Big Ideas Math Algebra 2 Answers Chapter 10 Probability

Big Ideas Math Algebra 2 Answers Chapter 10 Probability

Big Ideas Math Algebra 2 Answers Chapter 10 Probability is provided by subject experts adhering to the latest common core curriculum guidelines. Want to be Proficient in the Concepts of BIM Algebra 2 Ch 10 Probability? Then, start practicing constantly from our BigIdeas Math Book High School Algebra 2 Lesson 10 Probability Answer Key. Here, you can get the solutions in a detailed way for easy and better understanding to students so download the Ch 10 lesson-wise Big Ideas Math Algebra 2 Textbook Answers on Probability for free of charge & ace up your preparation.

Big Ideas Math Book Algebra 2 Answer Key Chapter 10 Probability

High School Candidates are recommended to access the below provided Topicwise Big Ideas Math Algebra 2 Ch 10 Probability Solutions and clear their doubts within no time. You can see step-by-step explained solutions in a simple and easy-to-understand language which are designed by the subject experts as per the latest common core 2019 curriculum. All you need to do is hit on the direct links available below and download the Big Ideas Math Textbook Solution Key for preparing the Algebra 2 Ch 10 Probability concepts efficiently in a free time. 

Probability Maintaining Mathematical Proficiency

Write and solve a proportion to answer the question.
Question 1.
What percent of 30 is 6?
Answer:

Question 2.
What number is 68% of 25?
Answer:

Question 3.
34.4 is what percent of 86?
Answer:

Display the data in a histogram.
Question 4.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 1
Answer:

Question 5.
ABSTRACT REASONING
You want to purchase either a sofa or an arm chair at a furniture store. Each item has the same retail price. The sofa is 20% off. The arm chair is 10% off, and you have a coupon to get an additional 10% off the discounted price of the chair. Are the items equally priced after the discounts are applied? Explain.
Answer:

Probability Mathematical Practices

Mathematically proficient students apply the mathematics they know to solve real-life problems.

Monitoring Progress

In Exercises 1 and 2, describe the event as unlikely, equally likely to happen or not happen, or likely. Explain your reasoning.
Question 1.
The oldest child in a family is a girl.
Answer:

Question 2.
The two oldest children in a family with three children are girls.
Answer:

Question 3.
Give an example of an event that is certain to occur.
Answer:

Lesson 10.1 Sample Spaces and Probability

Essential Question How can you list the possible outcomes in the sample space of an experiment?
The sample space of an experiment is the set of all possible outcomes for that experiment.

EXPLORATION 1

Finding the Sample Space of an Experiment
Work with a partner. In an experiment, three coins are flipped. List the possible outcomes in the sample space of the experiment.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 1

EXPLORATION 2

Finding the Sample Space of an Experiment
Work with a partner. List the possible outcomes in the sample space of the experiment.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 2

EXPLORATION 3

Finding the Sample Space of an Experiment
Work with a partner. In an experiment, a spinner is spun.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 3
a. How many ways can you spin a 1? 2? 3? 4? 5?
b. List the sample space.
c. What is the total number of outcomes?

EXPLORATION 4

Finding the Sample Space of an Experiment
Work with a partner. In an experiment, a bag contains 2 blue marbles and 5 red marbles. Two marbles are drawn from the bag.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 4
a. How many ways can you choose two blue? a red then blue? a blue then red? two red?
b. List the sample space.
c. What is the total number of outcomes?

Communicate Your Answer

Question 5.
How can you list the possible outcomes in the sample space of an experiment?
Answer:

Question 6.
For Exploration 3, find the ratio of the number of each possible outcome to the total number of outcomes. Then find the sum of these ratios. Repeat for Exploration 4. What do you observe?
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 5
Answer:

Monitoring Progress

Find the number of possible outcomes in the sample space. Then list the possible outcomes.
Question 1.
You flip two coins.
Answer:

Question 2.
You flip two coins and roll a six-sided die.
Answer:

Question 3.
You flip a coin and roll a six-sided die. What is the probability that the coin shows tails and the die shows 4?
Answer:

Find P(\(\bar{A}\)).
Question 4.
P(A) = 0.455.
Answer:

Question 5.
P(A) = \(\frac{1}{4}\)
Answer:

Question 6.
P(A) = 1
Answer:

Question 7.
P(A) = 0.03
Answer:

Question 8.
In Example 4, are you more likely to get 10 points or 5 points?
Answer:

Question 9.
In Example 4, are you more likely to score points (10, 5, or 2) or get 0 points?
Answer:

Question 10.
In Example 5, for which color is the experimental probability of stopping on the color greater than the theoretical probability?
Answer:

Question 11.
In Example 6, what is the probability that a pet-owning adult chosen at random owns a fish?
Answer:

Sample Spaces and Probability 10.1 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
A number that describes the likelihood of an event is the __________ of the event.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 1

Question 2.
WRITING
Describe the difference between theoretical probability and experimental probability.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, find the number of possible outcomes in the sample space. Then list the possible outcomes.
Question 3.
You roll a die and flip three coins.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 3.1
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 3.2

Question 4.
You flip a coin and draw a marble at random from a bag containing two purple marbles and one white marble.
Answer:

Question 5.
A bag contains four red cards numbered 1 through 4, four white cards numbered 1 through 4, and four black cards numbered 1 through 4. You choose a card at random.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 5

Question 6.
You draw two marbles without replacement from a bag containing three green marbles and four black marbles.
Answer:

Question 7.
PROBLEM SOLVING
A game show airs on television five days per week. Each day, a prize is randomly placed behind one of two doors. The contestant wins the prize by selecting the correct door. What is the probability that exactly two of the five contestants win a prize during a week?
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 6
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 7.1
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 7.2

Question 8.
PROBLEM SOLVING
Your friend has two standard decks of 52 playing cards and asks you to randomly draw one card from each deck. What is the probability that you will draw two spades?
Answer:

Question 9.
PROBLEM SOLVING
When two six-sided dice are rolled, there are 36 possible outcomes. Find the probability that (a) the sum is not 4 and (b) the sum is greater than 5.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 9

Question 10.
PROBLEM SOLVING
The age distribution of a population is shown. Find the probability of each event.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 7
a. A person chosen at random is at least 15 years old.
b. A person chosen at random is from 25 to 44 years old.
Answer:

Question 11.
ERROR ANALYSIS
A student randomly guesses the answers to two true-false questions. Describe and correct the error in finding the probability of the student guessing both answers correctly.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 8
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 11

Question 12.
ERROR ANALYSIS
A student randomly draws a number between 1 and 30. Describe and correct the error in finding the probability that the number drawn is greater than 4.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 9
Answer:

Question 13.
MATHEMATICAL CONNECTIONS
You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. What is the probability your dart lands in the yellow region?
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 13

Question 14.
MATHEMATICAL CONNECTIONS
The map shows the length (in miles) of shoreline along the Gulf of Mexico for each state that borders the body of water. What is the probability that a ship coming ashore at a random point in the Gulf of Mexico lands in the given state?
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 11
a. Texas
b. Alabama
c. Florida
d. Louisiana
Answer:

Question 15.
DRAWING CONCLUSIONS
You roll a six-sided die 60 times. The table shows the results. For which number is the experimental probability of rolling the number the same as the theoretical probability?
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 12
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 15

Question 16.
DRAWING CONCLUSIONS
A bag contains 5 marbles that are each a different color. A marble is drawn, its color is recorded, and then the marble is placed back in the bag. This process is repeated until 30 marbles have been drawn. The table shows the results. For which marble is the experimental probability of drawing the marble the same as the theoretical probability?
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 13
Answer:

Question 17.
REASONING
Refer to the spinner shown. The spinner is divided into sections with the same area.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 14
a. What is the theoretical probability that the spinner stops on a multiple of 3?
b. You spin the spinner 30 times. It stops on a multiple of 3 twenty times. What is the experimental probability of stopping on a multiple of 3?
c. Explain why the probability you found in part (b) is different than the probability you found in part (a).
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 17

Question 18.
OPEN-ENDED
Describe a real-life event that has a probability of 0. Then describe a real-life event that has a probability of 1.
Answer:

Question 19.
DRAWING CONCLUSIONS
A survey of 2237 adults ages 18 and over asked which sport is their favorite. The results are shown in the figure. What is the probability that an adult chosen at random prefers auto racing?
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 15
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 19

Question 20.
DRAWING CONCLUSIONS
A survey of 2392 adults ages 18 and over asked what type of food they would be most likely to choose at a restaurant. The results are shown in the figure. What is the probability that an adult chosen at random prefers Italian food?
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 16
Answer:

Question 21.
ANALYZING RELATIONSHIPS
Refer to the board in Exercise 13. Order the likelihoods that the dart lands in the given region from least likely to most likely.
A. green
B. not blue
C. red
D. not yellow
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 21

Question 22.
ANALYZING RELATIONSHIPS
Refer to the chart below. Order the following events from least likely to most likely.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 17
A. It rains on Sunday.
B. It does not rain on Saturday.
C. It rains on Monday.
D. It does not rain on Friday.
Answer:

Question 23.
USING TOOLS
Use the figure in Example 3 to answer each question.
a. List the possible sums that result from rolling two six-sided dice.
b. Find the theoretical probability of rolling each sum.
c. The table below shows a simulation of rolling two six-sided dice three times. Use a random number generator to simulate rolling two six-sided dice 50 times. Compare the experimental probabilities of rolling each sum with the theoretical probabilities.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 18
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 23

Question 24.
MAKING AN ARGUMENT
You flip a coin three times. It lands on heads twice and on tails once. Your friend concludes that the theoretical probability of the coin landing heads up is P(heads up) = \(\frac{2}{3}\). Is your friend correct? Explain your reasoning.
Answer:

Question 25.
MATHEMATICAL CONNECTIONS
A sphere fits inside a cube so that it touches each side, as shown. What is the probability a point chosen at random inside the cube is also inside the sphere?
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 19
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 25

Question 26.
HOW DO YOU SEE IT?
Consider the graph of f shown. What is the probability that the graph of y = f (x) +c intersects the x-axis when c is a randomly chosen integer from 1 to 6? Explain.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 20
Answer:

Question 27.
DRAWING CONCLUSIONS
A manufacturer tests 1200 computers and finds that 9 of them have defects. Find the probability that a computer chosen at random has a defect. Predict the number of computers with defects in a shipment of 15,000 computers. Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 27

Question 28.
THOUGHT PROVOKING
The tree diagram shows a sample space. Write a probability problem that can be represented by the sample space. Then write the answer(s) to the problem.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 21
Answer:

Maintaining Mathematical Proficiency

Find the product or quotient.
Question 29.
\(\frac{3 x}{y} \cdot \frac{2 x^{3}}{y^{2}}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 29

Question 30.
\(\frac{4 x^{9} y}{3 x^{3}} \cdot \frac{2 x y}{8 y^{2}}\)
Answer:

Question 31.
\(\frac{x+3}{x^{4}-2}\) • (x2 − 7x + 6)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 31

Question 32.
\(\frac{2 y}{5 x} \div \frac{y}{6 x}\)
Answer:

Question 33.
\(\frac{3 x}{12 x-11} \div \frac{x+1}{5 x}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.1 a 33

Question 34.
\(\frac{3 x^{2}+2 x-13}{x^{4}}\) ÷ (x2 + 9)
Answer:

Lesson 10.2 Independent and Dependent Events

Essential Question How can you determine whether two events are independent or dependent?
Two events are independent events when the occurrence of one event does not affect the occurrence of the other event. Two events are dependent events when the occurrence of one event does affect the occurrence of the other event.

EXPLORATION 1

Identifying Independent and Dependent Events
Work with a partner. Determine whether the events are independent or dependent. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 1
a. Two six-sided dice are rolled.
b. Six pieces of paper, numbered 1 through 6, are in a bag. Two pieces of paper are selected one at a time without replacement.
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 2

EXPLORATION 2

Finding Experimental Probabilities
Work with a partner.
a. In Exploration 1(a), experimentally estimate the probability that the sum of the two numbers rolled is 7. Describe your experiment.
b. In Exploration 1(b), experimentally estimate the probability that the sum of the two numbers selected is 7. Describe your experiment.

EXPLORATION 3

Finding Theoretical Probabilities
Work with a partner.
a. In Exploration 1(a), find the theoretical probability that the sum of the two numbers rolled is 7. Then compare your answer with the experimental probability you found in Exploration 2(a).
b. In Exploration 1(b), find the theoretical probability that the sum of the two numbers selected is 7. Then compare your answer with the experimental probability you found in Exploration 2(b).
c. Compare the probabilities you obtained in parts (a) and (b).

Communicate Your Answer

Question 4.
How can you determine whether two events are independent or dependent?
Answer:

Question 5.
Determine whether the events are independent or dependent. Explain your reasoning.
a. You roll a 4 on a six-sided die and spin red on a spinner.
b. Your teacher chooses a student to lead a group, chooses another student to lead a second group, and chooses a third student to lead a third group.
Answer:

Monitoring Progress

Question 1.
In Example 1, determine whether guessing Question 1 incorrectly and guessing Question 2 correctly are independent events.
Answer:

Question 2.
In Example 2, determine whether randomly selecting a girl first and randomly selecting a boy second are independent events.
Answer:

Question 3.
In Example 3, what is the probability that you spin an even number and then an odd number?
Answer:

Question 4.
In Example 4, what is the probability that both bills are $1 bills?
Answer:

Question 5.
In Example 5, what is the probability that none of the cards drawn are hearts when (a) you replace each card, and (b) you do not replace each card? Compare the probabilities.
Answer:

Question 6.
In Example 6, find (a) the probability that a non-defective part “passes,” and (b) the probability that a defective part “fails.”
Answer:

Question 7.
At a coffee shop, 80% of customers order coffee. Only 15% of customers order coffee and a bagel. What is the probability that a customer who orders coffee also orders a bagel?
Answer:

Independent and Dependent Events 10.2 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Explain the difference between dependent events and independent events, and give an example of each.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 1

Question 2.
COMPLETE THE SENTENCE
The probability that event B will occur given that event A has occurred is called the __________ of B given A and is written as _________
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, tell whether the events are independent or dependent. Explain your reasoning.
Question 3.
A box of granola bars contains an assortment of flavors. You randomly choose a granola bar and eat it. Then you randomly choose another bar.
Event A: You choose a coconut almond bar first.
Event B: You choose a cranberry almond bar second.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 3

Question 4.
You roll a six-sided die and flip a coin.
Event A: You get a 4 when rolling the die.
Event B: You get tails when flipping the coin.
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 3
Answer:

Question 5.
Your MP3 player contains hip-hop and rock songs. You randomly choose a song. Then you randomly choose another song without repeating song choices.
Event A: You choose a hip-hop song first.
Event B: You choose a rock song second.
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 5

Question 6.
There are 22 novels of various genres on a shelf. You randomly choose a novel and put it back. Then you randomly choose another novel.
Event A: You choose a mystery novel.
Event B: You choose a science fiction novel.
Answer:

In Exercises 7–10, determine whether the events are independent.
Question 7.
You play a game that involves spinning a wheel. Each section of the wheel shown has the same area. Use a sample space to determine whether randomly spinning blue and then green are independent events.
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 7

Question 8.
You have one red apple and three green apples in a bowl. You randomly select one apple to eat now and another apple for your lunch. Use a sample space to determine whether randomly selecting a green apple first and randomly selecting a green apple second are independent events.
Answer:

Question 9.
A student is taking a multiple-choice test where each question has four choices. The student randomly guesses the answers to the five-question test. Use a sample space to determine whether guessing Question 1 correctly and Question 2 correctly are independent events.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 9

Question 10.
A vase contains four white roses and one red rose. You randomly select two roses to take home. Use a sample space to determine whether randomly selecting a white rose first and randomly selecting a white rose second are independent events.
Answer:

Question 11.
PROBLEM SOLVING
You play a game that involves spinning the money wheel shown. You spin the wheel twice. Find the probability that you get more than $500 on your first spin and then go bankrupt on your second spin.
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 11

Question 12.
PROBLEM SOLVING
You play a game that involves drawing two numbers from a hat. There are 25 pieces of paper numbered from 1 to 25 in the hat. Each number is replaced after it is drawn. Find the probability that you will draw the 3 on your first draw and a number greater than 10 on your second draw.
Answer:

Question 13.
PROBLEM SOLVING
A drawer contains 12 white socks and 8 black socks. You randomly choose 1 sock and do not replace it. Then you randomly choose another sock. Find the probability that both events A and B will occur.
Event A: The first sock is white.
Event B: The second sock is white.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 13

Question 14.
PROBLEM SOLVING
A word game has 100 tiles, 98 of which are letters and 2 of which are blank. The numbers of tiles of each letter are shown. Yourandomly draw 1 tile, set it aside, and then randomly draw another tile. Find the probability that both events A and B will occur.
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 7
Answer:

Question 15.
ERROR ANALYSIS
Events A and B are independent. Describe and correct the error in finding P(A and B).
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 15

Question 16.
ERROR ANALYSIS
A shelf contains 3 fashion magazines and 4 health magazines. You randomly choose one to read, set it aside, and randomly choose another for your friend to read. Describe and correct the error in finding the probability that both events A and B occur.
Event A: The first magazine is fashion.
Event B: The second magazine is health.
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 9
Answer:

Question 17.
NUMBER SENSE
Events A and B are independent. Suppose P(B) = 0.4 and P(A and B) = 0.13. Find P(A).
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 17

Question 18.
NUMBER SENSE
Events A and B are dependent. Suppose P(B | A) = 0.6 and P(A and B) = 0.15. Find P(A).
Answer:

Question 19.
ANALYZING RELATIONSHIPS
You randomly select three cards from a standard deck of 52 playing cards. What is the probability that all three cards are face cards when (a) you replace each card before selecting the next card, and (b) you do not replace each card before selecting the next card? Compare the probabilities.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 19

Question 20.
ANALYZING RELATIONSHIPS
A bag contains 9 red marbles, 4 blue marbles, and 7 yellow marbles. You randomly select three marbles from the bag. What is the probability that all three marbles are red when (a) you replace each marble before selecting the next marble, and (b) you do not replace each marble before selecting the next marble? Compare the probabilities.
Answer:

Question 21.
ATTEND TO PRECISION
The table shows the number of species in the United States listed as endangered and threatened. Find (a) the probability that a randomly selected endangered species is a bird, and (b) the probability that a randomly selected mammal is endangered.
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 10
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 21

Question 22.
ATTEND TO PRECISION
The table shows the number of tropical cyclones that formed during the hurricane seasons over a 12-year period. Find (a) the probability to predict whether a future tropical cyclone in the Northern Hemisphere is a hurricane, and (b) the probability to predict whether a hurricane is in the Southern Hemisphere.
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 11
Answer:

Question 23.
PROBLEM SOLVING
At a school, 43% of students attend the homecoming football game. Only 23% of students go to the game and the homecoming dance. What is the probability that a student who attends thefootball game also attends the dance?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 23

Question 24.
PROBLEM SOLVING
At a gas station, 84% of customers buy gasoline. Only 5% of customers buy gasoline and a beverage. What is the probability that a customer who buys gasoline also buys a beverage?
Answer:

Question 25.
PROBLEM SOLVING
You and 19 other students volunteer to present the “Best Teacher” award at a school banquet. One student volunteer will be chosen to present the award. Each student worked at least 1 hour in preparation for the banquet. You worked for 4 hours, and the group worked a combined total of 45 hours. For each situation, describe a process that gives you a “fair” chance to be chosen, and find the probability that you are chosen.
a. “Fair” means equally likely.
b. “Fair” means proportional to the number of hours each student worked in preparation.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 25

Question 26.
HOW DO YOU SEE IT?
A bag contains one red marble and one blue marble. The diagrams show the possible outcomes of randomly choosing two marbles using different methods. For each method, determine whether the marbles were selected with or without replacement.
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 12
Answer:

Question 27.
MAKING AN ARGUMENT
A meteorologist claims that there is a 70% chance of rain. When it rains, there is a 75% chance that your softball game will be rescheduled. Your friend believes the game is more likely to be rescheduled than played. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 27

Question 28.
THOUGHT PROVOKING
Two six-sided dice are rolled once. Events A and B are represented by the diagram. Describe each event. Are the two events dependent or independent? Justify your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 13
Answer:

Question 29.
MODELING WITH MATHEMATICS
A football team is losing by 14 points near the end of a game. The team scores two touchdowns (worth 6 points each) before the end of the game. After each touchdown, the coach must decide whether to go for 1 point with a kick (which is successful 99% of the time) or 2 points with a run or pass (which is successful 45% of the time).
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 14
a. If the team goes for 1 point after each touchdown, what is the probability that the team wins? loses? ties?
b. If the team goes for 2 points after each touchdown, what is the probability that the team wins? loses? ties?
c. Can you develop a strategy so that the coach’s team has a probability of winning the game that is greater than the probability of losing? If so, explain your strategy and calculate the probabilities of winning and losing the game.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 29

Question 30.
ABSTRACT REASONING
Assume that A and B are independent events.
a. Explain why P(B) = P(B|A) and P(A) =P(A|B).
b. Can P(A and B) also be defined as P(B) • P(A|B)? Justify your reasoning.
Answer:

Maintaining Mathematical Proficiency

Solve the equation. Check your solution.
Question 31.
\(\frac{9}{10}\)x = 0.18
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 31

Question 32.
\(\frac{1}{4}\)x + 0.5x = 1.5
Answer:

Question 33.
0.3x − \(\frac{3}{5}\)x + 1.6 = 1.555
Answer:
Big Ideas Math Algebra 2 Answers Chapter 10 Probability 10.2 a 33

Lesson 10.3 Two-Way Tables and Probability

Essential Question How can you construct and interpret a two-way table?

EXPLORATION 1

Completing and Using a Two-Way Table
Work with a partner. A two-way table displays the same information as a Venn diagram. In a two-way table, one category is represented by the rows and the other category is represented by the columns.
The Venn diagram shows the results of a survey in which 80 students were asked whether they play a musical instrument and whether they speak a foreign language. Use the Venn diagram to complete the two-way table. Then use the two-way table to answer each question.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 1
a. How many students play an instrument?
b. How many students speak a foreign language?
c. How many students play an instrument and speak a foreign language?
d. How many students do not play an instrument and do not speak a foreign language?
e. How many students play an instrument and do not speak a foreign language?

EXPLORATION 2

Two-Way Tables and Probability
Work with a partner. In Exploration 1, one student is selected at random from the 80 students who took the survey. Find the probability that the student
a. plays an instrument.
b. speaks a foreign language.
c. plays an instrument and speaks a foreign language.
d. does not play an instrument and does not speak a foreign language.
e. plays an instrument and does not speak a foreign language.

EXPLORATION 3

Conducting a Survey
Work with your class. Conduct a survey of the students in your class. Choose two categories that are different from those given in Explorations 1 and 2. Then summarize the results in both a Venn diagram and a two-way table. Discuss the results.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 2

Communicate Your Answer

Question 4.
How can you construct and interpret a two-way table?
Answer:

Question 5.
How can you use a two-way table to determine probabilities?
Answer:

Monitoring Progress

Question 1.
You randomly survey students about whether they are in favor of planting a community garden at school. Of 96 boys surveyed, 61 are in favor. Of 88 girls surveyed, 17 are against. Organize the results in a two-way table. Then find and interpret the marginal frequencies.
Answer:

Question 2.
Use the survey results in Monitoring Progress Question 1 to make a two-way table that shows the joint and marginal relative frequencies.
Answer:

Question 3.
Use the survey results in Example 1 to make a two-way table that shows the conditional relative frequencies based on the column totals. Interpret the conditional relative frequencies in the context of the problem.
Answer:

Question 4.
Use the survey results in Monitoring Progress Question 1 to make a two-way table that shows the conditional relative frequencies based on the row totals. Interpret the conditional relative frequencies in the context of the problem.
Answer:

Question 5.
In Example 4, what is the probability that a randomly selected customer who is located in Santa Monica will not recommend the provider to a friend?
Answer:

Question 6.
In Example 4, determine whether recommending the provider to a friend and living in Santa Monica are independent events. Explain your reasoning.
Answer:

Question 7.
A manager is assessing three employees in order to offer one of them a promotion. Over a period of time, the manager records whether the employees meet or exceed expectations on their assigned tasks. The table shows the manager’s results. Which employee should be offered the promotion? Explain.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 3
Answer:

Two-Way Tables and Probability 10.3 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
A(n) _____________ displays data collected from the same source that belongs to two different categories.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 1

Question 2.
WRITING
Compare the definitions of joint relative frequency, marginal relative frequency, and conditional relative frequency.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, complete the two-way table.
Question 3.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 4
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 3

Question 4.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 5
Answer:

Question 5.
MODELING WITH MATHEMATICS
You survey 171 males and 180 females at Grand Central Station in New York City. Of those, 132 males and 151 females wash their hands after using the public rest rooms. Organize these results in a two-way table. Then find and interpret the marginal frequencies.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 6
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 5

Question 6.
MODELING WITH MATHEMATICS
A survey asks 60 teachers and 48 parents whether school uniforms reduce distractions in school. Of those, 49 teachers and 18 parents say uniforms reduce distractions in school. Organize these results in a two-way table. Then find and interpret the marginal frequencies.
Answer:

USING STRUCTURE In Exercises 7 and 8, use the two-way table to create a two-way table that shows the joint and marginal relative frequencies.
Question 7.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 7
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 7

Question 8.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 8
Answer:

Question 9.
MODELING WITH MATHEMATICS
Use the survey results from Exercise 5 to make a two-way table that shows the joint and marginal relative frequencies.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 9

Question 10.
MODELING WITH MATHEMATICS
In a survey, 49 people received a flu vaccine before the flu season and 63 people did not receive the vaccine. Of those who receive the flu vaccine, 16 people got the flu. Of those who did not receive the vaccine, 17 got the flu. Make a two-way table that shows the joint and marginal relative frequencies.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 9
Answer:

Question 11.
MODELING WITH MATHEMATICS
A survey finds that 110 people ate breakfast and 30 people skipped breakfast. Of those who ate breakfast, 10 people felt tired. Of those who skipped breakfast, 10 people felt tired. Make a two-way table that shows the conditional relative frequencies based on the breakfast totals.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 11

Question 12.
MODELING WITH MATHEMATICS
Use the survey results from Exercise 10 to make a two-way table that shows the conditional relative frequencies based on the flu vaccine totals.
Answer:

Question 13.
PROBLEM SOLVING
Three different local hospitals in New York surveyed their patients. The survey asked whether the patient’s physician communicated efficiently. The results, given as joint relative frequencies, are shown in the two-way table.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 10
a. What is the probability that a randomly selected patient located in Saratoga was satisfied with the communication of the physician?
b. What is the probability that a randomly selected patient who was not satisfied with the physician’s communication is located in Glens Falls?
c. Determine whether being satisfied with the communication of the physician and living in Saratoga are independent events.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 13

Question 14.
PROBLEM SOLVING
A researcher surveys a random sample of high school students in seven states. The survey asks whether students plan to stay in their home state after graduation. The results, given as joint relative frequencies, are shown in the two-way table.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 11
a. What is the probability that a randomly selected student who lives in Nebraska plans to stay in his or her home state after graduation?
b. What is the probability that a randomly selected student who does not plan to stay in his or her home state after graduation lives in North Carolina?
c. Determine whether planning to stay in their home state and living in Nebraska are independent events.
Answer:

ERROR ANALYSIS In Exercises 15 and 16, describe and correct the error in finding the given conditional probability.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 12

Question 15.
P(yes|Tokyo)
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 13
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 15

Question 16.
P(London|no)
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 14
Answer:

Question 17.
PROBLEM SOLVING
You want to find the quickest route to school. You map out three routes. Before school, you randomly select a route and record whether you are late or on time. The table shows your findings. Assuming you leave at the same time each morning, which route should you use? Explain.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 15
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 17.1
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 17.2

Question 18.
PROBLEM SOLVING
A teacher is assessing three groups of students in order to offer one group a prize. Over a period of time, the teacher records whether the groups meet or exceed expectations on their assigned tasks. The table shows the teacher’s results. Which group should be awarded the prize? Explain.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 16
Answer:

Question 19.
OPEN-ENDED
Create and conduct a survey in your class. Organize the results in a two-way table. Then create a two-way table that shows the joint and marginal frequencies.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 19

Question 20.
HOW DO YOU SEE IT?
A research group surveys parents and coaches of high school students about whether competitive sports are important in school. The two-way table shows the results of the survey.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 17
a. What does 120 represent?
b. What does 1336 represent?
c. What does 1501 represent?
Answer:

Question 21.
MAKING AN ARGUMENT
Your friend uses the table below to determine which workout routine is the best. Your friend decides that Routine B is the best option because it has the fewest tally marks in the “Does Not Reach Goal” column. Is your friend correct? Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 18
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 21

Question 22.
MODELING WITH MATHEMATICS
A survey asks students whether they prefer math class or science class. Of the 150 male students surveyed, 62% prefer math class over science class. Of the female students surveyed,74% prefer math. Construct a two-way table to show the number of students in each category if 350 students were surveyed.
Answer:

Question 23.
MULTIPLE REPRESENTATIONS
Use the Venn diagram to construct a two-way table. Then use your table to answer the questions.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 19
a.What is the probability that a randomly selected person does not own either pet?
b. What is the probability that a randomly selected person who owns a dog also owns a cat?
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 23

Question 24.
WRITING
Compare two-way tables and Venn diagrams. Then describe the advantages and disadvantages of each.
Answer:

Question 25.
PROBLEM SOLVING
A company creates a new snack, N,and tests it against its current leader, L. The table shows the results.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 20
The company is deciding whether it should try to improve the snack before marketing it, and to whom the snack should be marketed. Use probability to explain the decisions the company should make when the total size of the snack’s market is expected to (a) change very little, and (b) expand very rapidly.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 25

Question 26.
THOUGHT PROVOKING
Bayes’ Theorem is given by
P(A|B) = \(\frac{P(B \mid A) \cdot P(A)}{P(B)}\).
Use a two-way table to write an example of Bayes’ Theorem.
Answer:

Maintaining Mathematical Proficiency

Draw a Venn diagram of the sets described.
Question 27.
Of the positive integers less than 15, set A consists of the factors of 15 and set B consists of all odd numbers.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 27

Question 28.
Of the positive integers less than 14, set A consists of all prime numbers and set B consists of all even numbers.
Answer:

Question 29.
Of the positive integers less than 24, set A consists of the multiples of 2 and set B consists of all the multiples of 3.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 a 29

Probability Study Skills: Making a Mental Cheat Sheet

10.1–10.3 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 21

Core Concepts
Section 10.1
Theoretical Probabilities, p. 538
Probability of the Complement of an Event, p. 539
Experimental Probabilities, p. 541

Section 10.2
Probability of Independent Events, p. 546
Probability of Dependent Events, p. 547
Finding Conditional Probabilities, p. 549

Section 10.3
Making Two-Way Tables, p. 554
Relative and Conditional Relative Frequencies, p. 555

Mathematical Practices
Question 1.
How can you use a number line to analyze the error in Exercise 12 on page 542?
Answer:

Question 2.
Explain how you used probability to correct the flawed logic of your friend in Exercise 21 on page 560.
Answer:

Study Skills: Making a Mental Cheat Sheet

  • Write down important information on note cards.
  • Memorize the information on the note cards, placing the ones containing information you know in one stack and the ones containing information you do not know in another stack. Keep working on the information you do not know.
    Big Ideas Math Algebra 2 Solutions Chapter 10 Probability 10.3 22

Probability 10.1–10.3 Quiz

Question 1.
You randomly draw a marble out of a bag containing 8 green marbles, 4 blue marbles, 12 yellow marbles, and 10 red marbles. Find the probability of drawing a marble that is not yellow.
Answer:

Find P(\(\bar{A}\)).
Question 2.
P(A) = 0.32
Answer:

Question 3.
P(A) = \(\frac{8}{9}\)
Answer:

Question 4.
P(A) = 0.01
Answer:

Question 5.
You roll a six-sided die 30 times. A 5 is rolled 8 times. What is the theoretical probability of rolling a 5? What is the experimental probability of rolling a 5?
Answer:

Question 6.
Events A and B are independent. Find the missing probability.
P(A) = 0.25
P(B) = _____
P(A and B) = 0.05
Answer:

Question 7.
Events A and B are dependent. Find the missing probability.
P(A) = 0.6
P(B|A) = 0.2
P(A and B) = _____
Answer:

Question 8.
Find the probability that a dart thrown at the circular target shown will hit the given region. Assume the dart is equally likely to hit any point inside the target.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability q 1
a. the center circle
b. outside the square
c. inside the square but outside the center circle
Answer:

Question 9.
A survey asks 13-year-old and 15-year-old students about their eating habits. Four hundred students are surveyed, 100 male students and 100 female students from each age group. The bar graph shows the number of students who said they eat fruit every day.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability q 2
a. Find the probability that a female student, chosen at random from the students surveyed, eats fruit every day.
b. Find the probability that a 15-year-old student, chosen at random from the students surveyed, eats fruit every day.
Answer:

Question 10.
There are 14 boys and 18 girls in a class. The teacher allows the students to vote whether they want to take a test on Friday or on Monday. A total of 6 boys and 10 girls vote to take the test on Friday. Organize the information in a two-way table. Then find and interpret the marginal frequencies.
Answer:

Question 11.
Three schools compete in a cross country invitational. Of the 15 athletes on your team, 9 achieve their goal times. Of the 20 athletes on the home team, 6 achieve their goal times. On your rival’s team, 8 of the 13 athletes achieve their goal times. Organize the information in a two-way table. Then determine the probability that a randomly selected runner who achieves his or her goal time is from your school.
Answer:

Lesson 10.4 Probability of Disjoint and Overlapping Events

Essential Question How can you find probabilities of disjoint and overlapping events?
Two events are disjoint, or mutually exclusive, when they have no outcomes in common. Two events are overlapping when they have one or more outcomes in common.

EXPLORATION 1

Disjoint Events and Overlapping Events
Work with a partner. A six-sided die is rolled. Draw a Venn diagram that relates the two events. Then decide whether the events are disjoint or overlapping.
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 1
a. Event A: The result is an even number.
Event B: The result is a prime number.

b. Event A: The result is 2 or 4.
Event B: The result is an odd number.
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 2

EXPLORATION 2

Finding the Probability that Two Events Occur
Work with a partner. A six-sided die is rolled. For each pair of events, find (a) P(A), (b) P(B), (c) P(A and B), and (d) P(A or B).
a. Event A: The result is an even number.
Event B: The result is a prime number.

b. Event A: The result is 2 or 4.
Event B: The result is an odd number.
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 3

EXPLORATION 3

Discovering Probability Formulas
Work with a partner.
a. In general, if event A and event B are disjoint, then what is the probability that event A or event B will occur? Use a Venn diagram to justify your conclusion.
b. In general, if event A and event B are overlapping, then what is the probability that event A or event B will occur? Use a Venn diagram to justify your conclusion.
c. Conduct an experiment using a six-sided die. Roll the die 50 times and record the results. Then use the results to find the probabilities described in Exploration 2. How closely do your experimental probabilities compare to the theoretical probabilities you found in Exploration 2?

Communicate Your Answer

Question 4.
How can you find probabilities of disjoint and overlapping events?
Answer:

Question 5.
Give examples of disjoint events and overlapping events that do not involve dice.
Answer:

Monitoring Progress

A card is randomly selected from a standard deck of 52 playing cards. Find the probability of the event.
Question 1.
selecting an ace or an 8
Answer:

Question 2.
selecting a 10 or a diamond
Answer:

Question 3.
WHAT IF?
In Example 3, suppose 32 seniors are in the band and 64 seniors are in the band or on the honor roll. What is the probability that a randomly selected senior is both in the band and on the honor roll?
Answer:

Question 4.
In Example 4, what is the probability that the diagnosis is incorrect?
Answer:

Question 5.
A high school basketball team leads at halftime in 60% of the games in a season. The team wins 80% of the time when they have the halftime lead, but only 10% of the time when they do not. What is the probability that the team wins a particular game during the season?
Answer:

Probability of Disjoint and Overlapping Events 10.4 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Are the events A and \(\bar{A}\) disjoint? Explain. Then give an example of a real-life event and its complement.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 a 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 4
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, events A and B are disjoint. Find P(A or B).
Question 3.
P(A) = 0.3, P(B) = 0.1
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 a 3

Question 4.
P(A) = 0.55, P(B) = 0.2
Answer:

Question 5.
P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{4}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 a 5

Question 6.
P(A) = \(\frac{2}{3}\), P(B) = \(\frac{1}{5}\)
Answer:

Question 7.
PROBLEM SOLVING
Your dart is equally likely to hit any point inside the board shown. You throw a dart and pop a balloon. What is the probability that the balloon is red or blue?
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 5
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 a 7

Question 8.
PROBLEM SOLVING
You and your friend are among several candidates running for class president. You estimate that there is a 45% chance you will win and a 25% chance your friend will win. What is the probability that you or your friend win the election?
Answer:

Question 9.
PROBLEM SOLVING
You are performing an experiment to determine how well plants grow under different light sources. Of the 30 plants in the experiment, 12 receive visible light, 15 receive ultraviolet light, and 6 receive both visible and ultraviolet light. What is the probability that a plant in the experiment receives visible or ultraviolet light?
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 a 9

Question 10.
PROBLEM SOLVING
Of 162 students honored at an academic awards banquet, 48 won awards for mathematics and 78 won awards for English. There are 14 students who won awards for both mathematics and English. A newspaper chooses a student at random for an interview. What is the probability that the student interviewed won an award for English or mathematics?
Answer:

ERROR ANALYSIS In Exercises 11 and 12, describe and correct the error in finding the probability of randomly drawing the given card from a standard deck of 52 playing cards.
Question 11.
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 6
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 a 11

Question 12.
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 7
Answer:

In Exercises 13 and 14, you roll a six-sided die. Find P(A or B).
Question 13.
Event A: Roll a 6.
Event B: Roll a prime number.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 a 13

Question 14.
Event A: Roll an odd number.
Event B: Roll a number less than 5.
Answer:

Question 15.
DRAWING CONCLUSIONS
A group of 40 trees in a forest are not growing properly. A botanist determines that 34 of the trees have a disease or are being damaged by insects, with 18 trees having a disease and 20 being damaged by insects. What is the probability that a randomly selected tree has both a disease and is being damaged by insects?
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 8
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 a 15

Question 16.
DRAWING CONCLUSIONS
A company paid overtime wages or hired temporary help during 9 months of the year. Overtime wages were paid during 7 months, and temporary help was hired during 4 months. At the end of the year, an auditor examines the accounting records and randomly selects one month to check the payroll. What is the probability that the auditor will select a month in which the company paid overtime wages and hired temporary help?
Answer:

Question 17.
DRAWING CONCLUSIONS
A company is focus testing a new type of fruit drink. The focus group is 47% male. Of the responses, 40% of the males and 54% of the females said they would buy the fruit drink. What is the probability that a randomly selected person would buy the fruit drink?
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 a 17

Question 18.
DRAWING CONCLUSIONS
The Redbirds trail the Bluebirds by one goal with 1 minute left in the hockey game. The Redbirds’ coach must decide whether to remove the goalie and add a frontline player. The probabilities of each team scoring are shown in the table.
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 9
a. Find the probability that the Redbirds score and the Bluebirds do not score when the coach leaves the goalie in.
b. Find the probability that the Redbirds score and the Bluebirds do not score when the coach takes the goalie out.
c. Based on parts (a) and (b), what should the coach do?
Answer:

Question 19.
PROBLEM SOLVING
You can win concert tickets from a radio station if you are the first person to call when the song of the day is played, or if you are the first person to correctly answer the trivia question. The song of the day is announced at a random time between 7:00 and 7:30 A.M. The trivia question is asked at a random time between 7:15 and 7:45 A.M. You begin listening to the radio station at 7:20. Find the probability that you miss the announcement of the song of the day or the trivia question.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 a 19

Question 20.
HOW DO YOU SEE IT?
Are events A and B disjoint events? Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 10
Answer:

Question 21.
PROBLEM SOLVING
You take a bus from your neighborhood to your school. The express bus arrives at your neighborhood at a random time between 7:30 and 7:36 A.M. The local bus arrives at your neighborhood at a random time between 7:30 and 7:40 A.M. You arrive at the bus stop at 7:33 A.M. Find the probability that you missed both the express bus and the local bus.
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 11
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 a 21

Question 22.
THOUGHT PROVOKING
Write a general rule for finding P(A or B or C) for (a) disjoint and (b) overlapping events A, B, and C.
Answer:

Question 23.
MAKING AN ARGUMENT
A bag contains 40 cards numbered 1 through 40 that are either red or blue. A card is drawn at random and placed back in the bag. This is done four times. Two red cards are drawn, numbered 31 and 19, and two blue cards are drawn, numbered 22 and 7. Your friend concludes that red cards and even numbers must be mutually exclusive. Is your friend correct? Explain.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 a 23

Maintaining Mathematical Proficiency

Write the first six terms of the sequence.
Question 24.
a1 = 4, an = 2an-1 + 3
Answer:

Question 25.
a1 = 1, an = \(\frac{n(n-1)}{a_{n-1}}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 10 Probability 10.4 a 25

Question 26.
a1 = 2, a2 = 6, an = \(\frac{(n+1) a_{n-1}}{a_{n-2}}\)
Answer:

Lesson 10.5 Permutations and Combinations

Essential Question How can a tree diagram help you visualize the number of ways in which two or more events can occur?

EXPLORATION 1

Reading a Tree Diagram
Work with a partner. Two coins are flipped and the spinner is spun. The tree diagram shows the possible outcomes.
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 1
a. How many outcomes are possible?
b. List the possible outcomes.

EXPLORATION 2

Reading a Tree Diagram
Work with a partner. Consider the tree diagram below.
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 2
a. How many events are shown?
b. What outcomes are possible for each event?
c. How many outcomes are possible?
d. List the possible outcomes.

EXPLORATION 3

Writing a Conjecture
Work with a partner.
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 3
a. Consider the following general problem: Event 1 can occur in m ways and event 2 can occur in n ways. Write a conjecture about the number of ways the two events can occur. Explain your reasoning.
b. Use the conjecture you wrote in part (a) to write a conjecture about the number of ways more than two events can occur. Explain your reasoning.
c. Use the results of Explorations 1(a) and 2(c) to verify your conjectures.

Communicate Your Answer

Question 4.
How can a tree diagram help you visualize the number of ways in which two or more events can occur?
Answer:

Question 5.
In Exploration 1, the spinner is spun a second time. How many outcomes are possible?
Answer:

Monitoring Progress

Question 1.
In how many ways can you arrange the letters in the word HOUSE?
Answer:

Question 2.
In how many ways can you arrange 3 of the letters in the word MARCH?
Answer:

Question 3.
WHAT IF?
In Example 2, suppose there are 8 horses in the race. In how many different ways can the horses finish first, second, and third? (Assume there are no ties.)
Answer:

Question 4.
WHAT IF?
In Example 3, suppose there are 14 floats in the parade. Find the probability that the soccer team is first and the chorus is second.
Answer:

Question 5.
Count the possible combinations of 3 letters chosen from the list A, B, C, D, E.
Answer:

Question 6.
WHAT IF?
In Example 5, suppose you can choose 3 side dishes out of the list of 8 side dishes. How many combinations are possible?
Answer:

Question 7.
WHAT IF?
In Example 6, suppose there are 20 photos in the collage. Find the probability that your photo and your friend’s photo are the 2 placed at the top of the page.
Answer:

Question 8.
Use the Binomial Theorem to write the expansion of (a) (x + 3)5 and (b) (2p − q)4.
Answer:

Question 9.
Find the coefficient of x5 in the expansion of (x − 3)7.
Answer:

Question 10.
Find the coefficient of x3 in the expansion of (2x + 5)8.
Answer:

Permutations and Combinations 10.5 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
An arrangement of objects in which order is important is called a(n) __________.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 4
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, find the number of ways you can arrange (a) all of the letters and (b) 2 of the letters in the given word.
Question 3.
AT
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 3

Question 4.
TRY
Answer:

Question 5.
ROCK
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 5

Question 6.
WATER
Answer:

Question 7.
FAMILY
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 7

Question 8.
FLOWERS
Answer:

In Exercises 9–16, evaluate the expression.
Question 9.
5P2
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 9

Question 10.
7P3
Answer:

Question 11.
9P1
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 11

Question 12.
6P5
Answer:

Question 13.
8P6
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 13

Question 14.
12P0
Answer:

Question 15.
30P2
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 15

Question 16.
25P5
Answer:

Question 17.
PROBLEM SOLVING
Eleven students are competing in an art contest. In how many different ways can the students finish first, second, and third?
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 17

Question 18.
PROBLEM SOLVING
Six friends go to a movie theater. In how many different ways can they sit together in a row of 6 empty seats?
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 5
Answer:

Question 19.
PROBLEM SOLVING
You and your friend are 2 of 8 servers working a shift in a restaurant. At the beginning of the shift, the manager randomly assigns one section to each server. Find the probability that you are assigned Section 1 and your friend is assigned Section 2.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 19

Question 20.
PROBLEM SOLVING
You make 6 posters to hold up at a basketball game. Each poster has a letter of the word TIGERS. You and 5 friends sit next to each other in a row. The posters are distributed at random. Find the probability that TIGERS is spelled correctly when you hold up the posters.
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 6
Answer:

In Exercises 21–24, count the possible combinations of r letters chosen from the given list.
Question 21.
A, B, C, D; r = 3
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 21

Question 22.
L, M, N, O; r = 2
Answer:

Question 23.
U, V, W, X, Y, Z; r = 3
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 23.1
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 23.2

Question 24.
D, E, F, G, H; r = 4
Answer:

In Exercises 25–32, evaluate the expression.
Question 25.
5C1
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 25

Question 26.
8C5
Answer:

Question 27.
9C9
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 27

Question 28.
8C6
Answer:

Question 29.
12C3
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 29

Question 30.
11C4
Answer:

Question 31.
15C8
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 31

Question 32.
20C5
Answer:

Question 33.
PROBLEM SOLVING
Each year, 64 golfers participate in a golf tournament. The golfers play in groups of 4.How many groups of 4 golfers are possible?
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 7
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 33

Question 34.
PROBLEM SOLVING
You want to purchase vegetable dip for a party. A grocery store sells 7 different flavors of vegetable dip. You have enough money to purchase 2 flavors. How many combinations of 2 flavors of vegetable dip are possible?
Answer:

ERROR ANALYSIS In Exercises 35 and 36, describe and correct the error in evaluating the expression.
Question 35.
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 8
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 35

Question 36.
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 9
Answer:

REASONING In Exercises 37–40, tell whether the question can be answered using permutations or combinations. Explain your reasoning. Then answer the question.
Question 37.
To complete an exam, you must answer 8 questions from a list of 10 questions. In how many ways can you complete the exam?
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 37

Question 38.
Ten students are auditioning for 3 different roles in aplay. In how many ways can the 3 roles be filled?
Answer:

Question 39.
Fifty-two athletes are competing in a bicycle race. In how many orders can the bicyclists finish first, second, and third? (Assume there are no ties.)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 39

Question 40.
An employee at a pet store needs to catch 5 tetras in an aquarium containing 27 tetras. In how many groupings can the employee capture 5 tetras?
Answer:

Question 41.
CRITICAL THINKING
Compare the quantities 50C9 and 50C41 without performing any calculations. Explain your reasoning.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 41

Question 42.
CRITICAL THINKING
Show that each identity is true for any whole numbers r and n, where 0 ≤ r ≤ n.
a. nCn = 1
b. nCr = nCn-r − r
c. n+1Cr = nCr + nCr-1
Answer:

Question 43.
REASONING
Consider a set of 4 objects.
a. Are there more permutations of all 4 of the objects or of 3 of the objects? Explain your reasoning.
b. Are there more combinations of all 4 of the objects or of 3 of the objects? Explain your reasoning.
c. Compare your answers to parts (a) and (b).
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 43

Question 44.
OPEN-ENDED
Describe a real-life situation where the number of possibilities is given by 5P2. Then describe a real-life situation that can be modeled by 5C2.
Answer:

Question 45.
REASONING
Complete the table for each given value of r. Then write an inequality relating nPr and nCr. Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 10
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 45

Question 46.
REASONING
Write an equation that relates nPr and nCr. Then use your equation to find and interpret the value of \(\frac{182^{P_{4}}}{182^{C_{4}}}\).
Answer:

Question 47.
PROBLEM SOLVING
You and your friend are in the studio audience on a television game show. From an audience of 300 people, 2 people are randomly selected as contestants. What is the probability that you and your friend are chosen?
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 11
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 47

Question 48.
PROBLEM SOLVING
You work 5 evenings each week at a bookstore. Your supervisor assigns you 5 evenings at random from the 7 possibilities. What is the probability that your schedule does not include working on the weekend?
Answer:

REASONING In Exercises 49 and 50, find the probability of winning a lottery using the given rules. Assume that lottery numbers are selected at random.
Question 49.
You must correctly select 6 numbers, each an integer from 0 to 49. The order is not important.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 49

Question 50.
You must correctly select 4 numbers, each an integer from 0 to 9. The order is important.’
Answer:

In Exercises 51–58, use the Binomial Theorem to write the binomial expansion.
Question 51.
(x + 2)3
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 51

Question 52.
(c − 4)5
Answer:

Question 53.
(a + 3b)4
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 53

Question 54.
(4p − q)6
Answer:

Question 55.
(w3 − 3)4
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 55

Question 56.
(2s4 + 5)5
Answer:

Question 57.
(3u + v2)6
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 57

Question 58.
(x3 − y2)4
Answer:

In Exercises 59–66, use the given value of n to find the coefficient of xn in the expansion of the binomial.
Question 59.
(x − 2)10, n = 5
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 59

Question 60.
(x − 3)7, n = 4
Answer:

Question 61.
(x2 − 3)8, n = 6
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 61

Question 62.
(3x + 2)5, n = 3
Answer:

Question 63.
(2x + 5)12, n = 7
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 63

Question 64.
(3x − 1)9, n = 2
Answer:

Question 65.
(\(\frac{1}{2}\)x − 4 )11, n = 4
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 65

Question 66.
(\(\frac{1}{4}\)x + 6 )6, n = 3
Answer:

Question 67.
REASONING
Write the eighth row of Pascal’s Triangle as combinations and as numbers.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 67

Question 68.
PROBLEM SOLVING
The first four triangular numbers are 1, 3, 6, and 10.
a. Use Pascal’s Triangle to write the first four triangular numbers as combinations.
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 12
b. Use your result from part (a) to write an explicit rule for the nth triangular number Tn.
Answer:

Question 69.
MATHEMATICAL CONNECTIONS
A polygon is convex when no line that contains a side of the polygon contains a point in the interior of the polygon. Consider a convex polygon with n sides.
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 13
a. Use the combinations formula to write an expression for the number of diagonals in an n-sided polygon.
b. Use your result from part (a) to write a formula for the number of diagonals of an n-sided convex polygon.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 69

Question 70.
PROBLEM SOLVING
You are ordering a burrito with 2 main ingredients and 3 toppings. The menu below shows the possible choices. How many different burritos are possible?
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 14
Answer:

Question 71.
PROBLEM SOLVING
You want to purchase 2 different types of contemporary music CDs and 1 classical music CD from the music collection shown. How many different sets of music types can you choose for your purchase?
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 15
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 71

Question 72.
PROBLEM SOLVING
Every student in your history class is required to present a project in front of the class. Each day, 4 students make their presentations in an order chosen at random by the teacher. You make your presentation on the first day.
a. What is the probability that you are chosen to be the first or second presenter on the first day?
b. What is the probability that you are chosen to be the second or third presenter on the first day? Compare your answer with that in part (a).
Answer:

Question 73.
PROBLEM SOLVING
The organizer of a cast party for a drama club asks each of the 6 cast members to bring 1 food item from a list of 10 items. Assuming each member randomly chooses a food item to bring, what is the probability that at least 2 of the 6 cast members bring the same item?
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 73

Question 74.
HOW DO YOU SEE IT?
A bag contains one green marble, one red marble, and one blue marble. The diagram shows the possible outcomes of randomly drawing three marbles from the bag without replacement.
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 16
a. How many combinations of three marbles can be drawn from the bag? Explain.
b. How many permutations of three marbles can be drawn from the bag? Explain.
Answer:

Question 75.
PROBLEM SOLVING
You are one of 10 students performing in a school talent show. The order of the performances is determined at random. The first 5 performers go on stage before the intermission.
a. What is the probability that you are the last performer before the intermission and your rival performs immediately before you?
b. What is the probability that you are not the first performer?
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 75

Question 76.
THOUGHT PROVOKING
How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3, and 4? Repetition of digits is allowed.
Answer:

Question 77.
PROBLEM SOLVING
Consider a standard deck of 52 playing cards. The order in which the cards are dealt for a “hand” does not matter.
a. How many different 5-card hands are possible?
b. How many different 5-card hands have all 5 cards of a single suit?
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 17
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 77

Question 78.
PROBLEM SOLVING
There are 30 students in your class. Your science teacher chooses 5 students at random to complete a group project. Find the probability that you and your 2 best friends in the science class are chosen to work in the group. Explain how you found your answer.
Answer:

Question 79.
PROBLEM SOLVING
Follow the steps below to explore a famous probability problem called the birthday problem. (Assume there are 365 equally likely birthdays possible.)
a. What is the probability that at least 2 people share the same birthday in a group of 6 randomly chosen people? in a group of 10 randomly chosen people?
b. Generalize the results from part (a) by writing a formula for the probability P(n) that at least 2 people in a group of n people share the same birthday. (Hint: Use nPr notation in your formula.)
c. Enter the formula from part (b) into a graphing calculator. Use the table feature to make a table of values. For what group size does the probability that at least 2 people share the same birthday first exceed 50%?
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 79.1
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 79.2

Maintaining Mathematical Proficiency

Question 80.
A bag contains 12 white marbles and 3 black marbles. You pick 1 marble at random. What is the probability that you pick a black marble?
Answer:

Question 81.
The table shows the result of flipping two coins 12 times. For what outcome is the experimental probability the same as the theoretical probability?
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 18
Answer:
Big Ideas Math Answers Algebra 2 Chapter 10 Probability 10.5 a 81

Lesson 10.6 Binomial Distributions

Essential Question How can you determine the frequency of each outcome of an event?

EXPLORATION 1

Analyzing Histograms
Work with a partner. The histograms show the results when n coins are flipped.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 1
a. In how many ways can 3 heads occur when 5 coins are flipped?
b. Draw a histogram that shows the numbers of heads that can occur when 6 coins are flipped.
c. In how many ways can 3 heads occur when 6 coins are flipped?

EXPLORATION 2

Determining the Number of Occurrences
Work with a partner.
a. Complete the table showing the numbers of ways in which 2 heads can occur when n coins are flipped.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 2
b. Determine the pattern shown in the table. Use your result to find the number of ways in which 2 heads can occur when 8 coins are flipped.

Communicate Your Answer

Question 3.
How can you determine the frequency of each outcome of an event?
Answer:

Question 4.
How can you use a histogram to find the probability of an event?
Answer:

Monitoring Progress

An octahedral die has eight sides numbered 1 through 8. Let x be a random variable that represents the sum when two such dice are rolled.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 3
Question 1.
Make a table and draw a histogram showing the probability distribution for x.
Answer:

Question 2.
What is the most likely sum when rolling the two dice?
Answer:

Question 3.
What is the probability that the sum of the two dice is at most 3?
Answer:

According to a survey, about 85% of people ages 18 and older in the U.S. use the Internet or e-mail. You ask 4 randomly chosen people (ages 18 and older) whether they use the Internet or e-mail.
Question 4.
Draw a histogram of the binomial distribution for your survey.
Answer:

Question 5.
What is the most likely outcome of your survey?
Answer:

Question 6.
What is the probability that at most 2 people you survey use the Internet or e-mail?
Answer:

Binomial Distributions 10.6 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
What is a random variable?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 a 1

Question 2.
WRITING
Give an example of a binomial experiment and describe how it meets the conditions of a binomial experiment.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, make a table and draw a histogram showing the probability distribution for the random variable.
Question 3.
x = the number on a table tennis ball randomly chosen from a bag that contains 5 balls labeled “1,” 3 balls labeled “2,” and 2 balls labeled “3.”
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 a 3

Question 4.
c = 1 when a randomly chosen card out of a standard deck of 52 playing cards is a heart and c = 2 otherwise.
Answer:

Question 5.
w = 1 when a randomly chosen letter from the English alphabet is a vowel and w = 2 otherwise.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 a 5

Question 6.
n = the number of digits in a random integer from 0 through 999.
Answer:

In Exercises 7 and 8, use the probability distribution to determine (a) the number that is most likely to be spun on a spinner, and (b) the probability of spinning an even number.
Question 7.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 a 7

Question 8.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 5
Answer:

USING EQUATIONS In Exercises 9–12, calculate the probability of flipping a coin 20 times and getting the given number of heads.
Question 9.
1
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 a 9

Question 10.
4
Answer:

Question 11.
18
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 a 11

Question 12.
20
Answer:

Question 13.
MODELING WITH MATHEMATICS
According to a survey, 27% of high school students in the United States buy a class ring. You ask 6 randomly chosen high school students whether they own a class ring.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 6
a. Draw a histogram of the binomial distribution for your survey.
b. What is the most likely outcome of your survey?
c. What is the probability that at most 2 people have a class ring?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 a 13.1
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 a 13.2

Question 14.
MODELING WITH MATHEMATICS
According to a survey, 48% of adults in the United States believe that Unidentified Flying Objects (UFOs) are observing our planet. You ask 8 randomly chosen adults whether they believe UFOs are watching Earth.
a. Draw a histogram of the binomial distribution for your survey.
b. What is the most likely outcome of your survey?
c. What is the probability that at most 3 people believe UFOs are watching Earth?
Answer:

ERROR ANALYSIS In Exercises 15 and 16, describe and correct the error in calculating the probability of rolling a 1 exactly 3 times in 5 rolls of a six-sided die.
Question 15.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 a 15

Question 16.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 8
Answer:

Question 17.
MATHEMATICAL CONNECTIONS
At most 7 gopher holes appear each week on the farm shown. Let x represent how many of the gopher holes appear in the carrot patch. Assume that a gopher hole has an equal chance of appearing at any point on the farm.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 9
a. Find P(x) for x= 0, 1, 2, . . . , 7.
b. Make a table showing the probability distribution for x.
c. Make a histogram showing the probability distribution for x.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 a 17.1
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 a 17.2

Question 18.
HOW DO YOU SEE IT?
Complete the probability distribution for the random variable x. What is the probability the value of x is greater than 2?
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 10
Answer:

Question 19.
MAKING AN ARGUMENT
The binomial distribution shows the results of a binomial experiment. Your friend claims that the probability p of a success must be greater than the probability 1 −p of a failure. Is your friend correct? Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 11
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 a 19

Question 20.
THOUGHT PROVOKING
There are 100 coins in a bag. Only one of them has a date of 2010. You choose a coin at random, check the date, and then put the coin back in the bag. You repeat this 100 times. Are you certain of choosing the 2010 coin at least once? Explain your reasoning.
Answer:

Question 21.
MODELING WITH MATHEMATICS
Assume that having a male and having a female child are independent events, and that the probability of each is 0.5.
a. A couple has 4 male children. Evaluate the validity of this statement: “The first 4 kids were all boys, so the next one will probably be a girl.”
b. What is the probability of having 4 male children and then a female child?
c. Let x be a random variable that represents the number of children a couple already has when they have their first female child. Draw a histogram of the distribution of P(x) for 0 ≤ x ≤ 10. Describe the shape of the histogram.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 a 21

Question 22.
CRITICAL THINKING
An entertainment system has n speakers. Each speaker will function properly with probability p, independent of whether the other speakers are functioning. The system will operate effectively when at least 50% of its speakers are functioning. For what values of p is a 5-speaker system more likely to operate than a 3-speaker system?
Answer:

Maintaining Mathematical Proficiency

List the possible outcomes for the situation.
Question 23.
guessing the gender of three children
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 a 23

Question 24.
picking one of two doors and one of three curtains
Answer:

Probability Performance Task: A New Dartboard

10.4–10.6 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 12

Core Concepts
Section 10.4
Probability of Compound Events, p. 564
Section 10.5Permutations, p. 571Combinations, p. 572
The Binomial Theorem, p. 574

Section 10.6
Probability Distributions, p. 580
Binomial Experiments, p. 581

Mathematical Practices
Question 1.
How can you use diagrams to understand the situation in Exercise 22 on page 568?
Answer:

Question 2.
Describe a relationship between the results in part (a) and part (b) in Exercise 74 on page 578.
Answer:

Question 3.
Explain how you were able to break the situation into cases to evaluate the validity of the statement in part (a) of Exercise 21 on page 584.
Answer:

Performance Task A New Dartboard

You are a graphic artist working for a company on a new design for the board in the game of darts. You are eager to begin the project, but the team cannot decide on the terms of the game. Everyone agrees that the board should have four colors. But some want the probabilities of hitting each color to be equal, while others want them to be different. You offer to design two boards, one for each group. How do you get started? How creative can you be with your designs?
To explore the answers to these questions and more, go to BigIdeasMath.com.
Big Ideas Math Algebra 2 Answer Key Chapter 10 Probability 10.6 13

Probability Chapter Review

10.1 Sample Spaces and Probability (pp. 537–544)

Question 1.
A bag contains 9 tiles, one for each letter in the word HAPPINESS. You choose a tile at random. What is the probability that you choose a tile with the letter S? What is the probability that you choose a tile with a letter other than P?
Answer:

Question 2.
You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. Are you most likely to get 5 points, 10 points, or 20 points?
Big Ideas Math Algebra 2 Answers Chapter 10 Probability cr 1
Answer:

10.2 Independent and Dependent Events (pp. 545–552)

Find the probability of randomly selecting the given marbles from a bag of 5 red, 8 green, and 3 blue marbles when (a) you replace the first marble before drawing the second, and (b) you do not replace the first marble. Compare the probabilities.
Question 3.
red, then green
Answer:

Question 4.
blue, then red
Answer:

Question 5.
green, then green
Answer:

10.3 Two-Way Tables and Probability (pp. 553–560)

Question 6.
What is the probability that a randomly selected resident who does not support the project in the example above is from the west side?
Answer:

Question 7.
After a conference, 220 men and 270 women respond to a survey. Of those, 200 men and 230 women say the conference was impactful. Organize these results in a two-way table. Then find and interpret the marginal frequencies.
Answer:

10.4 Probability of Disjoint and Overlapping Events (pp. 563–568)

Question 8.
Let A and B be events such that P(A) = 0.32, P(B) = 0.48, and P(A and B) = 0.12. Find P(A or B).
Answer:

Question 9.
Out of 100 employees at a company, 92 employees either work part time or work 5 days each week. There are 14 employees who work part time and 80 employees who work 5 days each week. What is the probability that a randomly selected employee works both part time and 5 days each week?
Answer:

10.5 Permutations and Combinations (pp. 569–578)

Evaluate the expression.
Question 10.
7P6
Answer:

Question 11.
13P10
Answer:

Question 12.
6C2
ans;

Question 13.
8C4
Answer:

Question 14.
Use the Binomial Theorem to write the expansion of (2x + y2)4.
Answer:

Question 15.
A random drawing will determine which 3 people in a group of 9 will win concert tickets. What is the probability that you and your 2 friends will win the tickets?
Answer:

10.6 Binomial Distributions (pp. 579–584)

Question 16.
Find the probability of flipping a coin 12 times and getting exactly 4 heads.
Answer:

Question 17.
A basketball player makes a free throw 82.6% of the time. The player attempts 5 free throws. Draw a histogram of the binomial distribution of the number of successful free throws. What is the most likely outcome?
Answer:

Probability Chapter Test

You roll a six-sided die. Find the probability of the event described. Explain your reasoning.
Question 1.
You roll a number less than 5.
Answer:

Question 2.
You roll a multiple of 3.
Answer:

Evaluate the expression.
Question 3.
7P2
Answer:

Question 4.
8P3
Answer:

Question 5.
6C3
Answer:

Question 6.
12C7
Answer:

Question 7.
Use the Binomial Theorem to write the binomial expansion of (x + y2)5.
Answer:

Question 8.
You find the probability P(A or B) by using the equation P(A or B) = P(A) + P(B) − P(A and B). Describe why it is necessary to subtract P(A and B) when the events A and B are overlapping. Then describe why it is not necessary to subtract P(A and B) when the events A and B are disjoint.
Answer:

Question 9.
Is it possible to use the formula P(A and B) =P(A) • P(B|A) when events A and B are independent? Explain your reasoning.
Answer:

Question 10.
According to a survey, about 58% of families sit down for a family dinner at least four times per week. You ask 5 randomly chosen families whether they have a family dinner at least four times per week.
a. Draw a histogram of the binomial distribution for the survey.
b. What is the most likely outcome of the survey?
c. What is the probability that at least 3 families have a family dinner four times per week?
Answer:

Question 11.
You are choosing a cell phone company to sign with for the next 2 years. The three plans you consider are equally priced. You ask several of your neighbors whether they are satisfied with their current cell phone company. The table shows the results. According to this survey, which company should you choose?
Big Ideas Math Algebra 2 Answers Chapter 10 Probability ct 11
Answer:

Question 12.
The surface area of Earth is about 196.9 million square miles. The land area is about 57.5 million square miles and the rest is water. What is the probability that a meteorite that reaches the surface of Earth will hit land? What is the probability that it will hit water?
Answer:

Question 13.
Consider a bag that contains all the chess pieces in a set, as shown in the diagram.
Big Ideas Math Algebra 2 Answers Chapter 10 Probability ct 13
a. You choose one piece at random. Find the probability that you choose a black piece or a queen.
b. You choose one piece at random, do not replace it, then choose a second piece at random. Find the probability that you choose a king, then a pawn.
Answer:

Question 14.
Three volunteers are chosen at random from a group of 12 to help at a summer camp.
a. What is the probability that you, your brother, and your friend are chosen?
b. The first person chosen will be a counselor, the second will be a lifeguard, and the third will be a cook. What is the probability that you are the cook, your brother is the lifeguard, and your friend is the counselor?
Answer:

Probability Cumulative Assessment

Question 1.
According to a survey, 63% of Americans consider themselves sports fans. You randomly select 14 Americans to survey.
a. Draw a histogram of the binomial distribution of your survey.
b. What is the most likely number of Americans who consider themselves sports fans?
c. What is the probability at least 7 Americans consider themselves sports fans?
Answer:

Question 2.
Order the acute angles from smallest to largest. Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability ca 1
Answer:

Question 3.
You order a fruit smoothie made with 2 liquid ingredients and 3 fruit ingredients from the menu shown. How many different fruit smoothies can you order?
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability ca 2
Answer:

Question 4.
Which statements describe the transformation of the graph of f(x) = x3 − x represented by g(x) = 4(x − 2)3 − 4(x − 2)?
A. a vertical stretch by a factor of 4
B. a vertical shrink by a factor of \(\frac{1}{4}\)
C. a horizontal shrink by a factor of \(\frac{1}{4}\)
D. a horizontal stretch by a factor of 4
E. a horizontal translation 2 units to the right
F. a horizontal translation 2 units to the left
Answer:

Question 5.
Use the diagram to explain why the equation is true. P(A) + P(B) = P(A or B) + P(A and B)
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability ca 5
Answer:

Question 6.
For the sequence \(-\frac{1}{2},-\frac{1}{4},-\frac{1}{6},-\frac{1}{8}, \ldots\) describe the pattern, write the next term, graph the first five terms, and write a rule for the nth term.
Answer:

Question 7.
A survey asked male and female students about whether they prefer to take gym class or choir. The table shows the results of the survey.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability ca 7
a. Complete the two-way table.
b. What is the probability that a randomly selected student is female and prefers choir?
c. What is the probability that a randomly selected male student prefers gym class?
Answer:

Question 8.
The owner of a lawn-mowing business has three mowers. As long as one of the mowers is working, the owner can stay productive. One of the mowers is unusable 10% of the time, one is unusable 8% of the time, and one is unusable 18% of the time.
a. Find the probability that all three mowers are unusable on a given day.
b. Find the probability that at least one of the mowers is unusable on a given day.
c. Suppose the least-reliable mower stops working completely. How does this affect the probability that the lawn-mowing business can be productive on a given day?
Answer:

Question 9.
Write a system of quadratic inequalities whose solution is represented in the graph.
Big Ideas Math Algebra 2 Solutions Chapter 10 Probability ca 8
Answer:

Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers

Big Ideas Math Answers Grade 4 Chapter 4

Check out all the topics covered in the Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers from the below section. You can guide your students by making our Big Ideas Math Book 4th Grade Solution Key Chapter 4 Multiply by Two-Digit Numbers as reference. Students who are seriously practicing to score good marks in exams are suggested to practice the problems given in our Big Ideas Math Answers Grade 4 Chapter 4 Answer Key. Access the links given below and Download Big Ideas 4th Grade Solution Key pdf.

Big Ideas 4th Grade Chapter 4 Multiply by Two-Digit Numbers Solution Key

Students can easily understand the concepts in-depth with the help of Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers. We have provided all the questions, answers, along with explanations. Also, most of the explanations are given with images for the best learning for students. In order to become a math expert, you must refer to the Big Ideas Multiply by Two-Digit Numbers chapter. Verify the step-by-step procedure to solve a problem to make your preparation easy.

Lesson 1: Multiply by Tens

Lesson 2: Estimate Products

Lesson 3: Use Area Models to Multiply Two-Digit Numbers

Lesson 4: Use the Distributive Property to Multiply Two-Digit Numbers

Lesson 5: Use Partial Products to Multiply Two-Digit Numbers

Lesson 6: Multiply Two-Digit Numbers

Lesson 7: Practice Multiplication Strategies

Lesson 8: Problem Solving: Multiplication with Two-Digit Numbers

Performance Task

Lesson 4.1 Multiply by Tens

Explore and Grow
Model each product. Draw each model.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 1
What pattern do you notice in the products?

Answer:
The Place- value method. From the above pattern, we can conclude that the result has different place-values of 6

Explanation:
The position of 3 is different in the given 4 multiplications.
So,
2 × 3 = 6
2 × 30 = 60
2 × 300 = 600
2 × 3000 = 6,000
From the above pattern, we can conclude that the result has different place-values of 6
Repeated Reasoning
How can the pattern above help you find 20 × 30?

Answer:
20 × 30 = 600

Explanation:
You can think of 20 as two tens and 30 as Three tens.
So,
20 × 30 = 2 × 1 ten × 3 × 1 ten = 2 tens × 2 × 3 = 6 × 2 tens = 600
Think and Grow: Multiply by Multiples of Tens
You can use place value and properties to multiply two-digit numbers by multiples of ten.
Example
Find 40 × 20.
One Way: Use place value.
40 × 20 = 40 × ____ tens
= ____ tens
= _____
So, 40 × 20 = _____.

Answer:
800

Explanation:
By using the Place-value method,
40 × 20 = 40 × 2 tens
= 80 tens
= 800
So, 40 × 20 = 800
Another Way: Use the Associative Property of Multiplication.
40 × 20 = 40 × (2 × 10) Rewrite 20 as 2 × 10.
= (40 × 2) × 10 Associative Property of Multiplication
= ____ × 10
= ____
So, 40 × 20 = _____.

Answer:
800

Explanation:
By using the Associative Property of Multiplication,
40 × 20 = 40 × (2 × 10)
= (40 × 2) × 10
= 80 × 10
= 800
Note: Associative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c
Find 12 × 30.
One Way: Use place value
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 2
12 × 30 = 12 × _____ tens
= ____ tens
= _____
So, 12 × 30 = _____.

Answer:
360

Explanation:
Using the Place-value method,
12 × 30 = 12 × 3 tens
= 36 tens
= 360
So, 12 × 30 = 360
Another Way: Use the Associative Property of Multiplication
12 × 30 = 12 × (3 × 10) Rewrite 30 as 3 × 10.
= (12 × 3) × 10 Associative Property of Multiplication
= ____ × 10
= _____
So, 12 × 30 = ____.

Answer: 360

Explanation:
By using the Associative Property of Multiplication,
12 × 30 = 12 × (3 × 10)
= (12 × 3) × 10
= 36 × 10
= 360
Note: Associative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c
Show and Grow
Find the product.

Question 1.
70 × 40 = _____

Answer:
2800

Explanation:
By using the Associative Property of Multiplication,
70 × 40 = 70 × (4 × 10)
= (70 × 4) × 10
= 280 × 10
= 2800
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 2.
50 × 80 = ____

Answer:
4,000

Explanation:
By using the Associative Property of Multiplication,
50 × 80 = 50 × (8 × 10)
= (50 × 8) × 10
= 400 × 10
= 4000
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 3.
24 × 90 = _____

Answer:
2160

Explanation:
By using the Associative Property of Multiplication,
24 × 90 = 24 × (9 × 10)
= (24 × 9) × 10
= (8 × 3 × 9) × 10
= 216 × 10
= 2160
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 4.
45 × 60 = _____

Answer:
2700

Explanation:
By using the Associative Property of Multiplication,
45 × 60 = 45 × (6 × 10)
= (45 × 6) × 10
= (5 × 9 × 6) × 10
= 270 × 10
= 2700
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c
Apply and Grow: Practice
Find the product.

Question 5.
90 × 10 = _____

Answer:
900

Explanation:
By using the place-value method,
90 × 10 = 10 × 9 tens
= 90 tens
= 900
So,
90 × 10 = 900

Question 6.
40 × 60 = ____

Answer:
2400
Explaination:
By using the place-value method,
40 × 60 = 40 × 6 tens
= 4 tens × 6 tens
= 24 × tens × tens
= 2400
So,
40 × 60 = 2400

Question 7.
20 × 70 = _____

Answer: 1400

Explanation:
By using the place-value method,
70 × 20 = 70 × 2 tens
= 7 tens × 2 tens
= 14 × tens × tens
= 1400
So,
70 × 20 = 1400

Question 8.
11 × 30 = ____

Answer: 330

Explanation:
By using the place-value method,
11 × 30 = 11 × 3 tens
= 33 tens
= 330
So,
11 × 30 = 330

Question 9.
12 × 40 = ____

Answer: 480

Explanation:
By using the place-value method,
12 × 40 = 12 × 4 tens
= 48 tens
= 480
So,
12 × 40 = 480

Question 10.
15 × 50 = _____

Answer: 750

Explanation:
By using the place-value method,
15 ×50 = 15 × 5 tens
= 75 tens
= 750
So,
15 × 50 = 750

Question 11.
30 × 13 = _____

Answer: 390

Explanation:
By using the place-value method,
13 × 30 = 13 × 3 tens
= 39 tens
= 390
So,
13 × 30 = 390

Question 12.
10 × 76 = _____

Answer: 760

Explanation:
By using the place-value method,
10 × 76 = 76 × 1 ten
= 76 tens
= 760
So,
76 × 10 = 760

Question 13.
40 × 25 = ____

Answer: 1,000

Explanation:
By using the place-value method,
25 × 40 = 25 × 4 tens
= 5 × 5× 4 tens
= 100 tens
= 1,000
So,
25 × 40 = 1,000
Find the missing factor.

Question 14.
50 × ____ = 1,500

Answer: 30
Explanation;
Let the missing number be X
So, 50 × X = 1,500
X = 1,500 / 50 = 30
Hence, the value of X is: 30

Question 15.
20 × ____ = 1,800

Answer: 90
Explanation;
Let the missing number be X
So, 20 × X = 1,800
X = 1,800 /20 = 90
Hence, the value of X is: 90

Question 16.
60 × ___ = 4,200

Answer: 70
Explanation;
Let the missing number be X
So, 60 × X = 4,200
X = 4,200 / 60 = 70
Hence, the value of X is: 70

Question 17.
____ × 80 = 6,400

Answer: 80
Explanation;
Let the missing number be X
So, X × 80 = 6,400
X = 6,400 / 80 = 80
Hence, the value of X is: 80

Question 18.
____ × 90 = 3,600

Answer: 40
Explanation;
Let the missing number be X
So, X × 90 = 3,600
X = 3,600 / 90 = 40
Hence, the value of X is: 40

Question 19.
____ × 70 = 3,500

Answer: 50
Explanation;
Let the missing number be X
So, X × 70 = 3,500
X = 3,500 / 70 = 50
Hence, the value of X is: 50
Compare.

Question 20.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 3

Answer: 60 × 30 is equal to 1,800

Explanation:
60 × 30 = 1,800
Given numbers are: 1,800 and 1,800
By comparing 2 values, we can conclude that 1,800 is equal to 1,800

Question 21.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 4

Answer: 480 is greater than 460

Explanation:
40 × 12 = 480
Given numbers are: 480 and 460
By comparing 2 values, we can conclude that480 is greater than 460

Question 22.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 5

Answer: 2,250 is less than 2,340

Explanation:
25 × 90 = 2,250
Given numbers are 2,250 and 2,340
By comparing 2 values, we can conclude that 2,250 is less than 2,340

Question 23.
It takes 10 days to film 1 episode of a television show. How many days will it take to film a 20-episode season?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 6

Answer: 200 days

Explanation:
Given that it takes 10 days to film 1 episode of a Television show.
So,
The number of days it will take to film a 20-episode season is: 20 × 10 = 200 days

Question 24.
Reasoning
What is Descartes’s number? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 7

Answer:

Question 25.
YOU BE THE TEACHER
Newton says that the product of two multiples of ten will always have exactly two zeros. Is he correct? Explain.

Answer: He is correct

Explanation:
Let us suppose 2 numbers 10 and 20 which are the two multiples of 10.
Now,
10 × 20 = 200
According to Newton, the product of two multiples of ten will always have exactly two zeroes.
So, from the above multiplication, we can say that Newton is correct.
Think and Grow: Modeling Real Life

Example
Food drive volunteers collect 1,328 cans of food. The volunteers have50 boxes. Each box holds 20 cans. How many cans will fit in the boxes?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 8
Multiply to find how many cans will fit in the boxes.
20 × 50 = 50 × (5 × 10)
Rewrite 50 as 5 × 10.
= (20 × 5) × 10 Associative Property of Multiplication
= 100 × 10
= 1,000
So, 1,000 cans fit in the boxes.
Subtract the number of cans that will fit in the boxes from the total number of cans collected.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 9
The cans that can not fit in the boxes = 1,328 – 1,000 = 328
So,
328 cans will not fit in the boxes.
Show and Grow

Question 26
A library has 2,124 new books. The library has 40 empty shelves. Each shelf holds 35 books. How many not books
will fit on the empty shelves?

Answer: 724

Explanation:
Multiply to find how many books will hold on the shelves.
40 × 35 = 35 × (4 × 10)
Rewrite 40 as 5 × 10.
= (35 × 4) × 10 Associative Property of Multiplication
= 140 × 10
= 1,400
So, 1,400 cans fit in the boxes.
Subtract the number of books that will hold on the shelves from the total number of books collected.
The number of books that will not hold on the shelves = 2,124 – 1,400 = 724
So,
724 books will not hold on the shelves

Question 27.
An apartment building has 15 floors. Each floor is 10 feet tall. An office building has 30 floors. Each floor is 13 feet tall. How much taller is the office building than the apartment building?

Answer: 240 feet

Explanation:
Given that an apartment has 15 floors and in that, each floor is 10 feet tall.
So, the height of the apartment = 15 × 10 = 150 feet
Given that an office building has 30 floors and in that, each floor is 13 feet tall.
So, the height of the office building = 30 × 13 = 390 feet
Now, to calculate how much taller an office building than the apartment, we have to subtract both the heights of the apartment and the office building.
So,
The difference in height between the office building and the apartment = 390 feet – 150 feet = 240 feet.
From the above, we can conclude that the office building is 240 feet taller than the apartment.

Question 28.
You burn 35 calories each hour you spend reading and 50 calories each hour you spend playing board games. In 2
weeks, you spend14 hours reading and 28 hours playing board games. How many calories do you burn reading
and playing board games?

Answer:
The calories burned during reading in 2 weeks = 490 calories
The calories burned during playing board games = 1400 calories

Explanation:
Given that,
The calories burned during reading is: 35 calories each hour
The calories burned during playing board games is: 50 calories each hour
It is also given that In 2 weeks,
the time spend on reading and Playing Board Games are 14 hours and 28 hours
So, to calculate the calorie consumption in these 2 weeks, we have to multiply the number of hours and the number of calories.
So,
The calories burned during reading in 2 weeks = 490 calories
The calories burned during playing board games in 2 weeks = 1400 calories

Multiply by Tens Homework & Practice 4.1

Find the product.

Question 1.
30 × 10 = _____

Answer: 300

Explanation:
The product of multiply by tens can be done in 2 ways. They are:
A) The place-value method  B) The Associative Property of Multiplication
A) By using the place-value method:
30 × 10 = 10 × 3 tens
= 1 ten × 3 tens
= 3 × 1 ten × 1 ten
= 3 × 100
= 300
B) The Associative Property of Multiplication:
30 × 10 = 3 × (10 × 10)
= (3 × 10) × 10
= 30 × 10
= 300
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 2.
20 × 90 = _____

Answer: 1800

Explanation:
The product of multiply by tens can be done in 2 ways. They are:
A) The place-value method  B) The Associative Property of Multiplication
A) By using the place-value method:
20 × 90 = 20 × 9 tens
= 2 tens × 9 tens
= 18 × 1 ten × 1 ten
= 18 × 100
= 1800
B) The Associative Property of Multiplication:
20 × 90 = 20 × (9 × 10)
= (20 × 9) × 10
=(5 × 4 × 9) × 10
= 180 × 10
= 1800
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 3.
50 × 70 = _____

Answer: 3500

Explanation:
The product of multiply by tens can be done in 2 ways. They are:
A) The place-value method  B) The Associative Property of Multiplication
A) By using the place-value method:
50 × 70 = 50 × 7 tens
= 5 tens × 7 tens
= 35 × 1 ten × 1 ten
= 35 × 100
= 3500
B) The Associative Property of Multiplication:
50 × 70 = 50 × (7 × 10)
= (50 × 7) × 10
=(5 × 10 × 7) × 10
= 350 × 10
= 3500
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 4.
40 × 13 = ______

Answer: 520

Explanation:
The product of multiply by tens can be done in 2 ways. They are:
A) The place-value method  B) The Associative Property of Multiplication
A) By using the place-value method:
40 × 13 = 13 × 4 tens
= 13 × 4 tens
= 52 × 1 ten
= 52 × 10
= 520
B) The Associative Property of Multiplication:
40 × 13 = 13 × (4 × 10)
= (13 × 4) × 10
=(13 ×2× 2) × 10
= 54 × 10
= 540
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 5.
27 × 60 = _____

Answer: 1620

Explanation:
The product of multiply by tens can be done in 2 ways. They are:
A) The place-value method  B) The Associative Property of Multiplication
A) By using the place-value method:
27 × 60 = 27 × 6 tens
= 3 × 9 × 6 tens
= 162 × 1 ten
= 162 × 10
= 1620
B) The Associative Property of Multiplication:
27 × 60 = 27 × (6 × 10)
= (27 × 6) × 10
=(6 × 3 × 9) × 10
= 162 × 10
= 1620
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 6.
80 × 56 = _____

Answer: 4480

Explanation:
The product of multiply by tens can be done in 2 ways. They are:
A) The place-value method  B) The Associative Property of Multiplication
A) By using the place-value method:
80 × 56 = 56 × 8 tens
= 7 × 8 × 8 tens
= 448 × 1 ten
= 448 × 10
= 4480
B) The Associative Property of Multiplication:
56 × 80 = 56 × (8 × 10)
= (56 × 8) × 10
=(8 × 7 × 8) × 10
= 448 × 10
= 4480
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c
Find the missing factor.

Question 7.
70 × ____ = 2,100

Answer: 30

Explanation:
Let the missing number be X
So, 70 × X = 2,100
X = 2,100 / 70 = 30
Hence, the value of X is: 30

Question 8.
____ × 10 = 900

Answer: 90
Let the missing number be X
So, X × 10 = 900
X = 900 / 10 =90
Hence, the value of X is: 90

Question 9.
40 × ____ = 1,600

Answer: 40
Let the missing number be X
So, 40 × X = 1,600
X = 1,600 / 40 = 40
Hence, the value of X is: 40

Question 10.
____ × 20 = 1,600

Answer: 80
Let the missing number be X
So, X× 20 = 1,600
X = 1,600 / 20 = 80
Hence, the value of X is: 80

Question 11.
30 × ____ = 1,800

Answer: 60
Let the missing number be X
So, 30 × X = 1,800
X = 1,800 / 30 = 60
Hence, the value of X is: 60

Question 12.
____ × 50 = 3,000

Answer: 60
Let the missing number be X
So, X × 50 = 3,000
X = 3,000 / 50 = 60
Hence, the value of X is: 60
Compare.

Question 13.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 10

Answer: 7,200 is less than 8,100

Explanation:
90 × 80 = 7,200
Given numbers are: 7,200 and 8,100
By comparing 2 values, we can conclude that 7,200 is less than 8,100

Question 14.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 11

Answer: 1,200 is greater  than 1,020

Explanation:
60 ×17 = 1,020
Given numbers are: 1,200 and 1,020
By comparing 2 values, we can conclude that 1,200 is greater than 1,020

Question 15.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 12

Answer: 2,380 is equal to 2,380

Explanation:
34 × 70 = 2,380
Given numbers are:2,380 and 2,380
By comparing the 2 values, we can conclude that 2,380 is equal to2,380

Question 16.
A shallow moonquake occurs 20 kilometers below the moon’s surface. A deep moonquake occurs 35 times deeper than a shallow moonquake. How many kilometers below the surface does the deep moonquake occur?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 13

Answer: 15 kilometers

Explanation:
Given that a shallow moonquake occurred 20 kilometers below the moon’s surface and a deep moonquake occurs 35 meters deeper than a shallow moonquake.
Now, to calculate how much distance the deep moonquake occurred from the surface, we have to subtract the distance that a shallow moonquake occurred from the deep moonquake occurred.
Hence,
The distance below the surface the deep moonquake occurred = 30 – 25 = 15 kilometers

Question 17.
Structure
Write the multiplication equation represented by the number line.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 14

Answer:

Question 18.
Writing
Explain how you can use 20 × 10 = 200 to find 20 × 12.

Answer: By using the Associative Property of Multiplication,
20 × 10 = 10 × (2 × 10)
= (10 × 2) × 10
= 20 × 10
= 200
By using the same method, we can also find the value of 20 × 12.
Now,
By using the Associative Property of Multiplication,
20 × 12 = 12 × (2 × 10)
= (12 × 2) × 10
= 24 × 10
= 240
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 19.
DIG DEEPER!
The product of a number and twice that number is 800. What are the numbers?

Answer: 20, 40

Explanation:
Let the number be X
Given that,
X × 2X = 800
Take X as a common multiple.
Hence,
X × X ( 1 × 2) = 800
X × X × 2 = 800
X × X = 800/2 = 400
X × X = 20 × 20
From the above, we can conclude that the value of X is 20
Hence, the value of twice the X is  20 × 2 = 40
So,
The numbers that can give the product 800 are 20, 40

Question 20.
Modeling Real Life
There are 506 new plants in a greenhouse. A worker programs a robot to arrange the plants into14 rows with 30 plants in each row. How many plants will fit in the rows?

Answer: 420

Explanation:
Given that there are 506 new plants in a greenhouse and a robot can arrange the plants into 14 rows with 30 plants each.
So, to find how many plants will fit in the row, we have to multiply 14 and 30( Since the robot arranges the plants in rows)
Now, By using the Associative Property of Multiplication,
14 × 30 =  14 × (3 × 10)
= (14 × 3) × 10
= (2 × 7 × 3) × 10
= 42 × 10
= 420
From the above, we can conclude that 420 plants will fit in the row.

Question 21.
Modeling Real Life
The world’s largest pool is 13 meters longer than the total length of 20 Olympic pools. An Olympic pool is 50 meters long. How long is the world’s largest pool?

Answer: 1013 meters

Explanation: Given that an Olympic pool has a length of 50 meters. But, there are 20 Olympic pools.
So, to find the total length of the Olympic Pool, we have to multiply the number of pools and the length of each pool.
By using the Associative Property of Multiplication,
50 × 20 = 50 × (2 × 10)
= (50 × 2) × 10
= ( 5 × 10 × 2) × 10
= 100 × 10
= 1,000 meters
So, the total length of the 20 Olympic pools = 1,000 meters
The
Question also mentions that the world’s largest pool is 13 meters longer than the total length of 20 Olympic pools.
Hence,
The length of the World’s largest pool = 1,000 + 13 = 1,013 meters.
So, the length of the World’s largest pool is 1,013 meters
Review & Refresh
Find the value of the underlined digit.

Question 22.
52,618

Answer: The place-value of 8 in the given number is: 8

Explanation:
We can find the position of any given number by using the place- value method.
From this, we can conclude that the place-value of 8 is: 8

Question 23.
379,021

Answer: The place-value of 7 in the given number is: 70,000

Explanation:
We can find the position of any given number by using the place- value method.
From this, we can conclude that the place-value of 7 in the given number is: 70,000

Explanation:

Question 24.
203,557

Answer: The place-value of 2 in the given number is: 200,000

Explanation:
We can find the position of any given number by using the place- value method.
From this, we can conclude that the place-value of 2 in the given number is: 200,000

Question 25.
497,384

Answer: The place-value of 3 in the given number is: 300

Explanation:
We can find the position of any given number by using the place- value method.
From this, we can conclude that the place-value of 3 in the given number is: 300

Lesson 4.2 Estimate Products

Explore and Grow
Choose an expression to estimate each product. Write the expression. You may use an expression more than once.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 1
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 2
Compare your answers with a partner. Did you choose the same expressions?

Answer:
Let your Expression be 20 ×25.
The given Partner Expressions are:
A) 21 × 24  B) 26 × 38  C) 23 × 17  D) 42 × 23
By Comparing your Expression and your Partner Expression,
A) 500 is less than 504.

Explanation:
Let your Expression be 20 ×25.
By using the Associative Property of Multiplication,
20 × 25 = 25 × ( 2 × 10)
= (25 × 2) × 10
= (5 × 5 × 2) × 10
= 50 × 10
= 500
The given Partner Expressions are:
A) 21 × 24  B) 26 × 38  C) 23 × 17  D) 42 × 23
We can calculate the partner Expressions by Simplifying the given Expressions.
A)
21 × 24 = 3 × 7 × 3 × 8
= 9 × 56
= 504
By comparing your Expression with your Partner Expression, 500 is less than 504.
B)
26 × 38 = 13  × 2 × 19 × 2
= 4× 247
= 988
By comparing your Expression with your Partner Expression, 500 is less than 988.
C)
23 × 17
=391
By comparing your Expression with your Partner Expression, 500 is greater than 391.
D)
42 × 23 = 7  ×  2 ×3 ×  23
= 966
By comparing your Expression with your Partner Expression, 500 is less than 966.
Construct Arguments
Which estimated product do you think will be closer to the product of 29 and 37? Explain your reasoning.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 3

Answer: 1,000 will be closer to 1,073

Explanation:
Given Product is
29 × 37 = 1,073
Given Expressions are:
25 × 40 = 1,000
30 × 40 = 1,200
Compare the given Product and the Expressions.
By comparison, we can conclude that 1,073 is close to 1,000.
Think and Grow: Estimate Products
You can estimate products using rounding or compatible numbers. Compatible numbers are numbers that are easy to multiply and are close to the actual numbers.
Example
Use rounding to estimate 57 × 38.
Step 1: Round each factor to the nearest ten.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 4
Step 2: Multiply.
60 × 40 = 60 × 4 tens
= 240 tens
= 2400
So, 57 × 38 is about 2400.
Example
Use compatible numbers to estimate 24 × 31.
Step 1: Choose compatible numbers.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 5
Step 2: Multiply.
25 × 30 = 25 × 3 tens
= 75 tens
= 750
So, 24 × 31 is about 750.
Show and Grow
Use rounding to estimate the product.

Question 1.
27 × 50

Answer: 1500

Explanation:
Let 27 be Rounded to 30.
Now, we have to find the result of 30 × 50.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
30 × 50 = 30 × (5 × 10)
= (30 × 5) × 10
= ( 3 × 10 × 5) × 10
= 150 × 10
= 1,500
B) By using the place-value method,
30 × 50 = 30 × 5 tens
= 3 tens × 5 tens
= 15 × 1 ten × 1ten
= 15 × 10 × 10
= 1500
So,
27 × 50 can be rounded to 1,500

Question 2.
42 × 14

Answer: 600

Explanation:
Let 42 be Rounded to 40
Let 14 be Rounded to 15
Now, we have to find the result of 40 × 15.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
40 × 15 = 15 × (4 × 10)
= (15 × 4) × 10
= ( 3 × 4 × 5) × 10
= 60 × 10
= 600
B) By using the place-value method,
40 × 15 = 15 × 4 tens
= 60 tens
= 60 × 10
= 600
So,
42 × 16 can be rounded to 600

Question 3.
61 × 73

Answer: 4,200

Explanation:
Let 61 be Rounded to 60
Let 73 be Rounded to 70
Now, we have to find the result of 60 × 70.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
60 × 70 = 60 × (7 × 10)
= (60 × 7) × 10
= ( 6 × 10 × 7) × 10
= 420 × 10
= 4,200
B) By using the place-value method,
60 × 70 = 60 × 7 tens
= 6 tens ×7 tens
= 42 × 1 ten × 1ten
= 42 × 10 × 10
= 4,200
So,
61 ×73 can be rounded to 4,200
Use compatible numbers to estimate the product.

Question 4.
19 × 26

Answer: 500

Explanation:
Let 19 be Rounded to 20
Let 26 be Rounded to 25
Now, we have to find the result of 20 × 25.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
20 × 25 = 25 × (2 × 10)
= (25 × 2) × 10
= ( 5 × 5 ×2) × 10
= 50 × 10
= 500
B) By using the place-value method,
20 × 25 = 25 × 2 tens
= 50 tens
= 50 × 1 ten
= 50 × 10
= 500
So,
19 ×26 can be rounded to 500

Question 5.
23 × 78

Answer: 2,000

Explanation:
Let 23 be Rounded to 25
Let 78 be Rounded to 80
Now, we have to find the result of 25 × 80.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
25 × 80 = 25 × (8 × 10)
= (25 × 8) × 10
= ( 5× 5 × 8) × 10
= 200 × 10
= 2,000
B) By using the place-value method,
25 × 80 = 25 × 8 tens
= 200 tens
= 200 × 10
= 2,000
So,
23 ×78 can be rounded to 2,000

Question 6.
74 × 20

Answer: 1,500

Explanation:
Let 74 be Rounded to 75
Now, we have to find the result of 75 × 20.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
75 × 20 = 75 × (2 × 10)
= (75 × 2) × 10
= ( 5 × 5 × 3 × 2) × 10
= 150 × 10
= 1,500
B) By using the place-value method,
75× 20 = 75 ×2 tens
= 150 tens
= 150× 10
= 1,500
So,
74 ×20 can be rounded to 1,500
Apply and Grow: Practice
Estimate the product.

Question 7.
41 × 73

Answer:  2,800

Explanation:
Let 41 be Rounded to 40
Let 73 be Rounded to 70
Now, we have to find the result of 40 × 70.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
40 × 70 = 40 × (7 × 10)
= (40 × 7) × 10
= ( 4 × 10 × 7) × 10
= 280 × 10
= 2,800
B) By using the place-value method,
40 × 70 = 40 × 7 tens
= 4 tens ×7 tens
= 28 × 1 ten × 1ten
= 28 × 10 × 10
= 2,800
So,
41 ×73 can be rounded to 2,800

Question 8.
52 × 84

Answer: 4,250

Explanation:
Let 52 be Rounded to 50
Let 84 be Rounded to 85
Now, we have to find the result of 50 × 85.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
50 × 85 = 85 × (5× 10)
= (85 × 5) × 10
= ( 17× 5 × 5) × 10
= 425 × 10
= 4,250
B) By using the place-value method,
50 × 85 = 85 × 5 tens
= 425 tens
= 425 × 10
= 4,250
So,
52 ×84 can be rounded to 4,250

Question 9.
26 × 68

Answer: 1,750

Explanation:
Let 26 be Rounded to 25
Let 68 be Rounded to 70
Now, we have to find the result of 25 × 70.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
25 × 70 = 25 × (7 × 10)
= (25 × 7) × 10
= ( 5 × 5 × 7) × 10
= 175 × 10
= 1,750
B) By using the place-value method,
25 × 70 = 25 × 7 tens
= 175  tens
= 175 × 10
= 1,750
So,
26 ×68 can be rounded to 1,750

Question 10.
38 × 17

Answer: 600

Explanation:
Let 38 be Rounded to 40
Let 17 be Rounded to 15
Now, we have to find the result of 40 × 15.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
15 × 40 = 15 × (4× 10)
= (15 × 4) × 10
= ( 5 × 3 × 4) × 10
= 60 × 10
= 600
B) By using the place-value method,
15 × 40 = 15 × 4 tens
= 60  tens
= 60 × 10
= 600
So,
38 ×17 can be rounded to 600

Question 11.
75 × 24

Answer: 1,875

Explanation:
Let 24 be Rounded to 25
Now, we have to find the result of 25 × 75.
We can find the Product by using the simplification method.
25 × 75
= 5 × 5 × 25 × 3
= 5 × 5× 5 × 5 × 3
= 25 × 25 × 3
= 625 × 3
= 1,875
So,
38 × 17 can be Rounded to 1,875

Question 12.
93 × 53

Answer: 4,500

Explanation:
Let 93 be Rounded to 90
Let 53 be Rounded to 50
Now, we have to find the result of 90 × 50.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
90 × 50 = 90 × (5 × 10)
= (90 × 5) × 10
= ( 5 × 2 × 9 × 5) × 10
= 450 × 10
= 4,500
B) By using the place-value method,
90 × 50 = 90 × 5 tens
= 9 tens × 5 tens
= 45 × 1 ten × 1 ten
= 45 × 10 × 10
= 4,500
So,
93 ×53 can be rounded to 4,500

Question 13.
44 × 78

Answer: 3,600

Explanation:
Let 44 be Rounded to 45
Let 78 be Rounded to 80
Now, we have to find the result of 45 × 80.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
45 × 80 = 45 × (8 × 10)
= (45 × 8) × 10
= ( 5 × 2 × 9 × 4) × 10
= 360 × 10
= 3,600
B) By using the place-value method,
45 × 80 = 45 × 8 tens
= 5 × 9 × 8 tens
= 45 × 8 × 10
= 360 × 10
= 3,600
So,
44 ×78 can be rounded to 4,500

Question 14.
21 × 33

Answer: 600

Explanation:
Let 21 be Rounded to 20
Let 33 be Rounded to 30
Now, we have to find the result of 20 × 30.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
20 × 30 = 20 × (3× 10)
= (20 × 3) × 10
= ( 5× 4 × 3) × 10
= 60 × 10
= 600
B) By using the place-value method,
20 × 30 = 30 × 2 tens
= 3 tens × 2 tens
= 6 × 10 × 10
= 600
So,
21 ×33 can be rounded to 600

Question 15.
45 × 45

Answer: 2,500

Explanation:
Let 45 be Rounded to 50
Now, we have to find the result of 50 × 50.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
50 × 50 = 50 × (5× 10)
= (50 × 5) × 10
= ( 10× 5 × 5) × 10
= 250 × 10
= 2,500
B) By using the place-value method,
50 × 50 = 50 × 5 tens
= 250 tens
= 25 × 10 × 10
= 2,500
So,
45 ×45 can be rounded to 2,500
Open-Ended
Write two possible factors that can be estimated as shown.

Question 16.
2,400
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 6

Answer:

Explanation:
The given number is: 2,400
The Products of 24 are:
4 × 6 = 24
6 × 4 =24
From the above two products, we can conclude that the two possible numbers that can give the product 2,400 are: 40, 60

Question 17.
1,200
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 7

Answer:              

Explanation:

Explanation:
The given number is: 1,200
The Products of 12 are:
3 × 4 =12
4 × 3 =12
6 ×2 =12
2 ×6 =12
From the above two products, we can conclude that the two possible numbers that can give the product 2,400 are: 40, 30, and 20, 60

Question 18.
DIG DEEPER!
You use 50 × 30 to estimate 46 × 29. Will your estimate be greater than or less than the actual product? Explain.

Answer: We will Estimate the Product greater than the actual Product

Explanation:
Given Product is: 46 × 29

Explanation:
Let 46 be Rounded to 50
Let 29 be Rounded to 30
Now, we have to find the result of 50 × 30.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
50 × 30 = 50 × (3 × 10)
= (50 × 3) × 10
= ( 5 × 10 × 3) × 10
= 150 × 10
= 1,500
B) By using the place-value method,
50 × 30 = 30 × 5 tens
= 150  tens
= 150× 10
= 1,500
So,
46 ×29 can be rounded to 1,500

Question 19.
YOU BE THE TEACHER
Your friend uses rounding to estimate 15 × 72. She gets a product of 700. Is your friend’s estimate correct?
Explain.

Answer: No

Explanation:
Your friend is going to estimate 15 × 72 and she gets a product 700.
So, According to her Product, the Possible rounded off numbers to get the product 700 are 10 and 70
So, your Friend’s estimate is not correct.
Now,
Let 72 be Rounded to 70
Now, we have to find the result of 15 × 70.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
15 × 70 = 15 × (7 × 10)
= (15 × 7) × 10
= (3 × 5 × 7) × 10
= 105 × 10
= 1,050
B) By using the place-value method,
15 × 70 = 15 × 7 tens
= 105  tens
= 105 × 10
= 1,050
So,
15 ×72 can be rounded to 1,050
Think and Grow: Modeling Real Life
Example
About how much does 1 year of phone service cost?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 8
Think: What do you know? What do you need to find? How will you solve it?
There are 12 months in 1 year, so multiply the price per month by 12.
12 × 24 = 288
Estimate the product.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 9

Answer: 
Show and Grow
Question 20.
Use the table above. About how much does 1 year of Internet service cost? About how much does 1 year of cable television service cost?

Answer:
Internet Service cost: $540
Cable television service cost: $1068

Explanation:
In the given table,
The Internet cost and cable television service costs are given per month.
For 1 year, there are 12 months.
So,
The cost of Internet service for a year is: 12 × $45 = $ 540
The cost of cable television service for a year is: 12 × $89 = 1068

Question 21.
A giant panda eats 28 pounds of food each day. An orca eats 17 times as much food as the panda eats each day. About how much food does the orca eat each day?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 10

Answer: 476 Pounds

Explanation:
Given that a giant panda eats 28 pounds of food each day and an Orca eats 17 times as much food as the panda eats each day.
So,
The amount of food eaten by an Orca = 17 × The amount of food eaten by panda
= 17 × 28 = 476 pounds

Estimate Products Homework & Practice 4.2

Use rounding to estimate the product.

Question 1.
42 × 13

Answer:  600

Explanation:
Let 42 be Rounded to 40
Let 13 be Rounded to 15
Now, we have to find the result of 40 × 15.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
40 × 15 = 15 × (4 × 10)
= (15 × 4) × 10
= ( 3 × 4 × 5) × 10
= 60 × 10
= 600
B) By using the place-value method,
40 × 15 = 15 × 4 tens
= 60 tens
= 60 × 10
= 600
So,
42 × 13 can be rounded to 600

Question 2.
56 × 59

Answer: 3,300

Explanation:
Let 56 be Rounded to 55
Let 59 be Rounded to 60
Now, we have to find the result of 55 × 60.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
55 × 60 = 55 × (6 × 10)
= (55 × 6) × 10
= ( 5 × 11 × 6) × 10
= 330 × 10
= 3,300
B) By using the place-value method,
55 × 60 = 55 × 6 tens
= 330 tens
= 330 × 10
= 3,300
So,
42 × 16 can be rounded to 600

Question 3.
19 × 91

Answer: 1,800

Explanation:
Let 19 be Rounded to 20
Let91 be Rounded to 90
Now, we have to find the result of 20 × 90.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
20 × 90 = 20 × (9× 10)
= (20 × 9) × 10
= ( 5 × 4 × 9) × 10
= 180 × 10
= 1,800
B) By using the place-value method,
20 × 90 = 20 × 9 tens
= 2 tens × 9 tens
= 18 × 1 ten × 1 ten
=18 × 10 × 10
= 18 × 100
= 1,800
So,
19 × 91 can be rounded to 1,800
Use compatible numbers to estimate the product.

Question 4.
23 × 78

Answer: 2,000

Explanation:
Let 23 be Rounded to 25
Let 78 be Rounded to 80
Now, we have to find the result of 25 × 80.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
25 × 80 = 25 × (8 × 10)
= (25 × 8) × 10
= ( 5 × 5 × 8) × 10
= 200 × 10
= 2,000
B) By using the place-value method,
25 × 80 = 25 × 8 tens
= 200 tens
= 200 × 10
= 2,000
So,
23 × 78 can be rounded to 2,000

Question 5.
67 × 45

Answer:3,150

Explanation:
Let 67 be Rounded to 70
Now, we have to find the result of 70 × 45.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
45 × 70 = 45 × (7 × 10)
= (45 × 7) × 10
= ( 5 × 9 × 7) × 10
= 315 × 10
= 3,150
B) By using the place-value method,
45 × 70 = 45 × 7 tens
= 315 tens
= 315 × 10
= 3,150
So,
67 × 45 can be rounded to 3,150

Question 6.
19 × 24

Answer: 500
Explanation;
Let 19 be Rounded to 20
Let 24 be Rounded to 25
Now, we have to find the result of 25 × 20.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
20 × 25 =25 × (2 × 10)
= (25 × 2) × 10
= ( 5 × 5 × 2) × 10
= 50 × 10
= 500
B) By using the place-value method,
25 × 20 = 25 × 2 tens
= 50 tens
= 50 × 10
= 500
So,
19 × 24 can be rounded to 500
Estimate the product.

Question 7.
84 × 78

Answer: 6,800

Explanation:
Let 84 be Rounded to 85
Let 78 be Rounded to 80
Now, we have to find the result of 85 × 80.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
85 × 80 = 85 × (8 × 10)
= (85 × 8) × 10
= ( 5 × 17 × 8) × 10
= 680 × 10
= 6,800
B) By using the place-value method,
85 × 80 = 85 × 8 tens
= 680 tens
= 680 × 10
= 6,800
So,
84 × 78 can be rounded to 600

Question 8.
92 × 34

Answer: 3,150

Explanation:
Let 92 be Rounded to 90
Let 34 be Rounded to 35
Now, we have to find the result of 90 × 35.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
90 × 35 = 35 × (9 × 10)
= (35 × 9) × 10
= ( 5 × 7 × 9) × 10
= 315 × 10
= 3,150
B) By using the place-value method,
90 × 35 = 35 × 9 tens
= 315 tens
= 315 × 10
= 3,150
So,
92 × 34 can be rounded to 3,150

Question 9.
57 × 81

Answer: 4,800

Explanation:
Let 57 be Rounded to 60
Let 81 be Rounded to 80
Now, we have to find the result of 80 × 60.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
80 × 60 = 80 × (6 × 10)
= (80 × 6) × 10
= ( 8 × 10 × 6) × 10
= 480 × 10
= 4,800
B) By using the place-value method,
80 × 60 = 80 × 6 tens
= 480 tens
= 480 × 10
= 4,800
So,
57 × 81 can be rounded to 600
Open-Ended
Write two possible factors that could be estimated as shown.

Question 10.
6,400
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 11

Answer:    

Explanation:
The Products of 64 are:
8 × 8 = 64
16 × 4 =64
From the above two products, we can conclude that the two possible numbers that can give the product 6,400 are: 80,80 and. 160,40

Question 11.
1,600
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 12

Answer:

Explanation:
The Products of 16 are:
4 × 4 = 16
8 × 2 =16
From the above two products, we can conclude that the two possible numbers that can give the product 1,600 are: 40, 40 and, 80,20

Question 12.
Reasoning
Are both Newton’s and Descartes’s estimates reasonable? Explain.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 13

Answer: Both Newton’s and Descartes’s estimates are reasonable

Explanation:
According to Newton,
The estimated values of 27 and 68 are 30 and 70
According to Descartes,
The estimated values of 27 and 68 are 25 and 70
According to Newton:
27 × 68 = 2,100
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
30 × 70 = 30 × (7 × 10)
= (30 × 7) × 10
= ( 3 × 10 × 7) × 10
= 210 × 10
= 2,100
B) By using the place-value method,
30× 70 = 30 ×7 tens
= 210 tens
= 210 × 10
= 2,100
So,
27 × 68 can be rounded to 2,100 ( According to Newton)
According to Descartes:
27 × 68 = 1,750
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
25 × 70 = 25 × (7 × 10)
= (25 × 7) × 10
= ( 5 × 5 × 7) × 10
= 175 × 10
= 1,750
B) By using the place-value method,
25× 70 = 25 ×7 tens
= 175 tens
= 175 × 10
= 1,750
So,
27 × 68 can be rounded to 1,750 ( According to Descartes)

Question 13.
DIG DEEPER!
You use 90 × 30 to estimate 92 × 34. Will your estimate be greater than or less than the actual product? Explain.

Answer: No
Your friend is going to estimate 92 × 34 and she gets a product  2,700.
So, your Friend’s estimate is not correct.
Now,
Let 34 be Rounded to 35
Let 92 be Rounded to 90
Now, we have to find the result of 35 × 90.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
35 × 90 = 35 × (9 × 10)
= (35 × 9) × 10
= (9× 5 × 7) × 10
= 315 × 10
= 3,150
B) By using the place-value method,
35 × 90 = 35 × 9 tens
= 315  tens
= 315 × 10
= 3,150
So,
92 ×34 can be rounded to 3,150

Question 14.
Modeling Real Life
About how many hours of darkness does Barrow, Alaska have in December?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 14

Answer: 744 hours

Explanation:
From the above table,
The days of darkness in Barrow, Alaska are 31 days.
We know that there are 24 hours in a day.
So,
The number of hours of darkness does Barrow, Alaska have in December = 31 × 24 = 744 hours
Review & Refresh

Question 15.
Round 253,490 to the nearest ten thousand.

Answer: 250,000
Explanation;
The position of a given number is dependent on the place-value of that number.
So,
When 253,490 rounded off to the nearest ten thousand, the result ts 250,000

Question 16.
Round 628,496 to the nearest hundred thousand.

Answer: 630,000
Explanation;
The position of a given number is dependent on the place-value of that number.
So,
When 628,496 rounded off to the nearest ten thousand, the result ts 630,000

Lesson 4.3 Use Area Models to Multiply Two-Digit Numbers

Explore and Grow
Draw an area model that represents 15 × 18. Then break apart your model into smaller rectangles.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 1
What is the total area of your model? Explain how you found your answer.

Answer: The Total Area of your Model = 400

Explanation:

Count the number of boxes in both the vertical and horizontal directions.
Since all the sides of the figure look the same, take any 1 row in the vertical direction and horizontal direction and count the number of boxes in it.
After counting, you get
The number of boxes in a row present in the vertical direction is: 20
The number of boxes in a row present in the horizontal direction is: 20
So, we can find the total area of your model by multiplying the number of boxes in both the vertical and horizontal directions. ( Remember, there is no need to count all the boxes in both directions since the figure has an equal number of boxes on all sides)
So,
The total area of your model = 20 × 20 = 400
Reasoning
Compare with a partner. Do you get the same answer? Explain.

Answer: Yes

Explanation:
My Partner counted the number of boxes in the middle of the model in both vertical and horizontal directions.
He got,
The number of boxes in a row present in the vertical direction is: 20
The number of boxes in a row present in the horizontal direction is: 20
So,
The total area of my partner’s model = 20 × 20 = 400
So,
My partner and I got the same answer.
Think and Grow: Use Area Models to Multiply
Example
Use an area model and partial products to find 12 × 14.
Model the expression. Break apart 12 as 10 + 2 and 14 as 10 + 4.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 2
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 3

Answer: 168

Explanation:
So, 12 × 14 = 168
Show and Grow
Use the area model to find the product.

Question 1.
17 × 15 = _____
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 4

Answer: 255

Explanation:
So, 17 × 15 = 255

Question 2.
34 × 22 = _____

Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 5

Answer: 748

Explanation:
  So, 34 × 22 = 748
Apply and Grow: Practice
Use the area model to find the product.

Question 3.
13 × 19 = _____
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 6

Answer: 247

Explanation:
So, 13 × 19 = 247
Now,
100 + 90 + 30 + 27 = 247
So,
                 13 × 19 = 247

Question 4.
25 × 39 = ____
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 7

Answer: 975

Explanation:

Now,
600 + 150 + 180 + 45 = 975
So,
25 × 39 = 975
Draw an area model to find the product.

Question 5.
11 × 13 = ______
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 8

Answer:  143

Explanation:
By using the Partial Products method,
10 × 6 + 3 × 6 + 10 × 5 + 3 × 5
= 60 + 18 + 50 + 15 = 143
So,
11 × 13 = 143

Question 6.
23 × 26 = ______

Answer: 598

Explanation:

By using the Partial products method,
20 × 20 + 3 × 20 + 20 × 6 + 3 × 6
= 400 + 60 + 120 + 18 = 598
So,  23 × 26 = 598

Question 7.
27 × 45 = ______

Answer: 1,215

Explanation:

By using the partial products method,
20 × 20 + 7 × 20 + 20 × 25 + 7 ×25
= 400 + 140 +500 + 175 = 1,215
So, 27 × 45 = 1,215

Question 8. Perseid meteors travel 59 kilometers each second. How far does a perseid meteor travel in 15 seconds?
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 9

Answer: 885 kilometers

Explanation:
Given that the Perseid meteors travel 59 kilometers each second.
So,
The distance traveled by the Perseid meteors in 15 seconds = 59 × 15 = 885 kilometers
From the above,
We can conclude that the Perseid meteors travel 885 kilometers in 15 seconds.

Question 9.
DIG DEEPER!
Write the multiplication equation represented by the area model.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 10

Answer:

Explanation:
Using the Partial Products Model,
40 × 30 + 40 × 2 + 3 × 30 + 3 × 2
= 1,200 + 80 + 90 + 6 = 1,376
So, the multiplication equation represented by area model = 43 × 32
Think and Grow: Modeling Real Life
Example
A wind farm has 8 rows of new wind turbines and 3 rows of old wind turbines. Each row has 16 turbines. How many turbines does the wind farm have?
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 11

Answer: Add the number of rows of new turbines to the number of rows of old turbines
8 + 3 =11
So,
There are11 rows of turbines.
Multiply the number of rows by the number in each row.

So, the wind farm has 176  turbines.
Show and Grow

Question 10.
You can type 19 words per minute. Your cousin can type 33 words per minute. How many more words can your cousin type in 15 minutes than you?

Answer: 210 words

Explanation:
Given that you can type 19 words per minute and your cousin can type 33 words per minute.
So,
The number of words you can type in 15 minutes = 19 × 15 = 285
The number of words your cousin can type in 15 minutes = 495
Hence,
The number of words you have to type more than your cousin in 15 minutes = 495 – 285 =  310 words

Question 11.
A store owner buys 24 packs of solar eclipse glasses. Each pack has 12 glasses. The store did not sell 18 of the glasses. How many of the glasses did the store sell?
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 13

Answer: 270 glasses

Explanation:
Given that a store Owner buys 24 packs of solar eclipse glasses and each pack has 12 glasses.
So,
the total number of glasses that a store owner buy = 24 × 12 = 288 glasses
But,
It is also given that the store owner did not sell 18 glasses.
Hence, the total number of glasses the store owner sold = 288 – 18 = 270 glasses

Use Area Models to Multiply Two-Digit Numbers Homework & Practice 4.3

Use the area model to find the product.

Question 1.
12 × 13 = ______
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 14

Answer: 156

Explanation:
By using the partial products method,
10 × 10 + 2 ×10 + 10 × 3 + 2 ×3
= 100 + 20 +30 + 6 = 156
So, 12 × 13 = 156

Question 2.
38 × 24 = _____
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 15

Answer: 912

Explanation:
By using the Partial products method,
30 × 20 + 8 × 20 + 30 × 4 + 8 × 4
= 600 + 160 + 120 + 32 = 912
So,  38 × 24 = 912
Use the area model to find the product.

Question 3.
19 × 18 = ____
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 16

Answer: 342

Explanation:

By using the partial products method,
10 × 10 +8 × 10 + 10 × 9 +9 ×8
= 100 + 80 +90 + 72 = 342
So, 19 × 18 = 342

Question 4.
23 × 25 = _____
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 17

Answer: 575

Explanation:
By using the partial products method,
20 × 20 +5 × 20 + 3 × 20 +3 ×5
= 400 + 100 +60 + 15 = 575
So, 23 × 25 = 575
Draw an area model to find the product.

Question 5.
26 × 31 = _____

Answer: 806

Explanation:

By using the partial products method,
20 × 30 +6× 30 + 1 × 20 +6 ×1
= 600 + 180 +20 + 6 = 806
So, 26 × 31 = 806

Question 6.
22 × 47 = ______

Answer: 1,034

Explanation:

By using the partial products method,
20 × 40 +2×40 + 7 × 20 +2 ×7
= 800 + 80 +140 + 14 = 1,034
So, 22 × 47 = 1,034

Question 7.
YOU BE THE TEACHER
Your friend finds 12 × 42. Is your friend correct? Explain.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 18

Answer: Yes, your friend is correct.

Explanation:

By using the partial products method,
10 × 40 +2×10 + 2 × 40 +2 ×2
= 400 + 20 +80 + 4 = 504
So, 12 × 42 = 504

Question 8.
Writing
Explain how to use an area model and partial products to multiply two-digit numbers.

Answer:
Let the Partial Products are: a, b, c, d
By using the Partial Products method,
a × c + b × c + a × d + b × d
= ac + bc + ad + bd
So, by using the above method, we can find the product of 2- digit numbers.

Question 9.
Modeling Real Life
A mega-arcade has 9 rows of single-player games and 5 rows of multi-player games. Each row has 24 games. How many games does the arcade have?
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 19

Answer:  The arcade has 336 games

Explanation:
Given that a mega arcade has 9 rows of single-player games and 5 rows of multi-player games.
So,
Total number of rows present in the arcade = 9 + 5 = 14
It is also given that each row has 24 games.
So,
The total number of games present in the arcade = 14 × 24 = 336

By using the partial products method,
10 × 20 +4 × 20 + 10 × 4 + 4 ×4
= 200 + 80 +40 + 16 = 336
So, 14 × 24 = 336
Review & Refresh
Find the sum. Check whether your answer is reasonable.

Question 10.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 20

Answer: 84,016

Explanation:
To find the sum, add the digits starting from the Right-most position. If there is “Carry”, then add that carry to the result of the next Position value.

Question 11.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 21

Answer: 71,585

Explanation:
To find the sum, add the digits starting from the Right-most position. If there is “Carry”, then add that carry to the result of the next Position value. (As shown in the above figure)

Question 12.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 22

Answer: 569,821

Explanation:

To find the sum, add the digits starting from the Right-most position. If there is “Carry”, then add that carry to the result of the next Position value. (As shown in the above figure)

Lesson 4.4 Use the Distributive Property to Multiply Two-Digit Numbers

Explore and Grow
Use as few base ten blocks as possible to create an area model for 13 × 24. Draw to show your model.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 1
Color your model to show four smaller rectangles. Label the partial products.

Answer:
By using the partial products method,
10 × 20 +3 × 20 + 4 × 3 + 10 ×4
= 200 + 60 +12 + 40 = 312
So, 13 × 24 = 312
Reasoning
How do you think the Distributive Property relates to your area model? Explain.

Answer:
Distributive Property of Multiplication:
Let there are 3 numbers a, b, c.
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)
SO, by using the above property, we can conclude that the Distributive Property relates to your area model.
Think and Grow: Use the Distributive Property to Multiply
Example
Find 17 × 25.
One Way: Use an area model and partial products.


Show and Grow

Question 1.
Use the area model and the Distributive Property to find 32 × 19.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 3

Answer: 608

Explanation:
Using the Distributive Property, we can find the product of 32 × 19
32 × 19 = 32 × ( 10 + 9)
= ( 32 × 10 ) + ( 32 × 9)
= ( 30 + 2 ) × 10 + ( 30 + 2 ) × 9
= ( 30 × 10 ) + ( 2 × 10 ) + ( 30 × 9 ) + ( 2 × 9)
= 300 + 20 + 270 + 18
= 608
So, 32 × 19 = 608
Apply and Grow: Practice

Question 2.
Use the area model and the Distributive Property to find 34 × 26.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 4

Answer: 884

Explanation:
Using the Distributive Property, we can find the product of 34 × 26
34 × 26 = 34 × ( 20 + 6)
= ( 34 × 20 ) + ( 34 × 6)
= ( 30 + 4 ) × 20 + ( 30 + 4 ) × 6
= ( 30 × 20 ) + ( 4 × 20 ) + ( 30 × 6 ) + ( 4 × 6)
= 600 + 80 + 180 + 24
= 884
So, 34 × 26 = 884
Use the Distributive Property to find the product.

Question 3.
28 × 47 = 28 × (40 + 7)
= (28 × 40) + (28 × 7)
= (20 + 8) × 40  + (20 + 8) × 7
= (20 × 40) + (8 × 40) + (20 × 7) + (8 × 7)
= 800 + 320 + 140 + 56
=1,316
So, 28 × 47 = 1,316

Answer: 28 × 47 = 1,316

Question 4.
39 × 41 = _____

Answer:
39 × 41 = 39 × (40 + 1)
= (39 × 40) + (39 × 1)
= (30 + 9) × 40  + (30 + 9) × 1
= (30 × 40) + (9 × 40) + (30 × 1) + (9 ×1)
=1,200 + 360 + 30 + 9
=1,599
So, 39 × 41 = 1,599

Question 5.
74 × 12 = ______

Answer:
74 × 12 = 74 × (10 + 2)
= (74 × 10) + (74 × 2)
= (70 + 4) × 10  + (70 + 4) × 2
= (70 × 10) + (4 × 10) + (70 × 2) + (4 ×2)
=700 + 40 + 140 + 8
=888
So, 74 × 12 = 888

Question 6.
83 × 65 = _____

Answer:
83 × 65 = 83 × (60 + 5)
= (83 × 60) + (83 × 5)
= (80 + 3) × 60  + (80 + 3) ×5
= (80 × 60) + (3 × 60) + (80 × 5) + (3 ×5)
=4,800 + 180 + 400 + 15
=5,395
So, 83 × 65 = 5,395

Question 7.
Which One Doesn’t Belong?
Which expression does not belong with the other three?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 5

Answer:
Let the given Expressions be ordered as A), B), C) and D)
From the Order, we can say that Expression C) does not belong to the other three.

Explanation:
The given Expressions are:
A) ( 40 + 7) × 52
B) ( 40 + 7) × (50 + 2)
C) ( 40 × 7) × ( 50 × 2)
D) 47 × ( 50 + 2)
The given Expressions are written using the Distributive Property of Multiplication.
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)
So, from the above Property, we can conclude that Expression C) does not belong to the other three.
Think and Grow: Modeling Real Life
Example
The dunk tank at a school fair needs 350 gallons of water. There are 27 students in a class. Each student pours13 gallons of water into the tank. Is there enough water in the dunk tank?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 6
Find how many gallons of water the students put in the dunk tank.

Given that there are 350 gallons at a school fair.
But, we got 351 gallons of water.
So, there is 1 gallon enough water in the dunk tank.
Show and Grow

Question 8.
An event coordinator orders 35 boxes of T-shirts to give away at a baseball game. There are 48 T-shirts in each box. If 2,134 fans attend the game, will each fan get a T-shirt?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 8

Answer: No, each fan will not get a T-shirt.

Explanation:
Given that an event coordinator orders 35 boxes of T-shirts to give away at a baseball game and there are 48 T-shirts in each box.
So, we will get the total number of T-shirts due to the Product of 35 × 48.
We will get the product by using the Distributive Property of Multiplication.
35 × 48 = 35 × (40 + 8)
= (35 × 40) + (35 × 8)
= (30 + 5) × 40  + (30 + 5) × 8
= (30 × 40) + (5 × 40) + (30 × 8) + (8 ×5)
=1,200 + 20 + 240 + 40
=1,680
So, 35× 48 = 1,680
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 9.
A horse owner must provide 4,046 square meters of pasture for each horse. Is the pasture large enough for 2 horses? Explain.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 9

Answer: Yes, the pasture is large enough for 2 horses.

Explanation:
Given that a horse owner must provide 4,046 square meters for each horse.
But, it is also given that the area of pasture is 86 × 96 square meters.
We have to find the Product of 86 × 96 by using the Distributive Property of Multiplication..
86 × 96 = 86 × (90 + 6)
= (86 × 90) + (86 × 6)
= (80 + 6) × 90  + (80 + 6) × 6
= (80 × 90) + (6 × 90) + (80 × 6) + (6 ×6)
=7,200 + 540 + 480 + 36
=8,256
So, 86× 96 = 8,256
By comparing the area of the pasture of each horse and the Product, we can conclude that the pasture is large enough for the 2 horses.
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Use the Distributive Property to Multiply Two-Digit Numbers Homework & Practice 4.4

 

Question 1.
Use the area model and the Distributive Property to find 45 × 21.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 10

Answer: 945

Explanation: By using the Distributive Property of Multiplication,
45 × 21 = 45 × (20 + 1)
= (45 × 20) + (45 × 1)
= (40 + 5) × 20  + (40 + 5) ×1
= (20 × 40) + (5 × 20) + (40 × 1) + (1 ×5)
=800 + 100 + 40 + 5
=945
So, 45× 21 = 945
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 2.
Use the Distributive Property to find the product.
34 × 49 = 34 × (40 + 9)
= (34 × 40) + (34 × 9)
= (30 + 4) × 40 + (30 + 4) × 9
= (30 × 40) + (4 × 40) + (30 × 9) + (4 × 9)
= 1,200 +160 + 270 + 36
= 1,666
So, 34 × 49 =1,666

Question 3.
14 × 27 = ______

Answer: 378

Explanation: Using the Distributive Property of Multiplication,
14 × 27 = 14 × (20 + 7)
= (14 × 20) + (14 × 7)
= (10 + 4) × 20  + (10 + 4) × 7
= (10 × 20) + (4 × 20) + (10 × 7) + (4 ×7)
=200 + 80 + 70 + 28
=378
So, 14× 27 = 378
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 4.
38 × 31 = ______

Answer: 1,178

Explanation: Using the Distributive Property of Multiplication,
38 × 31 = 38 × (30 + 1)
= (38 × 30) + (38 × 1)
= (30 + 8) × 30  + (30 + 8) ×1
= (30 × 30) + (8× 30) + (30 × 1) + (8 ×1)
=900 + 240 + 30 + 8
=1,178
So, 38× 31 = 1,178
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 5.
58 × 26 = ______

Answer: 1,508

Explanation: Using the Distributive Property of Multiplication,
58 × 26 = 58 × (20 + 6)
= (58 × 20) + (58 × 6)
= (50 + 8) × 20  + (50 + 8) × 6
= (50 × 20) + (8 × 20) + (50 × 6) + (8 ×6)
=1,000 + 160 + 300 + 48
=1,508
So, 58× 26 = 1,508
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 6.
56 × 32 = ______

Answer: 1,792

Explanation: Using the Distributive Property of Multiplication,
56 × 32 = 56 × (30 + 2)
= (56 × 30) + (56× 2)
= (50 + 6) × 30  + (50 + 6) × 2
= (30 × 50) + (6 × 30) + (50 × 2) + (6 ×2)
=1,500 + 180 + 100 + 12
=1,792
So, 56× 32 = 1,792
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 7.
87 × 23 = ______

Answer: 2,001

Explanation: Using the Distributive Property of Multiplication,
87 × 23 = 87 × (20 + 3)
= (87 × 20) + (87 × 3)
= (80 + 7) × 20  + (80 + 7) × 3
= (80 × 20) + (7 × 20) + (3 × 80) + (7 ×3)
=1,600 + 140 + 240 + 21
=2,001
So, 87× 23 = 2,001
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 8.
95 × 81 = ______

Answer: 7,695

Explanation: Using the Distributive Property of Multiplication,
95 × 81 = 95 × (80 + 1)
= (95 × 80) + (95 × 1)
= (90 + 5) × 80  + (90 + 5) × 1
= (90 × 80) + (5 × 80) + (90 × 1) + (1 ×5)
=7,200 + 400 + 90 + 5
=7,695
So, 95× 81 = 7,695
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 9.
DIG DEEPER!
Find 42 × 78 by breaking apart 42 first.

Answer: 3,276

Explanation: Using the Distributive Property of Multiplication,
42 × 78 = 78 × (40 + 2)
= (78 × 40) + (78 × 2)
= (70 + 8) × 40  + (70 + 8) × 2
= (70 × 40) + (8 × 40) + (70 × 8) + (8 ×2)
=2,800 + 320 + 560 + 16
=3,276
So, 42× 78 = 3,276
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 10.
Modeling Real Life
The Elephant Building is 335 feet high. A real Asian elephant is 12 feet tall. If 29 real elephants could stand on top of each other, would they reach the top of the building?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 11

Answer: The 29 real elephants would reach the top of the building even when they stand on top of each other.

Explanation:
Give that the Elephant Building is 335 feet high. It is also given that a real Asian Elephant is 12 feet tall and there are 29 real elephants.
So, we have to find the height of 29 real elephants. We can find it using the product of 29 × 12.
We find the product by using the Distributive Property of Multiplication.
29 × 12 = 12 × (20 + 9)
= (12 × 20) + (12 × 9)
= (10 + 2) × 20  + (10 + 2) × 9
= (10 × 20) + (2 × 20) + (10 × 9) + (2 ×9)
=200 + 40 + 90 + 18
=348
So, 29× 12 = 348
We get the result of the Product as 348 feet but given that the Elephant Building is 335 feet high.
From this, we can conclude that if 29 real elephants could stand on top of each other,  they would reach the top of the building
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)
Review & Refresh
Find the difference. Then check your answer.

Question 11.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 12

Answer: 25,259

Explanation:

The difference between the multi-digit numbers can be found by the difference taken from the left-most digit. If the number we want to subtract is less than the number to be subtracted, then we will take the carry from the Previous digit and the Previous digit contains 1 less number.

Question 12.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 13

Answer: 53,162

Explanation:
The difference between the multi-digit numbers can be found by the difference taken from the left-most digit. If the number we want to subtract is less than the number to be subtracted, then we will take the carry from the Previous digit and the Previous digit contains 1 less number.

Question 13.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 14

Answer: 140,938

Explanation:
The difference between the multi-digit numbers can be found by the difference taken from the left-most digit. If the number we want to subtract is less than the number to be subtracted, then we will take the carry from the Previous digit and the Previous digit contains 1 less number.

Lesson 4.5 Use Partial Products to Multiply Two-Digit Numbers

Explore and Grow

How can you use the rectangles to find 24 × 53? Complete the equation.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 1
24 × 53 = _____

Answer: 1,272

Explanation:
By using the partial products method,
24 × 53 = 50 × 20 +3 × 20 + 4 × 3 + 50 ×4
= 1,000 + 60 +12 + 200 = 1,272
So, 53 × 24 = 1,272
Reasoning
What does the area of each rectangle represent.

Answer:
Think and Grow: Use Partial Products to Multiply Two-DigitNumbers
Example
Use place value and partial products to find 27 × 48.
Estimate: 30 × 50 = 1,500

So, 27 × 48 =1296
Check: Because 1,296 is close to the estimate, 1,500, the answer is reasonable.
Show and Grow
Find the product. Check whether your answer is reasonable.

Question 1.
Estimate: ______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 3

Answer:  585

Explanation:
Using the Partial Products method,
39 × 15 = ( 30 + 9) × ( 10 + 5)
= 30 × 10 + 9 × 10 + 30 × 5 + 9 × 5
= 300 + 90 + 150 + 45
= 585
Estimate:
Let 39 be Rounded to 40.
So, 40 × 15 = 600
As the Estimate and the actual answer are near, the answer is reasonable.

Question 2.
Estimate: ______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 4

Answer: 5,166

Explanation:
Using the Partial Products method,
82 × 63 = ( 80 + 2) × ( 60 + 3)
= 80 × 60 + 2 × 60 + 80 × 3 + 2 × 3
= 4,800 + 120 + 240 + 6
= 5,166
Estimate:
Let 82 be Rounded to 80.
Let 63 be Rounded to 65.
So, 80 × 65 = 5,200
As the Estimate and the actual answer are near, the answer is reasonable.

Question 3.
Estimate: _______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 5

Answer: 3,976

Explanation:
Using the Partial Products method,
56 × 71 = ( 50 + 6) × ( 70 + 1)
= 50 × 70 + 1 × 50 + 70 × 6 + 6 × 1
= 3,500 + 50 + 420 + 6
= 3,976
Estimate:
Let 56 be Rounded to 55.
Let 71 be Rounded to 70.
So, 55 × 70 = 3,850
As the Estimate and the actual answer are not  near, the answer is not  reasonable.
Apply and Grow: Practice
Find the product. Check whether your answer is reasonable.

Question 4.
Estimate: _____
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 6

Answer: 364

Explanation:
Using the Partial Products method,
14 × 26 = ( 10 + 4) × ( 20 + 6)
= 10 × 20 + 10 × 6 + 20 × 4 + 4 × 6
= 200 + 60 + 80 + 24
= 364
Estimate:
Let 26 be Rounded to 25.
Let 14 be Rounded to 15.
So, 25 × 15 = 375
As the Estimate and the actual answer are near, the answer is reasonable.

Question 5.
Estimate: _____
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 7

Answer: 1,767

Explanation:
Using the Partial Products method,
57 × 31 = ( 50 + 7) × ( 30 + 1)
= 50 × 30 + 1 × 50 + 30 × 7 + 7 ×1
= 1,500 + 50 + 210 + 7
= 1,767
Estimate:
Let 57 be Rounded to 55. (or) Let 57 be Rounded to 60.
Let 31 be Rounded to 30.
So, 30 × 55 = 1,650 (or) 30 × 60 = 1,800
As the Estimate and the actual answer are near, the answer is reasonable. ( Depending on the Estimate value).

Question 6.
Estimate: _______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 8

Answer: 2,116

Explanation:
Using the Partial Products method,
23 × 92 = ( 20 +3) × ( 90 + 2)
= 20 × 90 + 2 × 20 + 90 × 3 + 3 ×2
= 1,800 + 40 + 270 + 6
= 2,116
Estimate:
Let 23 be Rounded to 25.
Let 92 be Rounded to 90.
So, 90 × 25 = 2,250 
As the Estimate and the actual answer are near, the answer is reasonable.

Question 7.
Estimate: ______
13 × 98 = ______

Answer: 1,274

Explanation:
Using the Partial Products method,
13 × 98 = ( 10 + 3) × ( 90 +8)
= 10 × 90 + 3 × 90 + 10 × 8 + 3 ×8
= 900 + 270 + 80 + 24
= 1,274
Estimate:
Let 13 be Rounded to 15.
Let 98 be Rounded to 100.
So, 15 × 100 = 1,500 
As the Estimate and the actual answer are not near, the answer is not  reasonable.

Question 8.
Estimate: ______
65 × 22 = ______

Answer: 1,430

Explanation:
Using the Partial Products method,
65 × 22 = ( 60 + 5) × ( 20 + 2)
= 60 × 20 + 60 × 2 + 20 × 5 + 5 ×2
= 1,200 + 120 + 100 + 10
= 1,430
Estimate:
Let 22 be Rounded to 20.
So, 20 × 65 = 1,300
As the Estimate and the actual answer are not  near, the answer is not reasonable.

Question 9.
Estimate: ______
72 × 81 = _____

Answer: 5,832

Explanation:
Using the Partial Products method,
72 × 81 = ( 70 + 2) × (80 + 1)
= 70 × 80 + 1 × 70 + 80 × 2 + 2 ×1
= 5,600 + 70 + 160 + 2
= 5,832
Estimate:
Let 72 be Rounded to 70.
Let 81 be Rounded to 80.
So, 70 × 80 = 5,600
As the Estimate and the actual answer are not near, the answer is not  reasonable.

Question 10.
A farmer has 58 cows. Each cow produces 29 liters of milk. How many liters of milk do the cows produce in all?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 9

Answer: 1,682 liters

Explanation:
Given that a farmer has 58 cows and each cow gives 29 liters of milk.
So, to find the total quantity of milk, we have to find the Product of 58 × 29 by using the Partial Products method.
Now,
Using the Partial Products method,
58 × 29 = ( 50 + 8) × ( 20 + 9)
= 50 × 20 + 9 × 50 + 20 ×8 + 8 ×9
= 1,000 + 450 + 160 + 72
= 1,682
From the above value, we can conclude that the total quantity of milk produces is 1,682 liters.

Question 11.
Number Sense
How much greater is the product of 12 and 82 than the product of 11 and 82? Explain how you know without multiplying.

Answer: The product of 12 and 82 is greater than 82 than the product of 11 and 82.

Explanation:
The given Products are
A) product of 12 and 82  B) product of 11 and 82
From the above, we can see that both the products have “82” as a Common number. So, find the difference between the remaining 2 numbers and we find that difference as ‘1’.
From this, we can conclude that the product of 12 and 82 is greater than 82 than the product of 11 and 82.
To verify this, we can find the products using the Partial Products method.
Now,
12 × 82 = ( 10 + 2) × ( 80 +2)
= 10 × 80 + 10 × 2 + 80 × 2 + 2 ×2
= 800 + 20 + 160 + 4
= 984
11 × 82 = ( 10 + 1) × ( 80 + 2)
= 10 × 80 + 10 × 2 + 80 ×1 + 2 × 1
= 800 + 20 + 80 + 2
= 902
Now, 984 – 902 = 82
By multiplication also, we can conclude that the product of 12 and 82 is greater than 82 than the product of 11 and 82.

Question 12.
DIG DEEPER!
Write the multiplication equation shown by the partial products.
200 + 60 + 50 + 15

Answer:  13 × 25

Explanation:
By using the Partial Products method,
200 + 60 + 50 + 15
= 20 × 10 + 20 × 3 + 5 ×10 + 5 × 3
= ( 10 + 3) × 20 +( 10 + 3) × 5
= ( 10 + 3) × ( 20 + 5)
= 13 × 25
Think and Grow: Modeling Real Life
Example
How many hours does a koala sleep in 2 weeks?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 10
Find how many hours a koala sleeps each day.

We know that,
1 week = 7 days
So,
2 weeks = 2 × 7 days = 14 days
Hence,
A koala sleeps 308 hours in 2 weeks.
Show and Grow

Question 13.
Use the table above. How many hours does a python sleep in 3 weeks?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 12

Answer: 378 hours

Explanation:
From the above table, we can conclude that the python sleeps for 18 hours a day.
We know that,
1 week = 7 days
So,
3 weeks = 3 × 7 = 21 days
Hence, to find how many hours a python sleep in 3 weeks, we have to find the product of 18 × 21.
Using the Partial Products method,
18 × 21 = ( 10 + 8) × ( 20 +1)
= 10 × 20 + 10 × 1 + 20 × 8 + 8 ×1
= 200 + 10 + 160 + 8
= 378
Hence, from the above,
we can conclude that the python sleeps for 378 hours in 3 weeks.

Question 14.
You have 12 packets of pea seeds and23 packets of cucumber seeds. How many fewer pea seeds do you have than cucumber seeds?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 13

Answer:  102 pea seeds are less than cucumber seeds

Explanation:
Given that there are 12 packets of pea seeds and 23 packets of cucumber seeds.
From the table,
The number of seeds in each packet of pea seeds = 12 × 6 = 72
The number of seeds in each packet of cucumber = 12 × 3 = 36 + 6 = 42
So,
The total number of seeds in pea = 72  × 12
The total number of seeds in cucumber = 42 × 23
By using the Partial Products method,
72 × 12 = ( 10 + 2) × ( 70 +2)
= 10 × 70 + 10 × 2 + 2 × 70 + 2 ×2
= 700 + 20 + 140 + 4
= 864
42 × 23 = ( 40 + 2) × ( 20 +3)
= 40 × 20 + 40 × 3 + 20 × 2 + 2 ×3
= 800 + 120 + 40 + 6
= 966
So,
The number of pea seeds less than the cucumber seeds = 966 – 864 = 102 seeds

Use Partial Products to Multiply Two-Digit Numbers Homework & Practice 4.5

Find the product. Check whether your answer is reasonable.

Question 1.
Estimate: _______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 14

Answer: 442

Explanation:
Using the Partial Products method,
17 × 26 = ( 10 + 7) × ( 20 + 6)
= 10 × 20 + 10 × 6 + 20 × 7 + 7 × 6
= 200 + 140 + 60 + 24
= 424
Estimate:
Let 17 be Rounded to 15.
Let 26 be Rounded to 25.
So, 25 × 15 = 375
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 2.
Estimate: _____
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 15

Answer: 2,356

Explanation:
Using the Partial Products method,
38 × 62 = ( 30 + 8) × ( 60 + 2)
= 30 × 60 + 30 × 2 + 60 × 8 + 8 × 2
= 1,800 + 60 + 480 + 16
= 2,356
Estimate:
Let 38 be Rounded to 40.
Let 62 be Rounded to 60.
So, 40 × 60 = 2,400
As the Estimate and the actual answer are near, the answer is reasonable.

Question 3.
Estimate: ______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 16

Answer: 3,913

Explanation:
Using the Partial Products method,
91 × 43 = ( 90 + 1) × ( 40 + 3)
= 90 × 40 + 90 × 3 + 40 × 1 + 3 × 1
= 3,600 + 270 + 40 + 3
= 3,913
Estimate:
Let 91 be Rounded to 90.
Let 43 be Rounded to 45.
So, 90 × 45 = 4,050
As the Estimate and the actual answer are not near, the answer is not reasonable.
Find the product. Check whether your answer is reasonable.

Question 4.
Estimate: _____
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 17

Answer: 3,774

Explanation:
Using the Partial Products method,
51 × 74 = ( 50 + 1) × ( 70 + 4)
= 50 × 70 + 50 × 4 + 70 × 1 + 4 × 1
= 3,500 + 200 + 70 + 4
= 3,774
Estimate:
Let 51 be Rounded to 50.
Let 74 be Rounded to 75.
So, 50 × 75 = 3,750
As the Estimate and the actual answer are near, the answer is reasonable.

Question 5.
Estimate: ______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 18

Answer: 532

Explanation:
Using the Partial Products method,
28 × 19 = ( 20 + 8) × ( 10 + 9)
= 10 × 20 + 20 × 9 + 10 × 8 + 9 × 8
= 200 + 180 + 80 + 72
= 532
Estimate:
Let 28 be Rounded to 30.
Let 19 be Rounded to 20.
So, 30 × 15 = 600
As the Estimate and the actual answer are near, the answer is reasonable.

Question 6.
Estimate: ______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 19

Answer: 1,575

Explanation:
Using the Partial Products method,
35 × 45 = ( 30 + 5) × ( 40 + 5)
= 30 × 40 + 30 × 5 + 40 × 5 + 5 × 5
= 1,200 + 150 + 200 + 25
= 1,575
There is no need for Estimate because they are already rounded numbers. So, the answer is reasonable.

Question 7.

Estimate: ______
82 × 63 = ______

Answer: 5,166

Explanation:
Using the Partial Products method,
82 × 63 = ( 80 + 2) × ( 60 + 3)
= 80 × 60 + 80 × 3 + 60 × 2 + 2 × 3
= 4,800 + 240 + 120 + 6
= 5,166
Estimate:
Let 82 be Rounded to 80.
Let 63 be Rounded to 65.
So, 80 × 65 = 5,200
As the Estimate and the actual answer are near, the answer is reasonable.

Question 8.
Estimate: ______
36 × 93 = ______

Answer: 3,348

Explanation:
Using the Partial Products method,
36 × 93 = ( 30 + 6) × ( 90 + 3)
= 30 × 90 + 30 × 3 + 90 × 6 + 3 × 6
= 2,700 + 90 + 540 + 18
= 3,348
Estimate:
Let 36 be Rounded to 35.
Let 93 be Rounded to 95.
So, 35 × 95 = 3,325
As the Estimate and the actual answer are near, the answer is reasonable.

Question 9.
Estimate: _______
57 × 22 = ______

Answer: 1,254

Explanation:
Using the Partial Products method,
57 × 22 = ( 50 + 7) × ( 20 + 2)
= 50 × 20 + 50 × 2 + 20 × 7 + 7 × 2
= 1,000 + 100 + 140 + 14
= 1,254
Estimate:
Let 57 be Rounded to 55.
Let 22 be Rounded to 20.
So, 55 × 20 = 1,100
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 10.
DIG DEEPER!
Find the missing digits. Then find the product.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 20

Answer: The missing digits are 1, 1

Explanation:
Using the Partial Products method,
100 + 50 + 60 + 30
= 10 × 10 + 10 × 5 + 10 × 6 + 5 × 6
= 10( 10 + 5) + 6( 10 + 5)
= ( 10 + 5) × ( 10 + 6)
= 15 × 16
So,
From the above, we can conclude that the missing digits are: 1, 1

Question 11.
Modeling Real Life
If Newton meets his goal each month, how many liters of water will he drink in 1 year?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 21

Answer: 396 liters of water
Explanation;
Given,
Each ♦ = 6 liters
From this,
Half ♦ = 3 liters
From the above table,
Monthly water Intake by Newton = 6 × 5 + 3 = 33 liters
So, The yearly intake by Newton can find out by the product of 33 × 12. ( Since a year has 12 months)
Using the Partial Products method,
33 × 12 = ( 30 + 3) × ( 10 + 2)
= 30 × 10 + 30 × 2 + 10 × 3 + 3 × 2
= 300 + 60 + 30 + 6
= 396 liters
From the above,
we can conclude that the yearly intake by Newton is: 396 liters

Question 12.
Modeling Real Life
Use the table in Exercise 11. If you and Descartes each meet your goal each month, how many more liters will you drink in 1 year than Descartes?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 22

Answer: 468 liters
Explanation;
From the above table,
Given that,
The monthly intake of water by you = 6 × 7 + 3 = 45 liters
The monthly intake of water by Descartes = 6 liters
We know that 1 year consists of 12 months.
So,
The yearly intake of water by Descartes = 12 × 6 = 72 liters
The yearly intake of water by you = 45 × 12
We have to find 45 × 12 using the Partial Products method.
45 × 12 = ( 40 + 5) × ( 10 +2)
= 10 × 40 + 40 × 2 + 5 × 10 + 5 ×2
= 400 + 80 + 50 + 10
= 540 liters
Hence,
The amount of water you drink more than Descartes = 540 – 72 = 468 liters
Review & Refresh
Add or subtract. Then check your answer.

Question 13.
512,006 + 318,071 = ______

Answer: 830,077

Explanation:
To find the sum, add the digits starting from the Right-most position. If there is “Carry”, then add that carry to the result of the next Position value.

Question 14.
746,620 – 529,706 = ______

Answer: 216,914

Explanation:
The difference between the multi-digit numbers can be found by the difference taken from the left-most digit. If the number we want to subtract is less than the number to be subtracted, then we will take the carry from the Previous digit and the Previous digit contains 1 less number.

Lesson 4.6 Multiply Two-Digit Numbers

Explore and Grow

Use base ten blocks to model each product. Draw each model.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 1

Answer:
Reasoning
How are the models related to the product 17 × 13?

Answer:
The above model shows the pattern of the Partial Products method.
According to the Partial Products method,
(10 +7) × (10 + 3 ) ( As in the above model)
= 17 × 13
Think and Grow: Multiply Two-Digit Numbers

Show and Grow
Find the product. Check whether your answer is reasonable.

Question 1.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 3

Answer: 1,312

Explanation: Using the Partial Products method,
41 × 32 = ( 40 + 1) × ( 30 +2)
= 30 × 40 + 40 × 2 + 1 × 30 + 1 ×2
= 1,200 + 80 + 30 + 2
= 1,312
Estimate:
Let 41 be Rounded to 40.
Let 32 be Rounded to 30.
So, 40 × 30 = 1,200
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 2.
Estimate: _____
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 4

Answer: 2,392

Explanation: Using the Partial Products method,
52 × 46 = ( 50 + 2) × ( 40 +6)
= 50 × 40 + 50 × 6 +2 × 40 + 6 ×2
= 2,000 + 300 + 80 + 12
= 2,392
Estimate:
Let 52 be Rounded to 50.
Let 46 be Rounded to 45.
So, 45 × 50 = 2,250
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 3.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 5

Answer: 2,730

Explanation: Using the Partial Products method,
78 × 35 = ( 70 + 8) × ( 30 +5)
= 30 × 70 + 70 × 5 + 8 × 30 + 8 ×5
= 2,100 + 350 + 240 + 40
= 2,730
Estimate:
Let 78 be Rounded to 80.
So, 35 × 80 = 2,800
As the Estimate and the actual answer are  near, the answer is reasonable.
Apply and Grow: Practice
Find the product. Check whether your answer is reasonable.

Question 4.
Estimate: _____
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 6

Answer: 516

Explanation: Using the Partial Products method,
12 × 43 = ( 10 + 2) × ( 40 +3)
= 10 × 40 + 10 × 3 + 2 × 40 + 3 ×2
= 400 + 30 + 80 + 6
= 516
Estimate:
Let 12 be Rounded to 10.
Let 43 be Rounded to 45.
So, 45 × 10 = 450
As the Estimate and the actual answer are near, the answer is  reasonable.

Question 5.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 7

Answer: 1,992

Explanation: Using the Partial Products method,
83 × 24 = ( 80 + 3) × ( 20 +4)
= 80 × 20 + 80 × 4 + 3 × 20 + 3 ×4
= 1,600 + 320 + 60 + 12
= 1,992
Estimate:
Let 83 be Rounded to 85.
Let 24 be Rounded to 25.
So, 85 × 25 = 2,152
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 6.
Estimate: _____
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 8

Answer: 4,484

Explanation: Using the Partial Products method,
59 × 76 = ( 50 + 9) × ( 70 +6)
= 50 × 70 + 50 × 6 + 9 × 70 + 9 ×6
= 3,500 + 300 + 630 + 54
= 4,484
EStimate:
Let 59 be Rounded to 60.
Let 76 be Rounded to 75.
So, 60 × 75 = 4,500
As the Estimate and the actual answer are near, the answer is  reasonable.

Question 7.
Estimate: ______
22 × 41 = ______

Answer: 902

Explanation: Using the Partial Products method,
41 × 22 = ( 40 + 1) × ( 20 +2)
= 20 × 40 + 40 × 2 + 1 × 20 + 1 ×2
= 800 + 80 + 20 + 2
= 902
Estimate:
Let 41 be Rounded to 40.
Let 22 be Rounded to 20.
So, 40 × 20 = 800
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 8.
Estimate: _____
94 × 32 = ______

Answer: 3,008

Explanation: Using the Partial Products method,
94 × 32 = ( 90 + 4) × ( 30 +2)
= 30 × 90 + 90 × 2 + 4 × 30 + 4 ×2
= 2,700 + 180 + 120 + 8
= 3,008
Estimate:
Let 94 be Rounded to 95.
Let 32 be Rounded to 30.
So, 95 × 30 = 2,850
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 9.
Estimate: _____
63 × 54 = _____

Answer: 3,402

Explanation: Using the Partial Products method,
63 × 54 = ( 60 + 3) × ( 50 +4)
= 60 × 50 + 60 × 4 + 3 × 50 + 3 ×4
= 3,000 + 240 + 150 + 12
= 3,402
Estimate:
Let 63 be Rounded to 65.
Let 54 be Rounded to 55.
So, 55 × 65 = 3,575
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 10.
Newton eats 14 treats each week. Each treat has 33 calories. How many treat calories does Newton eat each week?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 9

Answer: 462 treat calories

Explanation:
Given that Newton eats 14 treats each week and each treat has 33 calories.
So, to find the total number of treat calories, we have to find the product of 14 × 33 by using the Partial Products method.
14 × 33 = ( 10 + 4) × ( 30 +3)
= 10 × 30 + 10 × 3 + 30 × 4 + 3 ×4
= 300 + 30 + 120 + 12
= 462
From the above,
we can conclude that Newton eats 462 treat calories each week.

Question 11.
YOU BE THE TEACHER
Your friend finds 43 × 26. Is he correct? Explain.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 10

Answer: Your friend is not correct.

Explanation:
First, we will find 43 × 26 by using the Partial Products method.
43 × 26 = ( 40 + 3) × ( 20 +6)
= 40 × 20 + 40 × 6 + 20 × 3 + 3 ×6
= 800 + 240 + 60 + 18
= 860 + 258
= 1,118
Fro the above, we can conclude that 258 must be placed instead of 248

Question 12.
Open-Ended
Use 4 cards to write 2 two-digit numbers that have a product that is close to, but not greater than, 1,200.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 11
_______ × ______

Answer:
Think and Grow: Modeling Real Life
Example
There are 16 hours in 1 day on Neptune. There are 88 times as many hours in 1 day on Mercury as 1 day on Neptune. There are 5,832 hours in 1 day on Venus. Are there more hours in 1 day on Mercury or 1 day on Venus?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 12
Multiply to find how many hours Compare if there are in 1 day on Mercury.
Using the Multiplication method,

So, there are 1,408 hours on Mercury in 1 day
It is also given that there are 5,832 hours on venus in 1 day.
So, to compare how many hours are more on Mercury when compared to venus = 5,832 – 1,408 = 4,424 hours.
So, there are more hours in 1 day on Venus.
Show and Grow

Question 13.
A ninja lantern-shark is 18 inches longer than a whale. A whale-shark is 16 times as long as the ninja lantern-shark. A hammerhead shark is 228 inches long. Is the whale shark ? or the hammerhead shark longer?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 14

Answer: The whale-shark is longer when compared to the hammerhead shark.

Explanation:
Given that a ninja lantern-shark is 18 inches long and a whale shark is 16 times longer than the ninja lantern-shark
So, to find the total length of a whale shark, we have to find the product of 18 × 16 using the Partial Products method.
18 × 16 = ( 10 + 8) × ( 10 +6)
= 10 × 10 + 10 × 6 + 10 × 8 + 8 ×6
= 100 + 60 + 80 + 48
= 288 inches
It is also given that the hammerhead shark is 228 inches long.
From the above, we can conclude that the whale shark is longer than the hammerhead shark.

Question 14.
There are 24 science classrooms in a school district. Each classroom receives 3 hot plates. Each hot plate costs $56. How much do all of the hot plates cost?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 15

Answer: $4,032

Explanation:
Given that there are 24 science classrooms in a school district and each classroom receives 3 hot plates.
So, the total number of hot plates received in a school district = 24 × 3 = 72 hot plates
It is also given that each hot plate costs $56.
So, to find the total cost of the hot plates we have to find the product of 56 × 72 using the Partial Products method.
56 × 72 = (50 + 6) × ( 70 +2)
= 50 × 70 + 50 × 2 + 70 × 6 + 2 ×6
= 3,500 + 100 + 420 + 12
= $4,032
From the above, we can conclude that the cost of all the hotplates in a school district is: $4,032

Question 15.
Fourteen adults and 68 students visit the art museum. What is the total cost of admission?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 16

Answer: $1,656

Explanation:
From the above table,
we can see that
the admission price of an adult is: $26
the admission price of a student is: $19
It is also given that there are 14 adults and 68 students.
So,
the total admission cost of the adults is: 14 × 26
the total admission cost of the children is: 68 × 19
Using  the Partial Products method,
14 × 26 = (20 + 6) × ( 10 +4)
= 20 × 10 + 20 × 4 + 10 × 6 + 4 ×6
= 200 + 80 + 60 + 24
= $364
Using  the Partial Products method,
68 × 19 = (60 + 8) × ( 10 +9)
= 60 × 10 + 60 × 9 + 10 × 8 + 8 ×9
= 600 + 540 + 80 + 72
= $1,292
Hence,
The total cost of admission (Both adults and children) = 364 + 1,292 = $1,656

Multiply Two-Digit Numbers Homework & Practice 4.6

Find the product. Check whether your answer is reasonable.

Question 1.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 17

Answer: 1,196

Explanation: Using the Partial Products method,
31 × 92 = ( 30 + 1) × ( 90 +2)
= 30 × 90 + 30 × 2 + 1 × 90 + 1 ×2
= 2,700 + 60 + 90 + 2
= 2,852
Estimate:
Let 31 be Rounded to 30.
Let 92 be Rounded to 90.
So, 30 × 90 = 2,700
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 2.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 18

Answer: 3,456

Explanation: Using the Partial Products method,
48 × 72 = ( 40 + 8) × ( 70 +2)
= 70 × 40 + 40 × 2 + 8 × 70 + 8 ×2
= 2,800 + 80 + 560 + 16
= 3,456
Estimate:
Let 48 be Rounded to 50.
Let 72 be Rounded to 70.
So, 50 × 70 = 3,500
As the Estimate and the actual answer are near, the answer is  reasonable.

Question 3.
Estimate: _____
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 19

Answer: 1,290

Explanation: Using the Partial Products method,
15 × 86 = ( 10 + 5) × ( 80 +6)
= 10 × 80 + 10 × 6 + 5 × 80 + 5 × 6
= 800 + 60 + 400 + 30
= 1,290
Estimate:
Let 86 be Rounded to 85.
So, 85 × 15 = 1,275
As the Estimate and the actual answer are near, the answer is reasonable.

Question 4.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 20

Answer: 4,374

Explanation: Using the Partial Products method,
81 × 54 = ( 80 + 1) × ( 50 +4)
= 80 × 50 + 80 × 4 + 1 × 50 + 1 × 4
= 4,000 + 320 + 50 + 4
= 4,374
Estimate:
Let 81 be Rounded to 80.
Let 54 be Rounded to 55.
So, 55 × 80 = 4,400
As the Estimate and the actual answer are near, the answer is reasonable.

Question 5.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 21

Answer: 1,426

Explanation: Using the Partial Products method,
23 × 62 = ( 20 + 3) × ( 60 +2)
= 20 × 60 + 20 × 2 + 3 × 60 + 3 ×2
= 1,200 + 40 + 180 + 6
= 1,426
Estimate:
Let 23 be Rounded to 25.
Let 62 be Rounded to 60.
So, 25 × 60 = 1,500
As the Estimate and the actual answer are near, the answer is  reasonable.

Question 6.
Estimate: _____
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 22

Answer: 5,335

Explanation: Using the Partial Products method,
97 × 55 = ( 90 + 7) × ( 50 +5)
= 90 × 50 + 90 × 5 + 7 × 50 + 7 × 5
= 4,500 + 450 + 350 + 35
= 5,335
Estimate:
Let 97 be Rounded to 95.
So, 55 × 95 = 5,225
As the Estimate and the actual answer are near, the answer is  reasonable.
Find the product. Check whether your answer is reasonable.

Question 7.
Estimate: ______
51 × 62 = ______

Answer: 3,162

Explanation: Using the Partial Products method,
51 × 62 = ( 50 +1) × ( 60 +2)
= 60 × 50 + 50 × 2 + 1 × 60 + 1 × 2
= 3,000 + 100 + 60 + 2
= 3,162
Estimate:
Let 51 be Rounded to 50.
Let 62 be Rounded to 60.
So, 50 × 60 = 3,000
As the Estimate and the actual answer are near, the answer is  reasonable.

Question 8.
Estimate: ______
37 × 13 = ______

Answer: 481

Explanation: Using the Partial Products method,
37 × 13 = ( 10 +3) × ( 30 +7)
= 10 × 30 + 10 × 7 + 3 × 30 + 3 × 7
= 300 + 70 + 90 + 21
= 481
Estimate:
Let 37 be Rounded to 40.
Let 13 be Rounded to 15.
So, 40 × 15= 600
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 9.
Estimate: _______
49 × 78 = ______

Answer: 3,822

Explanation: Using the Partial Products method,
49 × 78 = ( 40 +9) × ( 70 +8)
= 40 × 70 + 40 × 8 + 9 × 70 + 9 × 8
= 2,800 + 320 + 630 + 72
= 3,822
Estimate:
Let 49 be Rounded to 50.
Let 78 be Rounded to 80.
So, 50 × 80 = 4,000
As the Estimate and the actual answer are near, the answer is  reasonable.

Question 10.
Newton plays 21 basketball games. He scores 12 points each game. How many points does he score in all?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 23

Answer: He scored 252 points.

Explanation:
Given that Newton plays 21 basketball games and he scores 12 points each game.
So, to find the total number of points in all games, we have to find the product of 21 × 12
By using the Partial Products method,
21 × 12 = ( 20 +1) × ( 10 + 2)
= 20 × 10 + 20 × 2 + 1 × 10 + 1 × 2
= 200 + 40 + 10 + 2
= 252 points
Hence,
From the above, we can conclude that Newton had scored 252 points in 21 basketball games.

Question 11.
DIG DEEPER!
When you use regrouping to multiply two-digit numbers, why does the second partial product always end in 0?

Answer: Because we divide the partial Products in terms of 10 only.
Ex:
13 × 15 = ( 10 + 3) × (10 + 5)

Question 12.
Number Sense
Find the missing digits.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 24

Answer: The missing numbers are: 6, 7, 0 and 1

Explanation:

By using the Partial Products method,
34 × 65 = ( 30 +4) × ( 60 + 5)
= 30 × 60 + 30 × 5 + 4 × 60 + 4 × 5
= 1,800 + 150 + 240 + 20
= 2,040 + 170
= 2,210
So,
From the above, we can conclude that the missing numbers are: 6, 7, 0 and 1

Question 13.
Modeling Real Life
A tiger dives 12 feet underwater. An otter dives 25 times deeper than the tiger. A walrus dives 262 feet underwater. Does the otter or walrus dive deeper?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 25

Answer:  Otter drives deeper than the Walrus.

Explanation:
Given that a tiger dives 12 feet underwater and an otter dives 25 times deeper than the tiger.
So,
The total distance dived by otter = 25 × 12 feet
By using the Partial Products method,
25 × 12 = ( 20 +5) × ( 10 + 2)
= 20 × 10 + 20 × 2 + 5 × 10 + 2 × 5
= 200 + 40 + 50 + 10
= 300
It is also given that a walrus dives 262 feet underwater.
From the above,
we can conclude that the otter dives deeper than the walrus.
Review & Refresh

Question 14.
Complete the table.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 26

Answer:
A) 6,835
Word Form: Six thousand, eight hundred thirty-five
Expanded Form: 6,000 + 800 + 30 + 5

Explanation:
Any number can be written in 3 forms. They are:
A) Standard Form B) Word Form C) Expanded Form
So, we can write the given form in the remaining 2 forms.
B) 70,000 + 4,000 + 100 + 2
Standard Form: 74,102
Word Form: Seventy-four thousand, One hundred two

Explanation:
Any number can be written in 3 forms. They are:
A) Standard Form B) Word Form C) Expanded Form
So, we can write the given form in the remaining 2 forms.
C) Five hundred one thousand, three hundred twenty-nine
Standard Form: 501,329
Expanded Form: 500,000 +0 + 1,000 + 100 + 0 + 2

Explanation:
Any number can be written in 3 forms. They are:
A) Standard Form B) Word Form C) Expanded Form
So, we can write the given form in the remaining 2 forms.

Lesson 4.7 Practice Multiplication Strategies

Explore and Grow
Choose any strategy to find 60 × 80.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 1

Answer: The strategy to find 60 × 80 is: Associative Property of Multiplication

Explanation: By using the Associative Property of Multiplication,
60 ×80 =60 × (8 × 10)
= (60 × 8) × 10
= (6 × 10 × 8) × 10
= 480 × 10
= 4,800
Choose any strategy to find 72 × 13.

Answer: The strategy to find 72 × 13 is: Distributive Property

Explanation:
72 × 13 = 72 × ( 10 + 3)
= ( 72 × 10 ) + ( 72 ×3)
= ( 70 + 2 ) × 10 + ( 70 + 2 ) × 3
= ( 70 × 10 ) + ( 2 × 10 ) + ( 70 × 3 ) + ( 2 × 3)
= 700 + 20 + 210 + 6
= 936
So, 72 × 13 = 936
Reasoning
Explain why you chose your strategies. Compare your strategies to your partner’s strategies. How are they the same or different?

Answer:
Think and Grow: Practice Multiplication Strategies
Example
Find 62 × 40.
One Way: Use place value.
62 × 40 = 62 × 4 tens
=248 tens
= 248 × 10
= 2,480
So, 62 × 40 = 2,480.
Another Way: Use an area model and partial products.

So, 62 × 40 =2,480 .
Example
Find 56 × 83.
One Way: Use place value and partial products.

So, 56 × 83 = 4,648.
Another Way: Use regrouping
Multiply 56 by 3 ones. Then multiply 56 by 8 tens. Regroup if necessary.

So, 56 × 83 = 4,648.
Show and Grow
Find the product.

Question 1.
90 × 37 = _____

Answer: 3,330

Explanation: Using the Associative Property of Multiplication,
37 × 90 = 37 × (9 × 10)
= (37 × 9) × 10
= (3 × 3 × 37) × 10
= 333 × 10
= 3,330

Question 2.
78 × 21 = ______

Answer: 1,638

Explanation: Using the Distributive Property,
78 × 21 = 78 × ( 20 + 1)
= ( 78 × 20 ) + ( 78 × 1)
= ( 70 + 8 ) × 20 + ( 70 + 8 ) × 1
= ( 70 × 20 ) + ( 8 × 20 ) + ( 70 × 1 ) + ( 8 × 1)
= 1,400 + 160 + 70 + 8
= 1,638
So, 78 × 21 = 1,638

Question 3.
14 × 49 = _____

Answer: 686

Explanation: Using the Associative Property of Multiplication,
14 × 49 = 14 × ( 40 + 9)
= ( 14 × 40 ) + ( 14 × 9)
= ( 10 + 4 ) × 40 + ( 10 + 4 ) × 9
= ( 40 × 10 ) + ( 4 × 40 ) + ( 10 × 9 ) + ( 4 × 9)
= 400 + 160 + 90 + 36
= 686
So, 14 × 49 = 686
Apply and Grow: Practice
Find the product.

Question 4.
74 × 30 = _____

Answer: 2,220

Explanation: Using the Associative Property of Multiplication,
74 × 30 = 74 ( 3 × 10)
= ( 74 × 3) × 10
= 222 × 10
= 2,220
So, 74 × 30 = 2,220

Question 5.
51 × 86 = _____

Answer: 4,386

Explanation: Using the Distributive Property ,
51 × 86 = 51 × ( 80 + 6)
= ( 51 × 80 ) + ( 51 × 6)
= ( 50 + 1 ) × 80 + ( 50 + 1 ) × 6
= ( 50 × 80 ) + ( 1 × 80 ) + ( 50 × 6 ) + ( 1 × 6)
= 4,000 + 80 + 300 + 6
= 4,386
So, 51 × 86 = 4,386

Question 6.
40 × 29 = ______

Answer: 1,160

Explanation: Using the Associative Property of Multiplication,
40 × 29 = 29 × (4 × 10)
= (29 × 4) × 10
= (29× 2 × 2) × 10
= 116 × 10
= 1,160

Question 7.
92 × 80 = _____

Answer: 7,360

Explanation: Using the Associative Property of Multiplication,
92 × 80 = 92 × (8 × 10)
= (92 × 8) × 10
= (92 × 2 × 4) × 10
= 736 × 10
= 7,360

Question 8.
41 × 17 = ______

Answer: 697

Explanation: Using the Distributive Property,
41 × 17 = 41 × ( 10 + 7)
= ( 41 × 10 ) + ( 41 × 7)
= ( 40 + 1 ) × 10 + ( 40 + 1 ) × 7
= ( 40 × 10 ) + ( 1 × 10 ) + ( 40 × 7 ) + ( 1 × 7)
= 400 + 10 + 280 + 7
= 697
So, 41 × 17 = 697

Question 9.
60 × 53 = _____

Answer: 3,180

Explanation: By using the Associative Property of Multiplication,
60 × 53 = 53 × (6 × 10)
= (53 × 6) × 10
= (53× 3 × 2) × 10
= 318 × 10
= 3,180
Logic
Find the missing factor.

Question 10.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 5

Answer: The missing number is: 41

Explanation:

In the given figure, the partial fractions are given.
The 1st partial fraction represents the multiplication with the unit digit and the 2nd partial fraction represents the multiplication with the tens digit.
The first partial fraction will come when 72 is multiplied with 1 and the 2nd partial fraction will come when 72 is multiplied with 4.
So, the missing number is: 41

Question 11.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 6

Answer: The missing number is: 34

Explanation:

In the given figure, the partial fractions are given.
The 1st partial fraction represents the multiplication with the unit digit and the 2nd partial fraction represents the multiplication with the tens digit.
The first partial fraction will come when 65 is multiplied with 4 and the 2nd partial fraction will come when 65 is multiplied with 3.
So, the missing number is: 34

Question 12.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 7

Answer: The missing number is: 87

Explanation:

In the given figure, the partial fractions are given.
The 1st partial fraction represents the multiplication with the unit digit and the 2nd partial fraction represents the multiplication with the tens digit.
The first partial fraction will come when 93 is multiplied with 7 and the 2nd partial fraction will come when 93 is multiplied with 8.
So, the missing number is: 87

Question 13.
Writing
Explain why you start multiplying with the one’s place when using regrouping to multiply.

Answer: When using ” Regrouping” to multiply, we start multiplying from the rightmost position i.e.., one’s position because the place value of that position is 1.

Question 14.
DIG DEEPER!
Find the missing digit so that both products are the same.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 8

Answer: The missing digit so that both products are the same is: 3

Explanation:

The product of 26 × 15 is:
26 × 15 = 26 × ( 10 + 5)
= ( 26 × 10 ) + ( 26 × 5)
= ( 20 + 6 ) × 10 + ( 20 + 6 ) × 5
= ( 20 × 10 ) + ( 6 × 10 ) + ( 20 × 5 ) + ( 6 × 5)
= 200 + 60 + 100 + 30
= 390
So, 26 × 15 = 390
To get the same result in the next product, we have to find the missing number.
So, 390 / 30 = 13
Hence, the missing digit so that both products are the same is: 3
Think and Grow: Modeling Real Life
Example
A swinging ship ride runs 50 times each afternoon. The ship has 10 rows of benches with 4 seats in each bench. If the ship is full each time it runs, how many people will ride the swinging ship in 1 afternoon?
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 9
Multiply to find how many people will ride the swinging ship ride each time.
10 × 4 = 40
So,
40 people will ride the swinging ship ride each time.
Multiply to find how many people will ride the swinging ship in 1 afternoon.
40 × 50 = 40 × (5 × 10)  (Rewrite 50 as 5 × 10)
= (40 × 5) × 10 ( By using Associative Property of Multiplication)
= 200 × 10
= 2,000
Hence,
2,000 people will ride the swinging ship in 1 afternoon.
Show and Grow

Question 15.
A teacher orders 25 rock classification kits. Each kit has 4 rows with 9 rocks in each row. How many rocks are there in all?

Answer: 900 rocks

Explanation:
Given that a teacher orders 25 rock classification kits and each kit has 4 rows with 9 rocks in each row.
So,
The total number of rows present in 25 rock classification kits = 25 × 4 = 100 rows
We have to find the number of rocks in 100 rows by multiplying 100 × 9.
Now, 100 × 9 = 900 rocks.
From the above,
We can conclude that there are 900 total rocks.

Question 16.
A hotel has 12 floors with 34 rooms on each floor. 239 rooms are in use. How many rooms are not in use?

Answer: The total number of rooms that are not in use: 169

Explanation:
Given that a hotel has 12 floors with 34 rooms on each floor.
So, the total number of rooms in a hotel = 34 × 12
By using the Distributive method,
34 × 12 = 34 ( 10 + 2 )
= ( 34 × 10 ) + ( 34 × 2 )
= ( 30 + 4 ) × 10 + ( 30 + 4 ) × 2
= ( 30 × 10 ) + ( 4 × 10 ) + ( 30 × 2 ) + ( 4 × 2)
= 300 + 40 + 60 + 8
= 408
But, it is also given that 239 rooms are in use.
So, the number of rooms that are not in use = 408 – 239 = 169 rooms

Question 17.
A child ticket for a natural history museum costs $13. An adult ticket costs twice as much as a child ticket. How much does it cost for 21 children and 14 adults to go to the museum?
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 10

Answer: $637

Explanation:
Given that a child ticket for a natural history museum costs $13 and an adult ticket costs twice as much as a child ticket.
So, the cost of an adult ticket = 2 × $13 = $26
So, to find the total cost for 21 children and 14 adults in a museum, we have to find
21 × $13  and 14 × $26
Now,
21 × 13 = 21 ( 10 + 3 )
= ( 21 × 10 ) + ( 21 × 3 )
= ( 20 + 1 ) × 10 + ( 20 +1 ) × 3
= ( 20 × 10 ) + ( 1 × 10 ) + ( 20 × 3 ) + ( 1 × 3)
= 200 + 10 + 60 + 3
= 273
14 × 26 = 14 ( 20 + 6 )
= ( 14 × 20 ) + ( 14 × 6 )
= ( 10 + 4 ) × 20 + ( 10 +4 ) × 6
= ( 10 × 20 ) + ( 4 × 20 ) + ( 10 × 6 ) + ( 4 × 6)
= 200 + 80 + 60 + 24
= 364
Hence,
The total cost for children and adults in a museum = 273 + 364 = $637

Practice Multiplication Strategies Homework & Practice 4.7

Find the product.

Question 1.
16 × 13 = _____

Answer: 208

Explanation: By using the Distributive method,
16 × 13 = 16 ( 10 + 3 )
= ( 16 × 10 ) + ( 16 × 3 )
= ( 10 + 6 ) × 10 + ( 10 + 6 ) × 3
= ( 10 × 10 ) + ( 6 × 10 ) + ( 10 × 3 ) + ( 6 × 3)
= 100 + 30 + 60 + 18
= 208

Question 2.
29 × 50 = _____

Answer: 1,450

Explanation: By using the Associative Property of Multiplication,
29 × 50 = 29 × ( 5 × 10 )
= ( 29 × 5 ) × 10
= 145 × 10
= 1,450

Question 3.
78 × 45 = _____

Answer: 3,510

Explanation: By using the Distributive method,
78 × 45 = 78 ( 40 + 5 )
= ( 78 × 40 ) + ( 78 × 5 )
= ( 70 + 8 ) × 40 + ( 70 + 8 ) × 5
= ( 70 × 40 ) + ( 8 × 40 ) + ( 70 × 5 ) + ( 8 × 5)
= 2,800 + 320 + 350 + 40
= 3,510

Question 4.
30 × 71 = ______

Answer: 2,130

Explanation: By using the Associative Property of Multiplication,
30 × 71 = 71 × ( 3 × 10 )
= ( 71 × 3 ) × 10
= 213 × 10
= 2,130

Question 5.
62 × 14 = _____

Answer: 868

Explanation: By using the Distributive method,
62 × 14 = 62 ( 10 + 4 )
= ( 62 × 10 ) + ( 62 × 4 )
= ( 60 + 2 ) × 10 + ( 60 + 2 ) × 4
= ( 60 × 10 ) + ( 2 × 10 ) + ( 60 × 4 ) + ( 2 × 4)
= 600 + 20 + 240 + 8
= 868

Question 6.
80 × 90 = _____

Answer: 7,200

Explanation: By using the Associative Property of Multiplication,
80 × 90 = 80 × ( 9 × 10 )
= ( 80 × 9 ) × 10
= 720 × 10
= 7,200
Find the product.

Question 7.
70 × 18 = _____

Answer: 1,260

Explanation: By using the Associative Property of Multiplication,
70 × 18 = 18 × ( 7 × 10 )
= ( 18 × 7 ) × 10
= 126 × 10
= 1,260

Question 8.
32 × 59 = _____

Answer: 1,888

Explanation: By using the Distributive method,
32 × 59 = 59 ( 30 + 2 )
= ( 59 × 30 ) + ( 59 × 2 )
= (50 + 9 ) × 30 + ( 50 + 9 ) × 2
= ( 50 × 30 ) + ( 9 × 30 ) + ( 50 × 2 ) + ( 2 × 9)
= 1,500 + 270 + 100 + 18
= 1,888

Question 9.
67 × 20 = ____

Answer: 1,340

Explanation: By using the Associative Property of Multiplication,
67 × 20 = 67 × ( 2 × 10 )
= ( 67 × 2 ) × 10
= 134 × 10
= 1,340

Question 10.
51 × 84 = _____

Answer: 4,284

Explanation: By using the Distributive method,
51 × 84 = 51 ( 80 + 4 )
= ( 51 × 80 ) + ( 51 × 4 )
= (50 + 1 ) × 80 + ( 50 + 1 ) × 4
= ( 50 × 80 ) + ( 1 × 80 ) + ( 50 × 4 ) + ( 1 × 4)
= 4,000 + 80 + 200 + 4
= 4,284

Question 11.
40 × 40 = _____

Answer: 1,600

Explanation: By using the Associative Property of Multiplication,
40 × 40 = 40 × ( 4 × 10 )
= ( 40 × 4 ) × 10
= 160 × 10
= 1,600

Question 12.
23 × 97 = ______

Answer: 2,231

Explanation: By using the Distributive method,
23 × 97 = 97 ( 20 + 3 )
= ( 97 × 20 ) + ( 97 × 3 )
= (90 + 7 ) × 20 + ( 90 + 7 ) × 3
= ( 90 × 20 ) + ( 7 × 20 ) + ( 90 × 3 ) + ( 7 × 3)
= 1,800 + 140 + 270 + 21
= 2,231

Question 13.
Writing
Which strategy do you prefer to use when multiplying two-digit numbers? Explain.

Answer: The strategy you have to prepare when multiplying 2-digit numbers must depend on the numbers.
some strategies to multiply 2-digit numbers are:
A) The place-value method  B) The Associative Property of Multiplication  C) Distributive Property D) Partial Products method E) Regrouping method

Question 14.
Patterns
What number can you multiply the number of tires by to find the total weight? Use this pattern to complete the table.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 11

Answer: The number you can multiply the number of tires by to find the total weight is: 20

Explanation:

From the above table,
to find the multiple of total weight, divide the number of tires by the total weight. ( Take any 2 values)
So, 80 /4 = 20 ( Multiple of total weight)
From the above,
We can conclude that the number you can multiply the number of tires by to find the total weight is: 20

Question 15.
Modeling Real Life
Each bag of popcorn makes 13 cups. A school has a movie day, and the principal brings 15 boxes of popcorn. Each box has 3 bags of popcorn. How many cups of popcorn does the principal bring?
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 12

Answer: 585 cups of popcorn

Explanation:
Given that each bag of popcorn makes 13 cups and the principal brings 15 boxes of popcorn. It is also given that each box has 3 bags of popcorn.
So, to find the total number of cups of popcorn the principal bring = 13 × 15 × 3
Now,
By using the Distributive method,
13 × 45 = 45 ( 10 + 3 )
= (45 × 10 ) + ( 45 × 3 )
= (40 +5 ) × 10 + ( 40 + 5 ) × 3
= ( 40 × 10 ) + ( 5 × 10 ) + ( 40 × 3 ) + ( 5 × 3)
= 400 + 50 + 120 + 15
= 585
From the above,
we can conclude that the number of cups the principal bring is: 585 cups of popcorn
Review & Refresh
Find the product.

Question 16.
8 × 200 = _____

Answer: 1,600

Explanation: By using the Associative Property of Multiplication,
8 × 200 = 8 × (20 × 10)
= (8 × 20) × 10
= (8× 5 × 4) × 10
= 160 × 10
= 1,600

Question 17.
7 × 300 = _____

Answer: 2,100

Explanation: By using the Associative Property of Multiplication,
7 × 300 = 7 × (30 × 10)
= (7 × 30) × 10
= (7× 5 × 6) × 10
= 210 × 10
= 2,100

Question 18.
6,000 × 5 = _____

Answer: 30,000

Explanation: By using the Associative Property of Multiplication,
5 × 6,000 = 5 × (600 × 10)
= (5 × 600) × 10
= (6× 5 × 100) × 10
= 300 × 10
= 3,000

Question 19.
9 × 90 = ____

Answer: 810

Explanation: By using the Associative Property of Multiplication,
9 × 90 = 9 × (9 × 10)
= (9 × 9) × 10
= 81 × 10
= 810

Question 20.
3,000 × 6 = _____

Answer: 18,000

Explanation: By using the Associative Property of Multiplication,
6 × 3,000 = 6 × (300 × 10)
= (6 × 300) × 10
= (6× 3 × 100) × 10
= 1800 × 10
= 18,000

Question 21.
5 × 500 = _____

Answer: 2,500

Explanation: By using the Associative Property of Multiplication,
5 × 500 = 5 × (50 × 10)
= (5 × 50) × 10
= (5× 5 × 10) × 10
= 250 × 10
= 2,500

Lesson 4.8 Problem Solving: Multiplication with Two-Digit Numbers

Explore & Grow
Explain, in your own words, what the problem below is asking. Then explain how you can use multiplication to solve the problem.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 1
A ferry can transport 64 cars each time it leaves a port. The ferry leaves a port 22 times in 1 day. How many cars can the ferry transport in 1 day?

Answer: 1,408 cars can ferry transport in 1 day

Explanation:
Given that a ferry can transport 64 cars each time it leaves a port and the ferry leaves a port 22 times in 1 day.
From this, the total number of cars that can ferry transport in 1 day is: 64 × 22
By using the Distributive method,
64 × 22 = 64 ( 20 + 2 )
= (64 × 20 ) + ( 64 × 2 )
= (60 + 4 ) × 20 + ( 60 + 4 ) × 2
= ( 60 × 20 ) + ( 4 × 20 ) + ( 60 × 2 ) + ( 4 × 2)
= 1,200 + 80 + 120 + 8
= 1,408
From the above,
we can conclude that the total number of cars that ferry transport in 1 day is 1,408 cars.
Construct Arguments
Make a plan to find how many cars the ferry can transport in 1 week.

Answer: 9,856 cars

Explanation:
From the above problem, we know that the total number of cars a ferry can transport in 1 day is: 1,408 cars
We know that,
1 week = 7 days
First, we have to find the total number of cars the ferry can transport in 1 day using the Product 64 × 22.
By using the Distributive method,
64 × 22 = 64 ( 20 + 2 )
= (64 × 20 ) + ( 64 × 2 )
= (60 + 4 ) × 20 + ( 60 + 4 ) × 2
= ( 60 × 20 ) + ( 4 × 20 ) + ( 60 × 2 ) + ( 4 × 2)
= 1,200 + 80 + 120 + 8
= 1,408
Now,
the total number of cars a ferry can transport in 1 week = 1,408 × 7 = 9,856 cars
Think and Grow: Problem Solving: Multiplication with Two-Digit Numbers
Example
A pet store receives a shipment of 8 boxes of dog treats. Each box is 2 feet high and has 18 bags of dog treats. How many ounces of dog treats does the pet store receive in the shipment?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 2
Understand the Problem
What do you know?
• The store receives 8 boxes.
• Each box is 2 feet high.
• Each box has 18 bags of dog treats.
• Each bag weighs 32 ounces.
What do you need to find?
• You need to find how many ounces of dog treats the pet store receives in the shipment.
Make a Plan
How will you solve it?
• Multiply 32 by 18 to find how many ounces of dog treats are in each box.
• Then multiply the product by 8 to find how many ounces of dog treats the pet store receives in the shipment.
• The height of each box is unnecessary information.
Solve
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 3

Answer: 4,608 ounces

Explanation:
Given that the store receives 8 boxes and Each box has 18 bags of dog treats. It is also given that each bag weighs 32 ounces.
The number of ounces of dog treats the pet store receives in the shipment = 18 × 8 × 32
It is also given that each box is 2 feet high. But it is not necessary for the calculation of a total number of ounces.
First, we will find the weight of 1 box = 18 × 32
By using the Distributive method,
18 × 32 = 18 ( 30 + 2 )
= (18 × 30 ) + ( 18 × 2 )
= (10 + 8 ) × 30 + ( 10 + 8 ) × 2
= ( 10 ×30 ) + ( 8 × 30 ) + ( 10 × 2 ) + ( 8 × 2)
= 300 + 240 + 20 + 16
= 576
So,
The total weight of dog treats the pet store receives in the shipment = 576 × 8 = 4,608 ounces
Hence,
The pet store receives 4,608 ounces of dog treats.
Show and Grow

Question 1.
Show how to solve the problem above using one equation.

Answer: 18 × 32 × 8
Apply and Grow: Practice
Understand the problem. What do you know? What do you need to find? Explain.

Question 2.
Thirteen students create a petition for longer recess. They need 5,000 signatures in all. So far, each student has 99 signatures. How many more signatures do they need?

Answer: 3,713 signatures

Explanation:
Given that there are 13 students that create a petition for longer recess and they need 5,000 signatures in all.
It is also given that each student has 99 signatures.
So, the total number of signatures that 13 students have are: 13 × 99
By using the Distributive method,
13 × 99 = 13 ( 90 + 9 )
= (13 × 90 ) + ( 13 × 9 )
= (10 +3 ) × 90 + ( 10 + 3 ) × 9
= ( 90 × 10 ) + ( 3 × 90 ) + ( 10 × 9 ) + ( 9 × 3)
= 900 + 270 + 90 + 27
= 1,287
But, the students required 5000 signatures in total.
So,
The remainin number of signatures required by students = 5,000 – 1,287 = 3,713 signatures

Question 3.
An activity book has 35 pages and costs $7. Each page has 4 puzzles. You have completed all of the puzzles on 16 of the pages. How many puzzles do you have left to complete?

Answer: 76 puzzles

Explanation:
Given that an activity book has 35 pages and each page has 4 puzzles.
So,
The total number of puzzles = 35 × 4 = 140 puzzles
But, it is also given that the puzzles completed in all of the 16 pages and we know that each page has 4 puzzles.
So,
The number of puzzles in 16 pages = 16 × 4 = 64 puzzles
Now,
The remaining number of puzzles in the remaining pages = 140 – 64 = 76 puzzles
Understand the problem. Then make a plan. How will you solve it? Explain.

Question 4.
Twelve classes provide items for a time capsule. There are 23 students in each class. Each student puts 2 small items in the time capsule. How many items are in the time capsule?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 4

Answer: 552 items

Explanation:
Given that there are 12 classes that provide items for a time capsule and there are 23 students in each class and each student puts 2 small items in the time capsule.
First, we have to find the total number of students by the product 23 × 12
By using the Distributive method,
12 × 23 = 23( 10 + 2 )
= (23 × 10 ) + ( 23 × 2 )
= (20 +3 ) × 10 + ( 20 + 3 ) × 3
= ( 20 × 10 ) + ( 3 × 10 ) + ( 20 × 3 ) + ( 3 × 3)
= 200 + 30 + 60 + 9
= 296 students
Now,
The total number of small items put by the total number of students = 296 × 2 = 552 small items

Question 5.
A craftsman cuts letters and numbers from license plates to make signs. He has 15 Florida plates and 25 Georgia plates. Each plate has a total of 7 letters and numbers. How many letters and numbers does the craftsman cut in all?

Answer: 280 letters and numbers
Explanation;
Given that a craftsman has 15 Florida plates and 25 Georgia plates and each plate has a total of 7 letters and numbers.
Hence,
The total number of plates that a craftsman have = 25 + 15 = 40 plates
The total number of letters and numbers each plate has = 40 × 7 = 280 letters and numbers

Question 6.
DIG DEEPER!
In 1 month, the solar panel and the wind turbine can produce the kilowatt-hours of electricity shown. How much electricity can 28 solar panels and1 small wind turbine produce each month?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 5

Answer: 1,673 kilowatt-hours of electricity
Explanation;
Given that each solar panel produces 30 kilowatt-hours and a wind turbine produces 833 kilowatt-hours.
Now, the electricity produced by 28 solar panels = 28 × 30
By using the Associative Property of Multiplication,
28 × 30 = 28 × (3 × 10)
= (28 × 3) × 10
= (7× 4 × 3) × 10
= 84 × 10
= 840 kilowatt-hours
Hence,
The total electricity produced by 28 solar panels and 1 wind turbine = 840 + 833 = 1,673 kilowatt-hours

Think and Grow: Modeling Real Life
Example
An adult ticket for a zip line course costs $48.A child ticket costs $19 less than an adult ticket. In 1 day, 86 adults and 42 children ride the zipline. How much more money was earned from adult tickets than from child tickets?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 6
Think: What do you know? What do you need to find? How will you solve?
Step 1: How much money was earned from adult tickets?
$48 × 86 = ‘a’
‘a’ is an unknown product.

So,
a = 4,128
Step 2: How much money was earned from child tickets?
A child ticket costs $48 – $19 = $29.
$29 × 42 = c
c is the unknown product.

So,
c = 1,218
Step 3: Use m to represent how much more money was earned from adult tickets.

So, $ 2910 more was earned from adult tickets.
Show and Grow

Question 7.
A theater has 45 rows of 72 seats on the floor level and 22 rows of 36 seats in the balcony. How many seats are there in all?
How many more seats are on the floor level than on the balcony?

Answer: 4,032 seats
Explanation;
Given that a theater has 45 rows of 72 seats on the floor level and 22 rows of 36 seats in the balcony.
So,
The total number of seats on the floor level = 45 × 72
The total number of seats on the balcony = 22 × 36
By using the Distributive method,
72 × 45 = 45 ( 70 + 2 )
= (45 × 70 ) + ( 45 × 2 )
= (40 +5 ) × 70 + ( 40 + 5 ) × 2
= ( 40 × 70 ) + ( 5 × 70 ) + ( 40 × 2 ) + ( 5 × 2)
= 2,800 + 350 + 80 + 6
= 3,240
By using the Distributive method,
22 × 36 = 36 ( 20 + 2 )
= (36 × 20 ) + ( 36× 2 )
= (30 +6 ) × 20 + ( 30 + 6) × 2
= ( 30 × 20 ) + ( 6 × 20 ) + ( 30 × 2 ) + ( 6 × 2)
= 600 + 120 + 60 + 12
= 792
So,
The total number of seats ( Floor level and balcony) = 3,240 + 792 = 4,032 seats

Problem Solving: Multiplication with Two-Digit Numbers Homework & Practice 4.8

Understand the problem. Then make a plan. How will you solve it? Explain.

Question 1.
Seventy-two mushers compete in a sled-dog race. Each musher has 16 dogs. How many dogs compete in the race than mushers?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 10

Answer: 1,152 dogs

Explanation:
Given that there are 72 mushers compete in a sled-dog race and each musher has 16 dogs.
So,
The total number of dogs that compete in the race = 16 × 72
By using the Distributive method,
16 × 72 = 72 ( 10 + 6 )
= (72 × 10 ) + ( 72× 6 )
= (70 +2 ) × 10 + ( 70 + 2) × 6
= ( 70 × 10 ) + ( 2 × 10 ) + ( 70 × 6 ) + ( 6 × 2)
= 700 + 20 + 420 + 12
= 1,152
Hence,
The total number of dogs that compete in the race are: 1,152 dogs

Question 2.
A photographer buys 3 USB drives that each cost $5. She puts 16 folders on each drive. Each folder has 75 photographs. How many photographs does the photographer put on the USB drives in all?

Answer: 3,600 photographs
Explanation;
Given that a photographer has 3 USB drives and there are 16 folders and 75 photographers on each drive.
So,
The total number of photographs on all the USB drives = 16 × 3 × 75 = 48 × 75
By using the Distributive method,
48 × 75 = 75 (40 + 8 )
= (75 × 40 ) + ( 75 × 8 )
= (70 +5 ) × 40 + ( 70 + 5) × 8
= ( 70 × 40 ) + ( 5 × 40 ) + ( 70 × 8 ) + ( 5 × 8)
= 2,800 + 200 + 560 + 40
= 3,600 photographs
Hence,
The total number of photographs on all the USB drives are: 3,600 photographs

Question 3.
A teacher has 68 students take 25
Question test. The teacher checks the answers for 9 of the tests. How many answers does the teacher have left to check?

Answer: 1,475 answers

Explanation:
Given that a teacher has 68 students take 25
Question test.
So,
The total number of answers the teacher have to check = 68 × 25
By using the Distributive method,
68 × 25 = 68 ( 20 + 5 )
= (68 × 20 ) + ( 68× 5 )
= (60 +8 ) × 20 + ( 60 + 8) × 5
= ( 60 × 20 ) + ( 8 × 20 ) + ( 60 × 5 ) + ( 8 × 5)
= 1,200 + 160 + 300 + 40
= 1,700
It is also given that the teacher checked 9 papers only.
So.
The number of answers checked = 9 × 25 = 225
Hence,
The number of answers remained unchecked = 1,700 – 225 = 1,475 answers

Question 4.
Each day, a cyclist bikes uphill for 17 miles and downhill for 18 miles. She drinks 32 fluid ounces of water after each bike ride. How many miles does the cyclist bike in 2 weeks?

Answer: 490 miles
Explanation;
Given that a cyclist bikes uphill for 17 miles and downhill for 18 miles.
SO,
The total distance that cyclist bikes each day = 17 + 18 = 35 miles
We know that,
1 week = 7 days
S0, 2 weeks = 2 × 7  = 14 days.
So, the distance traveled by a cyclist in 2 weeks = 35 × 14
By using the Distributive method,
35 × 14 = 35 ( 10 + 4 )
= (35 × 10 ) + ( 35× 4 )
= (30 +5 ) × 10 + ( 30 + 5) × 4
= ( 30 × 10 ) + ( 5 × 10 ) + ( 30 × 4 ) + ( 5 × 4)
= 300 + 50 + 120 + 20
= 490 miles
Hence,
The distance traveled by a cyclist in 2 weeks is: 490 miles

Question 5.
Precision
Which expressions can be used to solve the problem?
Twelve friends play a game that has 308 cards. Each player receives 16 cards. How many cards are left?
(308 – 12) × 16
308 – (16 × 12)
308 – (12 × 16)
(308 – 16) – 12

Answer: Let the Expressions be named as A),  B), C) and D)
Either B) or C) can be used to solve the problem.

Explanation:
The given Expressions are;
A) (308 – 12) × 16
B) 308 – (16 × 12)
C) 308 – (12 × 16)
D) (308 – 16) – 12
It is given that there are 12 friends that have 16 cards each and there are 308 cards in total.
So,
The number of cards that have left = 308 – ( 16 × 12 ) (or) 308 – ( 12 × 16 )
Hence,
From the above,
we can conclude that either B) or C) can be used to solve the problem.

Question 6.
Modeling Real Life
A child ticket costs $14 less than an adult ticket. What is the total ticket cost for 18 adults and 37 children?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 11

Answer: The total ticket cost is: $2,617

Explanation:
The given cost of an Adult ticket = $57
It is also given that a child ticket costs $14 less than an adult ticket.
So,
The given cost of a child ticket = 57 – 14 = $43
Now,
The cost for 18 adults = 18 × 57
The cost for 37 children = 37 × 43
By using the Distributive method,
18 × 57 = 57 ( 10 + 8 )
= (57 × 10 ) + ( 57× 8 )
= (50 + 7 ) × 10 + ( 50 + 7) × 8
= ( 50 × 10 ) + ( 7 × 10 ) + ( 50 × 8 ) + ( 7 × 8)
= 500 + 70 + 400 + 56
= 1,026
By using the Distributive method,
37 × 43 = 37 ( 40 + 3 )
= (37 × 40 ) + ( 37× 3 )
= (30 +7 ) × 40 + ( 30 + 7) × 3
= ( 30 × 40 ) + ( 7 × 40 ) + ( 30 × 3 ) + ( 7 × 3)
= 1,200 + 280 + 90 + 21
= 1,591
Hence,
The total cost of tickets = 1,591 + 1,026 = $2,617

Question 7.
Modeling Real Life
An artist creates a pattern by alternating square and rectangular tiles. The pattern has 14 square tiles and 13 rectangular tiles. How long is the pattern?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 12

Answer: 1,680 cm

Explanation:
Given that an artist creates a pattern by alternating square and rectangular tiles and the pattern 14 square tiles and 13 rectangular tiles.
From the given fig.,
Area of Square = 8 × 8 = 64 square cm
Area of Rectangle = 8 × 14 = 112 square cm
So,
Area of 14 squares = 14 × 64
Area of 13 rectangles = 112 × 13
By using the Distributive method,
14 × 64 = 64 ( 10 + 4 )
= (64 × 10 ) + ( 64× 4 )
= (60 +4 ) × 10 + ( 60 + 4) × 4
= ( 60 × 10 ) + ( 4 × 10 ) + ( 60 × 4 ) + ( 4 × 4)
= 600 + 40 + 240 + 16
= 896
By using the Distributive method,
112 × 13 = 13 ( 100 + 12 )
= (13 × 100 ) + ( 13× 12 )
= (10 +3 ) × 100 + ( 10 + 3) × 12
= ( 10 × 100 ) + ( 3× 100 ) + ( 10 × 12 ) + ( 12 × 3)
= 1,000 + 300 + 120 + 36
= 1,456
Hence,
The total length of the Pattern = 896 + 1,456 = 2,352 square cm

Question 8.
Modeling Real Life
A cargo ship has go ship has 34 rows of crates. Each row has 16 stacks of crates. There are 5 crates in each stack. The ship workers unload 862 crates. How many crates are still on the ship?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 13

Answer: 1,858 crates are still on the ship.

Explanation:
Given that a cargo ship has a go ship has 34 rows of crates and each row has 16 stacks of crates. It is also given that there are 5 crates in each stack.
So,
The total number of crates = 34 × 16 × 5 = 34 × 80
By using the Associative Property of Multiplication,
34 × 80 = 34 × (8 × 10)
= (34 × 8) × 10
= (17× 2 × 8) × 10
= 272 × 10
= 2,720
It is also given that the ship workers unload 862 crates.
Hence,
The number of crates that the ship had = 2,720 – 862 = 1,858 crates
Review & Refresh
Estimate the product.

Question 1.
4 × 85

Answer: 340

Explanation:
By using the partial products method,
4 × 85 = ( 80 + 5 ) × ( 2 + 2 )
= ( 80 × 2 ) + (80 × 2 ) + ( 5 × 2 ) + ( 5 × 2 )
= 160 + 160 + 10 + 10
= 340

Question 2.
6 × 705

Answer: 4,230

Explanation:
By using the partial products method,
6 × 705 = ( 700 + 5 ) × ( 2 + 4 )
= ( 700 × 2 ) + (700 × 4 ) + ( 5 × 2 ) + ( 5 × 4 )
= 1,400 + 2,800 + 10 + 20
= 4,230

Question 3.
8 × 7,923

Answer: 63,384

Multiply by Two-Digit Numbers Performance Task

Wind turbines convert wind to energy. Most wind or turbines have 3 blades. The blades rotate slower or faster depending on the speed of the wind. More energy is generated when the blades spin faster.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 1

Question 1.
A wind turbine rotates between 15 and 40 times in 1 minute.
a.What is the least number of times the turbine rotates in 1 hour?
b.What is the greatest number of times the turbine rotates in 1 hour?

Answer:
a) 900 times
b) 2,400 times

Explanation:
Given that a wind turbine rotates between 15 and 40 times in 1 minute.
We know that,
1 hour = 60 minutes
The wind turbine rotates minimum 15 times and maximum 40 times in a minute.
So,
The minimum number of times a wind turbine rotate in 1 hour = 15 × 60
The maximum number of times a wind turbine rotate in 1 hour = 40 × 60
Using the Associative Property of Multiplication,
15 × 60 = 15 ( 6 × 10)
= ( 15 × 6) × 10
= 90 × 10
= 900
Using the Associative Property of Multiplication,
40 × 60 = 40 ( 6 × 10)
= ( 40 × 6) × 10
= 240 × 10
= 2,400
Hence,
The minimum number of times a wind turbine rotate in 1 hour = 900 times
The maximum number of times a wind turbine rotate in 1 hour = 2,400 times

Question 2.
The tips of the turbine blades spin 5 times faster than the speed of the wind. The speed of the wind is 22 miles per hour. How fast do the blade tips spin?

Answer: 110 miles per hour
Explanation;
Given that the tips of the turbine blades spin 5 times faster than the speed of the wind. It is also given that the speed of the wind is 22 miles per hour.
So,
The speed of the blade tips = 22 × 5 = 110 miles per hour

Question 3.
A turbine farm has 7 large wind turbines. Each wind turbine can generate enough energy to power 1,485 houses. How many houses can the turbine farm power in all?

Answer: 10,395 houses

Explanation:
Given that a turbine farm has 7 large wind turbines and each wind turbine can generate enough energy to power 1,485 houses.
So,
The total number of houses that a turbine farm generates enough energy = 1,485 × 7 = 10,395 houses

Question 4.
Use the chart.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 2
a. How many times more houses are powered when the length of each blade is doubled?
b. A wind turbine has blades that are each 60 meters long. How many houses can the wind turbine power?

Answer:
a) 2 times
b) 33,000 houses
Explanation;
a) The given original length of each blade is 15 meters and 30 meters. It is also given that some houses are powered.
Now,
It is also given that the length of each blade is doubled.
So, 15 meters becomes 30 meters and 30 meters becomes 60 meters.
From this, we have to observe that the change in length also changes the energy powered in a directly proportional way.
From this, we can conclude that
The number of times more houses are powered when the length of each blade is doubled is: 2 times
b) Given that the length f each wind turbine is 60 meters long.
So, the number of houses that can be powered up = 60 × 550
Using the Associative Property of Multiplication,
550 × 60 = 550 × ( 10 × 6)
= ( 550 × 6 ) × 10
= 3300 × 10
= 33,000
Hence,
The number of houses can the wind turbine power is: 33,000 houses

Multiply by Two-Digit Numbers Activity

Multiplication Boss
Directions:
1. Each player flips 4 Number Cards and uses them in any order to create a multiplication problem with two-digit factors.
2. Each player finds the product of the two factors.
3. Players compare products. The player with the greater product takes all 8 cards.
4. If the products are equal, each player flips 4 more cards and plays again. The player with the greater product takes all 16 cards.
5. The player with the most cards at the end of the round wins
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 3

Multiply by Two-Digit Numbers Chapter Practice

4.1 Multiply by Tens

Find the product.

Question 1.
50 × 20 = _____

Answer: 1,000

Explanation:
By using the Place-value method,
50 × 20 = 50 × 2 tens
= 5 tens × 2 tens
= 10 × 10 × 10
= 1,000
So, 50 × 20 = 1,000

Question 2.
30 × 60 = _____

Answer: 1,800

Explanation:
By using the Place-value method,
30 × 60 = 30 × 6 tens
= 3 tens × 6 tens
= 18 × 10 × 10
= 1,800
So, 30 × 60 = 1,800

Question 3.
80 × 10 = _____

Answer: 800

Explanation:
By using the Place-value method,
80 × 10 = 10 × 8 tens
= 1 ten × 8 tens
= 8 × 10 × 10
= 800
So, 80 × 10 = 800

Question 4.
40 × 70 = _____

Answer: 2,800

Explanation:
By using the Place-value method,
70 × 40 = 70 × 4 tens
= 7 tens × 4 tens
= 28 × 10 × 10
= 2,800
So, 70 × 40 = 2,800

Question 5.
60 × 50 = _____

Answer: 3,000

Explanation:
By using the Place-value method,
50 × 60 = 50 × 6 tens
= 5 tens × 6 tens
= 30 × 10 × 10
= 3,000
So, 50 × 60 = 3,000

Question 6.
90 × 90 = _____

Answer: 8,100

Explanation:
By using the Place-value method,
90 × 90 = 90 × 9 tens
= 9 tens × 9 tens
= 81 × 10 × 10
= 8,100
So, 90 × 90 = 8,100

Question 7.
70 × 11 = _____

Answer: 770

Explanation:
By using the Place-value method,
70 × 11 = 11 × 7 tens
= 77 tens
= 77 × 10
= 770
So, 70 × 11 = 770

Question 8.
18 × 30 = _____

Answer: 540

Explanation:
By using the Place-value method,
30 × 18 = 18 × 3 tens
= 54 tens
= 54 × 10
= 540
So, 30 × 18 = 540

Question 9.
20 × 75 = _____

Answer: 1,500

Explanation:
By using the Place-value method,
20 × 75 = 75 × 2 tens
= 150 tens
= 150 × 10
= 1,500
So, 20 × 75 = 1,500
Find the missing factor.

Question 10.
40 × ____ = 3,200

Answer: The missing number is: 80
Explanation;
Let the missing number be X
So, 40 × X = 3,200
X = 3,200 /40 = 80
Hence, the value of X is: 80

Question 11.
_____ × 20 = 1,200

Answer:
The missing number is: 60
Explanation;
Let the missing number be X
So, 20 × X = 1,200
X = 1,200 /20 = 60
Hence, the value of X is: 60

Question 12.
30 × ____ = 2,100

Answer:
The missing number is: 70
Explanation;
Let the missing number be X
So, 30 × X = 2,100
X = 2,100 /30 = 70
Hence, the value of X is: 70

4.2 Estimate Products

Estimate the product.

Question 13.
25 × 74

Answer: 1,850

Explanation:
Let 74 be rounded to 75
By using the Partial Fraction method,
25 × 75 = ( 20 + 5) × ( 70 + 5)
= ( 20 × 70) + ( 20 × 5) + ( 5 × 70) + ( 5 × 5)
= 1,400 + 100 + 350 + 25
= 1,875
So,
25 × 74 = 1,875

Question 14.
16 × 28

Answer: 448

Explanation:
Let 16 be rounded to 15
Let 28 be rounded to 30
By using the Partial Fraction method,
15 × 30 = ( 10 + 5) × ( 25 + 5)
= ( 10 × 25) + ( 10 × 5) + ( 5 × 25) + ( 5 × 5)
= 200 + 50 + 125 + 25
= 400
So,
16 × 28 = 400

Question 15.
42 × 81

Answer: 3,402

Explanation:
Let 42 be rounded to 40
Let 81 be rounded to 80
By using the tens method,
40 × 80 = 40 × 8 tens
= 4 tens × 8 tens
= 32 × 1 ten × 1 ten
= 32 × 10 × 10
= 3,200
So,
42 × 81 = 3,200
Open-Ended
Write two possible factors that can be estimated as shown.

Question 16.
8,100
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 16

Answer:

Explanation:
The Products of 81 are:
9 × 9 = 81
From the above two products, we can conclude that the two possible numbers that can give the product 6,400 are: 90,90

Question 17.
400
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 17

Answer:

Explanation:
The Products of 64 are:
2 × 2 = 4
4 × 1 =4
From the above two products, we can conclude that the two possible numbers that can give the product 400 are: 20,20 and. 10,40

4.3 Use Area Models to Multiply Two-Digit Numbers

Draw an area model to find the product.

Question 18.
13 × 19 = _____

Answer: 247

Explanation:
By using the Partial Products method,
13 × 19 = ( 10 + 3) × ( 10 + 9)
10 × 10 +  10× 9 + 10 × 3 + 3 × 9
= 100 + 90 + 30 + 27
= 247
So,
19 × 13 = 247

Question 19.
21 × 36 = _____

Answer: 756

Explanation:
By using the Partial Products method,
21 × 36 = ( 20 + 1) × ( 30 + 6)
= 20 × 30 +  20× 6 + 30 × 1 + 1 × 6
= 600 + 120 + 30 + 6
= 756
So,
21 × 36 = 756

Question 20.
YOU BE THE TEACHER
Your friend finds 28 × 24. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 20

Answer: Your friend is not correct.
Explanation;

By using the partial Products method,
400 + 160 +80 + 32 = 672
From the above,
we can conclude that in place of 16, 160 must be placed.

4.4 Use Distributive Property to Multiply Two-Digit Numbers

Use Distributive Property to find a product.

Question 21.
27 × 34 = 27 × (30 + 4)
= (27 × 30) + (27 × 4)
= (20 + 7) × 30 + (20 + 7) × 4
= (20 ×30) + (7 × 30) + (20 ×4) + (7 × 4)
= 600 + 210 + 80 + 28
= 918
So, 27 × 34 = 918.

Question 22.
43 × 18 = _____

Answer: 774

Explanation: Using the Partial Products method,
43 × 18 = 43× (10 + 8)
= (43 × 10) + (43 × 8)
= (40 + 3) × 10 + (40 + 3) × 8
= (40 ×10) + (3 × 10) + (40 ×8) + (3 × 8)
= 400 + 30 + 320 + 24
= 774
So, 43 × 18 = 774

Question 23.
35 × 57 = _____

Answer: 1,995

Explanation: Using the Partial products method,
35 × 57 = 35 × (50 + 7)
= (35 × 50) + (35 × 7)
= (30 + 5) × 50 + (30 + 5) × 7
= (50 ×30) + (5 × 50) + (30 ×7) + (7 × 5)
= 1,500 + 250 + 210 + 35
= 1,995
So, 35 × 57 = 1,995

Question 24.
81 × 76 = _____

Answer: 6,156
81 × 76 = 81 × (70 + 6)
= (81 × 70) + (81 × 6)
= (80 + 1) × 70 + (80 + 1) × 6
= (80 ×70) + (70 × 1) + (80 ×6) + (1 × 6)
= 5,600 + 70 + 480 + 6
= 6,156
So, 81 × 76 = 6,156

4.5 Use Partial Products to Multiply Two-Digit Numbers

Find the product. Check whether your answer is reasonable.

Question 25.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 25

Answer: 396

Explanation: Using the Partial Products method,
18 × 22 = 22 × (10 + 8)
= (22 × 10) + (22 × 8)
= (20 + 2) × 10 + (20 + 2) × 8
= (20 ×10) + (2 × 10) + (20 ×8) + (2 × 8)
= 200 + 20 + 160 + 16
= 396
So, 18 × 22 = 396

Question 26.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 26

Answer: 3,358

Explanation: Using the Partial Products method,
73 × 46 = 73 × (40 + 6)
= (73 × 40) + (73 × 6)
= (70 + 3) × 40 + (70 + 3) × 6
= (70 ×40) + (3 × 40) + (70 ×6) + (3 × 6)
= 2,800 + 120 + 420 + 18
= 3,358
So, 73 × 46 = 3,358.

Question 27.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 27

Answer: 3,276

Explanation: By using the Partial Products method,
39 × 84 = 39 × (80 + 4)
= (39 × 80) + (39 × 4)
= (30 + 9) × 80 + (30 + 9) × 4
= (80 ×30) + (9 × 80) + (30 ×4) + (9 × 4)
= 2,400 + 720 + 120 + 36
= 3,276
So, 39 × 84 = 3,276

Question 28.
57 × 19 = _____

Answer: 1,083

Explanation: By using the Partial Products method,
57 × 19 = 57 × (10 + 9)
= (57 × 10) + (57 × 9)
= (50 + 7) × 10 + (50 + 7) × 9
= (50 ×10) + (7 × 10) + (50 ×9) + (7 × 9)
= 500 + 70 + 450 + 63
= 1,083
So, 57 × 19 = 1,083

Question 29.
38 × 65 = _____

Answer: 2,470

Explanation: By using the Partial Products method,
38 × 65 = 65 × (30 + 8)
= (65 × 30) + (65 × 8)
= (60 + 5) × 30 + (60 + 5) × 8
= (60 ×30) + (5 × 30) + (60 ×8) + (5 × 8)
= 1,800 + 150 + 480 + 40
= 2,470
So, 38 × 65 = 2,470

Question 30.
94 × 26 = _____

Answer: 2,444

Explanation: By using the Partial Products method,
94 ×26 = 94 × (20 + 6)
= (94 × 20) + (94 × 6)
= (90 + 4) × 20 + (90 + 4) × 6
= (90 ×20) + (4 × 20) + (90 ×6) + (4 × 6)
= 1,800 + 80 + 540 + 24
= 2,444
So, 94 × 26 = 2,444
Reasoning
Find the missing digits. Then find the product.

Question 31.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 31

Answer:

Question 32.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 32

Answer:

4.6 Multiply Two-Digit Numbers

Find the product. Check whether your answer is reasonable.

Question 33.
Estimate: _____
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 33

Answer: 2,108

Explanation:
Using the Partial Products method,
34 × 62 = ( 60 + 2) × ( 30 + 4)
= 30 × 60 + 4 × 60 + 30 × 2 + 2 × 4
= 1,800 + 240 + 60 + 8
= 2,108
Estimate:
Let 62 be Rounded to 60.
Let 34 be Rounded to 35.
So, 60 × 35 = 2,100
As the Estimate and the actual answer are near, the answer is reasonable.

Question 34.
Estimate: ____
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 34

Answer: 7,917

Explanation:
Using the Partial Products method,
87 × 91 = ( 80 + 7) × ( 90 + 1)
= 80 × 90 + 7 × 90 + 80 × 1 + 7 × 1
= 7,200 + 630 + 80 + 7
= 7,917
Estimate:
Let 87 be Rounded to 85.
Let 91 be Rounded to 90.
So, 90 × 85 = 7,650
As the Estimate and the actual answer are not near, the answer is  not reasonable.

Question 35.
Estimate: _____
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 35

Answer: 3,285

Explanation:
Using the Partial Products method,
73 × 45 = ( 70 + 3) × ( 40 + 5)
= 70 × 40 + 3 × 40 + 70 × 5 + 5 × 3
= 2,800 + 120 + 350 + 15
= 3,285
Estimate:
Let 73 be Rounded to 75.
So, 45 × 75 = 3,375
As the Estimate and the actual answer are near, the answer is reasonable.

Question 36.
Estimate: ____
13 × 21 = ______

Answer: 273

Explanation:
Using the Partial Products method,
13 × 21 = ( 20 + 1) × ( 10 + 3)
= 20 × 10 + 20 × 3 + 10 × 1 + 1 × 3
= 200 + 60 + 10 + 3
= 273
Estimate:
Let 13 be Rounded to 15.
Let 21 be Rounded to 20.
So, 20 × 15 = 300
As the Estimate and the actual answer are near, the answer is reasonable.

Question 37.
Estimate: _____
42 × 53 = _____

Answer: 2,226

Explanation:
Using the Partial Products method,
42 × 53 = ( 40 + 2) × ( 50 + 3)
= 40 × 50 + 3 × 40 + 50 × 2 + 2 × 3
= 2,000 + 120 + 100 + 6
= 2,226
Estimate:
Let 42 be Rounded to 40.
Let 53 be Rounded to 55.
So, 40 × 55 = 2,200
As the Estimate and the actual answer are near, the answer is reasonable.

Question 38.
Estimate: _____
29 × 66 = _____

Answer: 1,914

Explanation:
Using the Partial Products method,
29 × 66 = ( 20 + 9) × ( 60 + 6)
= 20 × 60 + 9 × 60 + 20 × 6 + 9 × 6
= 1,200 + 540 + 120 + 54
= 1,914
Estimate:
Let 29 be Rounded to 30.
Let 66 be Rounded to 65.
So, 30 × 65 = 1,950
As the Estimate and the actual answer are near, the answer is reasonable.

4.7 Practice Multiplication Strategies

Find the product.

Question 39.
80 × 30 = _____

Answer: 2,400

Explanation: Using the Place-value method,
80 × 30 = 80 × 3 tens
= 240 tens
= 2,400
So, 80 × 30 = 2,400

Question 40.
26 × 51 = _____

Answer: 1,326

Explanation: Using the Partial Products method,
26 × 51 = ( 20 + 6) × ( 50 + 1)
= ( 20 × 50) + ( 20 × 1) + ( 6 × 50) + (6 × 1)
= 1,000 + 20 + 300 + 6
= 1,326

Question 41.
94 × 70 = _____

Answer: 6,580

Explanation: Using the Place-value method,
94 × 70 = 94 × 7 tens
= 658 tens
= 6,580
So, 94 × 70 = 6,580

Question 42.
15 × 67 = _____

Answer: 1,005

Explanation: Using the Partial Products method,
15 × 67 = ( 10 + 5) × ( 60 + 7)
= ( 10 × 60) + ( 10 × 7) + ( 5 × 60) + ( 5 × 7)
= 600 + 70 + 300 + 35
=1,005

Question 43.
40 × 38 = _____

Answer: 1,520

Explanation: Using the Place-value method,
38 × 40 = 38 × 4 tens
= 152 tens
= 1,520
So, 38 × 40 = 1,520

Question 44.
29 × 92 = _____

Answer: 2,668

Explanation: Using the partial products method,
29 × 92 = ( 20 + 9) × ( 90 + 2)
= ( 20 × 90 ) + ( 20 × 2) + ( 9 × 90) + ( 9 × 2)
= 1,800 + 40 + 810 + 18
= 2,668

Question 45.
Modeling Real Life
A Ferris wheel runs 40 times each day. It has 16 cars with 4 seats in each car. If the Ferris wheel is full each time it runs, how many people will ride it in 1 day?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 45

Answer: 2,560 people will ride the Ferris wheel in 1 day

Explanation:
Given that a  Ferris wheel runs 40 times each day and it has 16 cars with 4 seats in each car.
So, the total number of people that can ride a Ferris wheel = 40 × 16 × 4 = 40 × 64
By using the place-value method,
64  × 40  = 64 × 4 tens
= 256 tens
= 2,560
From the above,
We can conclude that there are 2,560 people who will ride the Ferris wheel in 1 day
4.8 Problem Solving: Multiplication with Two-Digit Numbers

Question 46.
A music fan memorizes 59 songs for a concert. Her goal is to memorize all of the songs from 13 albums. There are 15 songs on each album. How many more songs does the music fan still need to memorize?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 46

Answer: 136 songs

Explanation:
Given that a music fan has to memorize all of the songs from 13 albums and there are 15 songs on each album.
So, the total number of songs the music fan has to memorize = 13 × 15
By using the Partial Products method,
13 × 15 = ( 10 + 3) × ( 10 + 5)
= ( 10 × 10 ) + ( 10 × 5) + ( 3 × 10 ) + ( 3 × 5)
= 100 + 50 + 30 + 15
= 195
It is also given that the music fan memorized 59 songs for the Concert.
Hence,
The number of songs that has to memorize by the music fan = 195 – 59 = 136 songs

Question 47.
Find the area of the Jamaican flag.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 47

Answer: 2,592 inches
Explanation;
The given Jamaican flag is in the shape of a rectangle.
We know that,
Area of the rectangle = length × breadth
So,
Area of the Jamaican flag = 36 × ( 36 + 36) = 36 × 72
Using the Partial Products method,
36 × 72 = ( 30 + 6) × ( 70 + 2)
= ( 30 × 70 ) + ( 30 × 2 ) + ( 6 × 70) + ( 6 × 2)
= 2,100 + 60 + 720 + 12
= 2,592 inches
From this,
We can conclude that the area of the Jamaican flag is: 2,592 inches

Final Words:

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Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers

Big Ideas Math Answers Grade 1 Chapter 7

Students who are unable to solve the textbook problems can go through the Big Ideas Math Answers 1st Grade Chapter 7 Compare Two-Digit Numbers. The provided solutions to all the questions asked from Compare Two-Digit Numbers concepts will help you in real-time calculations also. Refer to Grade 1 Bigideas Math Answer Key of Ch 7 Compare Two-Digit Numbers to understand the solving methods in depth. Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers was explained by the math professionals in a simple way.

Big Ideas Math Book 1st Grade Answer Key Chapter 7 Compare Two-Digit Numbers

A major goal of our discussion is to explore the knowledge required to teach effectively to students. Go through the below attached links and get the Big Ideas Math 1st Grade 1 Answer Key for Chapter 7 Compare Two-Digit Numbers for better preparation and score best marks in the exams. Thus make use of these topic-wise links and solve each and every question covered in this Bigideas Math Answers Grade 1 Ch 7 Compare Two-Digit Numbers.

Lesson: 1 – Compare Numbers 11 to 19

Lesson: 2 – Compare Numbers

Lesson: 3 – Compare Numbers Using Place Value

Lesson 4: Compare Numbers Using Symbols

Lesson: 5 – Compare Numbers Using a Number Line

Lesson 6: 1 – More, 1 Less; 10 More, 10 Less

Performance Task

Compare Two-Digit Numbers Vocabulary

Organize It

Review Words:
fewer
more

Use the review words to complete the graphic organizer.

Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 1
Answer:
4 strawberries are more, 2 strawberries are fewer.

Explanation:
By seeing the above figure we can say that the first set has more number of strawberries than the second set.

Define It:

Use your vocabulary cards to complete the puzzle.

Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 2
Answer:
1 ) Greater than
2 ) Compare
3 ) Less than

Explanation:
we have to write the signs in the word form in the puzzle.
Use the clues given in the question and solve the puzzle.
Across means, we have to write the answer in the horizontal.
Down means, we have to write the answer in the vertical.

Lesson 7.1 Compare Numbers 11 to 19

Explore and Grow

Model each number. Circle the greater number.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 3
15

17
Answer: 17 is greater.

Explanation:
Compare the number of blocks of two whole numbers.
15 is smaller than 17.
Thus 17 is greater

Show and Grow

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 4
Answer:
15 is greater than 12 and 12 is less than 15.

Explanation:
Compare the number of blocks of two whole numbers.
15 is greater than 12.
12 is smaller than 15

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 5
Answer:
18 is greater than 14 and 14 is less than 18.

Explanation:
Compare the number of blocks of two whole numbers.
18 is greater than 14
14 is smaller than 18

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 6
Answer:
13 is greater than 11 and 11 is less than 13.

Explanation:
Compare the number of blocks of two whole numbers.
13 is greater than 11
11 is smaller than 13

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 7
Answer:
19 is greater than 17 and 17 is less than 19.

Explanation:
Compare the number of blocks of two whole numbers.
19 is greater than 17
17 is smaller than 19

Apply and Grow: Practice

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 8
Answer:
18 is greater than 16 and 16 is less than 18.

Explanation:
Compare the number of blocks of two whole numbers.
18 is greater than 16
16 is smaller than 18

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 9
Answer:
19 is greater than 15 and 15 is less than 19.

Explanation:
Compare the number of blocks of two whole numbers.
19 is greater than 15
15 is smaller than 19

Make quick sketches to compare the numbers.

Question 7.

Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 10
Answer:
11 < 12 and 12 > 11

Explanation:
Compare the number of blocks of two whole numbers.
12 is greater than 11
11 is smaller than 12

Question 8.

Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 11
Answer:
15 > 13 and 13 < 15

Explanation:
Compare the number of blocks of two whole numbers.
15 is greater than 13
13 is smaller than 15.

Question 9.
DIG DEEPER
Choose two numbers to complete the sentences.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 12
__________ is greater than ________ _________ is less than ________ .

Answer:
13 is greater than 12 and 12 is less than 13
16 is greater than 15 and 15 is less than 16
19 is greater than 18 and 18 is less than 19.

Explanation:
There are group of numbers
comparing each number with the other
and representing greater numbers and lesser numbers.

Think and Grow: Modeling Real Life

You have 16 tickets. Your friend has 11 tickets and wins 8 more. Who has more tickets?
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 13
You                Friend
Number of tickets:
Models:
Compare: _________ is greater than _________.
Who has more tickets? You             Friend

Answer:
Number of tickets: 35
Models: Tickets
Compare:
19 is greater than 16.

Explanation:
I have 16 tickets and my friend has 11 tickets and wins 8 more.
My friend has more tickets.

Show and Grow

Question 10.
You have 7 feathers and find 6 more. Your friend has 12 feathers. Who has more feathers?
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 14
You            Friend
Number of feathers:
Models:
Compare: _________ is greater than _________.
Who has more feathers? You             Friend

Answer:
Number of feathers: 25
Models: Feathers
Compare: 13 is greater than 12. I have more feathers.

Explanation: I have 7 feathers and found 6 more
6 + 7 = 13
and my friend has 12 featherswhich is lesser than me.

Compare Numbers 11 to 19 Practice 7.1

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 15
Answer:
16 is greater than 12 and 12 is less than 16.

Explanation:
Compare the number of blocks of two whole numbers.
16 is greater than 12
12 is smaller than 16.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 16
Answer:
18 is greater than 13 and 13 is less than 18.

Explanation:
Compare the number of blocks of two whole numbers.
18 is greater than 13
13 is smaller than 18.

Make quick sketches to compare the numbers.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 17
Answer:
19 > 12 and 12 < 19.

Explanation:
Compare the number of blocks of two whole numbers.
19 is greater than 12
12 is smaller than 19.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 18
Answer:
11 < 17 and 17 > 11

Explanation:
Compare the number of blocks of two whole numbers.
17 is greater than 11
11 is smaller than 17.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 19
Answer:
17 > 13 and 13 < 17

Explanation:
Compare the number of blocks of two whole numbers.
17 is greater than 13
13 is smaller than 17.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 20
Answer:
12 < 14 and 14 > 12.

Explanation:
Compare the number of blocks of two whole numbers.
14 is greater than 12
12 is smaller than 14.

Question 7.
DIG DEEPER
Choose two numbers to complete the sentences.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 21
__________ is greater than ________ .

__________ is less than ________ .
Answer:
12 is greater than 11 and 11 is less than 12.
14 is less than 15 and 15 is greater than 14.
17 is less than 18 and 18 is greater than 17.

Explanation:
There are group of numbers
comparing each number with the other
and representing greater numbers and lesser numbers.

Question 8.
Modeling Real Life
Your tower has 16 red blocks. Your friend’s tower has 10 red blocks and 4 blue blocks. Who uses more blocks?
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 22
__________ is greater than ________ .
Who has more? You Friend
Answer:
Me has 16 and my friend has 14.
So, 16 is greater than 14. Me has more.

Explanation:
My towel has 16 red blocks.
my friend has 10 red 4 blue in total 14 blocks
so, me has more blocks than my friend.

Review & Refresh

Question 9.
Make a quick sketch to complete the sentence.
59
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 23
________ is ________ tens and ________ ones.
Answer:
59 is 5 tens and 9 ones.

Explanation:
My friend has 59 choclates
there are sepereted with tens and ones.

Lesson 7.2 Compare Numbers

Explore and Grow

Model each number. Circle the greater number.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 24
34

26
Answer:
34 is greater.

Explanation:
Compare the number of blocks of two whole numbers.
34 is greater than 26
thus 34 is greater.

Show and Grow

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 25
Answer:
50 is greater than 48 and 48 is less than 50.

Explanation:
Compare the number of blocks of two whole numbers.
50 is greater than 48
48 is smaller than 50

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 26
Answer:
37 is greater than 34 and 34 is less than 37.

Explanation:
Compare the number of blocks of two whole numbers.
37 is greater than 34
34 is smaller than 37

Apply and Grow: Practice

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 27
Answer:
30 is greater than 28 and 28 is less than 30.

Explanation:
Compare the number of blocks of two whole numbers.
30 is greater than 28
28 is smaller than 30.

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 28
Answer:
45 is greater than 41 and 41 is less than 45.

Explanation:
Compare the number of blocks of two whole numbers.
45 is greater than 41
41 is smaller than 45.

Question 5.
Make quick sketches to compare the numbers.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 29
Answer:
80 > 63 and 63 < 80.

Explanation:
Compare the number of blocks of two whole numbers.
80 is greater than 63
63 is smaller than 80

Question 6.
DIG DEEPER!
Write a number that is greater than 90 but less than 94. Show how you know.
____________
Answer:
92

Explanation:
Is less than 94 the number which is greater than 90 is 91,92,93.
in 120 chart the number line wise we can see as 90, 91, 92, 93

Think and Grow: Modeling Real Life

Newton collects 61 acorns. Descartes collects 75 acorns. Who collects more acorns?
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 30
Models: Newton              Descartes

Compare: ___________ is greater than ___________ .
Who collects more acorns? Newton               Descartes
Answer:
75 is greater than 61. Descartes collects more acorns.

Explanation:
Newton collects 61 acorns.
Descartes collects 75 acornes.
75 is greater than 61
so, descartes has more.

Show and Grow

Question 7.
You pick 57 blueberries. Your friend picks 53 blueberries. Who picks more blueberries?
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 31
Models: You              Friend

Compare: ____________ is greater than ____________ .
Who pics more blueberries? You            Friend
Answer:
57 is greater than 53.
You picks more blueberries.

Explanation:
I pick 57 blueberries.
My friend has 53 blueberries
57 > 53 so, I collected more blueberries.

Compare Numbers Practice 7.2

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 32
Answer:
64 is greater than 46 and 46 is less than 64.

Explanation:
Compare the number of blocks of two whole numbers.
64 is greater than 46
46 is smaller than 64

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 33
Answer:
39 is greater than 35 and 35 is less than 39.

Explanation:
Compare the number of blocks of two whole numbers.
39 is greater than 35
35 is smaller than 39

Question 3.
Make quick sketches to compare the numbers.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 34
Answer:
91 > 70 and 70 < 91

Explanation:
Compare the number of blocks of two whole numbers.
91 is greater than 70
70 is smaller than 91

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 35
Answer:
86 > 68 and 68 < 86.

Explanation:
Compare the number of blocks of two whole numbers.
86 is greater than 68
68 is smaller than 86

Question 5.
MP Number Sense
Choose 2 numbers between 50 and 99. Write a sentence to compare the numbers.

___________                              ____________
Answer:
51 and 98.
These two numbers are in between 50 and 99.

Explanation:
There are apples on the tree from 50 to 90
A monkey came over there and plucked 51 and 98
It is monkeys choice it can pluck any of the fruit from 50 to 99.

Question 6.
Modeling Real Life
You collect 37 stamps. Your friend collects 27 stamps. Who collects more stamps?
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 36

Who collects more stamps? You               Friend
Answer:
You collect more stamps.

Explanation:
I have collected 37 stams
and my friend collected 27 stamps
37 > 27
so, I collected more stamps.

Review & Refresh

Circle the taller object.

Question 7.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 37
Answer:

The second lamp is taller.

Explanation:
Cubes are the measuring attribute
one lamp is compared with the other lamp
2 is taller than the 1

Question 8.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 38
Answer:
The first building is taller.

Explanation:
Cubes are the measuring attribute
one building is compared with the other building
1 is taller than the 2

Lesson 7.3 Compare Numbers Using Place Value

Explore and Grow

Model each number. What is the same about the models? What is different? Circle the greater number.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 39

32
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 40

23
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 40
Answer:

Explanation:
3 tens and 2 ones
2 tens and 3 ones. 32 is greater. This about tens and ones.

Show and Grow

Compare. Which digits help you decide?

Question 1.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 41
__________ tens __________ one                          __________ tens __________ one
Answer:
6 tens and 1 ones, 5 tens and 3 ones.
61 is greater than 53 and 53 is less than 61.

Explanation:
Compare the number of two whole numbers.
61 is greater than 53
53 is smaller than 61.

Question 2.

Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 42
__________ tens __________ ones                        __________ tens __________ one
Answer:
7 tens and 0 ones, 7 tens and 4 ones.

74 is greater than 70 and 70 is less than 74.

Apply and Grow: Practice

Compare. Which digits help you decide?

Question 3.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 43
__________ tens __________ ones                    __________ tens __________ one
Answer:
3 tens and 9 ones, 4 tens and 8 ones.
48 is greater than 39 and 39 is less than 48.
48 > 39 and 39 < 48

Question 4.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 44
__________ tens __________ ones                    __________ tens __________ one
Answer:
8 tens and 0 ones, 6 tens and 2 ones.
80 > 62 and 62 < 80.

Question 5.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 45
Answer:
26 > 23 and 23 < 26.

Explanation:
Compare the number of blocks of two whole numbers.
26 is greater than 23
23 is smaller than 26.

Question 6.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 46
Answer:
51 < 86 and 86 > 51.

Explanation:
Compare the number of blocks of two whole numbers.
81 is greater than 51
51 is smaller than 86.

Question 7.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 47
Answer:
17 < 71 and 71 > 17.

Explanation:
Compare the number of blocks of two whole numbers.
71 is greater than 17
17 is smaller than 71.

Question 8.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 48
Answer:
97 > 92 and 92 < 97

Explanation:
Compare the number of blocks of two whole numbers.
97 is greater than 92
92 is smaller than 97.

Question 9.
MP Precision
Match each ball with its bucket.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 49
Answer:
less than 65 is 58 and 62 and 64,
Greater than 65 is 67, 68 and 73.

Explanation:
There are 2 buckets first bucket collects less than 65
and second bucket collects greater than 65
There are some ballons with 58, 67, 62, 64, 73, 68.
They are seperated according to the representaion.

Think and Grow: Modeling Real Life

Who has more points?
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 50

Compare: _________ is greater than ________ .
_________ has more points.
Answer:
8 is greater than 7,
Descartes has more points.

Explanation:
Newton has 7 points and descartes has 8 points
descartes has more

Show and Grow:

Question 10.
Who has more points?
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 51
Compare: _________ is greater than ________ .
_________ has more points.
Answer:
8 is greater than 2,
Newton has more points.

Explanation:
Newton has 8 points and descartes has 7 points
so, newton has more points

Compare Numbers Using Place Value Practice 7.3

Compare. Which digits help you decide?

Question 1.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 52
__________ ten __________ ones                   __________ tens __________ ones
Answer:
1 ten and 6 ones,1 tens and 2 ones.
16 > 12 and 12 < 16.

Explanation:
16 can be written as one ten and six ones
12 can be written as one ten and two ones.
16 is greater than 12
12 is less than 16

Question 2.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 53
Answer:
32 < 38 and 38 > 32.

Explanation:
Compare the number of blocks of two whole numbers.
38 is greater than 32
32 is smaller than 38.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 54
Answer:
7 < 25 and 25 > 7.

Explanation:
Compare the number of blocks of two whole numbers.
25 is greater than 7
7 is smaller than 25.

Question 4.
MP Precision
Match each card with its pile.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 55
Answer:
Less than 85 are 79 and 81 and 84,
Greater than 85 are 87,89 and 93.

Explanation:
There are two walls first wall represents less than 85 and
second wall represents greater than 85
there are stickers numbered 93, 79, 84, 81, 87, 89
and placed according to the walls required.

Question 5.
Modeling Real Life
Who earns more points?
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 56
____________ is greater than ___________ .
Who earns more points? You            Friend
Answer: 41 is greater than 14

Explanation:
In you first basket 1 ball = 10 that is 4 x 10 = 40
second basket 1 ball = 1 total 40+1 = 41
In friend first basket 1 ball= 10 that is 1 x 10 = 10
second basket 4 balls= 4 x 1= 4 total 10+ 4 = 14
41 is greaterthan 14

Review & Refresh

Is the equation true or false?

Question 6.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 57
Answer:
True
15 – 9 = 6 and 11 – 5 = 6

Explanation:
jim has 15 choclates in that 9 has given to his sister
so 6 remaining
rim has 11 choclates in that 5 has given to his sister
so 6 is remaining
so both sisters have 6 each.

Question 7.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 58
Answer:
True
4 + 7 = 11 and 16 – 4 = 11

Explanation:
jim has 4 choclates and 7 has given to his sister
so 11 in all
rim has 16 choclates in that 4 has given to his sister
so 11 is remaining
so both sisters have 11 each.

Lesson 7.4 Compare Numbers Using Symbols

Explore and Grow

Make quick sketches. Complete the sentences.

Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 59
Answer:

25 < 27, 25 + 2 = 27, 28 > 27

Show and Grow

Make quick sketches to model each number. Compare.

Question 1.
53 ____________ 39.
53 ○ 39
Answer:
53 is greater than 39,
53 > 39.

Explanation:
Comaring the two whole numbers 53 and 39
53 is greater than 39.

Question 2.
28 _____________ 34.
28 ○ 34
Answer:
28 is less than 34,
28 < 34.

Explanation:
Comaring the two whole numbers 28 and 34
34 is greater than 28.

Apply and Grow: Practice

Question 3.
61 ______________ 72
61 ○ 72
Answer:
61 is Less than 72.
61 < 72

Explanation:
Comaring the two whole numbers 61 and 72
72 is greater than 71.

Question 4.
47 ______________ 47.
47 ○ 47
Answer:
47 is equal to 47.
47 = 47

Explanation:
Comaring the two whole numbers 41 and 41
41 is equal to 41.

Compare.

Question 5.
83 _______________ 49.
83 ○ 49
Answer:
83 is greater than 49,
83 > 49.

Explanation:
Comaring the two whole numbers 83 and 49
83 is greater than 49.

Question 6.
75 ______________ 99.
75 ○ 99
Answer:
75 is less than 99,
75 < 99.

Explanation:
Comaring the two whole numbers 75 and 99
99 is greater than 75.

Question 7.
54 ○ 70
Answer:
54 is less than 70,
54 < 70.

Explanation:
Comaring the two whole numbers 54 and 70
70 is greater than 54..

Question 8.
17 ○ 9
Answer:
17 is greater than 9,
17 > 9.

Explanation:
Comaring the two whole numbers 17 and 9
17 is greater than 9.

Question 9.
86 ○ 86
Answer:
86 is equal to 86,
86 = 86.

Explanation:
Comaring the two whole numbers 86 and 86
86 is equal to 86

Question 10.
DIG DEEPER!
19 Choose 2 cards for each problem. Compare the numbers. Use each card once.
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 60
Answer: 13 < 24 |  89 > 63 | 24 = 24.

Think and Grow: Modeling Real Life

You have 90 beads. Your friend has 75 beads. Who has more beads?
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 61
Models:

Compare: __________ ○ ___________
who has more beads? You Friend
Answer:
Me has more beads, 90 > 75.

Explanation:
I have 90 beads and my friend has 75 beads
90 > 75
so, my friend has more.

Show and Grow

Question 11.
You have 48 toy figures. Your friend has 54 toy figures. Who has fewer toy figures?
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 62
Models:

Compare: __________ ○ ___________
who has more beads? You Friend
Answer:
My friend has more toys, Me has fewer toys. 48 < 54.

Explanation:
I have 48 toy figures and my friend has 54 toy figures
48 < 54
so, i have fewer toys

Compare Numbers Using Symbols Practice 7.4

Make quick sketches to model each number. Compare.

Question 1.
37 ______________ 15.
37 ○ 15
Answer:
37 is greater than 35,
37 > 35.

Explanation:
Comaring the two whole numbers 37 and 35
37 is greater than 35.

Question 2.
22 ______________ 22.
22 ○ 22
Answer:
22 is equal to 22,
22 = 22.

Explanation:
Comaring the two whole numbers 22 and 22
22 is equal to 22.

Compare.

Question 3.
97 ○ 79
Answer:
97 is greater than 79,
97 > 79.

Explanation:
Comaring the two whole numbers 97 and 79
97 is greater than 79.

Question 4.
51 ○ 83
Answer:
51 is lesser than 83,
51 < 83.

Explanation:
Comaring the two whole numbers 51 and 83
83 is greater than 51.

Question 5.
39 ○ 62
Answer:
39 is less than 62,
39 < 62.

Explanation:
Comaring the two whole numbers 62 and 39
62 is greater than 39.

Compare.

Question 6.
19 ○ 91
Answer:
19 is less than 91,
19 < 91.

Explanation:
Comaring the two whole numbers 19 and 91
91 is greater than 19.

Question 7.
73 ○ 68
Answer:
73 is greater than 68,
73 > 68.

Explanation:
Comaring the two whole numbers 73 and 68
73 is greater than 68.

Question 8.
32 ○ 32
Answer:
32 is equal to 32,
32 = 32.

Explanation:
Comaring the two whole numbers 32 and 32
32 is equal to 32

Question 9.
DIG DEEPER!
Use each of the numbers once to complete the puzzle.
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 63
81 > __________
56 < __________
21 = __________
Answer:
81 > 45
56 < 74
21 = 21

Question 10.
Modeling Real Life
Who has more points?
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 64
__________ ○ _________
___________ has more points.
Answer:
37 < 68,
Descartes has more points.

Review & Refresh

Question 11.
6 + 8 = ___________
Answer: 14

Explanation:
To put two or more numbers or amounts together to get a total is adding
6 + 8 = 14

Question 12.
12 + 5 = ___________
Answer: 17

Explanation:
To put two or more numbers or amounts together to get a total is adding
12 + 5 = 17

Question 13.
___________ = 0 + 11
Answer: 11

Explanation:
To put two or more numbers or amounts together to get a total is adding
11= 0 + 11

Question 14.
___________ = 4 + 9
Answer: 13

Explanation:
To put two or more numbers or amounts together to get a total is adding
13 = 4 + 9.

Lesson 7.5 Compare Numbers Using a Number Line

Explore and Grow

Circle a number that is less than 45. Underline a number that is greater than 45. How do you know you are correct?

Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 65
Answer:
43 is circled and it is less than 45.
48 is underlined and it is greater than 45.
As the number line shows 40 to 50

Show and Grow

Compare.

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 66
43 ○ 48
Answer:
43 is less than 48, 43< 48.

Explanation:
compared the two whole numbers 43 and 48
and represented on the number line

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 66
44 ○ 36
Answer:
44 is greater than 36, 44 > 36.

Explanation:
compared the two whole numbers 44 and 36
and represented on the number line

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 66
39 ○ 39
Answer:
39 is equal to 39, 39 = 39.

Explanation:
compared the two whole numbers 39 and 39
and represented on the number line

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 66
31 ○ 50
Answer:
31 is less than 50, 31< 50.

compared the two whole numbers 31 and 50
and represented on the number line

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 66
37 ○ 33
Answer:
37 is greater than 33, 37 > 33.

Explanation:
compared the two whole numbers 37 and 33
and represented on the number line

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 66
38 ○ 42
Answer:
38 is less than 42, 38 < 42

Explanation:
compared the two whole numbers 42 and 38
and represented on the number line

Apply and Grow: Practice

Compare.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 71
65 ○ 66
Answer:
65 is less than 66, 65 < 66.

Explanation:
compared the two whole numbers 65 and 66
and represented on the number line

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 71
71 ○ 81
Answer:
71 is less than 81, 71 < 81.

Explanation:
compared the two whole numbers 71 and 81
and represented on the number line

Question 9.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 71
83 ○ 85
Answer:
83 < 85, 83 is less than 85.

Explanation:
compared the two whole numbers 83 and 85.
and represented on the number line

Question 10.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 71
74 ○ 69
Answer:
74 > 69, 74 is greater than 69.

Explanation:
compared the two whole numbers 74 and 69
and represented on the number line

Question 11.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 71
78 ○ 77
Answer:
78 is greater than 77, 78 > 77.

Explanation:
compared the two whole numbers 78 and 77
and represented on the number line

Question 12.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 71
72 ○ 72
Answer:
72 is equal to 72

Explanation:
compared the two whole numbers 72 and 72
and represented on the number line

Write a number that makes the statement true.

Question 13.
_________ > 47
Answer:
48 > 47.

Explanation:
compared the two whole numbers 47 and 48
and represented on the number line

Question 14.
_________ < 76
Answer:
75 < 76.

Explanation:
The below number line shows 75 is less than 76
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 71

Question 15.
81 = ________
Answer:
81 = 81.

Explanation:
The below number line shows 81 is equal to 81
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 71

Question 16.
DIG DEEPER!
Newton is thinking of a number that is less than 83 and greater than 74. His number has 6 ones. What is Newton’s number?
__________
Answer: 76,

Explanation:
Newton has 7 tens and 6 ones

Think and Grow: Modeling Real Life

The number on your bus is less than 91. Which buses can be yours?
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 67
Show how you know:
Answer:
88 and 90 can be mine.

Show and Grow

Question 17.
The number on your plane is greater than 58. Which planes can be yours?
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 68
Show how you know:
Answer:
It might be 60 or 63
By seeing the number on plane.

Compare Numbers Using a Number Line Practice 7.5

Compare.

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 69
53 ○ 56
Answer:
53 < 56, 53 is less than 56.

Explanation:
compared the two whole numbers 53 and 56
and represented on the number line

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 69
70 ○ 60
Answer:
70 is greater than 60, 70 > 60

Explanation:
compared the two whole numbers 70 and 60
and represented on the number line

Question 3.
63 ○ 67
Answer:
63 is less than 67, 63 < 67.

Explanation:
compared the two whole numbers 63 and 67
and represented on the number line

Question 4.
68 ○ 55
Answer:
68 is greater than 55, 68 > 55.

Explanation:
compared the two whole numbers 55 and 68
and represented on the number line

Write a number that makes the statement true.

Question 5.
55 < _________
Answer:
55 < 56, 55 is less than 56

Explanation:
The below number line shows 55 is less than 56
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 69

Question 6.
62 = __________
Answer:
62 = 62, 62 is equal to 62.

Explanation:
The below number line shows 62 is equal to 62
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 69

Question 7.
75 > _________
Answer:
75 > 74, 75 is greater than 74.

Explanation:
The below number line shows 75 is greater than 74.
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 71

Question 8.
MP Structure
How can you use a number line to tell whether 68 is greater than or less than 42?
Answer:
The below number line represents 68 is greater than 42

Question 9.
Modeling Real Life
The number on your taxi is greater than 40. Which taxis can be yours?
Big Ideas Math Answer Key Grade 1 Chapter 7 Compare Two-Digit Numbers 70
Show how you know:
Answer:
46 or 51 is mine as they are greater than 40.

Review & Refresh

Question 10.
You have 9 pencils. 7 are blue. The rest are orange. How many orange pencils do you have?

___________ orange pencils
Answer:
2 orange pencils

Lesson 7.6 1 More, 1 Less; 10 More, 10 Less

Explore and Grow

Model 43. Use your model to complete the sentences.

1 more than 43 is ___________ .
1 less than 43 is ___________ .
10 more than 43 is ___________ .
10 less than 43 is ___________ .
Answer:
1 more than 43 is 44 .
1 less than 43 is 42 .
10 more than 43 is 53 .
10 less than 43 is 33 .

Show and Grow

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 72
1 more than 47 is ___________ .
1 less than 47 is ___________ .
10 more than 47 is ___________ .
10 less than 47 is ___________ .
Answer:
1 more than 47 is 48 .
1 less than 47 is 46 .
10 more than 47 is 57 .
10 less than 47 is 37.

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 73
1 more than 61 is ___________ .
1 less than 61 is ___________ .
10 more than 61 is ___________ .
10 less than 61 is ___________ .
Answer:
1 more than 61 is 62 .
1 less than 61 is 60.
10 more than 61 is 71 .
10 less than 61 is 51 .

Apply and Grow: Practice

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 74
1 more than 39 is ___________ .
1 less than 39 is ___________ .
10 more than 39 is ___________ .
10 less than 39 is ___________ .
Answer:
1 more than 39 is 40 .
1 less than 39 is 38 .
10 more than 39 is 49 .
10 less than 39 is 29 .

Question 4.
1 more than 56 is ___________ .
1 less than 56 is ___________ .
10 more than 56 is ___________ .
10 less than 56 is ___________ .
Answer:
1 more than 56 is 57 .
1 less than 56 is 55.
10 more than 56 is 66 .
10 less than 56 is 46.

Question 5.
1 more than 75 is ___________ .
1 less than 75 is ___________ .
10 more than 75 is ___________ .
10 less than 75 is ___________ .
Answer:
1 more than 75 is 76 .
1 less than 75 is 74 .
10 more than 75 is 85 .
10 less than 75 is 65.

Question 6.
1 more than 33 is ___________ .
1 less than 33 is ___________ .
10 more than 33 is ___________ .
10 less than 33 is ___________ .
Answer:
1 more than 33 is 34 .
1 less than 33 is 32.
10 more than 33 is 43 .
10 less than 33 is 23 .

Question 7.
1 more than 80 is ___________ .
1 less than 80 is ___________ .
10 more than 80 is ___________ .
10 less than 80 is ___________ .
Answer:
1 more than 80 is 81 .
1 less than 80 is 79 .
10 more than 80 is 90 .
10 less than 80 is 70.

Question 8.
1 more than 12 is ___________ .
1 less than 12 is ___________ .
10 more than 12 is ___________ .
10 less than 12 is ___________ .
Answer:
1 more than 12 is 13 .
1 less than 12 is 11.
10 more than 12 is 22 .
10 less than 12 is 2 .

Question 9.
MP Number Sense
Make a quick sketch for the number that is 10 less than the model. What is the new number?
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 75
Answer:
32 is the number which is 10 less than the shown figure.

Think and Grow: Modeling Real Life

You have 25 markers. Newton has 10 more than you. Descartes has 1 fewer than Newton. How many markers does Descartes have?
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 76
Models:

___________ markers
Answer:
Models: Markers,
Descartes has 24.

Explanation:
You have 25 markers.
Newton has 10 more than you.
Descartes has 1 fewer than Newton.

Show and Grow

Question 10.
You have 42 party blowers. Descartes has 10 fewer than you. Newton has 1 more than Descartes. How many party blowers does Newton have?
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 77
Models:

____________ party blowers
Answer:
Models: party blowers
33 party blowers

1 More, 1 Less; 10 More, 10 Less Practice 7.6

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 78
1 more than 63 is ___________ .
1 less than 63 is ___________ .
10 more than 63 is ___________ .
10 less than 63 is ___________ .
Answer:
1 more than 63 is 64 .
1 less than 63 is 62.
10 more than 63 is 73 .
10 less than 63 is 53 .

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 79
1 more than 80 is ___________ .
1 less than 80 is ___________ .
10 more than 80 is ___________ .
10 less than 80 is ___________ .
Answer:
1 more than 80 is 81 .
1 less than 80 is 79 .
10 more than 80 is 90 .
10 less than 80 is 70.

Question 3.
1 more than 82 is ___________ .
1 less than 82 is ___________ .
10 more than 82 is ___________ .
10 less than 82 is ___________ .
Answer:
1 more than 82 is 83 .
1 less than 82 is 81 .
10 more than 82 is 92 .
10 less than 82 is 72 .

Question 4.
1 more than 16 is ___________ .
1 less than 16 is ___________ .
10 more than 16 is ___________ .
10 less than 16 is ___________ .
Answer:
1 more than 16 is 17 .
1 less than 16 is 15 .
10 more than 16 is 26 .
10 less than 16 is 6 .

Question 5.
1 more than 68 is ___________ .
1 less than 68 is ___________ .
10 more than 68 is ___________ .
10 less than 68 is ___________ .
Answer:
1 more than 68 is 69 .
1 less than 68 is 67 .
10 more than 68 is 78 .
10 less than 68 is 58 .

Question 6.
MP Number Sense
Make a quick sketch for the number that is 1 more than the model. What is the new number?
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 80
Answer:

The below figure shows 1 more than the model.

Question 7.
Modeling Real Life
You have 75 party cups. Descartes has lo more than you. Newton has I fewer than Descartes. How many party cups does Newton have?

_____________ party cups
Answer:
Descartes has 10 more than me is 85
Newton has 1 fewer than descartes that is 84

Review & Refresh

Question 8.
You have 17 erasers. Your friend takes some of them. You have 9 left. How many erasers did your friend take?

_____________ erasers
Answer:
My friend has taken 8 erasers
17 – 9 = 8

Compare Two-Digit Numbers Performance Task

Question 1.
Your school is having a toy drive. Each class wants to collect more than 100 toys.
Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 81

a. Your friend’s class collects 81 toys. Your cousin’s class collects 72 toys. Whose class collects more toys?

Your ____________ ‘s class collects more toys.
Answer:
Friends class collects toys = 81
Cousins class collects toys 72
My friends class collects more toys.

b. Your class collects 10 more toys than your friend’s class. Does your class reach the goal?
Yes                No
Answer:
If my friend class collect 10 more toys then the toys will be 82
Yes, they reach the goal.

Question 2.
Use the clues to match each class with the number of toys it collects.

  • Class A collects 10 fewer toys than Class C.
  • Class B collects the fewest number of toys.

Big Ideas Math Answers 1st Grade 1 Chapter 7 Compare Two-Digit Numbers 82
Answer:
class a = 78
class b = 68
Class c = 88.

Compare Two-Digit Numbers Chapter Practice

Compare Numbers 11 to 19 Homework & Practice 7.1

Question 1.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 83
Answer:
15 is greater than 14
14 is less than 15.

Explanation:
compare the blocks of two whole numbers 15 and 14
15 is greater than 14 and 14 is smaller than 15

Question 2.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 84
Answer:
16 is less than 19
19 is greater than 16.

Explanation:
compare the blocks of two whole numbers 16 and 19
19 is greater than 16 and 16 is smaller than 19

Compare Numbers Homework & Practice 7.2

Question 3.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 85
Answer:
47 is less than 52
52 is greater than 47

Question 4.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 86
Answer:
36 is greater than 33
33 is less than 36.

Question 5.
MP Structure
Write a number that is greater than 78 but less than 82. Show how you know.

______________
Answer:
79 is the number that is greater than 78 and less than 82

Compare Numbers Using Place Value Homework & Practice 7.3

Compare. Which digits help you decide?

Question 6.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 87
__________ tens ___________ ones                    __________ tens ___________ ones
Answer:
4 tens and 6 ones
5 tens and 5 ones.

Explanation:
compare the blocks of two whole numbers 46 and 55
55 is greater than 46 and 46 is smaller than 55

Question 7.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 88
__________ tens ___________ ones                   __________ tens ___________ ones
Answer:
8 tens and 9 ones
9 tens and 8 ones

Explanation:
compare the blocks of two whole numbers 8 and 9
9 is greater than 8 and 8 is smaller than 9

Question 8.
MP Precision
Match each chip with its box.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 89
Answer:
Less than 45 is 37,40 and 43.
Greater than 45 is 43,49 and 50.

Compare Numbers Using Symbols Homework & Practice 7.4

Make quick sketches to model each number. Compare.

Question 9.
84 ______________ 68.
84 ○ 68
Answer:
84 is greater than 68
84 > 68

Question 10.
29 ______________ 42.
29 ○ 42
Answer:
29 < 42
29 is less than 42

Compare Numbers Using a Number Line Homework & Practice 7.5

Compare.

Question 11.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 90
62 ○ 70
Answer:
62 <70

Explanation:
the numbers are represented on number line
62 is greater than 70

Question 12.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 90
78 ○ 69
Answer:
78>69.

Explanation:
they are represented on the number line
78 is greater than 69

Question 13.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 90
75 ○ 68
Answer: 75 > 68
Explanation:
75 is greater than 68
75>68

Question 14.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 90
64 ○ 73
Answer:
64 < 73

Explanation:
they are represented on the number line
64 is less than 73

Write a number that makes the statement true.

Question 15.
__________ < 36
Answer:
35 < 36

Explanation:
compare the blocks of two whole numbers 35 and 36
36 is greater than 35 and 35 is smaller than 36

Question 16.
28 = __________
Answer:
28 = 28

Explanation:
compare the blocks of two whole numbers 28 and 28
28 is equal to 28

Question 17.
____________ > 9
Answer:
10 > 9

Explanation:
compare the blocks of two whole numbers 10 and 9
10 is greater than 9 and 9 is smaller than 10

Question 18.
Modeling Real Life
The number on your train is less than 34. Which trains can be yours?
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 91
Show how you know:
Answer:
The number on my train is less than 34 which might be 29 or 30
and circled the numbers 29 and 30.

1 More, 1 Less; 10 More, 10 Less Homework & Practice 7.6

Question 19.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 92
1 more than 82 is ___________ .
1 less than 82 is ___________ .
10 more than 82 is ___________ .
10 less than 82 is ___________ .
Answer:
1 more than 82 is 83 .
1 less than 82 is 81 .
10 more than 82 is 92 .
10 less than 82 is 72 .

Question 20.
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 93
1 more than 29 is ___________ .
1 less than 29 is ___________ .
10 more than 29 is ___________ .
10 less than 29 is ___________ .
Answer:
1 more than 29 is 30 .
1 less than 29 is 28 .
10 more than 29 is 39 .
10 less than 29 is 19 .

Question 21.
Modeling Real Life
You have 34 oranges. Newton has I fewer than you. Descartes has 10 fewer than Newton. How many oranges does Descartes have?
Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers 94

________________ oranges
Answer:
Newton has one fewer than me is 34 – 1 = 33
Descrates has 10 fewer than newton 33 – 10 = 23
Descartes has 10 oranges.

Compare Two-Digit Numbers Cumulative Practice

Question 1.
6 people play basketball. Shade the circle next to the picture that shows how many more people need to join the group so there are 10 in all.
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 95

Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 96
Answer:
6 people play basketball, 4 people should be added to make 10.

Question 2.
Shade the circles next to the choices that match the model.
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 97
○ 6 tens and 3 ones
○ 3 tens and 6 ones
○ 63
○ 36
Answer:
Shaded the circle next to the choices that match the model

Question 3.
Newton has 8 toys. Shade the circle next to the picture that does not show a way he could share his toys with Descartes.
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 98
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 99
Answer:

7 + 2 is 9
Newton has 8 toys.

Question 4.
A group of students are at a carnival. 4 of them leave. There are 8 left. How many students were at the carnival to start?
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 100
__________ ○ __________ = __________ students
Answer:
4 + 8 = 12 students.

Question 5.
Shade the circle next to the number that completes the puzzle.
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 101
○ 70
○ 62
○ 64
○ 73
Answer:
73 is the number.

Question 6.
Circle the two numbers that complete the addition sentence.
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 102
Answer:
5 + 5 + 9 = 19

Question 7.
Shade the circle next to the number that tells how many ferrets are outside the cage.
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 103
○ 5
○ 4
○ 0
○ 1
Answer:
There are no ferrets are outside the cage.

Question 8.
Use the picture to complete the sentence.
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 104
___________ tens and __________ ones is __________ .
Answer:
11 tens and 7 ones.

Question 9.
Is each equation true or false?
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 105
Answer:
1 = 9 False
4 = 4 True
5 = 2 False
7 = 7 True

Question 10.
Write <, >, or = to compare the numbers.
21 ○ 45
98 ○ 97
67 ○ 60
36 ○ 36
Answer:
21 < 45
98 > 97
67 > 60
36 = 38

Question 11.
4 boats are docked. Some more join them. Now there are 6. Shade the circle next to the models that show how many more boats joined the first group.
Big Ideas Math Solutions Grade 1 Chapter 7 Compare Two-Digit Numbers 106
Answer:

Question 12.
Write two addition equations that can be solved using the double 7 + 7.
  7   + ________ = ___________               7   + ________ = ___________
Answer:
7 + 70 =77
7 + 7 = 14.

Explanation:
when 7 is added with 70 makes 77
and if 7 is added with another 7 then it makes 14.

Conclusion:

I hope the info presented in the Big Ideas Math Answers Grade 1 Chapter 7 Compare Two-Digit Numbers is satisfactory for all the students. The students of grade 1 can clarify their queries by posting their comments in the below comment section. Bookmark our Big Ideas Math Answer site to get the Answer key for all the BIM Grade 1 Chapters.

Big Ideas Math Answers Grade 7 Chapter 6 Percents

Big Ideas Math Answers Grade 7 Chapter 6 Percents

Download Big Ideas Math Book 7th Grade Answer Key Chapter 6 Percents Pdf for free of cost. If you are willing to be perfect in all concepts and math skills, then you must follow the syllabus given here. To score the best marks in the exam, students must learn all the conceptualized lessons on Simple Interest, Discounts, and also the Subject knowledge which will be helpful to improve practical skills.

Big Ideas Math Answers Grade 7 Chapter 6 Percents helps you to cross all the hurdle times while studying. You can easily grasp various concepts and problems on all the concepts given below. Get all the skills and different methods of solving problems easily and quickly. It helps to solve real-life calculations in a very dignified and smooth manner. Follow the below sections to get BIM Grade 7 Chapter 6 Percents problems and solutions.

Big Ideas Math Book 7th Grade Answer Key Chapter 6 Percents

Big ideas Math Book 7th Grade Answer Key Chapter 6 percents pdf is given here. It gives the most accurate answers to all the problems which are related to the given chapter. This chapter of percents contains various topics like Fractions, Decimals and Percents, The Percent Proportion, The Percent Equation, Percents of Increase and Decrease, Discounts and Markups, Simple Interest, and so on.

With the help of Big Ideas Math Book 7th Grade Solution Key Chapter 6 Percents, you can prepare the timetable. With the help of the timetable prepared, you can easily manage your syllabus and include all the topics into it as per the time remaining. In the further sections, you will know all the concepts that are present in Grade 7 Chapter 6 Percents. Check out all the information and kickstart your preparation.

STEAM Video/Performance Task

Getting Ready for Chapter 6

Lesson 1 : Fractions, Decimals, and Percents

Lesson 2 : The Percent Proportion

Lesson 3 : The Percent Equation

Lesson 4 : Percents of Increase and Decrease

Lesson 5 : Discounts and Markups

Lesson 6 : Simple Interest

Percents Connecting Concepts

STEAM Video/Performance Task

STEAM Video

Tornado!
More tornadoes occur each year in the United States than in any other country. How can you use a percent to describe the portion of tornadoes in the United States that occur in your state?
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 1
Watch the STEAM Video “Tornado!” Then answer the following questions.
1. The map below shows the average annual number of tornadoes in each state. Which regions have the most tornadoes? the fewest tornadoes?
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 2
2. Robert says that only Alaska, Hawaii, and Rhode Island average less than 1 tornado per year. What percent of states average more than 1 tornado per year?

Performance Task

Tornado Alley
After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given information about the average annual numbers of tornadoes in several states over a 25-year period. For example:
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 3
You will be asked to solve various percent problems about tornadoes. Why is it helpful to know the percent of tornadoes that occur in each state?

Getting Ready for Chapter 6

Chapter Exploration
Work with a partner. Write the percent of the model that is shaded. Then write the percent as a decimal.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 4

Answer : The Percent and decimal for the given models are
1. percent = 30%, decimal = 0.3
2. percent = 100%, decimal = 1
3. percent = 33%, decimal = 0.33
4. percent = 50%, decimal = 0.5
5. percent = 40%, decimal = 0.4
6. percent = 64%, decimal = 0.64
7. percent = 60%, decimal = 0.6

Explanation:
All the models given here are 10 by 10 grid forming a Square of 100 equal sections.
This entire square represents a whole and the shaded part is fraction.
Each of these shaded squares represents 1/100. So by using this data we have ,
1.    =
From the shaded part and the whole, we have the fraction of \(\frac{30}{100}\),
Then the percent will be \(\frac{30}{100}\) = 30%,
By rewriting it in decimal form we have 0.3

2.  =
From the shaded part and the whole, we have the fraction of \(\frac{100}{100}\),
Then the percent will be \(\frac{100}{100}\) = 100%,
By rewriting it in decimal form we have 1

3.  =
From the shaded part and the whole, we have the fraction of \(\frac{33}{100}\),
Then the percent will be \(\frac{33}{100}\) = 33%,
By rewriting it in decimal form we have 0.33

4.  =
From the shaded part and the whole, we have the fraction of \(\frac{50}{100}\),
Then the percent will be \(\frac{50}{100}\) = 50%,
By rewriting it in decimal form we have 0.5

5.  =
From the shaded part and the whole, we have the fraction of \(\frac{40}{100}\),
Then the percent will be \(\frac{40}{100}\) = 40%,
By rewriting it in decimal form we have 0.4

6.  =
From the shaded part and the whole, we have the fraction of \(\frac{64}{100}\),
Then the percent will be \(\frac{64}{100}\) = 64%,
By rewriting it in decimal form we have 0.64

7.  =
From the shaded part and the whole, we have the fraction of \(\frac{60}{100}\),
Then the percent will be \(\frac{60}{100}\) = 60%,
By rewriting it in decimal form we have 0.6

8. WRITE A PROCEDURE Work with a partner. Write a procedure for rewriting a percent as a decimal. Use examples to justify your procedure.

Answer:
Let us say that the fraction be \(\frac{44}{100}\),
Then its percentage will be 44%,
To rewrite it as decimal, we divide 44 by 100 we get 0.44 (a decimal number). So, to convert from percent to decimal divide by 100 and remove the “%” sign.
We get 0.44 as decimal of 44%.

Vocabulary
The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
percent of change
percent of decrease
discount
percent of increase
percent error
markup

Answer:
percent of change:
Percentage Change is all about comparing old values to new values.

percent of decrease:
percent of decrease is a measure of percent change, which is the extent to which something loses value. or
A negative percent of change indicates a decrease from the original value to the second value.

discount: A reduction of price  is known as discount .Sometimes discounts are in percent, such as a 10% discount, and then you need to do a calculation to find the price reduction.

percent of increase :
Percent increase is a measure of percent change, which is the extent to which something gains value. or
A positive percent of change indicates an increase from the original value to the second value.

percent error :
Percent error is the difference between estimated value and the actual value in comparison to the actual value and is expressed as a percentage.

markup:
Markup is all about how much a retailer increases the price over what they paid for it to buy the product or item in order to which is how they make money to pay for all their costs and hopefully make a profit.

Lesson 6.1 Fractions, Decimals, and Percents

EXPLORATION 1

Comparing Numbers in Different Forms
Work with a partner. Determine which number is greater. Explain your method.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.1 1
Answer:
a. 7% sales tax is greater than 5% sales tax
b. 0.37 cup of flour is greater than 0.33 cup of flour
c. 0.625 inch wrench is greater than 0.375 inch wrench
d. 12.6 dollars are greater than 12.56 dollars
e. 5.83 fluid ounces is greater than 5.6 fluid ounces

Explanation:
a. 7% sales tax or \(\frac{1}{20}\) sales tax
By using the method of converting fraction into Percent and comparing ,
We can rewrite \(\frac{1}{20}\) as  0.05 in decimal form,
To get get the percent , multiply 100 to 0.05, then we get 5%.
So, \(\frac{1}{20}\) can be write as 5%,
Finally, by comparing two values 7% sales tax is greater than 5% sales tax

b. 0.37 cup of flour or \(\frac{1}{3}\) cup for flour
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{1}{3}\) as  0.33 in decimal form
Finally, by comparing two values 0.37 cup of flour is greater than 0.33 cup of flour.

c. \(\frac{5}{8}\) inch wrench or 0.375 inch wrench
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{5}{8}\) by dividing 5 by 8, we have  0.625 in decimal form
Finally, by comparing two values 0.625 inch wrench is greater than 0.375 inch wrench.

d. \(12\) \({\Large\frac{3}{5}}\) dollars or 12.56 dollars
By using the method of converting fraction into decimal and comparing ,
We can rewrite fraction\(\frac{63}{5}\) by dividing 63 by 5, we have 12.6 in decimal form
Finally, by comparing two values 12.6 dollars are greater than 12.56 dollars

e.  \(5\) \({\Large\frac{5}{6}}\) fluid ounces or 5.6 fluid ounces
By using the method of converting fraction into decimal and comparing ,
We can rewrite fraction \(\frac{35}{6}\)  by dividing 35 by 6, we have 5.83 in decimal form
Finally, by comparing two values 5.83 fluid ounces is greater than 5.6 fluid ounces

 

EXPLORATION 2

Work with a partner and follow the steps below.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.1 2

  • Write five different numbers on individual slips of paper. Include at least one decimal, one fraction, and one percent.
  • On a separate sheet of paper, create an answer key that shows your numbers written from least to greatest.
  • Exchange slips of paper with another group and race to order the numbers from least to greatest. Then exchange answer keys to check your orders.

Answer: 78%, 0.95, \(\frac{83}{45}\), 6, 21

Explanation:
Let the numbers be 6,  21, 0.95, \(\frac{83}{45}\), 78%
In order to put them in an ascending order we have to calculate the fraction, percent, decimal in their respective form, So Fraction \(\frac{83}{45}\) can be rewrite as 1.84 in decimal form
Then 78% can be rewrite as 0.78 in decimal form,
As we can see 0.78 is less than 0.95 , 0.95 is less than \(\frac{83}{45}\),\(\frac{83}{45}\) is less than 6, 6 is less than 21,
Finally, we have the ascending order as 78%, 0.95, \(\frac{83}{45}\), 6, 21.

Try It

Write the percent as a decimal or the decimal as a percent. Use a model to represent the number.
Question 1.
39%
Answer: 0.39

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, 39% in decimal form is 0.39

Question 2.
12. 6 %
Answer: 0.126

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, 12.6% in decimal form is 0.126

Question 3.
0.05
Answer: 5%

Explanation:
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol.
So, 0.05 can be rewrite as 5%

Question 4.
1.25
Answer:  125%

Explanation:
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol.
So, 1.25 can be rewrite as 125%

Write the fraction as a decimal and a percent.
Question 5.
\(\frac{5}{8}\)
Answer: 0.625 or 62.5%

Explanation:
To get the percent or decimal from  fraction \(\frac{5}{8}\) we have to divide 5 by 8 ,
Then, we get 0.625,
To get the percent of 0.625 multiply by 100 , it will be 62.5%
So, \(\frac{5}{8}\) can be written as  0.625 or 62.5%

Question 6.
\(\frac{1}{6}\)
Answer: 0.166 or 16.6%

Explanation:
To get the percent or decimal from  fraction \(\frac{1}{6}\) we have to divide 1 by 6 ,
Then, we get 0.166,
To get the percent of 0.166 multiply by 100 , it will be 16.6%
So, \(\frac{1}{6}\) can be written as  0.166 or 16.6%

Question 7.
\(\frac{11}{3}\)
Answer:  3.66 or 366%

Explanation:
To get the percent or decimal from  fraction \(\frac{11}{3}\) we have to divide 11 by 3 ,
Then, we get 3.66,
To get the percent of 3.66 multiply by 100 , it will be 366%
So, \(\frac{11}{3}\) can be written as  3.66 or 366%

Question 8.
\(\frac{3}{1000}\)
Answer: 0.003 or 0.3%

Explanation:
To get the percent or decimal from  fraction \(\frac{3}{1000}\) we have to divide 3 by 1000 ,
Then, we get 0.003,
To get the percent of 0.003 multiply by 100 , it will be 0.3%
So, \(\frac{3}{1000}\) can be written as  0.003 or 0.3%

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

CONVERTING BETWEEN PERCENTS AND DECIMALS Write the percent as a decimal or the decimal as a percent. Use a model to represent the number.
Question 9.
46%
Answer: 0.46

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, 46% in decimal form is 0.46

Question 10.
\(66 . \overline{6} \%\)
Answer: \(0 .66 \overline{6}\)

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, \(66 . \overline{6} \%\) in decimal form is \(0 .66 \overline{6}\)

Question 11.
0.18
Answer: 18%

Explanation:
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol.
So, 0.18 can be rewrite as 18%

Question 12.
\(2 . \overline{3}\)
Answer: \(233 . \overline{3} \%\)

Explanation:
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol.
So, \(2 . \overline{3}\) can be rewrite as \(233 . \overline{3} \%\)

WRITING FRACTIONS AS DECIMALS AND PERCENTS Write the fraction as a decimal and a percent.
Question 13.
\(\frac{7}{10}\)
Answer: decimal = 0.7, percent = 70%

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{7}{10}\) as  0.7 in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then 0.7 can be rewrite as 70%
So, \(\frac{7}{10}\) in decimal = 0.7, percent = 70%

Question 14.
\(\frac{5}{9}\)
Answer: decimal = \(0 .\overline{5}\), percent = \(55 . \overline{5} \%\)

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{5}{9}\) as  \(0 .\overline{5}\) in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then \(0 .\overline{5}\) can be rewrite as \(55 . \overline{5} \%\)
So, \(\frac{5}{9}\) in decimal = \(0 .\overline{5}\), percent = \(55 . \overline{5} \%\)

Question 15.
\(\frac{7}{2000}\)
Answer: decimal = 0.0035, percent = 0.35%

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{7}{2000}\) as  0.0035 in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then 0.0035 can be rewrite as 0.35%
So, \(\frac{7}{2000}\) in decimal = 0.0035, percent = 0.35%

Question 16.
\(\frac{17}{15}\)
Answer: decimal = \(1.1 \overline{3}\) , percent = \(113 . \overline{3} \%\)

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{17}{15}\) as \(1.1 \overline{3}\) in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then \(1.1 \overline{3}\) can be rewrite as \(113 . \overline{3} \%\)
So, \(\frac{17}{15}\) in decimal = \(1.1 \overline{3}\), percent = \(113 . \overline{3} \%\)

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 17.
An astronaut spends 53% of the day working, 0.1 of the day eating, \(\frac{3}{10}\) of the day sleeping, and the rest of the day exercising. Order the events by duration from least to greatest. Justify your answer.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.1 3
Answer:  An astronaut spends the day 7% by exercising, 10% by eating, 30% by sleeping, 53% by working.

Explanation:
An astronaut spends 53% of the day working,
0.1 of the day eating, in terms of percent we can write it as 10%,
\(\frac{3}{10}\) of the day sleeping, in decimals we ca rewrite as 0.3 and in percent it will be 30%,
Let us say that the whole day be 100% , The sum of the works he is doing in percent we get,
53% + 10% + 30% = 93%, and
Given that the rest of the day exercising, so 100% – 93% = 7%, A whole day is completed with these works.
To put them in Order the events by duration from least to greatest, we have , 7%, 10%, 30%, 53%.
An astronaut spends the day 7% by exercising, 10% by eating, 30% by sleeping, 53% by working.

Question 18.
DIG DEEPER!
A band plays one concert in Arizona, one concert in California, and one concert in Georgia. In California, the band earned \(\frac{3}{2}\) the profit that they earned in Arizona. Of the total profit earned by the band, 32% is earned in Arizona. How many times more money did the band earn at the most profitable concert than at the least profitable concert? Justify your answer.
Answer:

Explanation:

Fractions, Decimals, and Percents Homework & Practice 6.1

Review & Refresh

Find the missing dimension. Use the scale 1 : 15.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.1 4
Answer: The model height of the Figure skater is 4.5inches and The Actual length of pipe is 75 feet .

Explanation:
Given , to use the scale oof 1 : 15 , Let the model height  be x
The model height of Figure skater is \(\frac{1}{15}\) = \(\frac{x}{67.5}\)
15x = 67.5
x = \(\frac{67.5}{15}\)
x = 4.5
So , The model height of the Figure skater is 4.5inches

Let the Actual length is y
The Actual length of pipe is \(\frac{1}{15}\) = \(\frac{5}{y}\)
y = 15 × 5 = 75
So , The Actual length of pipe is 75 feet .

Simplify the expression.
Question 3.
2(3p – 6) + 4p
Answer: p = 1.2

Explanation:
Let us say that whole expression is equal to 0
2(3p – 6) + 4p = 0
[2(3p) – 2(6)] + 4p = 0
6p – 12 + 4p = 0
10p – 12 = 0
p = 12/10 = 1.2
So, p = 1.2

Question 4.
5n – 3(4n + 1)
Answer: n = -0.42

Explanation:
Let us say that whole expression is equal to 0
5n – 3(4n + 1) = 0
5n – [ 3(4n) + 3(1) ] = 0
5n – 12n – 3 = 0
– 3 – 7n = 0
7n = – 3
n = -3/7 = -0.42
So, n = -0.42

Question 5.
What is the solution of 2n – 4 > – 12?
A. n < – 10
B. n < – 4
C. n > – 2
D. n > – 4
Answer:  D . n > -4

Explanation:
Given, 2n – 4 > – 12
add 4 in both sides,
2n – 4 + 4 > – 12 + 4
2n > – 8
divide both sides by 2
2n/2 > -8/2
n > – 4 .

Concepts, Skills, & Problem Solving
COMPARING NUMBERS IN DIFFERENT FORMS Determine which number is greater. Explain your method. (See Exploration 1, p. 235.)
Question 6.
4\(\frac{2}{5}\) tons or 4.3 tons
Answer: 4\(\frac{2}{5}\) tons is greater than 4.3 tons

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite fraction \(\frac{22}{5}\)  by dividing 22 by 5, we have 4.4 in decimal form
Finally, by comparing two values 4\(\frac{2}{5}\) tons is greater than 4.3 tons

Question 7.
82% success rate or \(\frac{5}{6}\) success rate
Answer: \(\frac{5}{6}\) success rate is greater than 82% success rate

Explanation:
By using the method of converting fraction into Percent and comparing ,
We can rewrite \(\frac{5}{6}\)  as  0.833 in decimal form,
To get get the percent , multiply 100 to 0.833, then we get 83.3%.
So, \(\frac{5}{6}\)  can be write as 83.3%,
Finally, by comparing two values \(\frac{5}{6}\) success rate is greater than 82% success rate

CONVERTING BETWEEN PERCENTS AND DECIMALS Write the percent as a decimal or the decimal as a percent. Use a model to represent the number.
Question 8.
26%
Answer: 0.26

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, 26% in decimal form is 0.26

Question 9.
0.63
Answer: 63%

Explanation:
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol.
So, 0.63 can be rewrite as 63%

Question 10.
9%
Answer: 0.09

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, 9% in decimal form is 0.09

Question 11.
0.6
Answer: 60%

Explanation:
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol.
So, 0.6 can be rewrite as 60%

Question 12.
44.7%
Answer: 0.447

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, 44.7% in decimal form is 0.447

Question 13.
55%
Answer: 0.55

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, 55% in decimal form is 0.55

Question 14.
\(39 . \overline{2} \%\)
Answer: \(0.39 \overline{2}\)

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, \(39 . \overline{2} \%\) in decimal form is \(0.39 \overline{2}\)

Question 15.
3.554
Answer: 355.4%

Explanation:
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol.
So, 3.554  can be rewrite as 355.4%

Question 16.
123%
Answer: 1.23

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, 123% in decimal form is 1.23

Question 17.
0.041
Answer: 4.1%

Explanation:
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol.
So, 0.041 can be rewrite as 4.1%

Question 18.
0.122
Answer: 12.2%

Explanation:
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol.
So, 0.122 can be rewrite as 12.2%

Question 19.
\(49 . \overline{92} \%\)
Answer: \(0.49 \overline{92}\)

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, \(49 . \overline{92} \%\) in decimal form is \(0.49 \overline{92}\)

Question 20.
YOU BE THE TEACHER
Your friend writes \(49 . \overline{8} \%\) as a decimal. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.1 5
Answer: He wrote the decimal for \(4. \overline{8} \%\) instead of \(49 . \overline{8} \%\)

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, \(49 . \overline{8} \%\)  in decimal form is \(0.49 \overline{8}\)

WRITING FRACTIONS AS DECIMALS AND PERCENTS Write the fraction as a decimal and a percent.
Question 21.
\(\frac{29}{100}\)
Answer: decimal = 0.29, percent = 29%

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{29}{100}\) as  0.29 in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then 0.29 can be rewrite as 29%
So, \(\frac{29}{100}\) in decimal = 0.29, percent = 29%

Question 22.
\(\frac{3}{4}\)
Answer: decimal = 0.75, percent = 75%

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite\(\frac{3}{4}\) as  0.75 in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then 0.75 can be rewrite as 75%
So, \(\frac{3}{4}\) in decimal = 0.75, percent = 75%

Question 23.
\(\frac{7}{8}\)
Answer: decimal = 0.875, percent = 87.5%

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{7}{8}\) as  0.875 in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then 0.875 can be rewrite as 87.5%
So, \(\frac{7}{8}\) in decimal = 0.875, percent = 87.5%

Question 24.
\(\frac{2}{3}\)
Answer: decimal = \(0. \overline{6}\), percent = \(66. \overline{6} \%\)

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{2}{3}\) as \(0. \overline{6}\) in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then\(0. \overline{6}\) can be rewrite as \(66. \overline{6} \%\)
So, \(\frac{2}{3}\) in decimal =\(0. \overline{6}\), percent = \(66. \overline{6} \%\)

Question 25.
\(\frac{7}{9}\)
Answer: decimal = 0.77, percent = 77.7%

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite\(\frac{7}{9}\) as  0.77 in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then 0.77 can be rewrite as 77.7%
So, \(\frac{7}{9}\) in decimal = 0.77, percent = 77.7%

Question 26.
\(\frac{12}{5}\)
Answer: decimal = 2.4, percent = 240%

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{12}{5}\) as  2.4 in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then 2.4 can be rewrite as 240%
So, \(\frac{12}{5}\) in decimal =2.4, percent = 240%

Question 27.
\(\frac{9}{2}\)
Answer: decimal = 4.5, percent = 450%

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite\(\frac{9}{2}\) as 4.5 in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then 4.5 can be rewrite as 450%
So, \(\frac{9}{2}\) in decimal = 4.5, percent = 450%

Question 28.
\(\frac{1}{1000}\)
Answer: decimal = 0.0010, percent = 0.10%

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{1}{1000}\) as  0.0010 in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then 0.0010 can be rewrite as 0.10%
So, \(\frac{1}{1000}\) in decimal = 0.0010, percent = 0.10%

Question 29.
\(\frac{17}{6}\)
Answer: decimal = \(2.8 \overline{3}\), percent = \(283 . \overline{3} \%\)

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{17}{6}\) as \(2.8 \overline{3}\) in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then\(2.8 \overline{3}\) can be rewrite as \(283 . \overline{3} \%\)
So, \(\frac{17}{6}\) in decimal = \(2.8 \overline{3}\), percent = \(283 . \overline{3} \%\)

Question 30.
\(\frac{3}{11}\)
Answer: decimal = 0.27, percent = 27%

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{3}{11}\) as  0.27 in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then 0.27 can be rewrite as 27%
So, \(\frac{3}{11}\) in decimal = 0.27, percent = 27%

Question 31.
\(\frac{1}{750}\)
Answer: decimal = \(0.001 \overline{3}\), percent =\(0.1 \overline{3} \%\)

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{1}{750}\) as  \(0.001 \overline{3}\) in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then \(0.001 \overline{3}\) can be rewrite as \(0.1 \overline{3} \%\)
So, \(\frac{1}{750}\) in decimal = \(0.001 \overline{3}\), percent = \(0.1 \overline{3} \%\)

Question 32.
\(\frac{22}{9}\)
Answer: decimal = \(2. \overline{4}\), percent = \(244 . \overline{4} \%\)

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{22}{9}\) as  \(2. \overline{4}\) in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then \(2. \overline{4}\) can be rewrite as \(244 . \overline{4} \%\)
So, \(\frac{22}{9}\) in decimal = \(2. \overline{4}\), percent = \(244 . \overline{4} \%\)

PRECISION Order the numbers from least to greatest.
Question 33.
66.1%, 0.66, \(\frac{2}{3}\), 0.667
Answer: 0.66, 66.1%, \(\frac{2}{3}\), 0.667

Explanation:
In order to put them in an ascending order we have to calculate the fraction, percent, decimal in their respective form, So Fraction \(\frac{2}{3}\) can be rewrite as \(0. \overline{6}\) in decimal form
Then 66.1% can be rewrite as 0.661 in decimal form,
As we can see 0.66 is less than 66.1% , 66.1% is less than \(\frac{2}{3}\), \(\frac{2}{3}\)  is less than 0.667,
Finally, we have the ascending order as 0.66, 66.1%, \(\frac{2}{3}\), 0.667.

Question 34.
\(\frac{2}{9}\), 21%, \(0.2 \overline{1}\), \(\frac{11}{50}\)
Answer: 21%, \(0.2 \overline{1}\), \(\frac{11}{50}\), \(\frac{2}{9}\)

Explanation:
In order to put them in an ascending order we have to calculate the fraction, percent, decimal in their respective form, So Fraction \(\frac{11}{50}\) can be rewrite as 0.22 in decimal form,
\(\frac{2}{9}\) can be rewrite as \(0.22 \overline{2}\)in decimal form,
Then 21% can be rewrite as 0.21 in decimal form,
As we can see 21%  is less than \(0.2 \overline{1}\) ,\(0.2 \overline{1}\) is less than \(\frac{11}{50}\),  \(\frac{11}{50}\)  is less than \(\frac{2}{9}\),
Finally, we have the ascending order as 21%, \(0.2 \overline{1}\), \(\frac{11}{50}\), \(\frac{2}{9}\)

MATCHING Tell which letter shows the graph of the number.
Question 35.
\(\frac{7}{9}\)
Answer: decimal = 0.777 , it is in the graph at the point A

Explanation:
By using the method of converting fraction into decimal ,
We can rewrite \(\frac{7}{9}\) as 0.777 in decimal form,
So, looking at the graph given, it is at point A.

Question 36.
0.812
Answer:  it is at the point C in the given graph.

Question 37.
\(\frac{5}{6}\)
Answer: decimal = 0.833 , it is in the graph at the point D

Explanation:
By using the method of converting fraction into decimal ,
We can rewrite \(\frac{5}{6}\) as 0.833 in decimal form,
So, looking at the graph given, it is at point D.

Question 38.
79.5%
Answer: 0.795, it is in the graph at the point B

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, 79.5% in decimal form is 0.795
0.795 is in the graph given , at the point B vb

Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.1 6

Question 39.
PROBLEM SOLVING
The table shows the portion of students in each grade that participate in School Spirit Week. Order the grades by portion of participation from least to greatest.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.1 7
Answer: The grades by portion of participation from least to greatest are 7 , 6 , 8.

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left. So, 65% in decimal form is 0.65.
By using the method of converting fraction into Percent  ,
We can rewrite \(\frac{5}{6}\) as 0.6 in decimal form,
Then grade 6 = 0.64 , grade 7 = 0.6 , grade 8 = 0.65 ,
So, The grades by portion of participation from least to greatest are 7 , 6 , 8.

Question 40.
MODELING REAL LIFE
The table shows the portion of gold medals that were won by the United States in five summer Olympic games. In what year did the United States win the least portion of gold medals? the greatest portion? Justify your answers.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.1 8
Answer: The United States win the least portion of gold medals in the year 2004 and the greatest portion of gold medals won in the year 2012

Explanation:
By using the method of converting fraction into Percent and comparing ,
We can rewrite \(\frac{36}{301}\)  as  0.119 in decimal form,
\(\frac{23}{150}\) as 0.153 in decimal form,
\(\frac{46}{307}\) as 0.149 in decimal form,
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
\(12. \overline{3} \%\) as \(0.12 \overline{3}\) in decimal form.
So, according to their years and portions of gold medals we have ,
year 2000 – \(0.12 \overline{3}\),
year 2004 – 0.119,
year 2008 – \(0. \overline{12}\),
year 2012 – 0.153,
year 2016 – 0.149,
Finally, The United States win the least portion of gold medals in the year 2004 and the greatest portion of gold medals won in the year 2012.

Question 41.
PROBLEM SOLVING
You, your friend, and your cousin have a basketball competition where each person attempts the same number of shots. You make 70% of your shots, your friend makes of her shots, \(\frac{7}{9}\) and your cousin makes \(0.7 \overline{2}\) of his shots. How many times more shots are made by the first place finisher than the third place finisher?
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.1 9
Answer: The first place finisher made 0.077 more shorts than the third place finisher.

Explanation:
The shorts made by 1st person are 70%, that is 0.7 in decimal,
The shorts made by 2nd person are \(\frac{7}{9}\) rewriting as 0.777 in decimal,
The shorts made by 3rd person are \(0.7 \overline{2}\)
The shorts made by  the first place finisher is  0.777 and by the Third place finisher is 0.7,
To know how many more shorts are made by first place finisher than third place finisher is the difference between 0.777 and 0.7 , that is 0.777 – 0.7 = 0.077.
So, the first place finisher made 0.077 more shorts than the third place finisher.

Question 42.
DIG DEEPER!
Three different mixtures contain small amounts of acetic acid. Mixture A is 0.036 acetic acid, Mixture B is 4.2% acetic acid, and Mixture C is \(\frac{1}{22}\) acetic acid. Explain how to use this information to determine which mixture contains the greatest amount of acetic acid.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.1 10
Answer: 0.045 , Mixture C has more amount of acetic acid as compared to Mixture A and Mixture B.

Explanation:
Mixture A is 0.036 acetic acid,
Mixture B is 4.2% acetic acid, in decimal we can write it as 0.042,
Mixture C is \(\frac{1}{22}\) acetic acid, rewriting as 0.045 in decimal form,
So, 0.036 is less than 0.042 , 0.042 is less than 0.045,
Finally,  0.045 Mixture c has the more amount of acetic acid as compared to Mixture A and Mixture B.

Question 43.
MODELING REAL LIFE
Over 44% of the 30 students in a class read a book last month. What are the possible numbers of students in the class who read a book last month? Justify your answer.
Answer: 13 number of students in the class read the book last month.

Explanation:
Given, 44% of 30 students
=(44%) × 30
= \(\frac{44}{100}\) × 30
= \(\frac{44 × 30}{100}\)
= \(\frac{1320}{100}\)
=13.2
So, 13 number of students in the class read the book last month.

Question 44.
NUMBER SENSE
Fill in the blanks using each of the numbers 0 – 7 exactly once, so that the percent, decimal, and fraction below are ordered from least to greatest. Justify your answer.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.1 11
Answer:  The percent, decimal, and fraction are ordered from least to greatest are 12.3%, 0.57, \(\frac{4}{6}\) by using 0 – 7 numbers only once .

Explanation:
Given , using each of the numbers 0 – 7 exactly once,
Estimating on the numbes
Using numbers 1 , 2 , 3 for percent gives 12.3% as shown and can be writen as 0.123
Using numbers 0 , 5 , 7 for decimal gives 0.57 as shown
Using numbers 4 and 6 for fraction gives \(\frac{4}{6}\) and can be writen as 0.66
The order from least to greatest is 0.123 , 0.57 , 0.66 .
So , The percent, decimal, and fraction are ordered from least to greatest are 12.3%, 0.57, \(\frac{4}{6}\)

Lesson 6.2 The Percent Proportion

EXPLORATION 1

Using Percent Models
Work with a partner.
a. Complete each model. Explain what each model represents.
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.2 1
b. Use the models in part (a) to answer each question.

  • What number is 50% of 30?
  • 15 is what percent of 75?
  • 96 is 133\(\frac{1}{3}\)% of what number?

c. How can you use ratio tables to check your answers in part(b)? How can you use proportions? Provide examples to support your reasoning.
d. Write a question different from those in part (b) that can be answered using one of the models in part(a). Trade questions with another group and find the solution.
Answer: All the answers are given below the explanation.

Explanation:
a.
This model represents the number 30 which is divided in to two equal halves ,which can be represented in percent and numbers,
that is 50% of 30 is 15.
50 percent × 30 =
(50:100) × 30 =
(50 × 30):100 =
1500:100 = 15


This model represents the number 75 which is divided in to five equal halves, which can be represented in percent and numbers ,That is
20% of 75 is 15,
40% of 75 is 30,
60% of 75 is 45,
80% of 75 is 60.


This model represents the number 96 which is divided in to four equal halves, which can be represented in percent and numbers ,That is
33\(\frac{1}{3}\%\) of 96 is 24,
66\(\frac{2}{3}\%\) of 96 is 48,
100% of 96 is 72.

b. As shown in the models in part (a) 15 is the number which is 50% of 30,
From the figure 2 of part (a) we know, 15 is the number which is 20% of 75,
From the figure 3 of part (a) we know, 96 is the number which is 133\(\frac{1}{3}\%\) of 96.

c. The models provided in the part (a) are representing the each of the number individually which are divided in to equal number of parts in respective of their numbers , Also the divided parts can be represented as percent and number form
Since the divided equal parts are equally represents the whole number given ,So that the ratios of the parts can be easily combined in to the whole number .
Thus, the proportion can be used to calculate the percent of each ratio and have the unchanged output .
For example,

In the picture shown The number 30 is divided into 1:1 ratio equally which is exactly the half of the number and percent that is 15 is 50% of 30.

d. From the figure 2 of part (a), we can assume the question as
What number is 80% of 75?

Answer: 60 is the number which is 80% of 75 , as shown in the model above.

Try It

Write and solve a proportion to answer the question.
Question 1.
What percent of 5 is 3?
Answer: 60%

Explanation:
3 : 5 × 100 =
(3 × 100): 5
300 : 5 = 60
So, 5 is 60% of 3.

Question 2.
24 is what percent of 20?
Answer: 120%

Explanation:
24 : 20 × 100 =
(24 × 100) : 20 =
2400 : 20 = 120
So, 24 is the 120% of 20.

Write and solve a proportion to answer the question.
Question 3.
What number is 80% of 60?
Answer:  48

Explanation:
80 % × 60 =
(80 :100) × 60 =
(80 × 60) :100 =
4800 : 100 = 48
So, 48 is 80% of 60.

Question 4.
10% of 40.5 is what number?
Answer: 4.05

Explanation:
10 % × 40.5 =
(10 : 100) × 40.5 =
(10 × 40.5) : 100 =
4.05 : 100 = 4.05
So, 4.05 is the 10% of 40.5

Write and solve a proportion to answer the question.
Question 5.
0.1% of what number is 4?
Answer: 4,000

Explanation:
Let the number be X
0.1% × X = 4
X = 4 ÷ 0.1%
= 4 ÷ (0.1 ÷ 100)
= (100 × 4 ) ÷ 0.1
= 400 ÷ 0.1
=4,000
So, 4 is the 0.1% of 4,000.

Question 6.
\(\frac{1}{2}\) is 25% of what number?
Answer:  2

Explanation:
To make calculation easier we can rewrite \(\frac{1}{2}\) as 0.5
let the number be X
25% × X = 0.5
X = 0.5 ÷ 25%
= 0.5 ÷ (25 ÷ 100)
= (100 × 0.5 ) ÷ 25
= 50 ÷ 25
=2
So, 0.5 is the 25% of 2.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
USING THE PERCENT PROPORTION
Write and solve a proportion to determine what percent of 120 is 54.
Answer:  45%

Explanation:
By using percent proportion, we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{54}{120}\) = \(\frac{p}{100}\)
\(\frac{54 × 100}{120}\) = p
p = \(\frac{5400}{120}\)
p = 45%
so, 54 is the 45% of 120.

Question 8.
CHOOSE TOOLS
Use a model to find 60% of 30.
Answer: 18

Explanation:
60% × 30
= (60 : 100) × 30
= (60 × 30) : 100
= 1800 : 100
= 18
So, 60% of 30 is 18.

Question 9.
WHICH ONE DOESN’T BELONG?
Which proportion at the left does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.2 2
Answer: second proportion doesn’t belong to other three

Explanation:
second proportion: 15/50 = p /100
p = (15 × 100) ÷50
p = 150 ÷ 5
p = 30
Here we got 30% , as the p values of the other three proportions are 50%
So, second proportion does not fit in to the other three proportions.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 10.
An arctic woolly-bear caterpillar lives for 7 years and spends 90% of its life frozen. How many days of its life is the arctic woolly-bear frozen?
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.2 3
Answer: The arctic woolly-bear frozen for 6.3 years in their life time.

Explanation:
Given, An arctic woolly-bear caterpillar lives for 7 years and spends 90% of its life frozen
We have, 90% × 7
= (90 : 100 ) × 7
= (90 × 7) : 100
= 630 : 100
= 6.3 years
So, The arctic woolly-bear frozen for 6.3 years in their life time.

Question 11.
DIG DEEPER!
The table shows the numbers of pictures you upload to a social media website for 5 days in a row. How many total pictures do you upload during the week when 32% of the total pictures are uploaded on Saturday and Sunday?
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.2 4
Answer: The total pictures uploaded during the week are 53.

Explanation:
Given,  32% of the total pictures are uploaded on Saturday and Sunday
And adding the photos that we uploaded during the 5 days are 2 + 2 + 4 + 1 + 8 = 17
Let the number we should find be X
32% × X = 17
X = 17 ÷ 32%
= 17 ÷ (32 ÷ 100)
= (100 × 17 ) ÷ 32
= 1700 ÷ 32
= 53.12
So, 17 is the 32% of 53.
Finally we have, The total pictures uploaded during the week are 53.

The Percent Proportion Homework & Practice 6.2

Review & Refresh

Write the fraction as a decimal and a percent.
Question 1.
\(\frac{42}{100}\)
Answer: decimal = 0.42 , percent = 42%

Explanation:
To get the percent or decimal from fraction \(\frac{42}{100}\) we have to divide 42 by 100 ,
Then, we get 0.42,
To get the percent of 0.42 multiply by 100 , it will be 42%
So, \(\frac{42}{100}\) can be written as  0.42 or 42%

Question 2.
\(\frac{7}{1000}\)
Answer: decimal = 0.007 , percent = 0.7%

Explanation:
To get the percent or decimal from  fraction \(\frac{7}{1000}\) we have to divide 7 by 1000 ,
Then, we get 0.007,
To get the percent of 0.007 multiply by 100 , it will be 0.7%
So, \(\frac{7}{1000}\) can be written as  0.007 or 0.7%

Question 3.
\(\frac{13}{9}\)
Answer: decimal = 1.444 , percent = 144.4%

Explanation:
To get the percent or decimal from \(\frac{13}{9}\)  fraction we have to divide 13 by 9 ,
Then, we get 1.444,
To get the percent of 1.444 multiply by 100 , it will be 144.4%
So, \(\frac{13}{9}\) can be written as  1.444 or 144.4%

Question 4.
\(\frac{41}{66}\)
Answer: decimal = \(0.62 \overline{12}\) , percent = \(62. \overline{12} \%\)

Explanation:
To get the percent or decimal from  fraction \(\frac{41}{66}\) we have to divide 41 by 66 ,
Then, we get \(0.62 \overline{12}\),
To get the percent of \(0.62 \overline{12}\) multiply by 100 , it will be \(62. \overline{12} \%\)
So, \(\frac{41}{66}\) can be written as  \(0.62 \overline{12}\) or \(62. \overline{12} \%\)

Evaluate the expression when a = – 15 and b = – 5.
Question 5.
a ÷ 5
Answer: -3

Explanation:
Given , a = – 15
Then , – 15 ÷ 5
= \(\frac{-15}{5}\)
= – 3.
so, a ÷ 5 = – 3.

Question 6.
\(\frac{b+14}{a}\)
Answer: \(\frac{9}{-15}\)

Explanation:
Given , a = -15 , b = -5 , by substituting the given values in the expression, we get
= \(\frac{(- 5)+14}{-15}\)
= \(\frac{9}{-15}\)
So, \(\frac{b+14}{a}\) = \(\frac{9}{-15}\)

Question 7.
\(\frac{b^{2}}{a+5}\)
Answer: \(\frac{25}{-10}\)

Explanation:
Given , a = -15 , b = -5, by substituting the given values in the expression, we get
= \(\frac{(-5)^{2}}{(-15)+5}\)
= \(\frac{25}{-10}\)
So, \(\frac{b^{2}}{a+5}\) = \(\frac{25}{-10}\)

What is the solution of 9x = 1.8?
A. x = – 5
B. x = – 0.2
C. x = 0.2
D. x = 5
Answer: C . x = 0.2

Explanation:
Given, 9x = 1.8
x = \(\frac{1.8}{9}\)
x = 0.2.

Concepts, Skills, &Problem Solving

CHOOSE TOOLS Use a model to answer the question. Use a proportion to check your answer. (See Exploration 1, p. 241.)
Question 9.
What number is 20% of 80?
Answer: 16

Explanation:
By using proportion , we have
\(\frac{a}{w}\)  = \(\frac{p}{100}\)
\(\frac{a}{80}\) = \(\frac{20}{100}\)
\(\frac{a}{80}\) = \(\frac{1}{5}\)
a = \(\frac{80}{5}\)
a = 16.
So, 16 is 20% of 80.

Question 10.
10 is what percent of 40?
Answer: 25%

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{10}{40}\) = \(\frac{p}{100}\)
\(\frac{1}{4}\)  = \(\frac{p}{100}\)
p = \(\frac{100}{4}\)
p = 25
So, 10 is 25% of 40.

Question 11.
15 is 30% of what number?
Answer: 50

Explanation:
By using proportion , we have
\(\frac{a}{w}\)  = \(\frac{p}{100}\)
\(\frac{15}{w}\)  = \(\frac{30}{100}\)
by cross multiplication we get,
30 × w = 15 × 100
w  = \(\frac{1500}{30}\)
w = \(\frac{150}{30}\)
w = 50.
So, 15 is 30% of 50.

Question 12.
What number is 120% of 70?
Answer: 84

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{70}\)  = \(\frac{120}{100}\)
\(\frac{a}{70}\)  = \(\frac{6}{5}\)
a = \(\frac{70 × 6}{5}\)
a = \(\frac{420}{5}\)
a = 84
So, 84 is 120% of 70.

Question 13.
20 is what percent of 50?
Answer: 40%

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{20}{50}\)  = \(\frac{p}{100}\)
p = \(\frac{20 × 100}{50}\)
p = \(\frac{200}{5}\)
p = 40
So, 20 is 40% of 50.

Question 14.
48 is 75% of what number?
Answer: 64

Explanation:
By using proportion , we have
\(\frac{a}{w}\)  = \(\frac{p}{100}\)
\(\frac{48}{w}\)  = \(\frac{75}{100}\)
w =\(\frac{48 × 100}{75}\)
w = \(\frac{4800}{75}\)
w = 64
So, 48 is 75% of 64.

USING THE PERCENT PROPORTION Write and solve a proportion to answer the question.
Question 15.
What percent of 25 is 12?
Answer: 48%

Explanation:
By using proportion , we have
\(\frac{a}{w}\)  = \(\frac{p}{100}\)
\(\frac{12}{25}\)  = \(\frac{p}{100}\)
p = \(\frac{12 × 100}{25}\)
p = \(\frac{1200}{25}\)
p = 48
So, 12 is 48% of 25.

Question 16.
14 is what percent of 56?
Answer: 25%

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{14}{56}\)  = \(\frac{p}{100}\)
p = \(\frac{14 × 100}{56}\)
p = \(\frac{1400}{56}\)
p = 25
So, 14 is 25% of 56.

Question 17.
25% of what number is 9?
Answer: 36

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{9}{w}\) = \(\frac{25}{100}\)
\(\frac{9}{w}\) = \(\frac{1}{4}\)
w = 9 × 4
w = 36.
So, 9 is 25% of 36.

Question 18.
36 is 0.9% of what number?
Answer: 4,000

Explanation:
By using proportion , we have
\(\frac{a}{w}\)  = \(\frac{p}{100}\)
\(\frac{36}{w}\) = \(\frac{0.9}{100}\)
w = \(\frac{36 × 100}{0.9}\)
w = \(\frac{3600}{0.9}\)
w = \(\frac{36,000}{9}\)
w = 4,000
So, 36 is 0.9% of 4,000.

Question 19.
75% of 124 is what number?
Answer: 93

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{124}\) = \(\frac{75}{100}\)
a = \(\frac{75 × 124}{100}\)
a  = \(\frac{9,300}{100}\)
a = 93
So, 93 is 75% of 124.

Question 20.
110% of 90 is what number?
Answer: 99

Explanation:
By using proportion , we have
\(\frac{a}{w}\)  = \(\frac{p}{100}\)
\(\frac{a}{90}\)  = \(\frac{110}{100}\)
a = \(\frac{110 × 90}{100}\)
a = \(\frac{9900}{100}\)
a = 99
So, 99 is 110% of 90.

Question 21.
What number is 0.4% of 40?
Answer: 0.16

Explanation:
By using proportion , we have
\(\frac{a}{w}\)  = \(\frac{p}{100}\)
\(\frac{a}{40}\)  = \(\frac{0.4}{100}\)
a = \(\frac{0.4 × 40}{100}\)
a = \(\frac{16}{100}\)
a = 0.16
So, 0.16 is 0.4% of 40.

Question 22.
72 is what percent of 45?
Answer:  160%

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{72}{45}\) = \(\frac{p}{100}\)
p = \(\frac{72 × 100}{45}\)
p = \(\frac{7200}{45}\)
p = 160
So, 72 is 160% of 45.

Question 23.
YOU BE THE TEACHER
Your friend uses the percent proportion to answer the question below. Is your friend correct? Explain your reasoning.
“40%of what number is 34?”
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.2 5
Answer:  yes, he used the correct percent proportion.

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{34}{w}\) = \(\frac{40}{100}\)
w = \(\frac{34 × 100}{40}\)
w = \(\frac{3400}{40}\)
w = 85
So, 34 is 40% of 85.

Question 24.
MODELING REAL LIFE
Of 140 seventh-grade students, 15% earn the Presidential Youth Fitness Award. How many students earn the award?
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.2 6
Answer: 21 students earn the Presidential Youth Fitness Award.

Explanation:
Given, Of 140 seventh-grade students, 15% earn the Presidential Youth Fitness Award.
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{140}\) = \(\frac{15}{100}\)
a = \(\frac{15 × 140}{100}\)
a = \(\frac{2100}{100}\)
a = 21
So, 21 is 15% of 140.
Totally ,21 students earn the Presidential Youth Fitness Award.

Question 25.
MODELING REAL LIFE
A salesperson receives a 3% commission on sales. The salesperson receives $180 in commission. What is the amount of sales?
Answer: The total amount of sale is $6,000.

Explanation:
Given, A salesperson receives a 3% commission on sales, The salesperson receives $180 in commission.
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{180}{w}\) = \(\frac{3}{100}\)
w = \(\frac{180 × 100}{3}\)
w = \(\frac{18,000}{3}\)
w = 6,000
So, $180 is 3% of $6,000.
Total amount of sale is $6,000.

USING THE PERCENT PROPORTION Write and solve a proportion to answer the question.
Question 26.
0.5 is what percent of 20?
Answer: 2.5%

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{0.5}{20}\) = \(\frac{p}{100}\)
p = \(\frac{0.5 × 100}{20}\)
p = 0.5 × 5
p = 2.5
So, 0.5 is 2.5% of 20.

Question 27.
14.2 is 35.5% of what number?
Answer: 40

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{14.2}{w}\) = \(\frac{35.5}{100}\)
w = \(\frac{14.2 × 100}{35.5}\)
w = \(\frac{142 × 10}{35.5}\)
w = 40
So, 14.2 is 35.5% of 40.

Question 28.
\(\frac{3}{4}\) is 60% of what number?
Answer: 1.25

Explanation:
\(\frac{3}{4}\) can be rewrite as 0.75 in decimal,
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{0.75}{w}\) = \(\frac{60}{100}\)
w = \(\frac{0.75 × 100}{60}\)
w = \(\frac{75}{60}\)
w = 1.25
So, 0.75 is 60% of 1.25.

Question 29.
What number is 25% of \(\frac{7}{8}\)?
Answer: 0.218

Explanation:
\(\frac{7}{8}\) can be rewrite as 0.875 in decimal,
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{0.875}\) = \(\frac{25}{100}\)
a = \(\frac{0.875 × 25}{100}\)
a = \(\frac{21.87}{100}\)
a = 0.218
So, 0.218 is 25% of 0.875.

Question 30.
MODELING REAL LIFE
You are assigned 32 math exercises for homework. You complete 75% of the exercises before dinner. How many exercises do you have left to do after dinner?
Answer: 24 exercises are left .

Explanation:
You are assigned 32 math exercises for homework. You complete 75% of the exercises before dinner.
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{32}\) = \(\frac{75}{100}\)
a = \(\frac{32 × 75}{100}\)
a = \(\frac{2,400}{100}\)
a = 24
So, 24 is 75% of 32.
Totally, 24 exercise are left .

Question 31.
MODELING REAL LIFE
Your friend earns $10.50 per hour, which is 125% of her hourly wage last year. How much did your friend earn per hour last year?
Answer: Friend earned $8.4 per hour last year

Explanation:
Your friend earns $10.50 per hour, which is 125% of her hourly wage last year,
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{10.5}{w}\) = \(\frac{125}{100}\)
w = \(\frac{10.5 × 100}{125}\)
w = \(\frac{1050}{125}\)
w = 8.4
So, 8.4 is 125% of 10.5.

Question 32.
MODELING REAL LIFE
The bar graph shows the numbers of reserved campsites at a campground for one week. What percent of the reservations were for Friday or Saturday?
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.2 7
Answer: The percent of reservations For Friday is 74.2% and for Saturday is  85%

Explanation:
As per the graph shown, The reservations made for the week are 35 ,
Friday reservations are 26, so
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{26}{35}\) = \(\frac{p}{100}\)
p = \(\frac{26 × 100}{35}\)
p = \(\frac{2600}{35}\)
p = 74.2
So, 26 is 74.2% of 35.

Saturday reservations are 30, so
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{30}{35}\) = \(\frac{p}{100}\)
p = \(\frac{30 × 100}{35}\)
p = \(\frac{3,000}{35}\)
p = 85
So, 30 is 85% of 35.

Totally, The percent of reservations For Friday is 74.2% and for Saturday is  85%.

Question 33.
PROBLEM SOLVING
Your friend displays the results of a survey that asks several people to vote on a new school mascot.
a. What is missing from the bar graph?
b. What percent of the votes does the least popular mascot receive? Explain your reasoning.
c. There are 124 votes total. How many votes does tiger receive?
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.2 8
Answer: a. The numerical values for the votes are missing from the graph

b. In order to calculate the percent of the votes does the least popular mascot received will be halted due to lack of clear mentioning of of the proportion of the votes or the ratio of the votes.

c. The votes acquired by the tiger are cannot be determined because of figure which does not contain proper information.

Question 34.
DIG DEEPER!
A quarterback completes 18 of 33 passes during the first three quarters of a football game. He completes every pass in the fourth quarter and 62.5% of his passes for the entire game. How many passes does the quarterback throw in the fourth quarter? Justify your answer.
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.2 9
Answer: Quarterback throw 20.6 passes in the fourth quarter

Explanation:
A quarterback completes 18 of 33 passes during the first three quarters of a football game,
He completes every pass in the fourth quarter and 62.5% of his passes for the entire game.
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{33}\) = \(\frac{62.5}{100}\)
a = \(\frac{33 × 62.5}{100}\)
a = \(\frac{2,062.5}{100}\)
a = 20.6
So, 20.6 is 62.5% of 33.

Hence, Quarterback thrown 20.6 passes in the fourth quarter

Question 35.
REASONING
20% of a number is x. What is 100% of the number? Assume x > 0.
Answer: 5x

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
substitute a = x and p = 20
\(\frac{x}{w}\) = \(\frac{20}{100}\)
by cross multiplication , we get
w × 20 = x × 100
20w = 100x
Divide both sides by 20,
\(\frac{20w}{20}\) = \(\frac{100x}{20}\)
w = 5x
So, 100% of the number is 5x.

Question 36.
STRUCTURE
Answer each question. Assume x > 0.
a. What percent of 8x is 5x?
b. What is 65% of 80x?
Answer: a. 62.5% , b. 52x

Explanation:
a. percent of 8x is 5x
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{5x}{8x}\) = \(\frac{p}{100}\)
\(\frac{5}{8}\) = \(\frac{p}{100}\)
by cross multiplying
8p = 500
p = \(\frac{500}{8}\)
p = 62.5
So, 5x is 62.5% of 8x.

b. By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{8x}\) = \(\frac{65}{100}\)
a = \(\frac{8x × 65}{100}\)
a = \(\frac{520x}{100}\)
a = 52x
So, 52x is 65% of 80x.

Lesson 6.3 The Percent Equation

EXPLORATION 1

Using Percent Equations
Work with a partner.
a. The circle graph shows the number of votes received by each candidate during a school election. So far, only half of the students have voted. Find the percent of students who voted for each candidate. Explain your method.
Big Ideas Math Answers Grade 7 Chapter 6 Percents 6.3 1
c. The circle graph shows the final results of the election after every student voted. Use the equation you wrote in part(b) to find the number of students who voted for each candidate.
d. Use a different method to check your answers in part(c). Which method do you prefer? Explain.
Answer:
a. The percent of students who voted for each candidate are
For Person A = 20% ,
For Person B = 25%
For Person C = 15%
For Person D = 40%

b. The equation is a = \(\frac{w × p}{100}\)

c. The number of students who voted for each candidate are
For Person A = 30
For Person B = 24
For Person C = 24
For Person D = 42

d. ratio proportion is used as another method.

Explanation:
a. Given, The circle graph shows the number of votes received by each candidate during a school election. So far, only half of the students have voted.
The number of votes received till now are 12 + 15 + 9 + 24 = 60,
To know the percent of students who voted for each candidate we have ,
For person A, w = 60 , a = 12 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{12}{60}\) = \(\frac{p}{100}\)
p = \(\frac{12 × 100}{60}\)
p = \(\frac{1200}{60}\)
p = 20
So, 12 is 20% of 60.
The percent of  students who voted for person A is 20%

For person B , w = 60 , a = 15 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{15}{60}\) = \(\frac{p}{100}\)
p = \(\frac{15 × 100}{60}\)
p = \(\frac{1500}{60}\)
p = 25
So, 15 is 25% of 60.
The percent of  students who voted for person B is 25%

For person C , w = 60 , a = 9 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{9}{60}\) = \(\frac{p}{100}\)
p = \(\frac{9 × 100}{60}\)
p = \(\frac{900}{60}\)
p = 15
So, 9 is 15% of 60.
The percent of  students who voted for person C is 15%

For person D , w = 60 , a = 24 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{24}{60}\) = \(\frac{p}{100}\)
p = \(\frac{24 × 100}{60}\)
p = \(\frac{2400}{60}\)
p = 40
So, 24 is 40% of 60.
The percent of  students who voted for person D is 40%

b. we know that percent proportion is \(\frac{a}{w}\) = \(\frac{p}{100}\)
where as a = part , w= whole , p = percent ,
To solve for a , we can write it as a = \(\frac{w × p}{100}\)

c. The figure showing the percent of all the candidates individually are after the final results ,
as shown in part (a)  half of the students are 60 ,
The half of the students voted in all the students are 60 and total strength of students are 60 + 60 = 120
To calculate the number of voting acquired by each candidate we have,
For person A , w  = 120 , p = 25 , a = ?
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{120}\) = \(\frac{25}{100}\)
a = \(\frac{120 × 25}{100}\)
a = \(\frac{120}{4}\)
a = 30
So, 30 is 25% of 120.
The number of students who voted for person A after final results are 30

For person B , w  = 120 , p = 20 , a = ?
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{120}\) = \(\frac{20}{100}\)
a = \(\frac{120 × 20}{100}\)
a = \(\frac{120}{5}\)
a = 24
So, 24 is 20% of 120.
The number of students who voted for person B after final results are 24

For person C , w  = 120 , p = 20 , a = ?
By using percent proportion , we have
same as for person B ,
So, 24 is 20% of 120.
The number of students who voted for person C after final results are 24

For person D , w  = 120 , p = 35 , a = ?
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{120}\) = \(\frac{35}{100}\)
a = \(\frac{120 × 35}{100}\)
a = \(\frac{4,200}{100}\)
a = 42
So, 42 is 35% of 120.
The number of students who voted for person D after final results are 42.

d. We are using ratio proportion method , To check the answers of  part (c),
For person A , 25 % × 120 =
(25 : 100) × 120 =
(25 × 120) : 100 =
3,000 : 100 = 30
So, 30 is the 25% of 120.

For person B , 20 % × 120 =
(20 : 100) × 120 =
(20 × 120) : 100 =
2,400 : 100 = 24
So, 24 is the 20% of 120.

For person C , 20% × 120 is same as person B
So, 24 is the 20% of 120.

For person D , 35 % × 120 =
(35 : 100) × 120 =
(35 × 120) : 100 =
4,200 : 100 = 42
So, 42 is the 35% of 120.
All the answers are verified with ratio proportion method.

Try It

Write and solve an equation to answer the question.
Question 1.
What number is 10% of 20?
Answer:  2

Explanation:
10 % × 20 =
(10 : 100) × 20 =
(10 × 20) : 100 =
200 : 100 = 2
So, 2 is the 10% of 20.

Question 2.
What number is 150% of 40?
Answer: 60

Explanation:
150 % × 40 =
(150 : 100) × 40 =
(150 × 40) : 100 =
6,000 : 100 = 60
So, 60 is the 150% of 40.

Write and solve an equation to answer the question.
Question 3.
3 is what percent of 600?
Answer: 0.5%

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{3}{600}\) = \(\frac{p}{100}\)
p = \(\frac{3 × 100}{600}\)
p = \(\frac{300}{600}\)
p = \(\frac{1}{2}\)
p = 0.5
So, 3 is 0.5% of 600.

Question 4.
18 is what percent of 20?
Answer:

By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{18}{20}\) = \(\frac{p}{100}\)
p = \(\frac{18 × 100}{20}\)
p = \(\frac{1800}{20}\)
p = \(\frac{180}{2}\)
p = 90
So, 18 is 90% of 20.

Write and solve an equation to answer the question.
Question 5.
8 is 80% of what number?
Answer: 10

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{8}{w}\) = \(\frac{80}{100}\)
w = \(\frac{8 × 100}{80}\)
w = \(\frac{800}{80}\)
w = 10
So, 8 is 80% of 10.

Question 6.
90 is 180% of what number?
Answer: 50

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{90}{w}\) = \(\frac{180}{100}\)
w = \(\frac{90 × 100}{180}\)
w = \(\frac{9,000}{180}\)
w = 50
So, 90 is 180% of 50.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
VOCABULARY
Write the percent equation in words.
Answer:
Percent Equation:
In this equation, the whole is the number of which we are taking a percentage and the part is the value that results from taking the percent of the whole. This means that in any percent problem, there are three basic values to be concerned about: the percent, the whole, and the resulting part.
we can represent percent = p , whole = w , part = a
So, we have the percent equation as,
\(\frac{a}{w}\) = \(\frac{p}{100}\).

USING THE PERCENT EQUATION Write and solve an equation to answer the question.
Question 8.
14 is what percent of 70?
Answer: 20%

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{14}{70}\) = \(\frac{p}{100}\)
p = \(\frac{14 × 100}{70}\)
p = \(\frac{1400}{70}\)
p = 20
So, 14 is 20% of 70.

Question 9.
What number is 36% of 85?
Answer: 30.6

Explanation:
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{85}\) = \(\frac{36}{100}\)
a = \(\frac{85 × 36}{100}\)
a = \(\frac{3,060}{100}\)
a = 30.6
So, 30.6 is 36% of 85.

Question 10.
9 is 12% of what number?
Answer: 75

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{9}{w}\) = \(\frac{12}{100}\)
w = \(\frac{9 × 100}{12}\)
w = \(\frac{900}{12}\)
w = 75
So, 9 is 12% of 75.

Question 11.
108 is what percent of 72?
Answer: 150%

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{108}{72}\) = \(\frac{p}{100}\)
p = \(\frac{108 × 100}{72}\)
p = \(\frac{10,800}{72}\)
p = 150
So, 108 is 150% of 72.

Question 12.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answers Grade 7 Chapter 6 Percents 6.3 2
Answer: 55 is 20% of what number ? , is different  from other three questions.

Explanation:
Given , 20% of 55 , we have to find the part of whole number
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{55}\) = \(\frac{20}{100}\)
a = \(\frac{55 × 20}{100}\)
a = \(\frac{1,100}{100}\)
a = 11
So, 11 is 20% of 55.

But this 55 is 20% of what number ? is different  from other three, because here we have to find out the whole number
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{55}{w}\) = \(\frac{20}{100}\)
w = \(\frac{55 × 100}{20}\)
w = \(\frac{5500}{20}\)
w = 275
So, 55 is 20% of 275.
Hence , 55 is 20% of what number ? , is different  from other three questions.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 13.
DIG DEEPER!
A school offers band and chorus classes. The table shows the percents of the 1200 students in the school who are enrolled in band, chorus, or neither class. How many students are enrolled in both classes? Explain.
Big Ideas Math Answers Grade 7 Chapter 6 Percents 6.3 3
Answer: The total number of students enrolled for both classes  are 744.

Explanation:
Given, The table shows the percent of the 1200 students in the school who are enrolled in band, chorus, or neither class.
For Band , w = 1200 , p = 34 , a = ?
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{1200}\) = \(\frac{34}{100}\)
a = \(\frac{1200 × 34}{100}\)
a = 12 × 34
a = 408
So, 408 is 34% of 1200.
The number of students enrolled for the Band are 408.

For Band , w = 1200 , p = 28 , a = ?
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{1200}\) = \(\frac{28}{100}\)
a = \(\frac{1200 × 28}{100}\)
a = 12 × 28
a = 336
So, 336 is 28% of 1200.
The number of students enrolled for the chorus are 336.

The total number of students enrolled for both classes  are 408 + 336 = 744.

Question 14.
Water Tank A has a capacity of 550 gallons and is 66% full. Water Tank B is 53% full. The ratio of the capacity of Water Tank A to Water Tank B is 11:15.
a. How much water is in each tank?
b. What percent of the total volume of both tanks is filled with water?
Answer:
a. The water tank A is filled with 363 gallons of water.
The water tank B is filled with 397.5 gallons of water.

b. The percent of the total volume of both tanks is filled with water is 58.5%.

Explanation:
a.  Given , Water Tank A has a capacity of 550 gallons and is 66% full.
w = 550 gallons , p = 66% , a = ?
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{550}\) = \(\frac{66}{100}\)
a = \(\frac{550 × 66}{100}\)
a = 550 × 0.66
a = 363
So, 363 is 66% of 550.
The water tank A is filled with 363 gallons of water.

Given, Water Tank B is 53% full.
The ratio of the capacity of Water Tank A to Water Tank B is 11:15.
The capacity of Water Tank A is 550 gallons
Let the capacity of tank B is x gallons
\(\frac{550}{x}\) = \(\frac{11}{15}\)
x = \(\frac{550 × 15}{11}\)
x =  \(\frac{8,250}{11}\)
x = 750.
The capacity of the water Tank B is 750 gallons,
To know the amount of water filled in the tank we have,
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{750}\) = \(\frac{53}{100}\)
a = \(\frac{750 × 53}{100}\)
a = 7.5 × 53
a = 397.5
So, 397.5 is 53% of 750.
The water tank B is filled with 397.5 gallons of water.

b. To know the percent of the total volume of both tanks is filled with water, we have
The total capacity of Water tank A and Water tank B = 550 + 750 = 1,300 gallons
The total amount of water filled in both tanks are 363 + 397.5 =  760.5 gallons
So, w = 1,300 , a = 760.5 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{760.5}{1,300}\) = \(\frac{p}{100}\)
p = \(\frac{760.5 × 100}{1,300}\)
p = \(\frac{760.5}{13}\)
p = 58.5
So, 760.5 is 58.5% of 1,300.
The percent of the total volume of both tanks is filled with water is 58.5%.

The Percent Equation Homework & Practice 6.3

Review & Refresh

Write and solve a proportion to answer the question.
Question 1.
30% of what number is 9?
Answer: 30

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{9}{w}\) = \(\frac{30}{100}\)
w = \(\frac{9 × 100}{30}\)
w = \(\frac{900}{30}\)
w = 30
So, 9 is 30% of 30.

Question 2.
42 is what percent of 80?
Answer:  52.5%

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{42}{80}\) = \(\frac{p}{100}\)
p = \(\frac{42 × 100}{80}\)
p = \(\frac{420}{8}\)
p = 52.5
So, 42 is 52.5% of 80.

Question 3.
What percent of 36 is 20?
Answer:

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{20}{36}\) = \(\frac{p}{100}\)
p = \(\frac{20 × 100}{36}\)
p = \(\frac{2,000}{36}\)
p = 5.55
So, 20 is55.5% of 36.

Question 4.
What number is 120% of 80?
Answer: 96

Explanation:
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{80}\) = \(\frac{120}{100}\)
a = \(\frac{80 × 120}{100}\)
a =  8 × 12
a = 96
So, 96 is 120% of 80.

Find the distance between the two numbers on a number line.
Question 5.
– 4 and 10
Answer: The distance between the two numbers  – 4 and 10  on a number line is 14 .

Explanation: 
For the number line above we have divided each segment in to equal parts ,
So, The distance between the two numbers  – 4 and 10  on a number line is 14 .

Question 6.
–\(\frac{2}{3}\) and \(\frac{4}{3}\)
Answer: The distance between the two numbers –\(\frac{2}{3}\) and \(\frac{4}{3}\) on a number line is 6 .

Explanation:
For the number line above we have divided each segment in to equal parts ,
So, The distance between the two numbers –\(\frac{2}{3}\) and \(\frac{4}{3}\) on a number line is 6 .

Question 7.
– 5\(\frac{2}{3}\) and – 1 \(\frac{3}{10}\)
Answer: The distance between the two numbers  – 5\(\frac{2}{3}\) and – 1 \(\frac{3}{10}\) on a number line is 6.

Explanation:
Given , – 5\(\frac{2}{3}\) and – 1 \(\frac{3}{10}\)
can be written as \(\frac{-13}{3}\) and  \(\frac{-7}{10}\)
converting into decimal form we get , -4.3 and -0.7
For the number line above we have divided each segment in to equal parts ,
So, The distance between the two numbers  – 5\(\frac{2}{3}\) and – 1 \(\frac{3}{10}\) on a number line is 6.

Question 8.
– 4.3 and 7.5
Answer: The distance between the two numbers  – 4.3 and 7.5  on a number line is 8 .

Explanation:
For the number line above we have divided each segment in to equal parts ,
So, The distance between the two numbers  – 4.3 and 7.5  on a number line is 8 .

Question 9.
There are 160 people in a grade. The ratio of boys to girls is 3 to 5. Which proportion can you use to find the number x of boys?
Big Ideas Math Answers Grade 7 Chapter 6 Percents 6.3 4
Answer:  A. \(\frac{3}{8}\) = \(\frac{x}{160}\)

Explanation:
Given, The ratio of boys to girls is\(\frac{3}{5}\)
The ratio of boys to the grade is \(\frac{3}{8}\)
to find the number of x boys, we have to
\(\frac{3}{8}\) = \(\frac{x}{160}\)
x = \(\frac{3 × 160}{8}\)
x = \(\frac{480}{8}\)
x = 60.
So, A =  \(\frac{3}{8}\) = \(\frac{x}{160}\) is the correct answer.

Concepts, Skills, & Problem Solving

USING PERCENT EQUATIONS The circle graph shows the number of votes received by each candidate during a school election. Find the percent of students who voted for the indicated candidate. Each Candidate(See Exploration 1, p. 247.)
Big Ideas Math Answers Grade 7 Chapter 6 Percents 6.3 5
Question 10.
Candidate A
Answer: The percent of students who voted for the candidate A is  36%

Explanation:
Given, the circle graph shows the number of votes received by each candidate during a school election.
The total number of students voted are 54 + 60 + 36 = 150
For candidate A we have , w = 150 , a = 54 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{54}{150}\) = \(\frac{p}{100}\)
p = \(\frac{54 × 100}{150}\)
p = \(\frac{540}{15}\)
p = 36
So, 54 is 36% of 150.
The percent of students who voted for the candidate A is 36%.

Question 11.
Candidate B
Answer: The percent of students who voted for the candidate B is 40%.

Explanation:
For candidate B we have , w = 150 , a = 60 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{60}{150}\) = \(\frac{p}{100}\)
p = \(\frac{60 × 100}{150}\)
p = \(\frac{600}{15}\)
p = 40
So, 60 is 40% of 150.
The percent of students who voted for the candidate B is 40%.

Question 12.
Candidate C
Answer: The percent of students who voted for the candidate C is 24%.

Explanation:
For candidate C we have , w = 150 , a = 36 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{36}{150}\) = \(\frac{p}{100}\)
p = \(\frac{36 × 100}{150}\)
p = \(\frac{360}{15}\)
p = 24
So, 36 is 24% of 150.
The percent of students who voted for the candidate C is 24%.

USING THE PERCENT EQUATION Write and solve an equation to answer the question.
Question 13.
20% of 150 is what number?
Answer:  30

Explanation:
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{150}\) = \(\frac{20}{100}\)
a = \(\frac{150 × 20}{100}\)
a = 15 × 2
a = 30
So, 30 is 20% of 150.

Question 14.
45 is what percent of 60?
Answer: 75%

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{45}{60}\) = \(\frac{p}{100}\)
p = \(\frac{45 × 100}{60}\)
p = \(\frac{4,500}{60}\)
p = 75
So, 45 is 75% of 60.

Question 15.
35% of what number is 35?
Answer: 35 is 35% of 100.

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{35}{w}\) = \(\frac{35}{100}\)
w = \(\frac{35 × 100}{35}\)
w = \(\frac{3500}{35}\)
w = 100
So, 35 is 35% of 100.

Question 16.
0.8% of 150 is what number?
Answer: 1.2 .

Explanation:
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{150}\) = \(\frac{0.8}{100}\)
a = \(\frac{150 × 0.8}{100}\)
a = \(\frac{120}{100}\)
a = 1.2
So, 1.2 is 0.8% of 150.

Question 17.
29 is what percent of 20?
Answer: 145%

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{29}{20}\) = \(\frac{p}{100}\)
p = \(\frac{29 × 100}{20}\)
p = \(\frac{2,900}{20}\)
p = \(\frac{2,90}{2}\)
p = 145
So, 29 is 145% of 20.

Question 18.
0.5% of what number is 12?
Answer:

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{12}{w}\) = \(\frac{0.5}{100}\)
w = \(\frac{12 × 100}{0.5}\)
w = \(\frac{1200}{0.5}\)
w = 2,400.
So, 12 is 0.5% of 2,400.

Question 19.
What percent of 300 is 51?
Answer: 17%

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{51}{300}\) = \(\frac{p}{100}\)
p = \(\frac{51 × 100}{300}\)
p = \(\frac{51}{3}\)
p = 17
So, 51 is 17% of 300.

Question 20.
120% of what number is 102?
Answer: 85

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{102}{w}\) = \(\frac{120}{100}\)
w = \(\frac{102 × 100}{120}\)
w = \(\frac{1020}{12}\)
w = 85
So, 102 is 120% of 85.

YOU BE THE TEACHER Your friend uses the percent equation to answer the question. Is your friend correct? Explain your reasoning.
Question 21.
What number is 35% of 20?
Big Ideas Math Answers Grade 7 Chapter 6 Percents 6.3 6
Answer: yes , He is correct .

Explanation:
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{20}\) = \(\frac{35}{100}\)
a = \(\frac{20 × 35}{100}\)
a = \(\frac{700}{100}\)
a = 7
So, 7 is 35% of 20.

Question 22.
30 is 60% of what number?
Big Ideas Math Answers Grade 7 Chapter 6 Percents 6.3 7
Answer: 30 is 60% of 50.

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{30}{w}\) = \(\frac{60}{100}\)
w = \(\frac{30 × 100}{60}\)
w = \(\frac{300}{6}\)
w = 50
So, 30 is 60% of 50.

Question 23.
MODELING REAL LIFE
A salesperson receives a 2.5% commission on sales. What commission does the salesperson receive for $8000 in sales?
Answer: He receives $200 for commission of the sale.

Explanation:
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{8000}\) = \(\frac{2.5}{100}\)
a = \(\frac{8000 × 2.5}{100}\)
a = 80 × 2.5
a = $200
So, $200 is 2.5% of $8000.
He receives $200 for commission of the sale.

Question 24.
MODELING REAL LIFE
Your school raised 125% of its fundraising goal. The school raised $6750. What was the goal?
Answer: The fundraising goal of the school is $5,400.

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{6750}{w}\) = \(\frac{125}{100}\)
w = \(\frac{6750 × 100}{125}\)
w =54 × 100
w = $5,400
So, $6750 is 125% of $5,400.
The fundraising goal of the school is $5,400.

Question 25.
MODELING REAL LIFE
The sales tax on the model rocket shown is $1.92. What is the percent of sales tax?
Big Ideas Math Answers Grade 7 Chapter 6 Percents 6.3 8
Answer: The percent of sales tax on the rocket model  is 8%.

Explanation:
Given, The sales tax on the model rocket shown is $1.92.
The tax on the rocket is $24 , we have ,
w = 24 , a = 1.92 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{1.92}{24}\) = \(\frac{p}{100}\)
p = \(\frac{1.92 × 100}{24}\)
p = \(\frac{192}{24}\)
p = 8
So, $1.92 is 8% of $24.
The percent of sales tax on the rocket model  is 8%.

PUZZLE There were n signers of the Declaration of Independence. The youngest was Edward Rutledge, who was x years old. The oldest was Benjamin Franklin, who was y years old.
Big Ideas Math Answers Grade 7 Chapter 6 Percents 6.3 9
Question 26.
x is 25% of 104. What was Rutledge’s age?
Answer: The age of Rutledge is 26.

Explanation:
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{104}\) = \(\frac{25}{100}\)
a = \(\frac{104 × 25}{100}\)
a = \(\frac{104}{4}\)
a = 26
x = 26.
So, 26 is 25% of 104.

Question 27.
7 is 10% of y. What was Franklin’s age?
Answer: The Franklin’s age is 70.

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{7}{w}\) = \(\frac{10}{100}\)
w = \(\frac{7 × 100}{10}\)
w = \(\frac{700}{10}\)
w = 70
y = 70.
So, 7 is 10% of 70.
The Franklin’s age is 70.

Question 28.
n is 80% of y. How many signers were there?
Answer: There are n = 56 members signers.

Explanation:
y = 70 , p = 80 ,
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{70}\) = \(\frac{80}{100}\)
a = \(\frac{70 × 80}{100}\)
a = 7 × 8
a =56
n = 56
So, 56 is 80% of 70.
There are n = 56 members signers.

Big Ideas Math Answers Grade 7 Chapter 6 Percents 6.3 10

Question 29.
LOGIC
How can you tell whether a percent of a number will be greater than, less than, or equal to the number? Give examples to support your answer.
Answer: The percent of a number is less than 100% , Then the percent of the number will be less than the number.
The percent of a number is 100% , Then the percent of the number will be equal the number.
The percent of a number is greater than 100% , Then the percent of the number will be greater than the number.

Explanation:
If the percent of a number is less than 100% , Then the percent of the number will be less than the number.
For example , 80% of 50
= 0.8 × 50
= 40
80% < 100% ,  so 40 < 50.

If the percent of a number is 100% , Then the percent of the number will be equal the number.
For example , 100% of 50
= 1 × 50
= 50.
100% = 100% , So, 50 = 50.

If the percent of a number is greater than 100% , Then the percent of the number will be greater than the number.
For example , 120% of 50
= 1.2 × 50
= 60.
120% > 100% , So, 60 > 50.

Question 30.
PROBLEM SOLVING
In a survey, a group of students is asked their favorite sport. Eighteen students choose “other” sports.
a. How many students participate in the survey?
b. How many choose football?
Answer:  a. The number of students participated  are 80.
b. The number of students chose football are 30.

Explanation:
a. 18 students chose ” other” sports So, a = 18 ,
The percent of the “other” sport = 100% – (40% + 37.5%) = 22.5%
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{18}{w}\) = \(\frac{22.5}{100}\)
w = \(\frac{18 × 100}{22.5}\)
w = \(\frac{1800}{22.5}\)
w =  80
So, 18 is 22.5% of 80.
The number of students participated  are 80.

b. 80 students are participated , so w = 80
The percent of the students who chose football is 37.5%
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{80}\) = \(\frac{37.5}{100}\)
a = \(\frac{80 × 37.5}{100}\)
a = \(\frac{3,000}{100}\)
a = 30
So, 30 is 37.5% of 80.
The number of students chose football are 30.

Question 31.
TRUE OR FALSE?
Tell whether the statement is true or false. Explain your reasoning.
If W is 25% of Z, then Z : W is 75 : 25.
Answer: The statement is False.

Explanation:
Given , W is 25% of Z
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{W}{Z}\) = \(\frac{25}{100}\)
So, \(\frac{Z}{W}\) = \(\frac{100}{25}\)
But given that Z : W is 75 : 25
So, The statement is False.

Question 32.
DIG DEEPER!
At a restaurant, the amount of your bill before taxes and tip is $19.83. A 6% sales tax is applied to your bill, and you leave a tip equal to 19% of the original amount. Use mental math to estimate the total amount of money you pay. Explain your reasoning. (Hint: Use 10% of the original amount.)
Answer: The total amount of the money to be paid is $22.18.

Explanation:
Given ,The amount of your bill before taxes and tip is $19.83.
A 6% sales tax is applied to your bill, and you leave a tip equal to 19% of the original amount.
bill before the tax = $19.83
sales tax = 6%
So, $19.83 – 6%
= $19.83 – 0.06
= $18.64
Tip = 19%
So, $18.64 + 19%
= $18.64 + 0.19
= $22.18
The total amount of the money to be paid is $22.18.

Question 33
REASONING
The table shows your test results in a math class. What score do you need on the last test to earn 90% of the total points on the tests?
Big Ideas Math Answers Grade 7 Chapter 6 Percents 6.3 11
Answer: Of all the total test points you need 720 points to earn 90%

Explanation:
Total point value = 100 + 250 + 150 + 300 = 800
Given p = 90% , w = 800 , a= ?
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{800}\) = \(\frac{90}{100}\)
a = \(\frac{800 × 90}{100}\)
a = 8 × 90
a = 720
So, 720 is 90% of 800.
Finally, of all the total test points you need 720 points to earn 90%

Lesson 6.4 Percents of Increase and Decrease

EXPLORATION 1

Exploring Percent of Change
Work with a partner. Each year in the Columbia River Basin, adult salmon swim upriver to streams to lay eggs.
To go up the river, the adult salmon use fish ladders. But to go down the river, the young salmon must pass through several dams.
Big Ideas Math Solutions Grade 7 Chapter 6 Percents 6.4 1
At one time, there were electric turbines at each of the eight dams on the main stem of the Columbia and Snake Rivers. About 88% of the young salmon pass through a single dam unharmed.
a. One thousand young salmon pass through a dam. How many pass through unharmed?
b. One thousand young salmon pass through the river basin. How many pass through all 8 dams unharmed?
c. By what percent does the number of young salmon decrease when passing through a single dam?
d. Describe a similar real-life situation in which a quantity increases by a constant percent each time an event occurs.
Answer: a. 880 salmon passed through single dam unharmed.
b. Totally , 358 salmon pass through all 8 dams unharmed.
c. The percent of the number of young salmon decrease when passing through a single dam is 12%
d. An example for, real-life situation in which a quantity increases by a constant percent each time an event occurs. is given below in explanation.

Explanation:
a. Given , One thousand young salmon pass through a dam. 88% of the young salmon pass through a single dam unharmed.
To know the number of salmon passed through unharmed we have,
88% of 1,000
= 88% × 1,000
= 0.88 × 1,000
= 880.
So, 880 salmon passed through single dam unharmed.

b. Given , One thousand young salmon pass through a dam. 88% of the young salmon pass through a single dam unharmed. To calculate number of salmon pass through all 8 dams unharmed are
number of  salmon passed through dam 1 , unharmed are 880 (as shown in part a)
number of  salmon passed through dam 2 , unharmed =  88%  of  880
= 0.88 × 880
= 774
number of  salmon passed through dam 3 , unharmed =  88%  of  774
= 0.88 × 774
= 681
number of  salmon passed through dam 4 , unharmed =  88%  of  681
= 0.88 × 681
= 599
number of  salmon passed through dam 5 , unharmed =  88%  of  599
= 0.88 × 599
= 527
number of  salmon passed through dam 6 , unharmed =  88%  of  527
= 0.88 × 527
= 463
number of  salmon passed through dam 7 , unharmed =  88%  of  463
= 0.88 × 463
= 407
number of  salmon passed through dam 8 , unharmed =  88%  of  407
= 0.88 × 407
= 358
Totally , 358 salmon pass through all 8 dams unharmed.

c. To calculate The percent of the number of young salmon decrease when passing through a single dam is
The total percent of salmon is 100%, The percent of salmon pass through a single dam is 88%
So, 100% – 88% = 12%
Finally , The percent of the number of young salmon decrease when passing through a single dam is 12%

d.  An Example of real-life situation in which a quantity increases by a constant percent each time an event occurs.  is ,  while we are filling the tank with water , The amount of water ingoing increases constantly with the speed of the motor power running the water , water levels in the tank increases by a constant percent each time until the tank is filled up with the water fully.

Big Ideas Math Solutions Grade 7 Chapter 6 Percents 6.4 2

Try It

Find the percent of change. Round to the nearest tenth of a percent if necessary.
Question 1.
10 inches to 25 inches
Answer:  percent of change is 150%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 25 and old value = 10 , because a change of 10 to 25 is a positive (increase) change
So, percent change = \(\frac{25 – 10}{10}\)
= \(\frac{15}{10}\)
= \(\frac{15}{10}\) × 100
= 150%
So, percent of change is 150%

Question 2.
57 people to 65 people
Answer:  percent of change is 14%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 65 and old value = 57 , because a change of 57 to 65 is a positive (increase) change
So, percent change = \(\frac{65 – 57}{57}\)
= \(\frac{8}{57}\)
= \(\frac{8}{57}\) × 100
= 14.03 %
Approximately We can write it as 14 %
So, percent of change is 14%

Question 3.
In Example 2, what was the percent of change from 2014 to 2015?
Answer: percent of change is  – 44%

Explanation:
In Example 2, change from 2014 to 2015 , that is 18 to 10
We know that , formula for percent change = \(\frac{New  value – old  value}{old  value}\)
where New value = 10 and old value = 18 , because a change of 18 to 10 is a negative (decrease) change
So, percent change = \(\frac{10 – 18}{18}\)
= \(\frac{-8}{18}\)
= – 0.444
=  – 0.444 × 100
=  – 44.4 %
Approximately We can write it as  – 44 %
So, percent of change is  – 44%

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 4.
VOCABULARY
What does it mean for a quantity to change by n%?
Answer: The meaning of a quantity to change is given by, which the amount of quantity increases or decreases of its original value to new value is called percent of change,
Then percent of change can be in positive or negative depending on the value of change , if it is for n% then it can be increase or decrease in the quantity of change.

Question 5.
NUMBER SENSE
Without calculating, determine which situation has a greater percent of change. Explain.

  • 5 bonus points added to 50 points
  • 5 bonus points added to 100 points

Answer: 5 bonus points added to 50 points has the greater percent of change.

Explanation:
5 bonus points added to 50 points
Then 50 points to 55 points
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 55 and old value = 50 , because a change of 50 to 55 is a positive (increase) change
So, percent change = \(\frac{55 – 50}{50}\)
= \(\frac{5}{50}\)
= \(\frac{1}{10}\) × 100
= 10%
So, percent of change is 10%

5 bonus points added to 100 points
Then 100 points to 105 points
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 105 and old value = 100 , because a change of 100 to 105 is a positive (increase) change
So, percent change = \(\frac{105 – 100}{100}\)
= \(\frac{5}{100}\)
= \(\frac{1}{20}\) × 100
= 5%
So, percent of change is 5%

Percent of change of 5 bonus points added to 50 points is 10%
Percent of change of 5 bonus points added to 100 points is 5%
So, 5 bonus points added to 50 points has the greater percent of change.

FINDING A PERCENT OF CHANGE Identify the percent of change as an increase or a decrease. Then find the percent of change.
Question 6.
8 feet to 24 feet
Answer: percent of change is increased that is 200%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 24 and old value = 8 , because a change of 8 to 24 is a positive (increase) change
So, percent change = \(\frac{24 -8}{8}\)
= \(\frac{16}{8}\)
= 2 × 100
= 200%
So, percent of change is 200%

Question 7.
300 miles to 210 miles
Answer: percent of change is decreased that is -30%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 210 and old value = 300 , because a change of300 to 210 is a negative (decrease) change
So, percent change = \(\frac{210 – 300}{300}\)
= \(\frac{-90}{300}\)
= \(\frac{-9}{30}\) × 100
= -0.3 × 100
= -30%
So, percent of change is -30%

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 8.
In one round of a game, you are asked how many bones are in a human body. If the percent error of your answer is at most 5%, you earn two points. If the percent error is at most 10%, but greater than 5%, you earn one point. You guess 195 bones. The correct answer is 206 bones. How many points do you earn?
Big Ideas Math Solutions Grade 7 Chapter 6 Percents 6.4 3
Answer: The percent Error we calculated  is at most of 5% so points we earned is 2 points

Explanation:
Given, You guess 195 bones. The correct answer is 206 bones.
The amount of error is 206 – 195 = 11
We know that , formula for percent error = \(\frac{Error value}{Actual value}\)
where Error value = 11 and Actual value = 206 ,
So, percent Error = \(\frac{11}{206}\)
= 0.053
= 0.053 × 100
=  5.3%
Approximately we can write as 5%
So, percent Error is 5%
Then , the percent Error we calculated  is at most of 5% so points we earned is 2 points

Question 9.
DIG DEEPER!
The manager of a restaurant offers a 20% decrease in price to tennis teams. A cashier applies a 10% decrease and then another 10% decrease. Is this the same as applying a 20% decrease? Justify your answer.
Answer: There is slight difference between these two methods but are approximately equal.

Explanation:
Given, The manager of a restaurant offers a 20% decrease in price to tennis teams.
Let the total price be 100
So, 100 decrease 20%
= 100 × (1 – 20%)
= 100 × (1 – 0.2)
= 80.

Given, A cashier applies a 10% decrease and then another 10% decrease.
Let the total price be 100
So, 100 decrease 10%
= 100 × (1 – 10%)
= 100 × (1 – 0.1)
= 90.

Again applying 10% decrease
90 decrease 10%
= 90 × (1 – 10%)
= 90 × (1 – 0.1)
= 81.

So, There is slight difference between these two methods but are approximately equal

Percents of Increase and Decrease Homework & Practice 6.4

Review & Refresh

Write and solve an equation to answer the question.
Question 1.
What number is 25% of 64?
Answer:  16

Explanation:
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{64}\) = \(\frac{25}{100}\)
a = \(\frac{64 × 25}{100}\)
a = \(\frac{64}{4}\)
a = 16
So, 16 is 25% of 64.

Question 2.
39.2 is what percent of 112?
Answer: 35%

Explanation:
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{39.2}{112}\) = \(\frac{p}{100}\)
p = \(\frac{39.2 × 100}{112}\)
p = \(\frac{3,920}{112}\)
p = 35
So, 39.2 is 35% of 112.

Find the sum. Write fractions in simplest form.
Question 5.
\(\frac{4}{7}\) + (- \(\frac{6}{7}\))
Answer:  – \(\frac{2}{7}\)

Explanation:
To find the sum of the fractions given, we have to do addition, that is
\(\frac{4}{7}\) + (- \(\frac{6}{7}\))
= \(\frac{4 – 6}{7}\)
= – \(\frac{2}{7}\)
So, The sum of \(\frac{4}{7}\) + (- \(\frac{6}{7}\)) is – \(\frac{2}{7}\)

Question 6.
– 4.621 + 3.925
Answer: – 0.696

Explanation:
To find the sum we have to add the given numbers , that is
– 4.621 + 3.925
=  3.925 – 4.621
= – 0.696
So, the sum of – 4.621 + 3.925 is – 0.696.

Question 7.
–\(\frac{5}{12}\) + \(\frac{3}{4}\)
Answer: \(\frac{1}{3}\)

Explanation:
To find the sum of the fractions given, we have to do addition, that is
Given , –\(\frac{5}{12}\) + \(\frac{3}{4}\)
\(\frac{3}{4}\) – \(\frac{5}{12}\)
Expand the fraction , multilpy the numerator and denominator by 3
We get , \(\frac{3 × 3}{3 × 4}\)
= \(\frac{3 × 3}{3 × 4}\) – \(\frac{5}{12}\)
= \(\frac{9}{12}\) – \(\frac{5}{12}\)
= \(\frac{9 – 5}{12}\)
= \(\frac{4}{12}\)
The simplest form of \(\frac{4}{12}\) is \(\frac{1}{3}\)
So, The sum of –\(\frac{5}{12}\) + \(\frac{3}{4}\) is \(\frac{1}{3}\)

Concepts, Skills, & Problem Solving

EXPLORING PERCENT CHANGE You are given the percent of salmon that pass through a single dam unharmed. By what percent does the number of salmon decrease when passing through a single dam? (See Exploration 1, p. 253.)
Question 8.
75%
Answer:  The percent of the number of salmon decrease when passing through a single dam is 25%

Explanation:
The percent of salmon that pass through a single dam unharmed is 75%
Let the percent of salmon coming to the dam are 100%
The percent of the number of salmon decrease when passing through a single dam is
100% – 75% = 25%
So, The percent of the number of salmon decrease when passing through a single dam is 25%

Question 9.
80%
Answer: The percent of the number of salmon decrease when passing through a single dam is 20%

Explanation:
The percent of salmon that pass through a single dam unharmed is 80%
Let the percent of salmon coming to the dam are 100%
The percent of the number of salmon decrease when passing through a single dam is
100% – 80% = 20%
So, The percent of the number of salmon decrease when passing through a single dam is 20%

Question 10.
62%
Answer: The percent of the number of salmon decrease when passing through a single dam is 38%

Explanation:
The percent of salmon that pass through a single dam unharmed is 62%
Let the percent of salmon coming to the dam are 100%
The percent of the number of salmon decrease when passing through a single dam is
100% – 62% = 38%
So, The percent of the number of salmon decrease when passing through a single dam is 38%

Question 11.
94%
Answer: The percent of the number of salmon decrease when passing through a single dam is 6%

Explanation:
The percent of salmon that pass through a single dam unharmed is 94%
Let the percent of salmon coming to the dam are 100%
The percent of the number of salmon decrease when passing through a single dam is
100% – 94% = 6%
So, The percent of the number of salmon decrease when passing through a single dam is 6%

FINDING A PERCENT OF CHANGE Identify the percent of change as an increase or a decrease. Then find the percent of change. Round to the nearest tenth of a percent if necessary.
Question 12.
12 inches to 36 inches
Answer: percent of change is 200%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 36 and old value = 12 , because a change of 12 to 36 is a positive (increase) change
So, percent change = \(\frac{36 – 12}{12}\)
= \(\frac{24}{12}\)
= 2 × 100
= 200%
So, percent of change is 200%

Question 13.
75 people to 25 people
Answer: percent of change is 66%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 25 and old value = 75 , because a change of 75 to 25 is a negative (decrease) change
So, percent change = \(\frac{25 – 75}{75}\)
= \(\frac{- 50}{75}\)
=  – 0.666
= – 0.666 × 100
= – 66.6%
Approximately we can write it as – 66%
So, percent of change is – 66%

Question 14.
50 pounds to 35 pounds
Answer: percent of change is – 30%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 35 and old value = 50 , because a change of 35 to 50 is a negative (decrease) change
So, percent change = \(\frac{35 – 50}{50}\)
= \(\frac{- 15}{50}\)
= – 0.3
= – 0.3 × 100
= – 30%
So, percent of change is – 30%

Question 15.
24 songs to 78 songs
Answer: percent of change is 225%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 78 and old value = 24 , because a change of 24 to 78 is a positive (increase) change
So, percent change = \(\frac{78 – 24}{24}\)
= \(\frac{54}{24}\)
= 2.25 × 100
= 225%
So, percent of change is 225%

Question 16.
10 gallons to 24 gallons
Answer: percent of change is 140%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 24 and old value = 10 , because a change of 10 to 24 is a positive (increase) change
So, percent change = \(\frac{24 – 10}{10}\)
= \(\frac{14}{10}\)
= 1.4
= 1.4 × 100
= 140%
So, percent of change is 140%

Question 17.
72 paper clips to 63 paper clips
Answer: percent of change is – 12.5%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 63 and old value = 72 , because a change of 72 to 63 is a negative (decrease) change
So, percent change = \(\frac{63 – 72}{72}\)
= \(\frac{- 9}{72}\)
= \(\frac{- 1}{8}\)
= – 0.125
= – 0.125 × 100
= – 12.5%
So, percent of change is – 12.5%

Question 18.
16 centimeters to 44.2 centimeters
Answer: percent of change is 176%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 44.2 and old value = 16 , because a change of 16 to 44.2 is a positive (increase) change
So, percent change = \(\frac{44.2 – 16}{16}\)
= \(\frac{28.2}{16}\)
= 1.76
= 1.76 × 100
= 176%
So, percent of change is 176%

Question 19.
68 miles to 42.5 miles
Answer: percent of change is – 37.5%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 42.5 and old value = 68 , because a change of 68 to 42.5 is a negative (decrease) change
So, percent change = \(\frac{42.5 – 68}{68}\)
= \(\frac{- 25.5}{68}\)
= – 0.375
= – 0.375 × 100
= – 37.5%
So, percent of change is – 37.5%

Question 20.
YOU BE THE TEACHER
Your friend finds the percent increase from 18 to 26. Is your friend correct? Explain your reasoning.
Big Ideas Math Solutions Grade 7 Chapter 6 Percents 6.4 4
Answer: No , The percent of change of 18 to 26 is positive (increase) that is 44.4%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 26 and old value = 18 , because a change of 18 to 26 is a positive (increase) change
So, percent change = \(\frac{26 – 18}{18}\)
= \(\frac{8}{18}\)
= 0.444
= 0.444 × 100
= 44.4%
So, percent of change is 44.4%

Question 21.
MODELING REAL LIFE
Last week, you finished Level 2 of a video game in 32 minutes. Today, you finish Level 2 in 28 minutes. What is the percent of change?
Answer: The percent of change from last week to today is – 12.5%

Explanation:
Given, you finished Level 2 of a video game in 32 minutes. Today, you finish Level 2 in 28 minutes.
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 28 and old value = 32 , because a change of 32 to 28 is a negative (decrease) change
So, percent change = \(\frac{28 – 32}{32}\)
= \(\frac{- 4}{32}\)
= \(\frac{- 1}{8}\)
= – 0.125
= – 0.125 × 100
= – 12.5%
So, percent of change is – 12.5%

Question 22.
MODELING REAL LIFE
You estimate that a baby pig weighs 20 pounds. The actual weight of the baby pig is 16 pounds. Find the percent error.
Big Ideas Math Solutions Grade 7 Chapter 6 Percents 6.4 5
Answer: The percent Error is 20%.

Explanation:
Given , You estimate that a baby pig weighs 20 pounds. The actual weight of the baby pig is 16 pounds.
The amount of error is 20 – 16 = 4
We know that , formula for percent error = \(\frac{Error value}{Actual value}\)
where Error value = 4 and Actual value = 20 ,
So, percent Error = \(\frac{4}{20}\)
= \(\frac{1}{5}\)
= 0.2
= 0.2 × 100
=  20%
So, percent Error is 20% .

Question 23.
PRECISION
A researcher estimates that a fossil is 3200 years old. Using carbon-14 dating, a procedure used to determine the age of an object, the researcher discovers that the fossil is 3600 years old.
Big Ideas Math Solutions Grade 7 Chapter 6 Percents 6.4 6.1
a. Find the percent error.
b. What other estimate gives the same percent error? Explain your reasoning.
Answer: a. The percent Error is 11.1%
b. The other estimate that gives the same percent error is 3,199 years old.

Explanation:
a. Given, A researcher estimates that a fossil is 3200 years old  the researcher discovers that the fossil is 3600 years old.
The amount of error is 3600 – 3200 = 400
We know that , formula for percent error = \(\frac{Error value}{Actual value}\)
where Error value = 400 and Actual value = 3600 ,
So, percent Error = \(\frac{400}{3600}\)
= \(\frac{1}{9}\)
= 0.111
= 0.111 × 100
=  11.1%
So, percent Error is 11.1% .

b. If The other estimate that gives the same percent error is 3,199 years old.
The amount of error is 3600 – 3199 = 401
We know that , formula for percent error = \(\frac{Error value}{Actual value}\)
where Error value = 401 and Actual value = 3600 ,
So, percent Error = \(\frac{401}{3600}\)
= 0.111
= 0.111 × 100
=  11.1%
So, percent Error is 11.1% .
So , The other estimate that gives the same percent error is 3,199 years old.

FINDING A PERCENT OF CHANGE Identify the percent of change as an increase or a decrease. Then find the percent of change. Round to the nearest tenth of a percent if necessary.
Question 24.
\(\frac{1}{4}\) to \(\frac{1}{2}\)
Answer: percent of change is 100%

Explanation:
The given fractions can be written in decimal form then we have,
\(\frac{1}{4}\) as 0.25
\(\frac{1}{2}\) as  0.5 , so , 0.25 to 0.5
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 0.5 and old value = 0.25 , because a change of 0.25 to 0.5 is a positive (increase) change
So, percent change = \(\frac{0.5 – 0.25}{0.25}\)
= \(\frac{0.25}{0.25}\)
= 1 × 100
= 100%
So, percent of change is 100%

Question 25.
\(\frac{4}{5}\) to \(\frac{3}{5}\)
Answer: percent of change is – 25%

Explanation:
The given fractions can be written in decimal form then we have,
\(\frac{4}{5}\) as 0.8
\(\frac{3}{5}\) as 0.6 , so , 0.8 to 0.6
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 0.6 and old value = 0.8 , because a change of 0.8 to 0.6 is a negative (decrease) change
So, percent change = \(\frac{0.6 – 0.8}{0.8}\)
= \(\frac{-0.2}{0.8}\)
= – 0.25
= – 0.25 × 100
= – 25%
So, percent of change is – 25%

Question 26.
\(\frac{3}{8}\) to \(\frac{7}{8}\)
Answer: percent of change is 135%

Explanation:
The given fractions can be written in decimal form then we have,
\(\frac{3}{8}\) as 0.37
\(\frac{7}{8}\) as 0.87 , So 0.37 to 0.87
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 0.87 and old value = 0.37 , because a change of 0.37 to 0.87 is a positive (increase) change
So, percent change = \(\frac{0.87 – 0.37}{0.37}\)
= \(\frac{0.5}{0.37}\)
= 1.35
= 1.35 × 100
= 135%
So, percent of change is 135%

Question 27.
\(\frac{5}{4}\) to \(\frac{3}{8}\)
Answer: percent of change is – 70.4%

Explanation:
The given fractions can be written in decimal form then we have,
\(\frac{5}{4}\) as 1.25
\(\frac{3}{8}\) as 0.37 , So, 1.25 to 0.37
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 0.37 and old value = 1.25 , because a change of 1.25 to 0.37 is a negative (decrease) change
So, percent change = \(\frac{0.37 – 1.25}{1.25}\)
= \(\frac{- 0.88}{1.25}\)
= – 0.704
= – 0.704 × 100
= – 70.4%
So, percent of change is – 70.4%

Question 28.
CRITICAL THINKING
Explain why a change from 20 to 40 is a 100% increase, but a change from 40 to 20 is a 50% decrease.
Answer: From 20 to 40 is a 100% increase because of increase in number value and from 40 to 20 is a 50% decrease because of decrease in number value.

Explanation:
Given , 20 to 40
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 40 and old value = 20 , because a change of 20 to 40 is a positive (increase) change
So, percent change = \(\frac{40 – 20}{20}\)
= \(\frac{20}{20}\)
= 1 × 100
= 100%
Then , percent of change is 100%
So, The percent of change from 20 to 40 is a 100% increase

Given , 40 to 20
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 20 and old value = 40 , because a change of 40 to 20 is a negative (decrease) change
So, percent change = \(\frac{20 – 40}{40}\)
= \(\frac{- 20}{40}\)
= \(\frac{- 1}{2}\)
= – 0.5
= – 0.5 × 100
= – 50%
Then , percent of change is – 50%
So , The percent of  change from 40 to 20 is a 50% decrease.

Finally , From 20 to 40 is a 100% increase because of increase in number value and from 40 to 20 is a 50% decrease because of decrease in number value.

Question 29.
MODELING REAL LIFE
The table shows population data for a community.
Big Ideas Math Solutions Grade 7 Chapter 6 Percents 6.4 6
a. What is the percent of change from 2011 to 2017?
b. Predict the population in 2023. Explain your reasoning.
Answer: a.  The percent of change from 2011 to 2017  is 169%
b. The estimated population of 2023 will be 158,000.

Explanation:
a. Given 2011 to 2017 , so from the table we know it as , 118,000 to 138,000
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 138,000  and old value = 118,000 , because a change of 118,000 to 138,000 is a positive (increase) change
So, percent change = \(\frac{138,000 – 118,000}{118,000}\)
= \(\frac{20,000}{118,000\)
= 0.169
=0.169 × 100
= 169%
So, percent of change is 169%

b. The population from 2011 to 2017 increased from 118,000 to 138,000,
The difference between 2011 to 2017 is 6 years ,So the population increase in numbers are
138,000 – 118,000 = 20,000.
If the population in 6 years is increased by 20,000.
Then from 2017 to 2023 is 6 years , So increase in population is 20,000
Then for 2023 The population will be 138,000 + 20,000 = 158,000
finally, The estimated population of 2023 will be 158,000.

Question 30.
GEOMETRY
Suppose the length and the width of the sandbox are doubled.
Big Ideas Math Solutions Grade 7 Chapter 6 Percents 6.4 7
a. Find the percent of change in the perimeter.
b. Find the percent of change in the area.
Answer: a. the percent of change in the perimeter of sandbox is 100%
b. the percent of change in the area of sandbox is

Explanation:
a. Given , length of sandbox = 10 ft , width of sandbox = 6 ft
The sandbox is in the form of a rectangle , So the perimeter of a rectangle is P = 2 ( l + w) , where l = length of the rectangle and  w = width of the rectangle,
Then P = 2( l + w )
= 2 ( 10 + 6 )
= 2 × 16
=18
So, the perimeter of sandbox is 32 ft

Given that , the length and the width of the sandbox are doubled.
Then l = 20 ft and w = 12 ft ,
P = 2( l + w )
= 2 ( 20 + 12 )
= 2 × 32
= 64
So, the perimeter of sandbox after the length and the width are doubled. is 64 ft .
The perimeter of sandbox changed  from 32 ft to 64 ft , Then
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 64 and old value = 32 , because a change of 32 to 64 is a positive (increase) change
So, percent change = \(\frac{64 – 32}{32}\)
= \(\frac{32}{32}\)
= 1 × 100
= 100%
So, percent of change is 100%
Finally , the percent of change in the perimeter of sandbox is 100%

b. Given , length of sandbox = 10 ft , width of sandbox = 6 ft
The sandbox is in the form of a rectangle , So the area of a rectangle is A = l × w ,
Then A = l × w
= 10 × 6 = 60
So , The Area of Sandbox is 60 ft

Given that , the length and the width of the sandbox are doubled.
Then , l = 20 ft and w = 12 ft ,
Then A = l × w
= 20 × 12 = 240
So , The Area of Sandbox after the length and the width are doubled is 240 ft
The area of sandbox changed from 60 ft to 240 ft
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 240 and old value = 60 , because a change of 60 to 240 is a positive (increase) change
So, percent change = \(\frac{240 – 60}{60}\)
= \(\frac{180}{60}\)
= 3
= 3 × 100
= 300%
So, percent of change is 300%
Finally , the percent of change in the area of sandbox is 300%.

Question 31.
MODELING REAL LIFE
A company fills boxes with about 21 ounces of cereal. The acceptable percent error in filling a box is 2.5%. Box A contains 20.4 ounces of cereal and Box B contains 21.5 ounces of cereal. Tell whether each box is an acceptable weight.
Answer: The percent error of Box A is greater than acceptable percent error that is 2.8% and The percent error of Box B is less than acceptable percent error that is 2.3%.

Explanation:
For Box A , Given , A company fills boxes with about 21 ounces of cereal and Box A contains 20.4 ounces of cereal
The amount of error is 21 – 20.4 = 0.6
We know that , formula for percent error = \(\frac{Error value}{Actual value}\)
where Error value = 0.6 and Actual value = 21 ,
So, percent Error = \(\frac{0.6}{21}\)
= 0.028
= 0.028 × 100
=  2.8%
So, percent Error is 2.8%
The percent error of Box A is greater than acceptable percent error that is 2.8%

For Box B A company fills boxes with about 21 ounces of cereal and  Box B contains 21.5 ounces of cereal.
The amount of error is 21.5 – 21 = 0.5
We know that , formula for percent error = \(\frac{Error value}{Actual value}\)
where Error value = 0.5 and Actual value = 21 ,
So, percent Error = \(\frac{0.5}{21}\)
= 0.023
= 0.023 × 100
=  2.3%
So, percent Error is 2.3%
The percent error of Box B is less than acceptable percent error that is 2.3%

Question 32.
PRECISION
Find the percent of change from June to September in the mile-run times shown.
Big Ideas Math Solutions Grade 7 Chapter 6 Percents 6.4 8
Answer: The percent of change from June to September in the mile-run times is -26%.

Explanation:
Given , change from 7.45 to 5.51
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 5.51 and old value = 7.45 , because a change of 7.45 to 5.51 is a negative (decrease) change
So, percent change = \(\frac{5.51 – 7.45}{7.45}\)
= \(\frac{- 1.94}{7.45}\)
= – 0.26
= – 0.26 × 100
= – 26%
So, percent of change is – 26%
Finally , The percent of change from June to September in the mile-run times is -26%.

Question 33.
CRITICAL THINKING
A number increases by 10% and then decreases by 10%. Will the result be greater than, less than or equal to, the original number? Explain.
Answer: A number increases by 10% and then decreases by 10% , the result will be less than the original number.

Explanation:
Let original number is 100
Given ,The original number increases by 10%
The new number = original number + The original number increases by 10%
= 100 + ( 100 × 10%)
= 100 + ( 100 × 0.1)
= 100 + 10
= 110
So, The new number is 110.

Let the original number is 110
Given ,The original number decreases by 10%
The new number = original number – The original number decreases by 10%
= 110 – ( 110 × 10%)
= 110 – ( 110 × 0.1)
= 110 – 11
= 99
So, The new number is 99

99 < 100 .
So , A number increases by 10% and then decreases by 10% , the result will be less than the original number.

Question 34.
PROBLEM SOLVING
You want to reduce your daily calorie consumption by about 9%. You currently consume about 2100 calories per day. Use mental math to estimate the number of calories you should consume in one week to meet your goal. Explain.
Answer: The estimated calories you should consume per week is 13,377.

Explanation:
Given ,You currently consume about 2100 calories per day, reduce your daily calorie consumption by about 9%.
so, The calories you have to consume after reduction of 9% is
The new number = original number – The original number decreases by 9%
= 2100 – ( 2100 × 9%)
= 2100 – (2100 × 0.09)
= 2100 – 189
= 1,911 .
The calories you have to consume after reduction of 9% is 1,911

The calories you should consume per day is 1,911.
The calories you should consume per week (7 days) = 1,911 × 7 = 13,377.

Finally , The estimated calories you should consume per week is 13,377.

Question 35.
DIG DEEPER!
Donations to an annual fundraiser are 15% greater this year than last year. Last year, donations were10% greater than the year before. The amount raised this year is $10,120. How much was raised two years ago?
Answer: The amount raised 2 years ago is $7,741.8.

Explanation:
Given , The amount raised this year is $10,120.
Let the amount raised last year = x
Donations are 15% greater than last year
The amount raised last year = The amount raised this year – ((The amount raised this year .15%)
x = 10,120 – ( 10,120 × 0.15)
x = 10,120 – 1,518
x = 8,602
The amount raised last year = $8,602.

We know that , The amount raised last year = $8,602.
Let the amount raised the year before = x
Donations are 10% greater than the year before
The amount raised the year before = The amount raised last year – ((The amount raised last year .10%)
x = 8,602 – ( 8,602 × 0.1)
x = 8,602 – 860.2
x = 7,741.8
The amount raised the year before = $7,741.8.

So, The amount raised 2 years ago is $7,741.8.

Question 36.
REASONING
Forty students are in the science club. Of those, 45% are girls. This percent increases to 56% after more girls join the club. How many more girls join?
Answer:  The number of new girls join the club is 10.

Explanation:
Let the number of new girls = x
The number of girls = x + 18
The number of students = x + 40
So, the number of girls = 56% The number of students
x + 18 = 0.56( x + 40 )
x + 18  = 0.56x + 22.4
0.56x – x = 22.4 – 18
0.44x = 4.4
x = \(\frac{4.4}{0.44}\)
x = 10.
So , The number of new girls is 10.

Lesson 6.5 Discounts and Markups

EXPLORATION 1

Comparing Discounts
Work with a partner.
a. The same pair of earrings is on sale at three stores. Which store has the best price? Use the percent models to justify your answer.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.5 1
b. You buy the earrings on sale for 30% off at a different store. You pay $22.40. What was the original price of the earrings? Use the percent model to justify your answer.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.5 2
c. You sell the earrings in part(b) to a friend for 60% more than what you paid. What is the selling price? Use a percent model to justify your answer.
Answer:
a. For store A ,The sales price is $27
For store B  ,The sales price is $24.5
For store C ,The sales price is $31.2

b. the original price  of the earrings is $32

c. the selling price is $35.84.

Explanation:
a. Given store A = $45 with 40% off

We know , The sales price be 100% – 40% = 60% of the original price
sales price = 60% of 45
= 0.6 × 45 = 27
So, The sales price is $27

For store B = $49 with 50% off

We know , The sales price be 100% – 50% = 50% of the original price
sales price = 50% of 49
= 0.5 × 49 = 24.5
So, The sales price is $24.5

For store c = $39 with 20% off

We know , The sales price be 100% – 20% = 80% of the original price
sales price = 80% of 39
= 0.8 × 39 = 31.2
So, The sales price is $31.2

b. Given , You buy the earrings on sale for 30% off at a different store. You pay $22.40
The saples price = 100% – 30% = 70%

we know  a = 22.4 , p = 70% , w = ?
a  =  p% × w
22.4 = 0.7 × w
w = \(\frac{22.4}{0.7}\)
w = 32.
So , the original price of the earrings is $32

c. Given , You sell the earrings in part(b) to a friend for 60% more than what you paid.
If the selling price is more than the buying price then it is called markup


Here , the markup is 60% of $22.4
a = p% × w
a = 0.6 × 22.4
a = 13.44
So, the markup is  $13.44
We know that selling price = cost of buying + markup
= 22.4 + 13.44
= 35.84
So , the selling price is $35.84.

Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.5 3

Try It

Question 1.
The original price of a skateboard is $50. The skateboard is on sale for 20% off. What is the sale price?
Answer: The sales price of skateboard is $40

Explanation:
Given , skateboard is $50 with 20% off
We know , The sales price be 100% – 20% = 80% of the original price
sales price = 80% of 50
= 0.8 × 50 = 40
So, The sales price is $40

Question 2.
The discount on a DVD is 50%. It is on sale for $10. What is the original price of the DVD?
Answer: The original price of DVD is $20.

Explanation:
Given , discount on a DVD is 50%. , It is on sale for $10
We know , The sales price be 100% – 50% = 50%
a = p% × w
10 = 0.5 × w
w = \(\frac{10}{0.5}\)
w = 20
So, The original price is $20.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 3.
WRITING
Describe how to find the sale price of an item that has a 15% discount.
Answer: To get the sales price , The discount percent must be cleared from the total percent of sales price , it gives the sales percentage of the original price , which is used to find the sales price.
So, The sales price = original price minus discount.

Explanation:
Let the original price be $50 with 15% off
We know , The sales price be 100% – 15% = 85% of the original price
sales price = 85% of 50
= 0.85 × 50 = 42.5
So, The sales price is $42.5

FINDING A SALE PRICE Find the sale price. Use a percent model to check your answer.
Question 4.
A portable table tennis set costs $30 before a 30% discount.
Answer: The sales price of portable table tennis set is $21.

Explanation:
Given , tennis set is $30 with 30% off
We know , The sales price be 100% – 30% = 70% of the original price
sales price = 70% of 30
= 0.7 × 30 = 21
So, The sales price is $21.

Question 5.
The original price of an easel is $70. The easel is on sale for 20% off.
Answer: The sales price  of an easel is $56.

Explanation:
Given , easel is $70 with 20% off
We know , The sales price be 100% – 20% = 80% of the original price
sales price = 80% of 70
= 0.8 × 70 = 56
So, The sales price is $56.

FINDING AN ORIGINAL PRICE Find the original price. Use a percent model to check your answer.
Question 6.
A bracelet costs $36 after a 25% discount.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.5 4
Answer: The original price of bracelet is $48.

Explanation: 
Given , discount on a bracelet is 25%. , It is cost  $36  after discount
We know , The sales price be 100% – 25% = 75%
a = p% × w
36 = 0.75 × w
w = \(\frac{36}{0.75}\)
w = 48
So, The original price is $48.

Question 7.
The discount on a toy robot is 40%. The toy robot is on sale for $54.
Answer: The sale price toy robot is $32.4 .

Explanation: 
Given , discount on a toy robot is 40%. The toy robot is on sale for $54.
We know , The sales price be 100% – 40% = 60%
a = p% × w
a = 0.6 × 54
a = 32.4
So, The sale price is $32.4 .

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 8.
DIG DEEPER!
You have two coupons for a store. The first coupon applies a $15 discount to a single purchase, and the second coupon applies a 10% discount to a single purchase. You can only use one coupon on a purchase. When should you use each coupon? Explain.
Answer: The first coupon of $15 discount is to be used on your highest cost of purchase  and the second coupon with 10% off is to be used on your least cost of purchase .

Explanation:
Given , The first coupon applies a $15 discount to a single purchase,
Let the purchase be $50
Then the first coupon applies = $50 – $15 = $35.
So, when the first coupon applies price will be $35.

Given , the second coupon applies a 10% discount to a single purchase
Let the original price be $50 and with 10% off
We know , The sales price be 100% – 10% = 90% of the original price
sales price = 90% of 50
= 0.9 × 50 = 45
So, The sales price is $45.
Here , we can see that $15 discount is offering the reduction of the original price and  discount with 10% of is offering to pay the 90% of its original price.

Finally , The first coupon of $15 discount is to be used on your highest cost of purchase  and the second coupon with 10% off is to be used on your least cost of purchase .

Question 9.
A store sells memory cards for $25 each.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.5 5
a. The markup for each memory card is 25%. How much did the store pay for 50 memory cards?
b. The store offers a discount when a customer buys two or more memory cards. A customer pays $47.50 for two memory cards. What is the percent of discount?
c. How much does a customer pay for three memory cards if the store increases the percent of discount in part (b) by 2%?
Answer: a. The store pay $937.5 for 50 memory cards
b. The discount offered by the store is 5%
c. The customer paid $69.75 for 3 memory cards.

Explanation:
a. The markup is 25% of $25
a = p% × w
= 25% × 25
= 0.25 × 25
= 6.25
So , the markup is $6.25.
To , find the cost to store , we have
cost to store = selling price – markup
= $25 – $6.25
= $18.75.
The cost to store for each memory card is $18.75.
Then for 50 memory cards = 50 × $18.75 = $937.5

b. A customer pays $47.50 for two memory cards.
Then for one memory card \(\frac{47.5}{2}\) = $23.75
we have , a = $23.75 , w = $25 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{23.75}{25}\) = \(\frac{p}{100}\)
p = \(\frac{23.75 × 100}{25}\)
p = \(\frac{2375}{25}\)
p = 95
So, 23.75 is 95% of 25.
To find the discount , = The percent of original price – the percent of selling price
=  100% – 95% = 5%
So, The discount offered by the store is 5%

c. If the discount is increased by 2% , Then the discount offered by store is 5% + 2% = 7%
The amount of selling price = 100% – 7% = 93%
So , 93% of $25
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{25}\) = \(\frac{93}{100}\)
a = \(\frac{25 × 93}{100}\)
a = \(\frac{2325}{100}\)
a = 23.25
So, The amount after  the discount is $23.25
Then for 3 memory cards = $23.25 × 3 = $69.75

The customer paid $69.75 for 3 memory cards.

Discounts and Markups Homework & Practice 6.5

Review & Refresh

Identify the percent of change as an increase or decrease. Then find the percent of change. Round to the nearest tenth of a percent if necessary.
Question 1.
16 meters to 20 meters
Answer: The percent of change is 25%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 20 and old value = 16 , because a change of 16 to 20 is a positive (increase) change
So, percent change = \(\frac{20 – 16}{16}\)
= \(\frac{4}{16}\)
= \(\frac{1}{4}\)
= 0.25
= 0.25 × 100
= 25%
So, percent of change is 25%

Question 2.
9 points to 4 points
Answer: The percent of change is – 55.5%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 4 and old value = 9 , because a change of 9 to 4 is a negative (decrease) change
So, percent change = \(\frac{4 – 9}{9}\)
= \(\frac{- 5}{9}\)
= – 0.555
= – 0.555 × 100
= – 55.5%
So, percent of change is – 55.5%

Question 3.
15 ounces to 5 ounces
Answer: The percent of change is – 66.6%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 5 and old value = 15 , because a change of 15 to 5 is a negative (decrease) change
So, percent change = \(\frac{5 – 15}{15}\)
= \(\frac{- 10}{15}\)
= – 0.666
= – 0.666 × 100
= – 66.6%
So, percent of change is – 66.6%

Question 4.
38 staples to 55 staples
Answer:

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 55 and old value = 38 , because a change of 38 to 55 is a positive (increase) change
So, percent change = \(\frac{55 – 38}{38}\)
= \(\frac{17}{38}\)
= 0.447
= 0.447 × 100
= 44.7%
So, percent of change is 44.7%

Find the product. Write fractions in simplest form.
Question 5.
\(\frac{4}{7}\left(-\frac{1}{6}\right)\)
Answer: The product of  \(\frac{4}{7}\left(-\frac{1}{6}\right)\) is 0.094.

Explanation:
The given fractions can be written in decimal form then we have,
\(\frac{4}{7}\left(-\frac{1}{6}\right)\) as 0.571 ( – 0.166)
To find the product we have to multiply the two numbers
= 0.571 × ( – 0.166)
= 0.094
So, The product of  \(\frac{4}{7}\left(-\frac{1}{6}\right)\) is 0.094.

Question 6.
– 1.58(6.02)
Answer: The product of – 1.58(6.02) is  – 9.51.

Explanation:
To find the product we have to multiply the two numbers
= – 1.58 × (6.02)
=  – 9.51
So, The product of – 1.58(6.02) is  – 9.51.

Question 7.
– 3(- 2\(\frac{1}{8}\))
Answer:

Explanation:
The given fractions can be written in decimal form then we have,
– 3(- 2\(\frac{1}{8}\)) as – 3 ( – 2.12)
To find the product we have to multiply the two numbers
= – 3 × ( – 2.12)
=  6.36
So, The product of  – 3 ( – 2.12) is 6.36

Concepts, Skills, & Problem Solving

COMPARING DISCOUNTS The same item is on sale at two stores. Which one is the better price? Use percent models to justify your answer. (See Exploration 1, p. 259.)
Question 8.
60% off $60 or 55% off $50
Answer: The item has better price at 55% off $50 .

Explanation:
a. Given , 60% off $60.
We know , The sales price be 100% – 60% = 40% of the original price
sales price = 40% of 60
= 0.4 × 60 = 24
So, The sales price is $24

b. Given , 55% off $50

We know , The sales price be 100% – 55% = 45% of the original price
sales price = 45% of 50
= 0.45 × 50 = 22.5
So, The sales price is $22.5

Question 9.
85% off $90 or 70% off $65
Answer: The item has better price at 85% off $90 .

Explanation:

a. Given , 85% off $90
We know , The sales price be 100% –85% = 15% of the original price
sales price = 15% of 90
= 0.15 × 90 = 13.5
So, The sales price is $13.5 .

b. Given , 70% off $65

We know , The sales price be 100% –70% = 30% of the original price
sales price = 30% of 65
= 0.3 × 65 = 19.5
So, The sales price is $19.5.

USING TOOLS Copy and complete the table.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.5 6
Answer:
10. Given , Original price of the item is $ 80 , percent of discount is 20% , Find sales price ?
Answer: The sales price of the is $64.

Explanation:
We know , The sales price be 100% – 20% = 80% of the original price
sales price = 80% of 80
= 0.8 × 80 = 64
So, The sales price is $64.

11. Given , Original price of the item is $42 , percent of discount is 15% , Find sales price ?
Answer: The sales price of the is $35.7.

Explanation:
We know , The sales price be 100% – 15% = 85% of the original price
sales price = 85% of 42
= 0.85 × 42 = 35.7
So, The sales price is $35.7.

12. Given , Original price of the item is $120 , percent of discount is 80% , Find sales price ?
Answer: The sales price of the is $24.

Explanation:
We know , The sales price be 100% – 80% = 20% of the original price
sales price = 20% of 120
= 0.2 × 120 = 24
So, The sales price is $24.

13. Given , Original price of the item is $112 , percent of discount is 32% , Find sales price ?
Answer: The sales price of the is $76.16.

Explanation:
We know , The sales price be 100% – 32% = 68% of the original price
sales price = 68% of 112
= 0.68 × 112 = 76.16
So, The sales price is $76.16.

14. Given , Original price of the item is $69.8 , percent of discount is 60% , Find sales price ?
Answer: The sales price of the is $27.92.

Explanation:
We know , The sales price be 100% – 60% = 40% of the original price
sales price = 40% of 69.8
= 0.4 × 69.8 = 27.92
So, The sales price is $27.92.

15. Given , sales price of the item is $40 , percent of discount is 25% , Find original price ?
Answer: The Original price of the is $53.

Explanation:
The sales price be 100% – 25% = 75%
we know  a = 40 , p = 75% , w = ?
a  =  p% × w
a = 75% × w
40 = 0.75 × w
w = \(\frac{40}{0.75}\)
w = 53
So , the original price of the earrings is $53.

16. Given , sales price of the item is $57 , percent of discount is 5% , Find original price ?
Answer: The Original price of the is $60.

Explanation:
The sales price be 100% – 5% = 95%
we know  a = 57 , p = 95% , w = ?
a  =  p% × w
a = 95% × w
57 = 0.95 × w
w = \(\frac{57}{0.95}\)
w = 60
So , the original price of the earrings is $60.

17. Given , sales price of the item is $90 , percent of discount is 80% , Find original price ?
Answer: The Original price of the is $450.

Explanation:
The sales price be 100% – 80% = 20%
we know  a = 90 , p = 20% , w = ?
a  =  p% × w
a = 20% × w
90 = 0.2 × w
w = \(\frac{90}{0.2}\)
w = 450
So , the original price of the earrings is $450.

18. Given , sales price of the item is $72 , percent of discount is 64% , Find original price ?
Answer: The Original price of the is $200.

Explanation:
The sales price be 100% – 64% = 36%
we know  a = 72 , p = 36% , w = ?
a  =  p% × w
a = 36% × w
72 = 0.36 × w
w = \(\frac{72}{0.36}\)
w = 200
So , the original price of the earrings is $200.

19. Given , sales price of the item is $146.54 , percent of discount is 15% , Find original price ?
Answer: The Original price of the is $172.4.

Explanation:
The sales price be 100% – 15% = 85%
we know  a = 146.54 , p = 85% , w = ?
a  =  p% × w
a = 85% × w
146.54 = 0.85 × w
w = \(\frac{146.54}{0.85}\)
w = 172.4
So , the original price of the earrings is $172.4.

20. Given , original price  of the item is $60 , sales price of the item is $45 , Find percent of discount ?
Answer: The percent of discount is 25% .

Explanation:
We have a = 45 , w = 60 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{45}{60}\) = \(\frac{p}{100}\)
p = \(\frac{45 × 100}{60}\)
p = \(\frac{4,500}{60}\)
p = 75
So, 45 is 75% of 60.
To get the percent of discount we have ,
The percent of discount = The percent of original price – the percent of sales price
= 100% – 75%
= 25%.
So, The percent of discount is 25% .

21. Given , original price  of the item is $82 , sales price of the item is $65.6 , Find percent of discount ?
Answer: The percent of discount is 20% .

Explanation:
We have a = 65.6 , w = 82 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{65.6}{82}\) = \(\frac{p}{100}\)
p = \(\frac{65.6 × 100}{82}\)
p = \(\frac{6,560}{82}\)
p = 80
So, 65.6 is 80% of 82.
To get the percent of discount we have ,
The percent of discount = The percent of original price – the percent of sales price
= 100% – 80%
= 20%.
So, The percent of discount is 20% .

22. Given , original price  of the item is $95 , sales price of the item is $61.75 , Find percent of discount ?
Answer: The percent of discount is 35% .

Explanation:
We have a = 61.75 , w = 95 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{61.75}{95}\) = \(\frac{p}{100}\)
p = \(\frac{61.75 × 100}{95}\)
p = \(\frac{6,175}{95}\)
p = 65
So, 61.75 is 65% of 90.
To get the percent of discount we have ,
The percent of discount = The percent of original price – the percent of sales price
= 100% – 65%
= 35%.
So, The percent of discount is 35% .

FINDING A SELLING PRICE .

Question 23.
Cost to store: $50
Markup: 10%
Answer:  The selling price is $55.

Explanation:
The markup is 10% of $50
a = p% × w
= 10% × 50
= 0.1 × 50
= 5
So , the markup is $5.
To , find the selling price , we have
selling price = cost to store + markup
= $50 + $5
= $55.
The selling price is $55.

Question 24.
Cost to store: $80
Markup: 60%
Answer: The selling price is $128.

Explanation:
The markup is 60% of $80
a = p% × w
= 60% × 80
= 0.6 × 80
= 48
So , the markup is $48.
To , find the selling price , we have
selling price = cost to store + markup
= $80 + $48
= $128.
The selling price is $128.

Question 25.
Cost to store: $140
Markup: 25%
Answer: The selling price is $175.

Explanation:
The markup is 25% of $140
a = p% × w
= 25% × 140
= 0.25 × 140
= 35
So , the markup is $35.
To , find the selling price , we have
selling price = cost to store + markup
= $140 + $35
= $175.
The selling price is $175.

Question 26.
YOU BE THE TEACHER
A store pays $60 for an item. Your friend finds the selling price when the markup is 20%. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.5 7
Answer: No , The selling price is $72.

Explanation:
Given , The markup is 20% of $60
a = p% × w
= 20% × 60
= 0.2 × 60
= 12
So , the markup is $12.
To , find the selling price , we have
selling price = cost to store + markup
= $60 + $12
= $72.
The selling price is $72.

Question 27.
STRUCTURE
The scooter is being sold at a 10% discount. The original price is shown. Which methods can you use to find the new sale price? Which method do you prefer? Explain.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.5 8
Answer: The sales price is $37.8. Used the method of Multiplying $42 by 0.9.

Explanation:
By Using the method of Multiplying $42 by 0.9
We know , The sales price be 100% – 10% = 90% of the original price
sales price = 90% of 42
= 0.9 × 42 = 37.8
So, The sales price is $37.8.

Question 28.
NUMBER SENSE
The original price of an item is P dollars. Is the price of the item with an 18% markup the same as multiplying the original price by 1.18? Use two expressions to justify your answer.
Answer: The selling price is $1.18P and it is same as multiplying the original price by 1.18 .

Explanation:
Given , The original price of an item is P dollars.
The markup is 18% of $P
a = p% × w
= 18% × P
= 0.18 × P
= $0.18P
So , the markup is $0.18P.
To , find the selling price , we have
selling price = cost to store + markup
= $P + $0.18P
= $P ( 1 + 0.18 )
= $1.18P.
The selling price is $1.18P.

The given method is multiplying the original price by 1.18,
The original price is $P = $P × 1018.
So , The selling price is $1.18P.

Finally , The selling price is $1.18P and it is same as multiplying the original price by 1.18 .

Question 29.
PROBLEM SOLVING
You are shopping for a video game system.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.5 9
a. At which store should you buy the system?
b. Store A has a weekend sale. What discount must Store A offer for you to buy the system there?
Answer: a. you should buy the system at store C has The selling price is $200.
b. To buy the system at store A  , it should have the discount of  28.6%.

a. Given , For store A cost to store is $162 , Markup is 40%
Then , The markup is 40% of $162
a = p% × w
= 40% × 162
= 0.4 × 162
= 64.8
So , the markup is $64.8.
To , find the selling price , we have
selling price = cost to store + markup
= $162 + $64.8
= $226.8.
So , For store A , The selling price is $226.8.

Given , For store B cost to store is $155 , Markup is 30%
Then , The markup is 30% of $155
a = p% × w
= 30% × 155
= 0.3 × 155
= 46.5
So , the markup is $46.5.
To , find the selling price , we have
selling price = cost to store + markup
= $155 + $46.5
= $201.5.
So , For store B , The selling price is $201.5.

Given , For store C cost to store is $160 , Markup is 25%
Then , The markup is 25% of $160
a = p% × w
= 25% × 160
= 0.25 × 160
= 40
So , the markup is $40.
To , find the selling price , we have
selling price = cost to store + markup
= $160 + $40
= $200.
So , For store C , The selling price is $200.

you should buy the system at store C has The selling price is $200.

b. Given , Store A has a weekend sale, to buy the system there , it should have the discount of ,
For store A , The selling price is $226.8 , cost to store is $162
We know a = $162 , w = $226.8
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{162}{226.8}\) = \(\frac{p}{100}\)
p = \(\frac{162 × 100}{226.8}\)
p = \(\frac{16,200}{226.8}\)
p = 71.4%
So, 162 is 71.4% of 226.8.
To get the percent of discount we have ,
The percent of discount = The percent of original price – the percent of sales price
= 100% – 71.4%
= 28.6%.
So, The percent of discount is 28.6% .
Hence , To buy the system at store A  , it should have the discount of  28.6%.

Question 30.
DIG DEEPER!
A pool manager balances the pH level of a pool. The price of a bucket of chlorine tablets is $90, and the price of a pH test kit is $11. The manager uses a coupon that applies a 40% discount to the total cost of the two items. How much money does the pool manager pay for each item?
Answer: The pool manager pays for each item $60.6.

Explanation:
Now, to find the money the pool manager pay for each item
The price of a bucket of  chlorine tablets is = $90.
The price of a pH test kit = $11.
So, to get the total price of both items we add both prices:
$90 + $11 = $101.
Total cost of two items = $101.
As, given a coupon that applies a 40% discount to the total cost of the two items.
Now, to get the total cost by applying discount:
101 – (40%  of 101)
= 101 – (0.4 × 101)
= 101 – 40.4
= 60.6
So, The pool manager pays for each item $60.6

 

 

Question 31.
PRECISION
You buy a pair of jeans at a department store.
Big Ideas Math Answer Key Grade 7 Chapter 6 Percents 6.5 10
a. What is the percent of discount to the nearest percent?
b. What is the percent of sales tax to the nearest tenth of a percent?
c. The price of the jeans includes a 60% markup. After the discount, what is the percent of markup to the nearest percent?
Answer: a. The percent of discount is to the nearest percent 25% .
b. The percent of sales tax to the nearest percent is 7%.
c. The percent of markup to the nearest percent is 54%

Explanation:
a. To find the percent of discount to the nearest percent , we have
a = 29.99 , w = 39.99 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{29.99}{39.99}\) = \(\frac{p}{100}\)
p = \(\frac{29.99 × 100}{39.99}\)
p = 0.749 × 100
p = 74.9
So, 29.99 is 74.9% of 39.99.
To get the percent of discount we have ,
The percent of discount = The percent of original price – the percent of sales price
= 100% – 74.9%
= 24.9%.
So, The percent of discount is 25% .

b. To find the percent of sales tax to the nearest tenth of a percent we have
sales tax = 1.95 , price = 29.99 , so , a = 1.95 , w = 29.99 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{1.95}{29.99}\) = \(\frac{p}{100}\)
p = \(\frac{1.95 × 100}{29.99}\)
p = 0.065 × 100
p = 6.5
So, 1.95 is 6.5% of 29.99.
Thus , The percent of sales tax is 7%

c. Given , The price of the jeans includes a 60% markup.
The markup is 60% of $39.99
a = p% × w
= 60% × 39.99
= 0.6 × 39.99
= 23.99
So , the markup is $23.99.
To , find the original price , we have
selling price = cost to store + markup
cost to store = selling price – markup
= $39.99 – $23.99
= $15.99.
The selling price is $15.99.
After the discount, The amount is $29.99
To know the mark up we have , a = 15.99 , w = 29.99
a = p% × w
p% = \(\frac{a}{w}\)
p% = \(\frac{15.99}{29.99}\)
p% = 0.533
p = 0.533 × 100
p = 53.3 .%
So , The percent of markup to the nearest percent is 54%

Question 32.
CRITICAL THINKING
You buy a bicycle helmet for $22.26, which includes 6% sales tax. The helmet is discounted 30% off the selling price. What is the original price?
Answer: The original price of the helmet is $30.

Explanation:
Given , You buy a bicycle helmet for $22.26, which includes 6% sales tax.
Then , 6% of $22.26 is $1.33 , by decreasing the tax amount from the buying price  we get $20.93
The helmet is discounted 30% off the selling price.
The sales price be 100% – 30% = 70%
we know  a = 20.93 , p = 70% , w = ?
a  =  p% × w
a = 70% × w
20.93 = 0.7 × w
w = \(\frac{20.93}{0.7}\)
w = 29.9 , approximately equal to 30
So , the original price of the helmet is $30.

Question 33.
REASONING
A drone that costs $129.50 is discounted 40%. The next month, the sale price is discounted an additional 60%. Is the drone now “free”? If so, explain. If not, find the sale price.
Answer: The sales price of the next month is $31.8 , The drone not for free.

Explanation:
Given , A drone that costs $129.50 is discounted 40%.
We know , The sales price be 100% – 40% = 60% of the original price
sales price = 60% of 129.50
= 0.6 × 129.5 = $77.7
So, The sales price is $77.7.

Given , The next month, the sale price is discounted an additional 60%.
The original price this time is $77.7
We know , The sales price be 100% – 60% = 40% of the original price
sales price = 40% of 77.7
= 0.4 × 77.7 = $31.8
So, The sales price of the next month is $31.8 .

Thus , The drone not for free.

Lesson 6.6 Simple Interest

EXPLORATION 1

Understanding Simple Interest
Work with a partner. You deposit $150 in an account that earns 6% simple interest per year. You do not make any other deposits or withdrawals. The table shows the balance of the account at the end of each year.
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.6 1
a. Describe any patterns you see in the account balance.
b. How is the amount of interest determined each year?
c. How can you find the amount of simple interest earned when you are given an initial amount, an interest rate, and a period of time?
d. You deposit $150 in a different account that earns simple interest. The table shows the balance of the account each year. What is the interest rate of the account? What is the balance after 10 years?
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.6 2
Answer: a. we observe the pattern of $9 increment in each year of the deposited amount.
b. the amount of interest each year is ,
For 0 years = $0
For 1 years  I = $9.54
For 2 years  I = $20.16
For 3 years  I = $31.86
For 4 years  I = $44.64
For 5 years  I = $58.5
For 6 years  I = $73.44
c. Formula for simple interest is I = Prt
d. The annual rate of interest is $ 2.60 .

Explanation:
a. As the table shows , we have a series of pattern among the years of amount deposited.
that is , For 0 years = $150
For 1 years = $150 + $9 = $159
For 2 years = $159 + $9 = $168
For 3 years = $168 + $9 = $177
For 4 years = $177 + $9 = $186
For 5 years = $186 + $9 = $195
For 6 years = $195 + $9 = $204 ,
Here, we observe the pattern of $9 increment in each year of the deposited amount.

b. From the table, we have , simple interest for each year as,
Principal = P , t = time in years , r = Annual interest rate
That is P = $150 , t = 0 years , r = 0.06
To find the simple interest  I  we know , I = Prt ,
For 0 years  I = 150 × 0 × 0.06 = $0 .
For 1 years , P = $159 , t = 1 year , r = 0.06
I = 159 × 1 × 0.06 = $9.54
For 2 years , P = $168 , t = 2 year , r = 0.06
I = 168 × 2 × 0.06 = $20.16
For 3 years , P = $177 , t = 3 year , r = 0.06
I = 177 × 3 × 0.06 = $31.86
For 4 years , P = $186 , t = 4 year , r = 0.06
I = 186 × 4 × 0.06 = $44.64
For 5 years , P = $195 , t = 5 year , r = 0.06
I = 195 × 5 × 0.06 = $58.5
For 6 years , P = $204 , t = 6 year , r = 0.06
I = 204 × 6 × 0.06 = $73.44

C. To find the simple interest , we have Principal = P , t = time in years , r = Annual interest rate ,
Formula for simple interest is I = Prt , So by using this formula we can find the simple interest of the deposit .

d. From the table given , we have the pattern of increment of $5 in each year ,
in this case the deposit in the 10 year will be $230 .
To find the interest rate of the account ,
r = \(\frac{I}{pt}\)
r = \(\frac{0.06}{230 × 10}\)
r = 2.60
So , the annual rate of interest is $ 2.60 .

Interest principal is money paid or earned for using or lending money. The is the amount of money borrowed or deposited.

Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.6 3

Try It

Question 1.
What is the balance of the account after 9 months?
Answer: The balance after 9 months = $500 + $1.66 = $501.66 .

Explanation:
From the example given , we know P = $500 , r = 0.03 , t =  9 months
I = prt
I = \(\frac{500 × 0.03}{9}\)
I = \(\frac{15}{9}\)
I = 1.66
The interest earned is $1.66 after 9 months ,
The balance after 9 months = $500 + $1.66 = $501.66 .

Question 2.
You deposit $350 in an account. The account earns $17.50 simple interest in 2.5 years. What is the annual interest rate?
Answer: The annual interest rate is 2%

Explanation:
Given , P = $350 , I = $17.5 , t = 2.5 ,
We know that  I = Prt
So , r = \(\frac{I}{Pt}\)
r = \(\frac{17.5}{350 × 2.5}\)
= \(\frac{17.5}{875}\)
= 0.02
we can write in percent as 2%
So , The annual interest rate is 2% .

Question 3.
In Example 3, how long does it take an account with a principal of $10,000 to earn $750 in interest?
Answer: It takes 3.75years to an account with a principal of $10,000 to earn $750 in interest .

Explanation:
Given , P = $10,000 , I = $750 , r = 0.02 , t = ? ,
We know that  I = Prt
t = \(\frac{I}{Pr}\)
= \(\frac{750}{10,000 × 0.02}\)
= \(\frac{750}{200}\)
= 3.75
So, It takes 3.75years to an account with a principal of $10,000 to earn $750 in interest .

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 4.
VOCABULARY
Explain the meaning of simple interest.
Answer:
Simple interest is interest calculated on the principal portion of a loan or the original contribution to a savings account. Simple interest does not compound, meaning that an account holder will only gain interest on the principal, and a borrower will never have to pay interest on interest already accrued.
So , To put in simplest form we can say that ,  simple interest money paid or earned only on the principal .

USING THE SIMPLE INTEREST FORMULA Use the simple interest formula.
Question 5.
You deposit $20 in a savings account. The account earns 4% simple interest per year. What is the balance after 4 years?
Answer: The balance of after 4 years is $23.2 .

Explanation:
To fin the principal , Given , P = $20 , r = 0.04 , t = 4 ,
We know that  I = Prt
I = 20 × 0.04 × 4
= 3.2
Simple interest I = $3.2
To find the balance after 4 years we have to add the simple interest to principal amount = $20 + $3.2 = $23.2 .
The balance of after 4 years is $23.2 .

Question 6.
You deposit $800 in an account. The account earns $360 simple interest in 3 years. What is the annual interest rate?
Answer: The annual interest rate is 15% .

Explanation:
Given , P = $800 , I = $360 , t = 3 years ,
We know that  I = Prt
So , r = \(\frac{I}{Pt}\)
r = \(\frac{360}{800 × 3}\)
= \(\frac{360}{2400}\)
= 0.15
we can write in percent as 15%
So , The annual interest rate is 15% .

Question 7.
You deposit $650 in a savings account. How long does it take an account with an annual interest rate of 5% to earn $178.25 in interest?
Answer: It takes 7.13 years an account with an annual interest rate of 5% to earn $178.25 in interest .

Explanation:
Given , P = $650 , I = $178.25 , r = 0.05 , t = ? ,
We know that  I = Prt
t = \(\frac{I}{Pr}\)
= \(\frac{178.25}{650 × 0.05}\)
= \(\frac{178.25}{32.5}\)
= 7.13
So, It takes 7.13 years an account with an annual interest rate of 5% to earn $178.25 in interest .

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 8.
You want to deposit $1000 in a savings account for 3 years. One bank adds a $100 bonus to your principal and offers a 2% simple annual interest rate. Another bank does not add a bonus, but offers 6% simple interest per year. Which bank should you choose? Explain.
Answer: Second bank offers the best deal .

Explanation:
Given , we know P = $1000 , t =  3 years ,
And one bank adds $100 bonus and offers  a 2% simple annual interest rate.
Then p = $1000 + $100 = $1100 , r = 0.02 ,
I = Prt
I = 1100 × 0.02 × 3
I = 11 × 2 × 3
I = 66
The simple interest of one bank = $66 .
The balance of the account will be = $1100 + $66 = $1166 .

Another bank offers no bonus, but  6% simple interest per year , r = 0.06  , p = $1000 , t = 3 years
I = prt
I = 1000 × 0.06 × 3
I = 10 × 6 × 3
I = 180
The simple interest of Another bank = $180 .
The balance of the account will be = $1000 + $180 = $1180 .

So, Second bank offers the best deal .

Question 9.
Your cousin borrows $1125 to repair her car. The simple annual interest rate is 10%. She makes equal monthly payments of $25. How many years will it take to pay off the loan?
Answer: It takes 2.2 years to pay off the loan.

Explanation:
Given , P = $1125 , r = 0.01 , I = $25 , t = ?
We know that  I = Prt
t = \(\frac{I}{Pr}\)
= \(\frac{25}{1125 × 0.01}\)
= \(\frac{25}{11.25}\)
= 2.2
So, It takes 2.2 years to pay off the loan.

Question 10.
DIG DEEPER!
You borrow$900 to buy a laptop. You plan to pay off the loan after 5 years of equal monthly payments. After 10 payments, you have $1200 left to pay. What is the simple annual interest rate of your loan?
Answer: The annual interest rate is 15% .

Explanation:
Given , P = $900 , I = $1200 , t = 5 years ,
We know that  I = Prt
So , r = \(\frac{I}{Pt}\)
r = \(\frac{900}{1200 × 5}\)
= \(\frac{900}{6000}\)
= 0.15
we can write in percent as 15%
So , The annual interest rate is 15% .

Simple Interest Homework & Practice 6.6

Review & Refresh

Find the selling price.
Question 1.
A store pays $8 for a pool noodle. The markup is 20%.
Answer: The selling price is $9.6

Explanation:
Then , The markup is 20% of $8
a = p% × w
= 20% × 8
= 0.2 × 8
= 1.6
So , the markup is $1.6 .
To , find the selling price , we have
selling price = cost to store + markup
= $8 + $1.6
= $9.6.
So ,The selling price is $9.6 .

Question 2.
A store pays $3 for a magazine. The markup is 5%.
Answer: The selling price is $3.15

Explanation:
Then , The markup is 5% of $3
a = p% × w
= 5% × 3
= 0.05 × 3
= 0.15
So , the markup is $0.15 .
To , find the selling price , we have
selling price = cost to store + markup
= $3 + $0.15
= $3.15.
So ,The selling price is $3.15 .

Solve the inequality. Graph the solution.
Question 3.
x + 5 < 2
Answer:  x  < -3

Explanation:
Given , x + 5 < 2
add  – 5 on both sides ,
x + 5 – 5 < 2 – 5
x  < -3

Question 4.
b – 2 ≥ – 1
Answer: b  ≥ 1

Explanation:
Given , b – 2 ≥ – 1
add  2 on both sides
b – 2 + 2  ≥ – 1 + 2
b  ≥ 1

Question 5.
w + 6 ≤ – 3
Answer: w  ≤ – 9

Explanation:
Given , w + 6 ≤ – 3
add -6 on both sides
w + 6 – 6  ≤ – 3 – 6
w  ≤ – 9

Concepts, Skills, & Problem Solving

UNDERSTANDING SIMPLE INTEREST The table shows the balance of an account each year. What is the interest rate of the account? What is the balance after 10 years? (See Exploration 1, p. 265.)
Question 6.
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.6 4
Answer: The annual interest rate is 3% .

Explanation:
Given , There is an increment of $2 each year , for 10 years it will be $60 .
Now we have , P = $60 , t = 10 years , I = $2 , r = ?
We know that  I = Prt
So , r = \(\frac{I}{Pt}\)
r = \(\frac{2}{60 × 10}\)
= \(\frac{2}{60}\)
= 0.03
we can write in percent as 3%
So , The annual interest rate is 3% .

Question 7.
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.6 5
Answer: The annual interest rate is 0.4% .

Explanation:
Given , There is an increment of $14 each year , for 10 years it will be $315 .
Now we have , P = $315 , t = 10 years , I = $14 , r = ?
We know that  I = Prt
So , r = \(\frac{I}{Pt}\)
r = \(\frac{14}{315 × 10}\)
= 0.004
we can write in percent as 0.4%
So , The annual interest rate is 0.4% .

FINDING INTEREST EARNED An account earns simple annual interest. (a) Find the interest earned. (b) Find the balance of the account.
Question 8.
$600 at 5% for 2 years
Answer: The simple interest of  bank is $60 .
The balance of the account will be $660 .

Explanation:
Given ,  P = $600 , t =  2 years , r = 0.05
I = prt
I = 600 × 0.05 × 2
I = 6 × 5 × 2
I = 60
The simple interest of bank is $60 .
The balance of the account will be = $600 + $60 = $660 .

Question 9.
$1500 at 4% for 5 years
Answer: The simple interest  is $300 .
The balance of the account will be  $1800 .

Explanation:
Given ,  P = $1500 , t =  5 years , r = 0.04
I = prt
I = 1500 × 0.04 × 5
I = 15 × 4 × 5
I = 300
The simple interest of  bank = $300 .
The balance of the account will be $1500 + $300 = $1800 .

Question 10.
$350 at 3 % for 10 years
Answer: The simple interest of  bank is $105 .
The balance of the account will be $455 .

Explanation:
Given , P = $350 , t =  10 years , r = 0.03
I = prt
I = 350 × 0.03 × 10
I = 35 × 3
I = 105
The simple interest of  bank is $105 .
The balance of the account will be $350 + $105 = $455 .

Question 11.
$1800 at 6.5% for 30 months
Answer:  The interest earned is $3.9 ,
The balance of the account will be $1,803.9 .

Explanation:
Given , P = $1800 , t =  30 months , r = 0.065
I = prt
I = \(\frac{1800 × 0.065}{30}\)
I = \(\frac{117}{30}\)
I = 3.9
The interest earned is $3.9 ,
The balance of the account will be $1800 + $3.9 = $1,803.9 .

Question 12.
$925 at2.3% for 2.4 years
Answer: The interest earned is $638.1 ,
The balance of the account will be $1,563.1 .

Explanation:
Given , P = $925 , t =  28 months , r = 0.023
I = prt
I = \(\frac{925 × 0.023}{28}\)
I = \(\frac{21.27}{30}\)
I = 638.1
The interest earned is $638.1 ,
The balance of the account will be $925 + $638.1 = $1,563.1 .

Question 13.
$5200 at 7.36% for 54 months
Answer: The interest earned is $7.08 ,
The balance of the account will be $5,207.

Explanation:
Given , P = $5200 , t = 54 months , r = 0.0736
I = prt
I = \(\frac{5200 × 0.0736}{54}\)
I = \(\frac{382.72}{54}\)
I = 7.08
The interest earned is $7.08 ,
The balance of the account will be $5200 + $7.08 = $5,207.

Question 14.
YOU BE THE TEACHER
Your friend finds the simple interest earned on $500 at 6% for 18 months. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.6 6
Answer: No , The simple interest is $1.6 .

Explanation:
Given , P = $500 , t =  18 months , r = 0.06
I = prt
I = \(\frac{500 × 0.06}{18}\)
I = \(\frac{30}{18}\)
I = 1.6
The simple interest is $1.6 .

FINDING AN ANNUAL INTEREST RATE Find the annual interest rate.
Question 15.
I = $24, P = $400, t = 2 years
Answer: The annual interest rate is 3% .

Explanation:
We know that  I = Prt
So , r = \(\frac{I}{Pt}\)
r = \(\frac{24}{400 × 2}\)
= \(\frac{24}{800}\)
= 0.03
we can write in percent as 3%
So , The annual interest rate is 3% .

Question 16.
I = $562.50, P = $1500, t = 5 years
Answer: The annual interest rate is 7.5% .

Explanation:
We know that  I = Prt
So , r = \(\frac{I}{Pt}\)
r = \(\frac{562.5}{1500 × 5}\)
= \(\frac{562.5}{7500}\)
= 0.075
we can write in percent as 7.5%
So , The annual interest rate is 7.5% .

Question 17.
I = $54, P = $900, t = 18 months
Answer: The annual interest rate is 108% .

Explanation:
We know that  I = Prt
So , r = \(\frac{I}{Pt}\)
r = \(\frac{54 × 18 }{900 }\)
= \(\frac{972}{900}\)
= 1.08
we can write in percent as 108%
So , The annual interest rate is 108% .

Question 18.
I = $160, P = $2000, t = 8 months
Answer: The annual interest rate is 64% .

Explanation:
We know that  I = Prt
So , r = \(\frac{I}{Pt}\)
r = \(\frac{160 × 8 }{2000 }\)
= \(\frac{1,280}{2,000}\)
= 0.64
we can write in percent as 64%
So , The annual interest rate is 64% .

FINDING AN AMOUNT OF TIME Find the amount of time.
Question 19.
I $30, P = $500, r = 3%
Answer: The amount of time is 1.6 years .

Explanation:
We know that  I = Prt
t = \(\frac{I}{Pr}\)
= \(\frac30}{500 × 0.03}\)
= \(\frac{30}{18}\)
=1.6
So , The amount of time is 1.6 years .

Question 20.
I = $720, P = $1000, r = 9%
Answer: The amount of time is 8 years .

Explanation:
We know that  I = Prt
t = \(\frac{I}{Pr}\)
= \(\frac{720}{1000 × 0.09}\)
= \(\frac{720}{90}\)
= 8
So , The amount of time is 8 years .

Question 21.
I = $54, P = $800, r = 4.5%
Answer: The amount of time is 1.5 years .

Explanation:
We know that  I = Prt
t = \(\frac{I}{Pr}\)
= \(\frac{54}{800 × 0.045}\)
= \(\frac{54}{36}\)
= 1.5
So , The amount of time is 1.5 years .

Question 22.
I = $450, P = $2400, r = 7.5%
Answer:  The amount of time is 2.5 years .

Explanation:
We know that  I = Prt
t = \(\frac{I}{Pr}\)
= \(\frac{450}{2400 × 0.075}\)
= \(\frac{450}{180}\)
= 2.5
So , The amount of time is 2.5 years .

Question 23.
FINDING AN ACCOUNT BALANCE
A savings account earns 5% simple interest per year. The principal is $1200. What is the balance after 4 years?
Answer: The bank balance is $1440 .

Explanation:
Given , P = $1200 , t =  4 years , r = 0.05
I = prt
I = 1200 × 0.05 × 4
I = 12 × 5 × 4
I = 240
The simple interest of  bank is $240 .
The bank balance is $1200 + $240 = $1440 .

Question 24.
FINDING AN ANNUAL INTEREST RATE
You deposit $400 in an account. The account earns $18 simple interest in 9 months. What is the annual interest rate?
Answer: The annual interest rate is 40.5% .

Explanation:
Given , P = $400 , I = $18 , t = 9 months ,
We know that  I = Prt
So , r = \(\frac{I}{Pt}\)
r = \(\frac{18 × 9}{400}\)
= \(\frac{162}{400}\)
= 0.405
we can write in percent as 40.5%
So , The annual interest rate is 40.5% .

Question 25.
FINDING AN AMOUNT OF TIME
You deposit $3000 in a CD (certificate of deposit) that earns 5.6% simple annual interest. How long will it take to earn $336 in interest?
Answer: It takes 2 years to earn $336 in interest .

Explanation:
Given , P = $3000  , r = 0.056 , I = $336
We know that  I = Prt
t = \(\frac{I}{Pr}\)
= \(\frac{336}{3000 × 0.056}\)
= \(\frac{336}{168}\)
= 2
So , It takes 2 years to earn $336 in interest .

FINDING AN AMOUNT PAID Find the amount paid for the loan.
Question 26.
$1500 at 9% for 2 years
Answer: The amount paid for the loan is $270 .

Explanation:
Given , P = $1500 , t =  2 years , r = 0.09
I = prt
I = 1500 × 0.09 × 2
I = 15 × 9 × 2
I = 270
So , The amount paid for the loan is $270 .

Question 27.
$2000 at 12% for 3 years
Answer: The amount paid for the loan is $72 .

Explanation:
Given , P = $2000 , t =  3 years , r = 0.012
I = prt
I = 2000 × 0.012 × 3
I = 2 × 12 × 3
I = 72
So , The amount paid for the loan is $72 .

Question 28.
$2400 at 10.5% for 5 years
Answer: The amount paid for the loan is $1,260 .

Explanation:
Given , P = $2400 , t =  5 years , r = 0.105
I = prt
I = 2400 × 0.105 × 5
I = 252 × 5
I = 1,260
So , The amount paid for the loan is $1,260 .

Question 29.
$4800 at 9.9% for 4 years
Answer: The amount paid for the loan is $1,900 .

Explanation:
Given , P = $4800 , t =  4 years , r = 0.099
I = prt
I = 4800 × 0.099 × 4
I = 475.2 × 4
I = 1,900
So , The amount paid for the loan is $1,900 .

USING THE SIMPLE INTEREST FORMULA Copy and complete the table.
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.6 7
Answer: 
30. The simple interest  is $1,530 .
31. The principal amount is $2,29,491 .
32. The time required is 4 years .
33. The annual interest rate is 1,275% .

30.  Explanation:
Given , P = $12,000 , t =  5 years , r = 0.0425
I = prt
I = 12,000 × 0.0425 × 5
I = 510 ×
I = 1,530
So , The simple interest  is $1,530 .

31.  Explanation:
Given , I = $828.75 , t =  18 months , r = 0.065
I = Prt
P = \(\frac{I}{rt}\)
= \(\frac{828.75 × 18}{0.065}\)
= \(\frac{14,916.96}{0.065}\)
= 2,29,491
So , The principal amount is $2,29,491 .

32. Explanation:
Given , P = $15,500 , I = $5425 , r = 0.0875
We know that  I = Prt
t = \(\frac{I}{Pr}\)
= \(\frac{5425}{15,500 × 0.0875}\)
= \(\frac{5425}{1,356.25}\)
= 4
So , The time required is 4 years .

33. Explanation:
Given , P = $18,000 , I = $4252.5 , t = 54 months ,
We know that  I = Prt
So , r = \(\frac{I}{Pt}\)
r = \(\frac{4252.5 × 54}{18,000}\)
= \(\frac{2,29,635}{18,000}\)
= 12.75
we can write in percent as 1,275%
So , The annual interest rate is 1,275% .

Question 34.
MODELING REAL LIFE
A family borrows money for a rainforest tour. The simple annual interest rate is 12%. The loan is paid after 3 months. What is the total amount paid for the tour?
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.6 8
Answer: The total amount paid for the tour will be $1,230 + $49.2 = $1,279.2 .

Explanation:
Given , P = $940 + $170 + $120 = 1,230
P = $1,230 , t =  3 months , r = 0.12
I = prt
I = \(\frac{1,230 × 0.12}{3}\)
I = \(\frac{147.6}{3}\)
I = 49.2
The simple interest  is $49.2 .
So , The total amount paid for the tour will be $1,230 + $49.2 = $1,279.2 .

Question 35.
MODELING REAL LIFE
You deposit $5000 in an account earning 7.5% simple interest per year. How long will it take for the balance of the account to be $6500?
Answer: It takes 4 years for the balance of the account to be $6500.

Explanation:
Given , the balance of the account  be $6500 ,  You deposit $5000
We know that . The total balance = deposit + simple interest
So, To get simple interest we have = The total balance – deposit = $6500 – $5000 = $1500 ,
P = $5000 , r = 0.075 , I = $1500 , t = ?
We know that  I = Prt
t = \(\frac{I}{Pr}\)
= \(\frac{1500}{5000 × 0.075}\)
= \(\frac{1500}{375}\)
= 4
So , It takes 4 years for the balance of the account to be $6500.

Question 36.
MODELING REAL LIFE
You borrow$1300 to buy a telescope. What is the monthly payment?
Big Ideas Math Answers 7th Grade Chapter 6 Percents 6.6 9
Answer: The monthly payment is $66.95 .

Explanation:
Given , P = $1300 , t =  2 years , r = 0.118
I = prt
I = 1300 × 0.118 × 2
I = 153.4 × 2
I = 306.8
The simple interest of  bank is $306.8 .
The balance of the account will be $1300 + $306.8 = $1,606.8 .
To get the monthly payment ,  convert 2 years into 24 months then , \(\frac{1,606.8}{24}\) = 66.95
So , The monthly payment is $66.95 .

Question 37.
REASONING
How many years will it take for $2000 to double at a simple annual interest rate of 8%? Explain how you found your answer.
Answer: It will take 12.5 years to make the principal amount double .

Explanation:
Given , it take for $2000 to double  , so the total balance will be $2000 + $2000 = $4000
We know that . The total balance = deposit + simple interest
So, To get simple interest we have = The total balance – deposit = $4000 – $2000 = $2000
P = $2000 , r = 0.08 , I = $2000 , t = ?
We know that  I = Prt
t = \(\frac{I}{Pr}\)
= \(\frac{2000}{2000 × 0.08}\)
= \(\frac{2000}{160}\)
= 12.5
So , It will take 12.5 years to make the principal amount double .

Question 38.
DIG DEEPER!
You take out two loans. After 2 years, the total interest for the loans is $138. On the first loan, you pay 7.5% simple annual interest on a principal of $800. On the second loan, you pay 3% simple annual interest. What is the principal for the second loan?
Answer: The principal amount of the second loan is $300 .

Explanation:
Given , For first loan , P = $800 , t =  2 years , r = 0.075
I = prt
I = 800 × 0.075 × 2
I = 120
The simple interest of  bank is $120 .

For Second loan , Given , the total interest for the loans is $138
So , I = $138 – $120 = $18 ,
Given r = 0.03 , t = 2 years , I = $18
I = Prt
p = \(\frac{I}{rt}\)
= \(\frac{18}{0.03 × 2}\)
= \(\frac{18}{0.06}\)
= 300
So, The principal amount of the second loan is $300 .

Question 39.
REPEATED REASONING
You deposit $500 in an account that earns 4% simple annual interest. The interest earned each year is added to the principal to create a new principal. Find the total amount in your account after each year for 3 years.
Answer:  The balance of the account for the first year will be $500 + $20 = $520
The balance of the account for the 2 year will be $520 + $41.6 = $561.6 .
The balance of the account for the 3 year will be $561.6 + $67.3 = $5683.9.

Explanation:
Given , P = $500 , t =  1 years , r = 0.04
I = prt
I = 500 × 0.04 × 1
I = 5 × 4
I = 20
The simple interest of  bank is $20 .
The balance of the account for the first year will be $500 + $20 = $520 .

For 2 years , p = $520 , t = 2 , r = 0.04
I = prt
I = 520 × 0.04 × 2
I = 20.8 × 2
I = 41.6
The simple interest of  bank is $41.6 .
The balance of the account for the 2 year will be $520 + $41.6 = $561.6 .

For 3 years , p = $561.6 , t = 3 , r = 0.04
I = prt
I = 561.6 × 0.04 × 3
I = 22.46× 3
I = 67.3
The simple interest of  bank is $67.3 .
The balance of the account for the 3 year will be $561.6 + $67.3 = $5683.9.

Question 40.
NUMBER SENSE
An account earns r% simple interest per year. Does doubling the initial principal have the same effect on the total interest earned as doubling the amount of time? Justify your answer.
Answer: Yes ,Doubling the initial principal have the same effect on the total interest earned as doubling the amount of time

Explanation:
Given , An account earns r% simple interest per year. Does doubling the initial principal
Let us say P = 2P , t = t years , r = r%
we know I = Prt ,
I = 2P × t × r
I = 2Prt
The simple interest of the doubling initial principal is 2Prt .

Now , doubling the amount of time , t = 2t years
we know I = Prt ,
I = P × 2t × r
I = 2Prt
The simple interest of the doubling the amount  is 2Prt .
So, Doubling the initial principal have the same effect on the total interest earned as doubling the amount of time

Percents Connecting Concepts

Using the Problem-Solving Plan
Question 1.
The table shows the percent of successful shots for each team in a hockey game. A total of 55 shots are taken in the game. The ratio of shots taken by the Blazers to shots taken by the Hawks is 6 : 5. How many goals does each team score?
Big Ideas Math Answers Grade 7 Chapter 6 Percents cc 1
Understand the problem.
You know that 55 shots are taken in a hockey game and that the Blazers take 6 shots for every 5 shots taken by the Hawks. You also know the percent of successful shots for each team.
Make a plan.
Use a ratio table to determine the number of shots taken by each team. Then use the percent equation to determine the number of successful shots for each team.
Solve and check.
Use the plan to solve the problem. Then check your solution.
Answer: The goals made by Hawks are 50. and The goals made by Blazers are 60

Explanation:
Given , The ratio of shots taken by the Blazers to shots taken by the Hawks is 6 : 5
the percent of successful shots for each team in a hockey game is  Blazers is 10% , Hawks is 16% ,
The goals made by Blazers are w = ? , a = 6 , p = 10%
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{6}{w}\) = \(\frac{10}{100}\)
w = \(\frac{6 × 100}{10}\)
w = \(\frac{600}{10}\)
w = 60
So, The goals made by Blazers are 60
The goals made by Hawks  are w= ? , a = 5 , p = 16%
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{5}{w}\) = \(\frac{10}{100}\)
w = \(\frac{5 × 100}{10}\)
w = \(\frac{500}{10}\)
w = 50
So , The goals made by Hawks are 50.

Question 2.
Fill in the blanks with positive numbers so that the sum of the fractions is 37.5% of the first fraction. Justify your answer.
Big Ideas Math Answers Grade 7 Chapter 6 Percents cc 2
Answer:  \(\frac{2}{5}\) + (- \(\frac{0.1}{4}\)) = 37.5%

Explanation:
Let us say that missing values be x and y
we have ,\(\frac{x}{5}\) + (- \(\frac{y}{4}\))
To get the simplified answer cross check the values of x and y
If x = 2 and y = 0.1 ,then
\(\frac{2}{5}\) + (- \(\frac{0.1}{4}\))
0.4 – 0.025 = 0.375
0.375 can be written as 37.5%
So ,  \(\frac{2}{5}\) + (- \(\frac{0.1}{4}\)) = 37.5%

Question 3.
The graph shows the distance traveled by a motorcycle on a dirt road. After turning onto a paved road, the motorcycle travels \(\frac{1}{5}\) mile every \(\frac{1}{4}\) minute. Find the percent of change in the speed of the motorcycle. Round to the nearest tenth of a percent if necessary.
Big Ideas Math Answers Grade 7 Chapter 6 Percents cc 3
Answer: The percent of change in the speed of the bike is 38%

Explanation:
speed of the bike on the dirty road  is  \(\frac{2.33}{4}\) = 0.58
speed of the bike on the paved road is \(\frac{0.2}{0.25}\) = 0.8
old value = 0.58 , new value = 0.8 ,
The amount of change = 0.8 – 0.58  = 0.22
So , percent of change is \(\frac{amount of change }{original amount}\)
= \(\frac{ 0.22}{0.58}\)
= 0.379
= 0.379 × 100
= 37.9 = 38%
So, the percent of change is 38%

Performance Task

Tornado Alley
At the beginning of this chapter, you watched a STEAM Video called “Tornado!” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answers Grade 7 Chapter 6 Percents cc 4
Answer:

Percents Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answers Grade 7 Chapter 6 Percents cr 1
percent of change : It is the percent that a quantity changes from the original amount =  \(\frac{amount of change }{original amount}\)
Example:
2 feet to 6 feet
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 6 and old value = 2 , because a change of 2 to 6 is a positive (increase) change
So, percent change = \(\frac{6 – 2}{2}\)
= \(\frac{4}{2}\)
= 2 × 100
= 200%
So, percent of change is 200%

percent of increase: When the original amount increases then the percent of change is called percent of increase = \(\frac{New value – old value}{old value}\)
Example:
5 cups to 10 cups
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 10 and old value = 5 , because a change of 5 to 10 is a positive (increase) change
So, percent change = \(\frac{10 – 5}{5}\)
= \(\frac{5}{5}\)
= 1 × 100
= 100%
So, percent of increase is 100%

percent of decrease:  When the original amount decrease then the percent of change is called percent of decrease =\(\frac{old value – New value}{old value}\)
Example:
15 inches to 12 inches
We know that , formula for percent change = \(\frac{old value – new value}{old value}\)
where New value = 12 and old value = 15 , because a change of 15 to 12 is a negative (decrease) change
So, percent change = \(\frac{15 – 12}{15}\)
= \(\frac{3}{15}\)
= 0.2 × 100
= 20%
So, percent of decrease is 20%

percent error: It is the percent that estimated value differs from Actual value \(\frac{Error value}{Actual value}\)
Example:
Estimated value = 40 , Actual value = 30
The amount of error is  40 – 30 = 10
We know that , formula for percent error = \(\frac{Error value}{Actual value}\)
where Error value = 10 and Actual value = 30 ,
So, percent Error = \(\frac{10}{30}\)
= 0.333
= 0.333 × 100
=  33.3%
So, percent Error is 33.3%

Discount:  It is a decrease in amount of original price of an item
Example:
A box is $5 with 20% off
We know , The sales price be 100% – 20% = 80% of the original price
sales price = 80% of 5
= 0.8 × 5 = 4
So, The sales price is $4.

Markup: The increase from what the stores pays to the selling price is called as Markup
Example:
The markup is 15% of $40
a = p% × w
= 15% × 40
= 0.15 × 40
= 6
So , the markup is $6.
To , find the selling price , we have
selling price = cost to store + markup
= $40 + $6
= $46.
The selling price is $46.

Interest: A money paid or earned for using or lending money is called interest
Example:
A bank offers a loan of $500 with an interest of $10 .
Here $10 is the interest amount to be paid for the loan

Principal: The amount of money borrowed or deposited
Example:
A bank offers a loan of $500 with an interest of $10 .
here the loan amount is the principal amount

Simple Interest: It is the money paid or earned only on the principal I = Prt , where I = simple interest , P = principal , r = rate of interest , t = time in years
Example:
A bank offers a loan of $500 with an interest rate of 3% for 2 years  . the simple interest is
p = $500 , r  = 0.03 , t = 2
I = 500 × 0.03 × 2
I = 30
So , The simple interest is $15 .

Graphic Organizers
You can use a Summary Triangle to explain a concept. Here is an example of writing a percent as a decimal a Summary Triangle for Writing a percent as a decimal.
Big Ideas Math Answers Grade 7 Chapter 6 Percents cr 2

Answer: Summary triangle for writing a decimal into percent is 
Choose and complete a graphic organizer to help you study the concept.
Big Ideas Math Answers Grade 7 Chapter 6 Percents cr 3
1. writing a decimal as a percent
2. comparing and ordering fractions, decimals, and percents
3. the percent proportion
4. the percent equation
5. percent of change
6. discount
7. markup

Writing a decimal as a percent

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answers Grade 7 Chapter 6 Percents cr 4

6.1 Fractions, Decimals, and Percents (pp. 235–240)
Learning Target: Rewrite fractions, decimals, and percents using different representations.

Write the percent as a decimal or the decimal as a percent. Use a model to represent the number.
Question 1.
74%
Answer: decimal form is 0.74

Explanation: 
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, 74% in decimal form is 0.74

Question 2.
2%
Answer: decimal form is 0.02

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, 2% in decimal form is 0.02

Question 3.
221%
Answer: decimal form is 2.21

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, 221% in decimal form is 2.21

Question 4.
0.17
Answer: 17%

Explanation:
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol.
So, 0.17 can be rewrite as 17%

Question 5.
\(4 . \overline{3}\)
Answer: \(433. \overline{3} \%\)

Explanation:
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol.
So, \(4 . \overline{3}\) can be rewrite as \(433. \overline{3} \%\)

Question 6.
0.079
Answer: 7.9%

Explanation:
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol.
So, 0.079  can be rewrite as 7.9%

Write the fraction as a decimal and a percent.
Question 7.
\(\frac{17}{20}\)
Answer: decimal = 0.85, percent = 85%

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{17}{20}\) as  0.85 in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then 0.85 can be rewrite as 85%
So,\(\frac{17}{20}\) in decimal = 0.85, percent = 85%

Question 8.
\(\frac{3}{8}\)
Answer: decimal = 0.375, percent = 37.5%

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{3}{8}\) as  0.375 in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then 0.375 can be rewrite as 37.5%
So, \(\frac{3}{8}\) in decimal = 0.375, percent = 37.5%

Question 9.
\(\frac{14}{9}\)
Answer: decimal = \(1 . \overline{5}\), percent = \(155. \overline{5} \%\)

Explanation:
By using the method of converting fraction into decimal and comparing ,
We can rewrite \(\frac{14}{9}\) as  \(1. \overline{5}\) in decimal form,
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol , then \(1 . \overline{5}\)can be rewrite as \(155. \overline{5} \%\)
So, \(\frac{14}{9}\) in decimal = \(1 . \overline{5}\), percent = \(155. \overline{5} \%\)

Question 10.
For school spirit day, 11.875% of your class wears orange shirts, \(\frac{5}{8}\) of your class wears blue shirts, 0.15625 of your class wears white shirts, and the rest of your class wears gold shirts. Order the portions of shirts of each color from least to greatest. Justify your answer.
Answer:  In ascending order we have 10.1% , 15.6% , 62.5% , 11.875%

Explanation:
Given , 11.875% of your class wears orange shirts,
\(\frac{5}{8}\) of your class wears blue shirts,
It can be written as 0.625 in decimals and 62.5% in percent
0.15625 of your class wears white shirts,
it can be written as 15.6%
And the rest of your class wears gold shirts.
Let the total be 100% , The percent in all colors is 11.875% + 62.5% +  15.6%  = 89.9%
So , the rest of the girls wears 100% – 89.9% = 10.1% ,
In ascending order we have 10.1% , 15.6% , 62.5% , 11.875%

6.2 The Percent Proportion (pp. 241–246)
Learning Target: Use the percent proportion to find missing quantities.

Write and solve a proportion to answer the question.
Question 11.
What percent of 60 is 18?
Answer: 18 is 30% of 60.

Explanation:
Given , a = 18 , w = 60 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{18}{60}\) = \(\frac{p}{100}\)
p = \(\frac{18 × 100}{60}\)
p = \(\frac{1800}{60}\)
p = 30
So, 18 is 30% of 60.

Question 12.
40 is what percent of 32?
Answer: 40 is 125% of 32.

Explanation:
Given , a = 40 , w = 32 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{40}{32}\) = \(\frac{p}{100}\)
p = \(\frac{40 × 100}{32}\)
p = \(\frac{4,000}{32}\)
p = 125
So, 40 is 125% of 32.

Question 13.
What number is 70% of 70?
Answer: 49 is 70% of 70.

Explanation:
Given , a = ? , w = 70 , p = 70%
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{70}\) = \(\frac{70}{100}\)
a = \(\frac{70 × 70}{100}\)
a = \(\frac{4900}{100}\)
a = 49
So, 49 is 70% of 70.

Question 14.
\(\frac{3}{4}\) is 75% of what number?
Answer: 0.75 is 755% of 1.

Explanation:
we can write \(\frac{3}{4}\) as 0.75 in decimal
Given , a = 0.75 , w = ? , p = 75%
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{0.75}{w}\) = \(\frac{75}{100}\)
w = \(\frac{0.75 × 100}{75}\)
w = \(\frac{75}{755}\)
w = 1
So, 0.75 is 755% of 1.

Question 15.
About 29% of the Earth’s surface is covered by land. The total surface area of the Earth is about 510 million square kilometers. What is the area of the Earth’s surface covered by land?
Big Ideas Math Answers Grade 7 Chapter 6 Percents cr 15
Answer: The area of the Earth’s surface covered by land is 147.9 million square kilometers .

Explanation:
Given , p = 29% , w = 510 , a = ?
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{510}\) = \(\frac{29}{100}\)
a = \(\frac{510 × 29}{100}\)
a = \(\frac{14,790}{100}\)
a = 147.9
So, 147.9 is 29% of 510.
The area of the Earth’s surface covered by land is 147.9 million square kilometers .

6.3 The Percent Equation (pp. 247–252)
Learning Target: Use the percent equation to find missing quantities. Write and solve an equation to answer the question.
Question 16.
What number is 24% of 25?
Answer: 6 is 24% of 25.

Explanation:
Given p = 24% , w = 25 , a = ?
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{25}\) = \(\frac{24}{100}\)
a = \(\frac{25 × 24}{100}\)
a = \(\frac{600}{100}\)
a = 6
So, 6 is 24% of 25.

Question 17.
9 is what percent of 20?
Answer: 9 is 45% of 20.

Explanation:
Given p = ? , w = 20 , a = 9
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{9}{20}\) = \(\frac{p}{100}\)
p = \(\frac{9 × 100}{20}\)
p = \(\frac{900}{20}\)
p = 45
So, 9 is 45% of 20.

Question 18.
60.8 is what percent of 32?
Answer: 60.8 is 190% of 32.

Explanation:
Given p = ? , w = 32, a = 60.8
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{60.8}{32}\) = \(\frac{p}{100}\)
p = \(\frac{60.8 × 100}{32}\)
p = \(\frac{6080}{32}\)
p = 190
So, 60.8 is 190% of 32.

Question 19.
91 is 130% of what number?
Answer: 91 is 130% of 70.

Explanation:
Given p = 130% , w = ? , a = 91
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{91}{w}\) = \(\frac{130}{100}\)
w = \(\frac{91 × 100}{130}\)
w = \(\frac{9100{130}\)
w = 70
So, 91 is 130% of 70.

Question 20.
85% of what number is 10.2?
Answer: 10.2 is 85% of 12.

Explanation:
Given p = 85% , w = ? , a = 10.2
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{10.2}{w}\) = \(\frac{85}{100}\)
w = \(\frac{10.2 × 100}{85}\)
w = \(\frac{1020}{85}\)
w = 12
So, 10.2 is 85% of 12.

Question 21.
83% of 20 is what number?
Answer: 16.6 is 83% of 20.

Explanation:
Given p = 83% , w = 20 , a = ?
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{20}\) = \(\frac{83}{100}\)
a = \(\frac{20 × 83}{100}\)
a = \(\frac{1,660}{100}\)
a = 16.6
So, 16.6 is 83% of 20.

Question 22.
15% of the parking spaces at a school are handicap spaces. The school has 18 handicap spaces. How many parking spaces are there in total?
Big Ideas Math Answers Grade 7 Chapter 6 Percents cr 22
Answer: Totally , there are 120 parking spaces .

Explanation:
Given , p = 15% , a = 18 , w = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{18}{w}\) = \(\frac{15}{100}\)
w = \(\frac{18 × 100}{15}\)
w = \(\frac{1800}{15}\)
w = 120
So, 18 is 15% of 120.
Totally , there are 120 parking spaces .

Question 23.
Of the 25 students on a field trip, 16 bring cameras. What percent of the students bring cameras?
Answer: 64% of students brought the cameras .

Explanation:
Given , a = 16 , w = 25 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{16}{25}\) = \(\frac{p}{100}\)
p = \(\frac{16 × 100}{25}\)
p = \(\frac{1600}{25}\)
p = 64
So, 16 is 64% of 25.
Thus, 64% of students brought the cameras .

6.4 Percents of Increase and Decrease (pp. 253–258)
Learning Target: Find percents of change in quantities.

Identify the percent of change as an increase or a decrease. Then find the percent of change. Round to the nearest tenth of a percent if necessary.
Question 24.
6 yards to 36 yards
Answer: percent of change is 500%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 36 and old value = 6 , because a change of 6 to 36 is a positive (increase) change
So, percent change = \(\frac{36 – 6}{6}\)
= \(\frac{30}{6}\)
= 5× 100
= 500%
So, percent of change is 500%

Question 25.
120 meals to 52 meals
Answer: percent of change is – 56.6%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 52 and old value = 120 , because a change of 120 to 52 is a negative (decrease) change
So, percent change = \(\frac{52 – 120}{120}\)
= \(\frac{- 68}{120}\)
= – 0.566
= – 0.566 × 100
= – 56.6%
So, percent of change is – 56.6%

Question 26.
You estimate that a jar contains 68 marbles. The actual number of marbles is 60. Find the percent error.
Answer: percent Error is 13%

Explanation:
The amount of error is 68 – 60 = 8
We know that , formula for percent error = \(\frac{Error value}{Actual value}\)
where Error value = 8 and Actual value = 60 ,
So, percent Error = \(\frac{8}{60}\)
= 0.133
= 0.133 × 100
=  13.3%
Approximately we can write as 13%
So, percent Error is 13%

Question 27.
The table shows the numbers of skim boarders at a beach on Saturday and Sunday. What was the percent of change in boarders from Saturday to Sunday?
Big Ideas Math Answers Grade 7 Chapter 6 Percents cr 27
Answer: percent of change is – 12.5%

Explanation:
Given , 12 to 9
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 63 and old value = 72 , because a change of 72 to 63 is a negative (decrease) change
So, percent change = \(\frac{63 – 72}{72}\)
= \(\frac{- 9}{72}\)
= \(\frac{- 1}{8}\)
= – 0.125
= – 0.125 × 100
= – 12.5%
So, percent of change is – 12.5%

6.5 Discounts and Markups (pp. 259–264)
Learning Target: Solve percent problems involving discounts and markups.

Find the sale price or original price.
Question 28.
Original price: $50
Discount: 15%
Sale price: ?
Answer: The sales price is $42.5

Explanation:
We know , The sales price be 100% – 15% = 85% of the original price
sales price = 85% of 50
= 0.85 × 50 = 42.5
So, The sales price is $42.5

Question 29.
Original price: ?
Discount: 20%
Sale price: $75
Answer: The original price is $93.75.

Explanation:
We know , The sales price be 100% – 20% = 80%
a = p% × w
75 = 0.8 × w
w = \(\frac{75}{0.8}\)
w = 93.75
So, The original price is $93.75.

Question 30.
What is the original price of the tennis racquet?
Big Ideas Math Answers Grade 7 Chapter 6 Percents cr 30
Answer: The original price is $30.

Explanation:
We know , The sales price be 100% – 30% = 70%
a = p% × w
21 = 0.7 × w
w = \(\frac{21}{0.7}\)
w = 30
So, The original price is $30.

Question 31.
A store pays $50 for a pair of shoes. The markup is 25%.
a. What is the selling price for the shoes?
b. What is the total cost for a person to buy the shoes including a 6% sales tax?
Answer: a. The selling price is $62.5.
b. Total cost for a person is  $66.25 .

Explanation:
The markup is 25% of $50
a = p% × w
= 25% × 50
= 0.25 × 50
= 12.5
So , the markup is $12.5.
To , find the selling price , we have
selling price = cost to store + markup
= $50 + $12.5
= $62.5.
The selling price is $62.5.

b. with 6% sales tax we have ,
6% of $62.5
= 0.06 × 62.5
= 3.75 , it is the tax rate
So , total cost for a person is $62.5 + $3.75 = $66.25

6.6 Simple Interest (pp. 265–270)
Learning Target: Understand and apply the simple interest formula.

An account earns simple interest. (a) Find the interest earned. (b) Find the balance of the account.
Question 32.
$300 at 4% for 3 years
Answer: a. The interest earned is $36
b. The balance of the account is $336 .

Explanation:
we know P = $300 , r = 0.04 , t = 3 years
I = prt
I = 300 × 0.04 × 3
I = 3 × 4 × 3
I = $36
The interest earned is $36
The balance of the account  = $300 + $36 = $336 .

Question 33.
$2000 at 3.5% for 4 years
Answer: a. The interest earned is $280
b. The balance of the account  is $2,280 .

Explanation:
we know P = $2000 , r = 0.035 , t = 4 years
I = prt
I = 2000 × 0.035 × 4
I = 2 × 35 × 4
I = $280
The interest earned is $280
The balance of the account  = $2000 + $280 = $2,280 .

Find the annual interest rate.
Question 34.
I = $17, P = $500, t = 2 years
Answer: The annual interest rate is 1.7% .

Explanation:
We know that  I = Prt
So , r = \(\frac{I}{Pt}\)
r = \(\frac{17}{500 × 2}\)
= \(\frac{17}{1000}\)
= 0.017
we can write in percent as 1.7%
So , The annual interest rate is 1.7% .

Question 35.
I = $426, P = $1200, t = 5 years
Answer: The annual interest rate is 7.1% .

Explanation:
We know that  I = Prt
So , r = \(\frac{I}{Pt}\)
r = \(\frac{426}{1200 × 5}\)
= \(\frac{426}{6000}\)
= 0.071
we can write in percent as 7.1%
So , The annual interest rate is 7.1% .

Find the amount of time.
Question 36.
I = $60, P = $400, r = 5%
Answer: The amount of time is 3 years .

Explanation:
We know that  I = Prt
t = \(\frac{I}{Pr}\)
= \(\frac{60}{400 × 0.05}\)
= \(\frac{60}{20}\)
= 3
So, The amount of time is 3 years .

Question 37.
I = $237.90, P = $1525, r = 2.6%
Answer: The amount of time is 6 years .

Explanation:
We know that  I = Prt
t = \(\frac{I}{Pr}\)
= \(\frac{237.9}{1525 × 0.026}\)
= \(\frac{237.9}{39.65}\)
= 6
So, The amount of time is 6 years .

Question 38.
You deposit $100 in an account. The account earns $2 simple interest in 6 months. What is the annual interest rate?
Answer: The annual interest rate is 12%

Explanation:
Given , p = $100 , I = $2 , t = 6 months ,
We know that  I = Prt
So , r = \(\frac{I}{Pt}\)
r = \(\frac{2 × 6}{100}\)
= \(\frac{12}{100}\)
= 0.12
we can write in percent as 12%
So , The annual interest rate is 12%

Question 39.
Bank A is offering a loan with a simple interest rate of 8% for 2 years. Bank B is offering a loan with a simple interest rate of 6.5% for 3 years.
Big Ideas Math Answers Grade 7 Chapter 6 Percents cr 39
a. Assuming the monthly payments are equal, what is the monthly payment for the four wheeler from Bank A? from Bank B?
b. Give reasons for why a person might choose Bank A and why a person might choose Bank B for a loan to buy the four wheeler. Explain your reasoning.
Answer: a. The monthly payment for bank A is $261 and for Bank B is $179.25 .
b. If a person chooses Bank A , it will be high in monthly payment but within 2 years loan will be cleared .
If a person chooses Bank B , it will be low monthly payment and can pay the loan a year extra in small amounts.

Explanation:
For bank A , Then p = $5400 , r = 0.08 , t = 2 years
I = Prt
I = 5400 × 0.08 × 2
I = 54 × 8 × 2
I = 864
The simple interest of one bank = $864 .
The balance of the account will be = $5400 + $864 = $6,264 .
The monthly payment = \(\frac{6,264}{24}\) = $261 .

For bank B,  Then p = $5400 , r = 0.065 , t = 3 years
I = Prt
I = 5400 × 0.065 × 3
I = 54 × 6.5 × 3
I = 1,053
The simple interest of one bank = $1,053 .
The balance of the account will be = $5400 + $1,053 = $6,453 .
The monthly payment = \(\frac{6,453}{36}\) = $179.25 .

b. If a person chooses Bank A , it will be high in monthly payment but within 2 years loan will be cleared .
If a person chooses Bank B , it will be low monthly payment and can pay the loan a year extra in small amounts.

Percents Practice Test

Write the percent as a decimal, or the decimal as a percent. Use a model to represent the number.
Question 1.
0.96%
Answer: 0.0096

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, 0.96% in decimal form is 0.0096

Question 2.
3%
Answer: 0.03

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, 3% in decimal form is 0.03

Question 3.
\(25 . \overline{5} \%\)
Answer: \(0 .25 \overline{5}\)

Explanation:   
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, \(25 . \overline{5} \%\) in decimal form is \(0 .25 \overline{5}\)

Question 4.
\(0 . \overline{6} \%\)
Answer: \(0 .006\overline{6}\)

Explanation:
To write percent into decimal we have to remove percent symbol and divide by 100, which moves the decimal point two places to the left.
So, \(0 . \overline{6} \%\)  in decimal form is\(0 .006 \overline{6}\)

Question 5.
7.88
Answer: 788%

Explanation: 
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol.
So, 7.88 can be rewrite as 788%

Question 6.
0.58
Answer: 58%

Explanation: 
To write decimal into percent we should multiply the number by 100, which moves the decimal point two places to the right and add a percent symbol.
So, 0.58 can be rewrite as 58%

Order the numbers from least to greatest.
Question 7.
86%, \(\frac{15}{18}\), 0.84, \(\frac{8}{9}\), \(0 . \overline{86} \%\)
Answer: The numbers from least to greatest are  \(0 .00 \overline{86}\) , \(0 .8 \overline{3}\) , 0.84 , 0.86 , \(0 .\overline{8}\)

Explanation:
86%,  in decimals as 0.86
\(\frac{15}{18}\), can be \(0 .8 \overline{3}\)
\(\frac{8}{9}\),  can be \(0 .\overline{8}\)
\(0 . \overline{86} \%\) can be written as \(0 .00 \overline{86}\)
So, The numbers from least to greatest are  \(0 .00 \overline{86}\) , \(0 .8 \overline{3}\) , 0.84 , 0.86 , \(0 .\overline{8}\)

Question 8.
91.6%, 0.91, \(\frac{11}{12}\), 0.917, 9.2%
Answer:  The numbers from least to greatest are 0.092 , 0.91 , 0.916 , \(0 .91\overline{6}\) , 0.917 .

Explanation:
91.6%, can be written as 0.916
\(\frac{11}{12}\), ca be written as \(0 .91\overline{6}\)
9.2% can be written as 0.092
So, The numbers from least to greatest are 0.092 , 0.91 , 0.916 , \(0 .91\overline{6}\) , 0.917 .

Write and solve a proportion or equation to answer the question.
Question 9.
What percent of 28 is 21?
Answer:  21 is 75% of 28.

Explanation:
Given , a = 21 , w = 28 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{21}{28}\) = \(\frac{p}{100}\)
p = \(\frac{21 × 100}{28}\)
p = \(\frac{2100}{28}\)
p = 75
So, 21 is 75% of 28.

Question 10.
64 is what percent of 40?
Answer: 64 is 160% of 40.

Explanation:
Given , a = 64 , w = 40 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{64}{40}\) = \(\frac{p}{100}\)
p = \(\frac{64 × 100}{40}\)
p = \(\frac{6400}{40}\)
p = 160
So, 64 is 160% of 40.

Question 11.
What number is 80% of 45?
Answer: 36 is 80% of 45.

Explanation:
Given , a = ? , w = 45 , p = 80%
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{45}\) = \(\frac{80}{100}\)
a = \(\frac{45 × 80}{100}\)
a = \(\frac{3600}{100}\)
a = 36
So, 36 is 80% of 45.

Question 12.
0.8% of what number is 6?
Answer: 6 is 0.8% of 750.

Explanation:
Given , a = 6 , w = ? , p = 0.8%
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{6}{w}\) = \(\frac{0.8}{100}\)
w = \(\frac{6 × 100}{0.8}\)
w = \(\frac{600}{0.8}\)
w = 750
So, 6 is 0.8% of 750.

Identify the percent of change as an increase or a decrease. Then find the percent of change. Round to the nearest tenth of a percent if necessary.
Question 13.
4 strikeouts to 10 strikeouts
Answer: percent of change is 150%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 10 and old value = 4 , because a change of 4 to 10 is a positive (increase) change
So, percent change = \(\frac{10 – 4}{4}\)
= \(\frac{6}{4}\)
= 1.5 × 100
= 150%
So, percent of change is 150%

Question 14.
$24 to $18
Answer: percent of change is – 25%

Explanation:
We know that , formula for percent change = \(\frac{New value – old value}{old value}\)
where New value = 18 and old value = 24 , because a change of 24 to 18 is a negative (decrease) change
So, percent change = \(\frac{18 – 24}{24}\)
= \(\frac{- 6}{24}\)
= – 0.25
= – 0.25 × 100
= – 25%
So, percent of change is – 25%

Find the sale price or selling price.
Question 15.
Original price: $15
Discount: 5%
Sale price: ?
Answer: The sales price is $14.25

Explanation:
We know , The sales price be 100% – 5% = 95% of the original price
sales price = 95% of 15
= 0.95 × 15 = 14.25
So, The sales price is $14.25

Question 16.
Cost to store: $5.50
Markup: 75%
Selling price: ?
Answer: The selling price is $9.625.

Explanation:
The markup is 75% of $5.50
a = p% × w
= 75% × 5.5
= 0.75 × 5.5
= 4.125
So , the markup is $4.125.
To , find the selling price , we have
selling price = cost to store + markup
= $5.5 + $4.125
= $9.625.
The selling price is $9.625.

An account earns simple interest. Find the interest earned or the principal.
Question 17.
Interest earned: ?
Principal: $450
Interest rate: 6%
Time: 8 years
Answer: The interest earned is $216 .

Explanation:
we know P = $450 , r = 0.06 , t = 8 years
I = prt
I = 450 × 0.06 × 8
I = 27 × 8
I = $216
The interest earned is $216 .

Question 18.
Interest earned: $27
Principal: ?
Interest rate: 1.5%
Time: 2 years
Answer: The principal is $900 .

Explanation:
Given , P = ? , I = $27 , t = 2 years , r = 0.015
We know that  I = Prt
So , p = \(\frac{I}{rt}\)
p = \(\frac{27}{0.015 × 2}\)
= \(\frac{27}{0.03}\)
= 900
So, The principal is $900 .

Question 19.
You spend 8 hours each weekday at school. (a) Write the portion of a weekday spent at school as a fraction, a decimal, and a percent. (b) What percent of a week is spent at school if you go to school 4 days that week? Round to the nearest tenth.
Answer: a. the portion of a weekday spent at school is 33% , It can be written as \(\frac{100}{3}\) and in decimal as 0.333
b . The 9% of a week is spent at school if you go to school 4 days that week.

Explanation:
Given , You spend 8 hours each weekday at school.
a. In a day we have 24 hours so , The percent of 8 hours in 24 hours is
a = 8 , w = 24 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{8}{24}\) = \(\frac{p}{100}\)
p = \(\frac{8 × 100}{24}\)
p = \(\frac{800}{24}\)
p = 33.3 %
It can be written as \(\frac{100}{3}\) and in decimal as 0.333
So, 8 is 33% of 24.

b. The number of hours in a week  , that is 7 days are 168 hours
The number of hours in 4 days to school are 32
So , we have a = 32 , w = 168 , p = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{32}{168}\) = \(\frac{p}{100}\)
p = \(\frac{32 × 100}{168}\)
p = \(\frac{3200}{168}\)
p = 9.04 %
p = 9%
So , The 9% of a week is spent at school if you go to school 4 days that week.

Question 20.
Research indicates that90% of the volume of an iceberg is below water. The volume of the iceberg above the water is 160,000 cubic feet. What is the volume of the iceberg below water?
Big Ideas Math Solutions Grade 7 Chapter 6 Percents pt 20
Answer: The volume of the iceberg below water is  1,77,777 cubic feet

Explanation:
Given , p = 90% , a = 160,000 , w = ?
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{160,000}{w}\) = \(\frac{90}{100}\)
w = \(\frac{160,000 × 100}{90}\)
w = \(\frac{1050}{125}\)
w = 1,77,777.7
So, the volume of the iceberg below water is  1,77,777 cubic feet

Question 21.
You estimate that there are 66 cars in a parking lot. The actual number of cars is 75.
a. Find the percent error.
b. What other estimate gives the same percent error? Explain your reasoning.
Answer: a. percent Error is 12%
b, If the estimated number is 66.1 then percent error will be 12%

Explanation:
The amount of error is 75 – 66 = 9
We know that , formula for percent error = \(\frac{Error value}{Actual value}\)
where Error value = 9 and Actual value = 75 ,
So, percent Error = \(\frac{9}{75}\)
= 0.12
= 0.12 × 100
=  12%
So, percent Error is 12%

b. if the estimation is 66.1
The amount of error is 75 – 66.1 = 8.9
We know that , formula for percent error = \(\frac{Error value}{Actual value}\)
where Error value = 8.9 and Actual value = 75 ,
So, percent Error = \(\frac{8.9}{75}\)
= 0.1198
= 0.1198 × 100
=  11.9%
Apporximately 12%
So, percent Error is 12%

Percents Cumulative Practice

Big Ideas Math Solutions Grade 7 Chapter 6 Percents cp 1
Question 1.
A movie theater offers 30% a movie ticket to students from your school. The regular price of a movie ticket is $8.50. What is the discounted price that you pay for a ticket?
A. $2.55
B. $5.50
C. $5.95
D. $8.20
Answer:  C. $5.95

Explanation:
Given , original price = $8.5 , with 30% off
We know , The sales price be 100% – 30% = 70% of the original price
sales price = 70% of 8.5
= 0.7 × 8.5 = 5.95
So, The sales price is $5.95.

Question 2.
What is the least value of x for which the inequality is true?
Big Ideas Math Solutions Grade 7 Chapter 6 Percents cp 2
16 ≥ – 2x
Answer: x = 1

Explanation:
Let us say X = 1
16 ≥ – 2(1)
16 ≥ – 2
So , x = 1 , for 16 ≥ – 2x

Question 3.
You are building a scale model of a park that is planned for a city. The model uses the scale 1 centimeter = 2 meters. The park will have a rectangular reflecting pool with a length of 20 meters and a width of 12 meters. In your scale model, what will be the area of the reflecting pool?
F. 60 cm2
G. 120 cm2
H. 480 cm2
I. 960 cm2
Answer:  F .  60 cm2

Explanation:
Given ,actual  l = 20 meters , The model uses the scale 1 centimeter = 2 meters.  let the model length be x
The model length x will be , \(\frac{1cm}{2m}\) = \(\frac{x cm}{20m}\)
2x = 20
x = 10 ,
Let the model width be y , The model width y will be \(\frac{1 cm}{2m}\) = \(\frac{y}{12}\)
y= 6 ,
So The model area of the reflecting pool , Area of rectangle = l × w
l = 10 , y = 6 , then A = 10 × 6 = 60 cm2
The model area of the reflecting pool is 60 cm2

Question 4.
Which proportion represents the problem?
Big Ideas Math Solutions Grade 7 Chapter 6 Percents cp 4
Answer: D . \(\frac{43}{n}\) = \(\frac{17}{100}\)

Explanation:
Given ,  p = 17% , a = 43
By using proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{43}{w}\) = \(\frac{17}{100}\)
w = \(\frac{43 × 100}{17}\)
w = \(\frac{4300}{17}\)
w = 252.9
So, 43 is 17% of 252.9.

Question 5.
Which list of numbers is in order from least to greatest?
F. 0.8, \(\frac{5}{8}\), 70%, 0.09
G. 0.09, \(\frac{5}{8}\), 0.8, 70%
H. \(\frac{5}{8}\), 70%, 0.8, 0.09
I. 0.09, \(\frac{5}{8}\), 70%, 0.8
Answer: H. \(\frac{5}{8}\), 70%, 0.8, 0.09

Explanation:
Given,
\(\frac{5}{8}\) can be written as 0.625 ,
70% can be written as 0.7
The order from least to greatest  is  \(\frac{5}{8}\) , 70% , 0.8 , 0.9

Question 6.
What is the value of \(\frac{9}{8}\) ÷ (-\(\frac{11}{4}\))?
Big Ideas Math Solutions Grade 7 Chapter 6 Percents cp 6
Answer: the value of \(\frac{9}{8}\) ÷ (-\(\frac{11}{4}\)) is – 0.413

Explanation:
\(\frac{9}{8}\)  can be written as  1.125
–\(\frac{11}{4}\) can be written as  – 2.75
then , \(\frac{1.125}{-2.75}\) = – 0.413

Question 7.
The number of calories you burn by playing basketball is proportional to the number of minutes you play. Which of the following is a valid interpretation of the graph?
Big Ideas Math Solutions Grade 7 Chapter 6 Percents cp 7
A. The unit rate is \(\frac{1}{9}\) calorie per minute.
B. You burn 5 calories by playing basketball for 45 minutes.
C. You do not burn any calories if you do not play basketball for at least 1 minute.
D. You burn an additional 9 calories for each minute of basketball you play.
Answer: C. You do not burn any calories if you do not play basketball for at least 1 minute.

Explanation:
As , the graph shows the coordinates of the minutes to calories by (0,0) at the initial stage of playing basketball for at least a minute .
So , the graph represents , You do not burn any calories if you do not play basketball for at least 1 minute.

Question 8.
A softball team is ordering uniforms. Each player receives one of each of the items shown in the table.
Big Ideas Math Solutions Grade 7 Chapter 6 Percents cp 8
Which expression represents the total cost (in dollars) when there are 15 players on the team?
F. x + 24.86
G. 15x + 372.90
H. x + 372.90
I. x + 387.90
Answer: G. 15x + 372.90

Explanation:
Given , jersey = x
So , the only option in the options with an x for 15 members is 15x ,
So, G. 15x + 372.90 is the correct option

Question 9.
Your friend solves the equation. What should your friend do to correct the error that he made?
Big Ideas Math Solutions Grade 7 Chapter 6 Percents cp 9
A. Multiply – 45 by – 3.
B. Add 3 to – 45.
C. Add 2 to – 15.
D. Divide – 45 by – 3.
Answer: A. Multiply – 45 by – 3.

Explanation:
Given , – 3(2 + w) = -45
By , Multiply – 45 by – 3.
we get 2 + w = 15
w = 15 – 2
w = 13.

Question 10.
You are comparing the costs of a certain model of ladder at a hardware store and at an online store.
Big Ideas Math Solutions Grade 7 Chapter 6 Percents cp 10
Part A What is the total cost of buying the ladder at each of the stores? Show your work and explain your reasoning.
Part B Suppose that the hardware store is offering 10% off the price of the ladder and that the online store is offering free shipping and handling. Which store offers the lower total cost for the ladder? by how much? Show your work and explain your reasoning.
Answer: Part A , The total cost at hardware store is $371 and The total cost at online store is $355.2
Part B ,The hardware store offers the best price $336 by $3.2 less than the online store $339.2

Explanation:
Part A , The ladder cost at hardware store is $350 , with 6% tax ,
So, 6% of 350
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{350}\) = \(\frac{6}{100}\)
a = \(\frac{350 × 6}{100}\)
a = \(\frac{2100}{100}\)
a = 21
So, 21 is 6% of 350.
The total cost at hardware store is $350 + $ 21 = $371

The ladder cost at online store is $320 , with 6% tax ,
So, 6% of 320
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{320}\) = \(\frac{6}{100}\)
a = \(\frac{320 × 6}{100}\)
a = \(\frac{1920}{100}\)
a = 19.2
So,19.2 is 6% of 320.
Additionally shipping cost is 5% of $320
By using percent proportion , we have
\(\frac{a}{w}\) = \(\frac{p}{100}\)
\(\frac{a}{320}\) = \(\frac{5}{100}\)
a = \(\frac{320 × 5}{100}\)
a = \(\frac{1600}{100}\)
a = 16
So,16 is 5% of 320.
The total cost at online store is $320 + $ 19.2 + $16 = $355.2

Part B, If the hardware store offers 10% off
Then , We know , The sales price be 100% – 10% = 90% of the original price
sales price = 90% of 350
= 0.9 × 350 = 315
So, The sales price is $315 .
The 6% of 320  is  $21
The total cost in hardware store after 10% off is $315 + $ 21 = $336

The online store offers free shipping and handling then the price will be $320 and 6% tax
So , 6% of 320  is $19.2
The total cost online store is $320 + $ 19.2 = $339.2

The hardware store offers the best price $336 by $3.2 less than the online store $339.2

Question 11.
Which graph represents the inequality – 5 – 3x ≥ – 11.
Big Ideas Math Solutions Grade 7 Chapter 6 Percents cp 11
Answer:

Explanation:11
Option F represents the x value from -4 to 2
So , if x = -4 ,
Then – 5 – 3(-4) ≥ – 11
-5 + 12 ≥ – 11
7 ≥ – 11
if x = 2
– 5 – 3(2) ≥ – 11
-5 – 6 ≥ – 11
-11 ≥ – 11
So, option F represents the inequality – 5 – 3x ≥ – 11.

Final Conclusion:

We have given Big Ideas Math Answers Grade 7 Chapter 6 Percents Pdf free links here. Download BIM Grade 7 Chapter 6 Pdf from the direct links given in the above section. We wish you make use of the given information and score better marks in the exam. If you have any doubts you can easily clear them within the below-given comment section. All the best to all the candidates!!

Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20

Big Ideas Math Answers Grade 1 Chapter 4

In Order to attempt the 1st Grade Exam, you need to have strong fundamentals. Download Big Ideas Math Book 1st Grade Answer Key Chapter 4 Add Numbers within 20 free of cost and get the homework help you need. Attempt the practice test to know your preparation level within the Chapter. Practice as much as you can using the Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 and clear the exam with better scores.

Big Ideas Math Book 1st Grade Answer Key Chapter 4 Add Numbers within 20

You can make the most out of the Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 through the quick links available. Simply tap on the topic you want and solve various questions involved in it easily. There are various models in addition and subtraction. We have listed all of them by considering enough examples and explained every problem step by step.

Vocabulary

Lesson: 1 Add Doubles from 6 to 10

Lesson: 2 Use Doubles within 20

Lesson: 3 Count on to Add within 20

Lesson 4: Add Three Numbers

Lesson 5 Add Three Numbers by Making a 10

Lesson: 6 Add 9

Lesson 7 Make a 10 to Add

Lesson: 8 Problem Solving: Addition within 20

Performance Task

Add Numbers within 20 Vocabulary

Organize It

Review Words:
addend
doubles
doubles minus 1
doubles plus 1
sum

Use the review words to complete the graphic organizer.

Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 1

Answer:

Explanation:
By Looking at graphic organizer
1.  Here we have 2 blocks and in second block we have one more added,
so we write the review words as doubles plus 1.
2. Here we have only two block so we write the review words as doubles.
3. Here also we have 2 blocks, but in second block on one dot it is crossed,
so we write the review words as doubles minus 1.

Define It

Match the review word to its definition.

Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 2
Answer:

Explanation:
Addend means any of the numbers that are added together.
Sum can be defined as the result or answer we get on adding two or more numbers or terms.
Here 1. 5 + 3 = 8  numbers 5, 3 are addends and 8 is sum
2. 4 + 3 = 7 number 4,3 are addends and 7 is sum.

Lesson 4.1 Add Doubles from 6 to 10

Explore and Grow

Use counters to model the story.

You have 7 marbles. Your friend has 7 marbles. How many marbles are there in all?

Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 3
Answer:
In all, there are total of 14 marbles.

Explanation:
Given I have 7 marbles and my friend has 7 marbles,
The total number of marbles available are ( my marbles and friend’s marble),
7 + 7 = 14 marbles. So 14 marbles are there in all.

Show and Grow

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 4
___8____ + ___8___ = __16_____
Answer:
8 + 8 = 16 dots

Explanation:
Given in block 1 we have 8 dots and in block 2 we have 8 dots,
So in total we have 8 + 8 = 16 dots.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 5
____7___ + __7____ = ___14____
Answer:
7 + 7 = 14 dots

Explanation:
Given in block 1 we have 7 dots and in block 2 we have 7 dots,
So in total we have 7 + 7 = 14 dots.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 6 =
Answer:
9 + 9 = 18 dots

Explanation:
Given in block 1 we have 9 dots and in block 2 we have 9 dots,
So in total we have 9 + 9 = 18 dots.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 7

Answer:
10 + 10 = 20 dots

Explanation:
Given in block 1 we have 10 dots and in block 2 we have 10 dots,
So in total we have 10 + 10 = 20 dots.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 8
____8___ + _8_____ = _______
Answer:
8 + 8 = 16 dots

Explanation:
Given in block 1 we have 8 dots and in block 2 we have 8 dots,
So in total we have 8 + 8 = 16 dots.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 9
___6____ + __6___ = ___12____
Answer:
6 + 6 = 12 dots

Explanation:
Given in block 1 we have 6 dots and in block 2 we have 6 dots,
So in total we have 6 + 6 = 12 dots.

Question 7.
7 + 7 = ___14____
Answer:
7 + 7 = 14 dots

Explanation:
Given first number is 7 and second number is 7,
So in total we have 7 + 7 = 14 dots.

Question 8.
10 + 10 = ___20_____
Answer:
10 + 10 = 20

Explanation:
Given first number is 10 and second number is 10,
So in total we have 10 + 10 = 20.

Question 9.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 10
Answer:

Explanation:
Given first number as 9 and second number as 9,
So in total we have 9 + 9 = 18.

Question 10.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 11
Answer:

Explanation:
Given first number as 8 and second number as 8,
So in total we have 8 + 8 = 16.

Question 11.
MP Reasoning
You and your friend each read 6 books. How many books do you and your friend read in all?
____12______ books
Answer:
Total number of books I and my friend read are 12.

Explanation:
Given I and my friend each read 6 books, that means I read 6 books,
my friend read 6 books, So in total we read 6 + 6 = 12.

Think and Grow: Modeling Real Life

You and your friend have the same number of video games. There are 16 in all. How many video games do you have?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 12
Draw a Picture:

Addition equation: X + X = 16,

___8_______ video games
Answer:
I have 8 number of video games.

Explanation:
Given I and my friend have same number of video games,
So in all both have 16 video games , Let us take I have X number of video games,
means friend will also have X number of video games,
Now addition equation is X + X = 16, 2 X = 16 , So X = 16 ÷ 2 = 8,
So I have 8 video games.

Show and Grow

Question 12.
2 friends give you the same number of pictures. You have 12 in all. How many pictures does each friend give you?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 13
Draw a Picture:

Addition equation:
X + X = 12,

____6______ pictures
Answer:
Each friend gives 6 pictures to me.

Explanation:
Given 2 friends gave me same number of pictures,
So in all both gave 12 pictures , Let us take each gave X number of pictures,
So we write addition equation as X + X = 12, 2 X = 12 , So X = 12 ÷ 2 = 6,
So each friend gave me 6 number of pictures.

Add Doubles from 6 to 10 Practice 4.1

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 14
____10______ + ____10______ = ___20______
Answer:
10 + 10 = 20 dots

Explanation:
Given in block 1 we have 10 dots and in block 2 we have 10 dots,
So in total we have 10 + 10 = 20 dots.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 15
_____6_____ + ____6______ = ____12_____
Answer:
6 + 6 = 12 dots

Explanation:
Given in block 1 we have 6 dots and in block 2 we have 6 dots,
So in total we have 6 + 6 = 12 dots.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 16
Answer:


Explanation:
Given first number as 9 and second number as 9,
So in total we have 9 + 9 = 18.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 17
Answer:

Explanation:
Given first number as 10 and second number as 10,
So in total we have 10 + 10 = 20.

Question 5.
MP Reasoning
You and your friend each do 7 jumping jacks. How many jumping jacks do you and your friend do in all?

____14______ jumping jacks
Answer:
14 jumping jacks I and my friend do in all .

Explanation:
Given I and my friend each do 7 jumping jacks,
means I did 7 jumping jacks, My friend did 7 jumping jacks,
So both overall did 7 + 7 = 14 jumping jacks.

Question 6.
Modeling Real Life
Newton and Descartes each have the same number of treats. They have 18 treats in all.
How many treats does Newton have?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 18

____9______ treats
Answer:
Newton have 9 treats.

Explanation:
Given Newton and Descartes each have the same number of treats,
Suppose if  Newton have  X treats means Descartes have same X treats
and in all both have 18 treats, So X + X = 18 , 2X = 18 , X = 18 ÷ 2 = 9,
So Newton has 9 treats.

Review & Refresh

Use the picture to write an equation.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 19
____3______ + ____2_____ = ____5_____
Answer:
3 + 2 = 5

Explanation:
By seeing the picture there are 3 children sitting and reading books on the chair and
2 children are having bags and going to school, So we write the equation as
3 + 2 = 5.

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 20
___5_______ + ____1_____ = ____6_____
Answer:
5 + 1 = 6

Explanation:
By seeing the picture there are 5 children sitting on the ground and reading books,
1 child having book is waving hand, So we write the equation as
5 + 1 = 6.

Lesson 4.2 Use Doubles within 20

Explore and Grow

Use counters to model the story.

You collect 8 leaves. Your friend collects 8 leaves. How many leaves are there in all?

____16_____ leaves
Explanation:
Both I and friend collect same leaves ,
8 + 8 = 16 leaves we call the sum as doubles as both addends are 8.

You collect 1 more leaf. How many leaves are there now?

_____17_____ leaves
Explanation:
We got doubles as 16, Now the required sum is one more than doubles,
That is 17,Therefore 8+8+1= 17 so we call as double plus 1.

Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 21

Show and Grow

Question 1.
Use the double 6 + 6 to find each sum.
6 + 7 = ___13____
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 22
Answer:
6 + 7= 13

Explanation:
We Know the double  6 + 6 = 12 , The number 7 is one more than 6,
So we can write 7 as 6 + 1, The required sum is one more than double,
12 + 1 =13, Sum is 13, therefore 6 + 7 = 13, Double plus 1.

6 + 5 = ___11____
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 23
Answer:
6+5 = 11

Explanation:
We Know the double of 6 + 6 = 12 , The number 5 is one less than 6,
So we can write 5 as 6 – 1, The required sum is one less than double,
12 – 1 = 11, Sum is 11, therefore 6 + 5 = 11, Double minus 1.

Apply and Grow: Practice

Use the double 7 + 7 to find each sum.

Question 2.
7 + 8 = ___15_____
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 24
Answer:
7 + 8 = 15

Explanation:
We know the double of 7 + 7 = 14 , The number 8 is one more than 7,
So we can write 8 as 7 + 1, The required sum is one more than double,
14 + 1 =15, Sum is 15, therefore 7 + 8 = 15, Double plus 1.

7 + 6 = ____13_____
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 25
Answer:
7 + 6 = 13

Explanation:
We know the double of 7+ 7 = 14 , The number 6 is one less than 7,
So we can write 6 as 7 – 1, The required sum is one less than double,
14 – 1 = 13, Sum is 13, therefore 7 + 6 = 13, Double minus 1.

Find the sum. Write the double you used.

Question 3.
5 + 6 = ___11______

___5_+5__+_ ○ ___1___ = ___11____
Answer:
5 + 6 = 11, we use double 5 + 5,

Explanation:
As the first addend is 5 and we know the double is 5 + 5 ,
So 5 + 5 = 10 , The number 6 is one more than 5,
So we can write 6 as 5 + 1, The required sum is one more than double,
10 + 1 =11, Sum is 11, therefore 5 + 6 = 11, Double plus 1.
We use double 5 + 5.

Question 4.
9 + 8 = ___17______

___9+9 -____ ○ ___1____ = __17_____
Answer:
9 + 8 = 17, we use double 9 + 9 ,

Explanation:
As the first addend is 9 and we know the double is 9 + 9,
9 + 9 = 18 , The number 8 is one less than 9,
So we can write 8 as 9 – 1, The required sum is one less than double,
18 – 1 = 17, Sum is 17, therefore 9 + 8 = 17, Double minus 1.
We use double 9 + 9

Question 5.
DIG DEEPER!
Circle two ways you can solve 9 + 10. Show how you know.
9 + 9 and 1 more
9 + 9 and 1 less
10 – 10 and 1 more
10 + 10 and 1 less
Answer:
To solve 9 + 10 the two ways  which we can solve are
9 + 9 and 1 more and other is 10 + 10 and 1 less.

Explanation:
We know 9 + 10 = 19 now we will see which way we can solve,
1. 9 + 9 and 1 more means , it is 18 + 1 which is equal to 19,
Similar to 9 + 10 = 19 , therefore we can solve with 9 + 9 and 1 more .
2. 9 + 9 and 1 less means it is 18 – 1 which is 17,
As 17 is not equal to 9 + 10 = 19, So 9 + 9 and 1 less way we will not solve 9 + 10 .
3. 10 – 10 and 1 more means 0 + 1 = 1 , here this is not equal to 9 + 10,
So by 10 – 10 and 1 more way we will not solve for 9 + 10.
4. 10 + 10 and 1 less means 10 + 9 = 19 which is equal to 9 + 10 ,
So 10 + 10 and 1 less way we will solve 9 + 10.
Therefore to solve 9 + 10 the two ways which we can solve are
9 + 9 and 1 more and other is 10 + 10 and 1 less.

Think and Grow: Modeling Real Life

A music room has 7 keyboards. There is 1 more recorder than keyboards. How many instruments are there?
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 26
Which doubles can you use to find the sum?
7 + 7          8 + 8            6 + 6
Equation:

__15_____ instruments
Answer:
There are 15 instruments in the music room.
We use doubles 7 + 7  to find the sum 7 + 8 = 15.

Explanation:
Given a music room has 7 keyboards and there are  1 more recorder than keyboards,
means number of recorders are 7 + 1 = 8, so total number of instruments in room are
7+ 8 = 7 + 7 + 1 = 15, Here we have used doubles 7 + 7.
Therefore  to find the sum we use doubles 7 + 7 and the sum is 7 + 8 = 15.

Show and Grow

Question 6.
A museum has lo paintings. There is 1 fewer sculpture than paintings. How many art pieces are there?
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 27
Which doubles can you use to find the sum?
8 + 8        9 + 9      10 + 10
Equation:

____19______ art pieces
Answer:
There are 19 art pieces, We use doubles 10 + 10  to find the sum 10 + 9 = 19.

Explanation:
Given a museum has 10 paintings and there is 1 fewer sculpture than paintings,
means number of sculptures are one less than paintings 10 – 1 = 9,
So total number of art pieces are 10 + 9 = 10 + 10 – 1 = 19,
Here we have used doubles 10 + 10.
Therefore  to find the sum we use doubles 10 + 10 and the sum is 10 + 9 = 19.

Use Doubles within 20 Practice 4.2

Question 1.
9 + 10 = ____19____
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 28
Answer:
9 + 10 = 19, We use double 9 + 9

Explanation:
As the first addend is 9 and we know the double is 9 + 9 ,
So 9 + 9 = 18 , The number 10 is one more than 9,
So we can write 10 as 9 + 1, The required sum is one more than double,
18 + 1 =19, Sum is 19, therefore 9 + 10 = 19, Double plus 1.
Here we have use double 9 + 9

9 + 8 = ____17____
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 29
Answer:
9 + 8 = 17, We use double 9 + 9

Explanation:
As the first addend is 9 and we know the double is 9 + 9,
9 + 9 = 18 , The number 8 is one less than 9,
So we can write 8 as 9 – 1, The required sum is one less than double,
18 – 1 = 17, Sum is 17, therefore 9 + 8 = 17, Double minus 1.
We have use double 9 + 9

Find the sum. Write the double you used.

Question 2.
10 + 9 = ____19_____

____10+10___ -○ ___1____ = __19_____
Answer:
10 + 9 = 19, we use double 10 + 10,

Explanation:
As the first addend is 10 and we know the double is 10 + 10,
10 + 10 = 20 , The number 9 is one less than 10,
So we can write 9 as 10 – 1, The required sum is one less than double,
20 – 1 = 19, Sum is 19, therefore 10 + 9 = 17, Double minus 1.
We have use double 10 + 10 .

Question 3.
7 + 8 = ____15_____

_7_+ 7_+____ ○ __1_____ = ___15___
Answer:
7 + 8 = 15, We use double 7 + 7
Explanation:
As the first addend is 7 and we know the double is 7 +7  ,
So 7 + 7 = 14 , The number 8 is one more than 7,
So we can write 8 as 7 + 1, The required sum is one more than double,
14 + 1 =15, Sum is 15, therefore 7 + 8 = 15, Double plus 1.
We use double 7 + 7.

Question 4.
DIG DEEPER!
Circle two ways you can solve 8 + 9. Show how you know.
8 + 8 and 1 more
8 + 8 and 1 less
9 + 9 and 1 more
9 + 9 and 1 less
Answer:
To solve 8 + 9 the two ways  which we can solve are
8 + 8 and 1 more and other is 9 + 9 and 1 less.

Explanation:
We know 8 + 9 = 17 now we will see which way we can solve,
1. 8 + 8 and 1 more means , it is 16 + 1 which is equal to 17,
Similar to 8 + 9 = 17 , therefore we can solve with 8 + 8 and 1 more .
2. 8 + 8 and 1 less means it is 16 – 1 which is 15,
As 15 is not equal to 8 + 9 = 17, So 8 + 8 and 1 less way we will not solve 8 + 9 .
3. 9 + 9 and 1 more means 18 + 1 = 19 , here this is not equal to 17,
So by 9 + 9 and 1 more way we will not solve for 8 + 9.
4. 9 + 9 and 1 less means 18 -1 = 17 which is equal to sum of 8 + 9  ,
So 9 + 9 and 1 less way we will solve 8 + 9.
Therefore to solve 8 + 9 the two ways which we can solve are
8 + 8 and 1 more and other is 9 + 9 and 1 less.

Question 5.
Modeling Real Life
There are 6 balls. There is 1 more toy hoop than balls. How many toys are there?
Which doubles can you use to find the sum?
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 30
7 + 7           8 + 8          6 + 6

___13_____ toys
Answer:
There are 13 toys in total,
We use doubles 6 + 6 to find the sum 6 + 7 = 13.

Explanation:
Given there are 6 balls and there is  1 more toy hoop than balls means
number of toy hoops =  6 + 1 = 7, so total number of toys are
6+ 7 = 6+ 6 + 1 = 13, Here we have used doubles 6 + 6.
Therefore to find the sum we use doubles 6 + 6 and the sum is 6 + 7 = 13.

Review & Refresh

Use the picture to write an equation.

Question 6.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 31
____7____ – ___1____ = ____6____
Answer:
Equation is 7 – 1 = 6

Explanation:
By seeing the picture there are total 7 children in the picture,
6 children are sitting and reading books and one boy with bag on shoulder’s is
walking away from other children, So we write the equation as 7 -1 = 6.

Question 7.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 32
____5____ – ___1____ = ___4_____
Answer:
Equation is 5 – 1 = 4

Explanation:
By seeing the picture there are total 5 children ,
in them four are sitting and playing with toy,
one boy is walking away from others,
So we write the equation as 5 – 1 =4.

Lesson 4.3 Count on to Add within 20

Explore and Grow

Model the story.

There are 8 coins in a piggy bank. You put in 5 more. How many coins are in the bank now?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 33
Answer:
There are 13 coins in the piggy bank.

Explanation:
Given there are 8 coins in a piggy bank and I have put 5 more means,
we start from 8 and move 5 units from 8 on number line we get 13, 8 + 5 = 13,
There are total 13 coins in the piggy bank.

Show and Grow

Question 1.
8 + 4 = _____12_____
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 34
Answer:
8 + 4 = 12

Explanation:
First we start from 8 and move 4 units we get 12 on the number line,
So sum is 8 + 4 = 12 .

Question 2.
5 + 10 = ___15______
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 35
Answer:
5 + 10 = 15

Explanation:
First we start from 5 and move 10 units we get 15 on the number line,
So sum is 5 + 10 = 15 .

Apply and Grow: Practice

Question 3.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 36
6 + 8 = _____14_____
Answer:
6 + 8 = 14

Explanation:
First we start from 6 and move 8 units we get 14 on the number line,
So sum is 6 + 8 = 14 .

Question 4.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 36
12 + 5 = __17______
Answer:
12 + 5 = 17

Explanation:
First we start from 12 and move 5 units we get 17 on the number line,
So sum is 12 + 5 = 17 .

Question 5.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 36
7 + 8 = ___15______
Answer:
7 + 8 = 15

Explanation:
First we start from 7 and move 8 units we get 15 on the number line,
So sum is 7 + 8 = 15 .

Question 6.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 36
10 + 9 = ___19_____
Answer:
10 + 9 = 19

Explanation:
First we start from 10 and move 9 units we get 19 on the number line,
So sum is  10 + 9 = 19 .

Question 7.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 36
____11___ = 0 + 11
Answer:
11 = 0 + 11 or 0 + 11 = 11

Explanation:
First we start from 0 and move 11 units we get 11 on the number line,
So sum is 11= 0 + 11 or 0 + 11 = 11.

Question 8.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 36
___13_____ = 4 + 9
Answer:
13 = 4 + 9

Explanation:
First we start from 4 and move 9 units we get 13 on the number line,
So sum is 13= 4 + 9  or 4 + 9 = 13.

Question 9.
MP Structure
Write the equation shown by the number line. Then write the equation another way.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 37
Answer:
7 + 4 = 11     or  4 + 7 = 11 or 11 = 7 + 4

Explanation:
As shown in the number line we are starting from 7
and moving 4 units we get 11,
So the equation is 7 + 4 = 11, we can also write  the equation as 4 + 7 =11,
there is no change in the value of sum.
We get sum 11 if we add 7 and 4.

Think and Grow: Modeling Real Life

You have 7 train cars. Newton has 5 more than you. Descartes has 4 more than you. Who has more train cars, Newton or Descartes?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 38
Model:
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 39

Equations:
Newton               Descartes
7 + 5 = 12          7 + 4 = 11

Who has more?         Newton             Descartes
Newton has more trains.

Explanation:
Given I have 7 trains, Newton has 5 more than you,
So the equation is 7 + 5 = 12, Newton has 12 trains,
and Descartes has 4 more than me , the equation is
7 + 4 = 11, Descartes has 11 trains. Now we compare with
Newton and Descartes so in between 12 and 11 we compare,
as 12 is more or greater than 11, means Newton has more trains = 12.

Show and Grow

Question 10.
You have 6 comic books. Newton has 5 more than you. Descartes has 6 more than you. Who has more comic books, Newton or Descartes?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 40
Model:
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 39

Equations:
Newton             Descartes
6 + 5 = 11          6 + 6 = 12

Who has more?         Newton             Descartes
Descartes has more comic books.

Explanation:
Given I have 6 comic books , Newton has 5 more than you,
So the equation is 6 + 5 = 11, Newton has 11 comic books,
and Descartes has 6 more than me , the equation is
6 + 6 = 12, Descartes has 12 comic books. Now we compare with
Newton and Descartes so in between 11 and 12 we compare,
as 12 is more or greater than 11, means Descartes has more comic books = 12.

Count on to Add within 20 Practice 4.3

Question 1.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 41
10 + 7 = ___17_______
Answer:
10 + 7 = 17

Explanation:
First we start from 10 and move 7 units we get 17 on the number line,
So sum is 10 + 7 = 17.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 41
8 + 9 = ____17_____
Answer:
8 + 9 = 17

Explanation:
First we start from 8 and move 9 units we get 17 on the number line,
So sum is 10 + 7 = 17.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 41
_____15_____ = 6 + 9
Answer:
6 + 9 = 15 or 15 = 6 + 9

Explanation:
First we start from 6 and move 9 units we get 15 on the number line,
So sum is 6 + 9 = 15 or 15 = 6 + 9.

Question 4.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 41
______16_____ = 12 + 4
Answer:
16= 12 + 4 or 12 + 4 = 16

Explanation:
First we start from 12 and move 4 units we get 16 on the number line,
So sum is  12 + 4 = 16 or 16 = 12 + 4.

Question 5.
MP Structure
Write the equation shown by the number line. Then write the equation another way.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 42
Answer:
5 + 8 = 13  or  8 + 5 = 13 or 13 = 5 + 8

Explanation:
As shown in the number line we are starting from 5
and moving 8 units we get 13,
So the equation is 5 + 8 = 13, we can also write  the equation as 8 + 5 =13,
there is no change in the value of sum.
We get sum 13 if we add 5 and 8.

Question 6.
Modeling Real Life
You have 11 toys. Newton has 3 more than you. Descartes has 6 more than you. Who has more toys, Newton or Descartes?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 43
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 44
Answer:
Equations:     Newton             Descartes
11 + 3 = 14          11 + 6 = 17
Descartes has more toys than Newton

Explanation:
Given I have 11 toys , Newton has 3 more than you,
So the equation is 11 + 3 = 14, Newton has 14 toys,
and Descartes has 6 more than me , the equation is
11 + 6 = 17, Descartes has 17 toys. Now we compare with
Newton and Descartes so in between 14 and 17 toys if  we compare
we get 17 is more or greater than 14, means Descartes has more toys = 17.

Review & Refresh

Question 7.
Circle the triangle. Draw a rectangle around the hexagon.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 45
Answer:

Explanation:
We know triangle means a closed, two-dimensional shape with three straight sides.
In shapes given in figure we have triangle in green so we will circle it,
and we know hexagon can be defined as a polygon with six sides.
In shapes given in the figure we have hexagon in yellow,
So we draw a rectangle around the hexagon.

Question 8.
Circle the cube. Draw a rectangle around the sphere.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 46
Answer:

Explanation:
Cube means a symmetrical three-dimensional shape,
either solid or hollow, contained by six equal squares,
In figure we have cube in purple so we circle the cube.
Sphere means a round solid figure, or its surface,
with every point on its surface equidistant from its center.
In the shapes given in figure we have sphere in yellow,
So we draw a rectangle around hexagon.

Lesson 4.4 Add Three Numbers

Explore and Grow

Use linking cubes to model the story.

You have 5 red pencils, 4 yellow pencils, and 3 blue pencils. How many pencils do you have in all?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 47
Answer:

Explanation:
Given 5 red pencils , 4 yellow pencils  and 3 blue pencils,
a. Now by using linking the cubes first we link with 5 red and 4 yellow pencils
we get 5 + 4 = 9 pencils now we link with 3 blue pencils 9 + 3 = 12 pencils.
b. Now by using linking the cubes first we link with 4 yellow pencils and 3 blue pencils
we get 4 + 3 = 7 pencils now we link with 5 red pencils 7 + 5 = 12 pencils.
c. Now by using linking the cubes first we link with 5 red and 3 blue pencils
we get 5 + 3 = 8 pencils now we link with 4 yellow pencils 8 + 4 = 12 pencils.
In all we have 12 pencils.

Show and Grow

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 48
Answer:

Explanation:
By using linking the cubes
a. First we add 2 + 5 = 7 and now we add 7+ 6 we get 13.
b. Here we add 5 + 6 =11, now we add 11 + 2 =13
c. Now here we add first 2 with 6 we get 8 to 8 we add 5
we will get 8 + 5 = 13, So we get sum as 13.

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 49
Answer:

Explanation:
By using linking the cubes
a. First we add 4 + 4 = 8 and now we add 8+ 3 we get 11.
b. Here we add 3 + 4 =7, now we add 7 + 4 =11
c. Now here we add first 4 with 3 we get 7 to 7 we add 4
we will get 7 + 4 = 11, So we get sum as 11.

Apply and Grow: Practice

Question 3.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 50
Answer:

Explanation:
By using linking the cubes
a. First we add 3 + 3 = 6 and now we add 6+ 5 we get 11.
b. Here we add 3 + 5 =8, now we add 8 + 3 =11
c. Now here we add first 5 with 3 we get 8 to 8 we add 3
we will get 8 + 3 = 11, So we get sum as 11.

Question 4.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 51
Answer:

Explanation:
By using linking the cubes
a. First we add 3 + 1 = 4 and now we add 4+ 2 we get 6.
b. Here we add 1 + 2 =3, now we add 3 + 3 =6,
c. Now here we add first 3 with 2 we get 5 to 5 we add 1
we will get 5 + 1 = 6, So we get sum as 6.

Question 5.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 52
Answer:

Explanation:
By using linking the cubes
a. First we add 7 + 5 = 12 and now we add 12+ 5 we get 17.
b. Here we add 5 + 5 =10, now we add 10 + 7 = 17,
c. Now here we add first 7 with 5 we get 12 to 12 we add 5
we will get 12 + 5 = 17, So we get sum as 17.

Question 6.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 53
Answer:

Explanation:
By using linking the cubes
a. First we add 4 + 5 = 9 and now we add 9 + 2 we get 11.
b. Here we add 5 + 2 = 7, now we add 7 + 4 = 11,
c. Now here we add first 4 with 2 we get 6 to 6 we add 5
we will get 6 + 5 = 11, So we get sum as 11.

Question 7.
7 + 8 + 1 = _________
Answer:
7 + 8 +1 = 16

Explanation:
By using linking the cubes
a. First we add 7 + 1 = 8 and now we add 8 + 8 we get 16.
b. Here we add 7 + 8 = 15, now we add 15 + 1 = 16,
c. Now here we add first 8 with 1 we get 9 to 9 we add 7
we will get 9 + 7 = 16, So we get sum as 16.

Question 8.
8 + 6 + 2 = _________
Answer:
8 + 6 + 2 = 16

Explanation:
By using linking the cubes
a. First we add 8 + 2 = 10 and now we add 10 + 6 we get 16.
b. Here we add 8 + 6 = 14, now we add 14 + 2 = 16,
c. Now here we add first 6 with 2 we get 8 to 8 we add
8 we will get 8 + 8 = 16, So we get sum as 16.

Question 9.
4 + 9 + 4 = __________
Answer:
4 + 9 + 4 = 17

Explanation:
By using linking the cubes
a. First we add 4 + 4 = 8 and now we add 8 + 9 we get 17.
b. Here we add 4 + 9 = 13, now we add 13 + 4 = 17,
c. Now here we add first 9 with 4 we get 13 to 13 we add 4
we will get 13 + 4 = 17, So we get sum as 17.

Question 10.
DIG DEEPER
Complete the number puzzle so that each branch has a sum of 14.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 54
Answer:

Explanation:
Let us take the missing number as X, Y Z, P
We take 6 + 1 + ___X___ = 14 , means X = 7 + X = 14,
therefore X = 14 – 7 = 7, So 6 + 1 + 7 = 14,
Next  Y, 5 + Y + 6 = 14 , 11 + Y = 14 , Y = 14 -11,
We get Y = 3,Making 5 + 3 + 6 = 14
Now Z, 6 + 7 + Z = 14, 13 + Z = 14, Z = 14 – 13,
Z = 1 Making 6 + 7 + 1 = 14
Now P, 4 + P + 3 = 14, 7 + P = 14 ,P = 14 – 7,
We get P = 7 making  4 + 7 + 3 = 14.

Think and Grow: Modeling Real Life

You have 6 goldfish, 7 minnows, and 6 guppies. Will the tank hold all of your fish?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 55
Equation:
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 56

Show how you know:
Answer:

Explanation:
Given I have 6 goldfish, 7 minnows and 6 guppies, Total Number of
Fishes with me are  6 + 7 + 6 = 19 , No the tank can not hold
all of fishes as given in the figure a tank can hold 15 Fish and
I have 19 which are more than 15 ,
No, it is not possible for the tank to hold 19 Fish.

Show and Grow

Question 11.
You have 2 lovebirds, 4 canaries, and 3 finches. Will the cage hold all of your birds?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 57
Equation:
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 58

Show how you know:
Answer:

Explanation:
Given I have 2 lovebirds, 4 canaries, and 3 finches ,
Total Number of birds I have are 2 + 4 + 3 = 9,
Now as shown in the picture a cage can holds 10 birds,
We have total 9 birds which is less than 10 means,
Yes, the cage can hold all of my birds .

Add Three Numbers Practice 4.4

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 59
Answer:

Explanation:
By using linking the cubes
a. First we add 2 + 2 = 4 and now we add 6 + 4 we will get 10.
b. Here we add first 6 + 2 = 8, now we add 8 + 2 = 10,
c. Now here we add 6 with 2 we get 8 to 8 we add 2,
we will get  8 + 2 = 10, So we get sum as 10.

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 60
Answer:

Question 3.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 61
Answer:

Explanation:
By using linking the cubes
a. First we add 8 + 7 = 15 and now we add 15 + 4 we will get 19.
b. Here we add first 7 + 4 = 11, now we add 8 + 11 = 19,
c. Now here we add 8 with 4 we get 12 to 12 we add 7,
we will get  12 + 7 = 19, So we get sum as 19.

Question 4.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 62
Answer:

Explanation:
By using linking the cubes
a. First we add 9 + 1 = 10 and now we add 10 + 10 we will get 20.
b. Here we add first 9 + 10 = 19, now we add 19 + 1 = 20,
c. Now here we add 10 with 1 we get 11 to 11 we add 9,
we will get  11 + 9 = 20, So we get sum as 20.

Question 5.
3 + 4 + 5 = _________
Answer:

Explanation:
By using linking the cubes
a. First we add 3 + 5 = 8 and now we add 8 + 4 we will get 12.
b. Here we add first 3 + 4 = 7, now we add  7 and 5 as 7 + 5 = 12,
c. Now here we add 4 with 5 we get 9 to 9 we add 3,
we will get  3 + 9 = 12, So we get sum as 12.

Question 6.
7 + 5 + 5 = ________
Answer:

Explanation:
By using linking the cubes
a. First we add 7 + 5 = 12 and now we add 12 + 5 we will get 17.
b. Here we add first 7 + 5 = 12, now we add  12 and 5 as 12 + 5 = 17,
c. Now here we add 5 and 5 we get 10, we add 7 to 10,
we will get  7 + 10 = 17, So we get sum as 17.

Question 7.
8 + 4 + 8 = ________
Answer:

Explanation:
By using linking the cubes
a. First we add 8 + 8 = 16 and now we add 16 + 4 we will get 20.
b. Here we add first 8 + 4 = 12, now we add 12 and 8 as 12 + 8 = 20,
c. Now here we add 4 with 8 we get 12 ,we add 8 to 12,
we will get  8 + 12 = 20, So we get sum as 20.

Question 8.
DIG DEEPER!
Complete the number puzzle so that each branch has the sum of 13.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 63
Answer:

Explanation:
Let us take the missing number as X, Y Z, P
We take 5 + 1 + ___X___ = 13 , means X = 6 + X = 13,
therefore X = 13 – 6 = 7, So 5 + 1 + 7 = 13,
Next  Y, 1 + Y + 6 = 13 , 7 + Y = 13, Y = 13 – 7,
We get Y = 6,Making 1 + 6 + 6 = 13
Now Z, 4 + 5 + Z = 13, 9 + Z = 13, Z = 13 – 9,
Z = 4 Making 4 + 5 + 4 = 13
Now P, 7 + P + 2 = 13, 9 + P = 13 ,P = 13 – 9,
We get P = 4 making  7 + 4 + 2 = 13.

Question 9.
Modeling Real Life
You have 7 white chickens, I black chicken, and 8 brown chickens. Will the chicken coop hold all of your chickens?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 64
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 65

Show how you know:
Answer:

Explanation:
Given I have 7 white chickens, 1 black chicken and 8 brown chickens,
Total Number of  chickens with me are  7 + 1 + 8 = 16 ,
Given in the figure a chicken coop can hold 15 chickens and
I have 16 which is more than 15.
No, it is not possible for the chicken coop to hold all 16 chickens.

Review & Refresh

Question 10.
5 + __5__ = 10
Answer:
5 + 5 = 10

Explanation:
Let us take the missing number as X,
It is given in the equation that 5 + X = 10,
So X = 10 – 5 = 5, means 5 + 5 = 10.

Question 11.
10 = 6 + ___4___
Answer:
10 = 6 + 4

Explanation:
Let us take the missing number as X,
It is given in the equation that 10 = 6 + X,
So X = 10 – 6 = 4, means 10 = 6 + 4 .

Question 12.
8 + __2___ = 10
Answer:
8 + 2 = 10

Explanation:
Let us take the missing number as X,
It is given in the equation that 8 + X = 10,
So X = 10 – 8 = 2, means 8 + 2 = 10.

Question 13.
10 = 3 + __7___
Answer:
10 = 3 + 7

Explanation:
Let us take the missing number as X,
It is given in the equation that 10 = 3 + 7,
So X = 10 – 3 = 7, means 10 = 3 + 7.

Lesson 4.5 Add Three Numbers by Making a 10

Explore and Grow

Show three ways to find a sum.
4 + 3 + 7 = ________

4 + 3 + 7 = ________

4 + 3 + 7 = ________
Answer:
4 + 3 + 7 = 14

Explanation:
By using the three ways we find sum,
Way 1. First we add 4 + 7 = 11 and now we add 11 + 3 we will get 14.
Way 2. Here we add first 4 + 3 = 7, now we add 7 and 7 as 7 + 7 = 14,
Way 3. Now here we add 3 with 7 we get 10 to 10 we add 4,
we will get  4 + 10 = 14, So we get sum as 14.

Show and Grow

Make a 10 to add.

Question 1.

Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 66

Answer:

Add 9,1 first to make 10 then add 3, sum = 9 + 1 + 3 = 13

Explanation:
To make 10 first we see adding 9 + 1 = 10 is equal to 10,
next we add 1 +3 =4 not equal to 10, 9 + 3 = 12 which
is not equal to 10 ,So to make 10 we add  9,1 first, then add 3,
So the sum is 9 + 1 + 3 = 13.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 67
Answer:

Add 2,8 first to make 10 then add 4, sum = 4 + 2 + 8 = 14

Explanation:
To make 10 first we see adding 4 + 2 = 6,
6 is not equal to 10, 2 + 8 = 10 which is equal to 10 ,
and now we add 4 + 8 = 12 which is not equal to 10,
So to make 10 we add 2,8 first then add 4
So the sum is 4 + 2 + 8 = 14.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 68
Answer:

Add 3,7 first to make 10 then add 2, sum = 2 + 3 + 7 = 12

Explanation:
To make 10 first we see adding 2 + 3 = 5,
5 is not equal to 10, 3 + 7 = 10 which is equal to 10 ,
and now we add 2 + 7 = 9 which is not equal to 10,
So to make 10 we add 3,7 first and then add 2
So the sum is 2 + 3 + 7 = 12.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 69
Answer:

Add 5,5 first to make 10 then add 9, sum = 5 + 5 + 9 = 19

Explanation:
To make 10 first we see adding 5 + 5 = 10,
so is equal to 10, now we add 5 + 9 = 14  which is not equal to 10 ,
and now we add other 5 + 9 = 14 which is also not equal to 10,
So to make 10 we add 5,5 first and then add 9,
So the sum is 5 + 5 + 9 = 19.

Apply and Grow: Practice

Make a 10 add.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 70
Answer:

Add 3,7 or 7,3 first to make 10 then add 3, sum = 3 + 7 + 3 = 13

Explanation:
To make 10 first we see adding 3 + 7 = 10,
so is equal to 10, now we add 7 + 3 = 10  which is also equal to 10 ,
and now we add other 3 + 3 = 6 which is not equal to 10,
So to make 10 we add 3,7 or 7,3 first and then add 3
So the sum is 3 + 7 + 3 = 13.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 71
Answer:

Add 8,2 first to make 10 then add 6, sum = 8 + 6 + 2 = 16

Explanation:
To make 10 first we see adding 8 + 6 = 14,
14 is not equal to 10, now we add 6 + 2 = 8  which is not equal to 10 ,
and now we add 8 + 2 = 10 which is equal to 10,
So to make 10 we add 8,2 first and then add 6
So the sum is 8 + 6 + 2 = 16.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 72
Answer:

Add 6,4 first to make 10 then add 7, sum = 6 + 7 + 4 = 17

Explanation:
To make 10 first we see adding 6 + 7 = 13,
It is not equal to 10, now we add 7 + 4 = 11  which is not equal to 10 ,
and now we add 6 + 4 = 10 which is equal to 10,
So to make 10 we add 6,4 first then add 7
So the sum is 6 + 7 + 4 = 17.

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 73
Answer:

Add 5,5 first to make 10 then add 8, sum = 8 + 5 + 5 = 18

Explanation:
To make 10 first we see adding 8 + 5 = 13,
It is not equal to 10, now we add 5 + 5 = 10  which equal to 10 ,
and now we add 8 + 5 = 13 which is not equal to 10,
So to make 10 we add 5,5 first then add 8
So the sum is 8 + 5 + 5 = 18.

Question 9.
4 + 6 + 2 = ________
Answer:

Add 4,6 first to make 10 then add 2, sum = 4 + 6 + 2 = 12

Explanation:
To make 10 first we see adding 4 + 6 = 10,
which is equal to 10, now we add 6 + 2 = 8  which is not equal to 10 ,
and now we add 4 + 2 = 6 which is not equal to 10,
So to make 10 we add 4,6 first then add 2
So the sum is 4 + 6 + 2 = 12.

Question 10.
5 + 9 + 1 = ________
Answer:

Add 9,1 first to make 10 then add 5, sum = 5 + 9 + 1 = 15

Explanation:
To make 10 first we see adding 5 + 9 = 14,
which is not equal to 10, now we add 9 + 1 = 10  which is equal to 10 ,
and now we add 5 + 1 = 6 which is not equal to 10,
So to make 10 we add 9,1 first then add 5,
So the sum is 5 + 9 + 1 = 15.

Question 11.
DIG DEEPER!
What do you know about the missing addends and the sum?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 74
Answer:
We know the sum of the missing addends is 10 and the total sum is 10 + 3 = 13.

Explanation:
From the picture it is showing that if we add the missing
addends we get sum as 10 and the total sum is 10 + 3 = 13.
So whenever we are adding two missing addends we should always check,
their sum should be always 10, Example if one number is 1
other numbers will be 10-1 = 9. So the missing numbers are (1,9).
It can be even (2,8), (3,7),(4,6),(5,5),(6,4),(7,3),(8,2) or (9,1).

Think and Grow: Modeling Real Life

You need to knock down 20 pins to win. Do you win?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 75
Equation:
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 76
Show how you know:
Answer:

6 + 3 + 7 = 16, No we will not win.

Explanation:
Given pins as 6,3,7 if we add all we get 6 + 3 + 7 = 16,
we need 20 pins to knock down to win and we got 16 which are less than 20,
So no we will not win.

Show and Grow

Question 12.
Your hockey team needs 12 goals to break a record. Does your team break the record?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 77
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 78
Equation:
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 79
Show how you know:
Answer:

5 + 6 + 4 = 15, Yes my hockey team will break the record.

Explanation:
Given the goals made are 5,6,4 so the total number of goals
made are 5 + 6 + 4 = 15, to break hockey record
we need 12 goals as my team made 15 goals which is more than 12,
Yes my hockey team will break the record.

Add Three Numbers by Making a 10 Practice 4.5

Make a 10 to add.

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 80
Answer:

Add 6,4 first to make 10 then add 8, sum = 6 + 8 + 4 = 18

Explanation:
To make 10 first we see adding 6 + 8 = 14,
which is not equal to 10, now we add 8 + 4 = 12
which is not equal to 10 ,
and now we add 6 + 4 = 10 which is equal to 10,
So to make 10 we add 6,4 first then add 8,
So the sum is 6 + 8 + 4 = 18.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 81
Answer:

Add 3,7 first to make 10 then add 4, sum = 3 + 7 + 4 = 14

Explanation:
To make 10 first we see adding 3 + 7 = 10,
which is equal to 10, now we add 7 + 4 = 11
which is not equal to 10 ,
and now we add 3 + 4 = 7 which is not equal to 10,
So to make 10 we add 3,7 first then add 4,
So the sum is 3 + 7 + 4 = 14.

Question 3.
5 + 8 + 2 = _______
Answer:

Add 8,2 first to make 10 then add 5, sum = 5 + 8 + 2 = 15

Explanation:
To make 10 first we see adding 5 + 8 = 13,
which is not equal to 10, now we add 8 + 2 = 10
which is equal to 10 ,
and now we add 5 + 2 = 7 which is not equal to 10,
So to make 10 we add 8,2 first then add 5,
So the sum is 5 + 8 + 2 = 15.

Question 4.
1 + 9 + 7 = _________
Answer:

Add 1,9 first to make 10 then add 7, sum = 1 + 9 + 7 = 17

Explanation:
To make 10 first we see adding 1 + 9 = 10,
which is equal to 10, now we add 9 + 7 = 16
which is not equal to 10 ,
and now we add 1 + 7 = 8 which is not equal to 10,
So to make 10 we add 1,9 first then add 7,
So the sum is 1 + 9 + 7 = 17.

Question 5.
DIG DEEPER!
What do you know about the missing addends and the sum?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 82
Answer:
we know the sum of the missing addends is 10 and the total sum is 10 + 2 = 12.

Explanation:
From the picture it is showing that if we add the missing
addends we get sum as 10 and the total sum is 10 + 2 = 12.
So whenever we are adding two missing addends we should always check,
their sum should be always 10, Example if one number is 2
other number will be 10-2 = 8. So the missing addends here are (2,8).
It can be (1,9),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2) or (9,1).

Question 6.
Modeling Real Life
Your baseball team needs 15 runs to break a record. Does your team break the record?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 83
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 84
Equation:
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 85
Show how you know:
Answer:

6 + 4 + 6 = 16, Yes my team will break the record.

Explanation:
Given my team makes 6,4,6 runs so total runs made are
6 + 4 + 6 = 16, to break the record we require 15 runs
but the team made 16 as 16 is more than 15,
Yes, my team will break the record.

Review & Refresh

Circle the heavier object.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 86
Answer:

Explanation:
By looking at the pictures book is heavier than a pencil.
So the book is circled.

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 87
Answer:

Explanation:
By looking at the pictures apple is heavier than the feather,
So apple is circled.

Lesson 4.6 Add 9

Explore and Grow

Use counters and the ten frames to find the sum. Show how you can make a lo to solve.

Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 88
Answer:
9 + 1 + 4 = 14

Explanation:
First we make sum 10 , For 9 to make 10 we need 10-9=1,
as we have add 1 to 9 now we will minus 1 from 5 we get 5 – 1 = 4,
Now we write the total sum as 9 + 1 + 4 = 14 .

Show and Grow

Make a 10 to add.

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 89
Answer:

Sum is 9 + 1 + 2 = 12  or  9 + 3 = 12.

Explanation:
First we make sum 10 , For 9 to make 10 we need 10-9=1,
as we have to  add 1 to 9 now  we will minus 1 from 3 we get 3 – 1 = 2,
Now we write the total sum as 9 + 1 + 2 = 12  or 9 + 3 = 12.

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 90
Answer:

Sum is 9 + 1 + 5 = 15 or 9 + 6 = 15.
Explanation:
First we make sum 10 , For 9 to make 10 we need 10-9=1,
as we have to  add 1 to 9 now  we will minus 1 from 6 we get 6 – 1 = 5,
Now we write the total sum as 9 + 1 + 5 = 15 or 9 + 6 = 15.

Apply and Grow: Practice

Make a 10 to add.

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 91
Answer:

Sum is 9 + 1 + 1 = 11  or 9 + 2 = 11.

Explanation:
First we make sum 10 , For 9 to make 10 we need 10-9=1,
as we have to  add 1 to 9 now  we will minus 1 from 2 we get 2 – 1 = 1,
Now we write the total sum as 9 + 1 + 1 = 11  or 9 + 2 = 11.

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 92
Answer:

Sum is 9 + 1 + 8 = 18  or 9 + 9 = 18.

Explanation:
First we make sum 10 , For 9 to make 10 we need 10-9=1,
as we have to add 1 to 9 now  we will minus 1 from 9 we get 9 – 1 = 8,
Now we write the total sum as 9 + 1 + 8 = 18  or 9 + 9 = 18.

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 93
Answer:

Sum is 9 + 1 + 4 = 14 or 9 + 5 = 14.

Explanation:
First we make sum 10 , For 9 to make 10 we need 10-9=1,
as we have to add 1 to 9 now  we will minus 1 from 5 we get 5 – 1 = 4,
Now we write the total sum as 9 + 1 + 4 = 14 or 9 + 5 = 14.

Question 6.
DIG DEEPER!
Use the ten frame to complete the equations.
9 + ? = 17
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 94
10 + _______ = 17
So, 9 + ________ = 17
Answer:
10 + 7 = 17,
So  9 + 1 + 7 = 17

Explanation:
Given 10 + ____ = 17, to find the missing number we take it as X,
10 + X = 17 , X = 17 – 10 =7. So the equation is 10 + 7 = 17.
Now we write 10 as 9 + 1 by using ten frame and
add 7 so the total sum is 9 + 1 + 7 = 17 or 10 + 7 = 17.

Think and Grow: Modeling Real Life

You have 9 stickers and earn 5 more. Your friend has 6 stickers and earns 10 more. Do you and your friend have the same number of stickers?
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 95
Addition equations:
you                Friend
9 + 5 = 14          6 + 10 = 16

Yes       No

Show how you know:
Answer:
My equation is 9 + 5 = 14 and friend has 6 + 10 =16
as 14 is not equal to 16, so I and my friend don’t
have the same number of stickers.

Explanation:
Given I have 9 stickers and earn 5 more means in total I have 9 + 5 = 14 stickers and friend has 6 stickers and earns 10 more means in total friend have 6 + 10 = 16 stickers,
Now comparing between me and my friend as 14 and 16 are not same, So No, I and my friend do not have the same number of stickers.

Show and Grow

Question 7.
Your friend has 9 magnets and finds 6 more. You have 5 magnets and find 10 more. Do you and your friend have the same number of magnets?
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 96
Addition equations:
 you             Friend
5 + 10 =15      9 + 6 = 15

Yes       No

Show how you know:

Answer:
My equation is 5 + 10 = 15 and friend equation is 9 + 6 = 15,
as both are equal, So Yes, I and my friend have the same number of magnets.

Explanation:
Given I have 5 magnets and find 10 more means in total, I have 5 + 10 = 15 magnets and friend has 9 magnets and finds 6 more means in total friend have 9 + 6 = 15 magnets, Now comparing between me and my friend as 15 and 15 are same, So Yes, I and my friend have the same number of magnets.

Add 9 Practice 4.6

Make a 10 to add

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 97
Answer:

Sum is 9 + 1 + 2 = 12 or 9 + 3 = 12.

Explanation:
First we make sum 10, For 9 to make 10 we need 10-9=1,
as we have to add 1 to 9 now we will minus 1 from 3 we get 3 – 1 = 2,
Now we write the total sum as 9 + 1 + 2 = 12 or 9 + 3 = 12.

Question 2.
DIG DEEPER!
Use the ten frame to complete the equations.
16 = ? + ?
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 98
10 + ___6_____ = ___16____
So, 16 = __9 +__1____ + __6___.
Answer:
We write the equation as 16 = 10 + 6
So 16 =  9 + 1 + 6 or 16 = 10 + 6

Explanation:
Given 10 + ____ = 16, to find the missing number we take it as X,
10 + X = 16 , X = 16 – 10 =6. So the equation is 10 + 6 = 16.
Now we write 10 as 9 + 1 by using the ten frame
and add 6 so the total sum is 9 + 1 + 6 = 16 or 10 + 6 = 16.

Question 3.
Modeling Real Life
You have 9 rocks and collect 7 more. Your friend has 8 rocks and collects 8 more. Do you and your friend have the same number of rocks?
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 99
Yes             No
Show how you know:
Answer:
I have 9 + 7 = 16 rocks,
Friend have 8 + 8 = 16 rocks,
Yes, I and my friend have the same number of rocks.

Explanation:
Given I have 9 rocks and collected 7 more,
So I have total 9 + 7 = 16 rocks and my friend have 8 rocks and collects 8 rocks total he have 16 rocks,
Now comparing both me and my friend both have the equal number of rocks, Yes, I and my friend have the same number of rocks.

Review & Refresh

Circle the longer object.

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 100
Answer:

Explanation:
By seeing the picture we can identify yellow spectacles are longer than the red spectacles, So circled the yellow spectacles.

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 101
Answer:

Explanation:
By seeing the picture we can identify tooth paste is longer than the toothbrush, So circled the toothpaste.

Lesson 4.7 Make a 10 to Add

Explore and Grow

Use counters and the ten frame to find the sum.
Show how you can make a lo to solve.

Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 102
Answer:

Explanation:
First we make sum 10, For 8 to make 10 we need 10-8=2,
as we have to add 2 to 8 now we will minus 2 from 6 we get 6 – 2 = 4,
Now we write the total sum as 8 + 2 + 4 = 14.

Show and Grow

Make a 10 to add.

Question 1.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 103
Answer:

Sum as 8 + 2 + 3 = 13 or 8 + 5 = 13.

Explanation:
First we make sum 10, For 8 to make 10 we need 10-8=2,
as we have to add 2 to 8 now we will minus 2 from 5 we get 5 – 2 = 3,
Now we write the total sum as 8 + 2 + 3 = 13 or 8 + 5 = 13.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 104
Answer:

Sum as 7 + 3 + 4  = 14 or 7 + 7 = 14.

Explanation:
First, we make sum 10, For 7 to make 10 we need 10-7=3,
as we have to add 3 to 7 now we will minus 3 from 7 we get 7 – 3 = 4,
Now we write the total sum as 7 + 3 + 4  = 14 or 7 + 7 = 14.

Apply and Grow: Practice

Make a 10 to add.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 105
Answer:

Sum as 7 + 3 + 3  = 13 or 7 + 6 = 13.

Explanation:
First we make sum 10, For 7 to make 10 we need 10-7=3,
as we have to add 3 to 7 now we will minus 3 from 6 we get 6 – 3 = 3,
Now we write the total sum as 7 + 3 + 3  = 13 or 7 + 6 = 13.

Question 4.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 106
Answer:

Sum as 7 + 3 + 1  = 11 or 7 + 4 = 11.

Explanation:
First we make sum 10, For 7 to make 10 we need 10-7=3,
as we have to add 3 to 7 now we will minus 3 from 4 we get 4 – 3 = 1,
Now we write the total sum as 7 + 3 + 1  = 11 or 7 + 4 = 11.

Question 5.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 107
Answer:

Sum as 8 + 2 + 5 = 15 or 8 + 7 = 15.

Explanation:
First we make sum 10, For 8 to make 10 we need 10-8=2,
as we have to add 2 to 8 now we will minus 2 from 7 we get 7 – 2 = 5,
Now we write the total sum as 8 + 2 + 5 = 15 or 8 + 7 = 15.

Question 6.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 108
Answer:

Sum as 6 + 4 + 2 = 12 or 6 + 6 = 12.
Explanation:
First we make sum 10, For 8 to make 10 we need 10-6=4,
as we have to add 4 to 6 now we will minus 4 from 6 we get 6 – 4 = 2,
Now we write the total sum as 6 + 4 + 2 = 12 or 6 + 6 = 12.

Question 7.
MP Number Sense
Use 4, 7, and 10 to complete the sentence.
7 + ___7_____ has the same sum as ___4___ + __10_____.
Answer:
7 + 7  has the same sum as  4 + 10

Explanation:
Given to use numbers 4,7,10 to make the sentence
7 +  _____ = ________ + _______ both sides sum is same
As left side 7 is given we substitute and check first we use 4, then 7 and 10
7 +  4 = 11 now we have numbers 7 and 10, sum is 7+10 = 17
So (7 + 4)= 11  ≠ (7 + 10) =17, next we take 7 and check 7 + 7 = 14
now we have numbers 4 , 10 and sum is 4 + 10 = 14
So (7 + 7) = 14 = (4 + 10) = 14 here if we take for 7 left side and 4,10 right
we get same sum both side, now we  even check taking 10 first,
7 + 10 = 17 in left side and we have numbers 7 and 4 for right as 7 + 4 =11,
So (7 + 10)= 17 ≠ (7 + 4) =11, So to make both the same sum
we take left side number as 7 and right side numbers 4, 10
making the sentence as (7 + 7) = (4 + 10) = 14.

Think and Grow: Modeling Real Life

There are 8 crabs. 7 more join them. There are 10 turtles. 5 more join them.
Is the number of crabs the same as the number of turtles?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 109
Addition equations:
Yes              No
Show how you know:
Answer:
Addition equations:
Crabs                 Turtles
8 + 7 = 15          10 + 5 =15
Yes, the number of crabs are the same as the number of turtles.

Explanation:
Given 8 crabs and 7 more joined them so the number of crabs are 8 + 7 = 15 crabs and there are 10 turtles and 5 more joined them
So the total number of turtles are 10 + 5 = 15 turtles, as both are 15,
Yes, the number of crabs are the same as the number of turtles.

Show and Grow

Question 8.
There are 6 squirrels. 6 more join them. There are 10 birds. 2 more join them. Is the number of squirrels the same as the number of birds?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 110
Addition equations:
Yes               No
Show how you know:
Answer:
Squirrels          Birds
6 + 6 = 12    10 + 2 = 12
Yes, the number of squirrels are the same as the number of birds.

Explanation:
Given 6 squirrels and 6 more joined them so total squirrels are 6 + 6 = 12 and 10 birds, 2 more birds joined making total as 10 + 2 = 12 birds as both are 12,
Yes, the number of squirrels are the same as the number of birds.

Make a 10 to Add Practice 4.7

Make a 10 to add.

Question 1.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 111
Answer:

Sum as 6 + 4 + 1  = 11 or 6 + 5 = 11.

Explanation:
First we make sum 10, For 6 to make 10 we need 10-6=4,
as we have to add 4 to 6 now we will minus 4 from 5 we get 5 – 4 = 1,
Now we write the total sum as 6 + 4 + 1  = 11 or 6 + 5 = 11.

Question 2.
MP Number Sense
Use 2, 6, and 10 to complete the sentence.
________ + ________ has the same sum as ______ + 6.
Answer:

2 + 10   has the same sum as  6 + 6

Explanation:
Given to use numbers 2,6,10 to make the sentence
_______ +  _____ = ________ +  6 both sides sum is same
As right side 6 is given we substitute and check first we use 2, then 6 and 10,
2 +  6 = 8 now we have numbers 6 and 10, sum is 6 + 10 = 16
So (6 + 10) = 16  ≠ (2 + 6) = 8, next we take 6 and check 6 + 6 = 12
now we have numbers 2 , 10 and sum is 2 + 10 = 12
So (2 + 10) = 12 = ( 6 + 6) = 12 here if we take for 6 right side and 2,10 left
we get same sum both side, now we  even check taking 10 first,
10 + 6 = 16 in right side and we have numbers 2 and 6 for left as 2 + 6 =8,
So (2 + 6)= 8 ≠ (10 + 6) =16, So to make both the same sum
we take left side numbers as 2,10 and right side number 6
making the sentence as (2 + 10) = (6 + 6) = 12.

Question 3.
Modeling Real Life
There are 7 monkeys. 4 more join them. There are 6 birds. 5 more join them. Is the number of monkeys the same as the number of birds?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 112
Yes            No
Show how you know:
Answer:
Equations are:   Monkeys       Birds
7 + 4 = 11     6 + 5 = 11
Yes, the number of monkeys are the same as the number of birds.

Explanation:
Given 7 monkeys and 4 more joined them so total
monkeys are 7 + 4 = 11 and 6 birds, 5 more birds joined
making total as 6 + 5 = 11 birds as both are 11,
Yes, the number of monkeys are the same as the number of birds.

Review & Refresh

Question 4.
There are 3 Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 113
2 more Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 113 join them.
How many Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 113 are there now?
____3____ + ___2____ = ____5____ Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 113
Answer:
3 + 2 = 5, there are 5 aero planes now.

Explanation:
Given there are 3 aero planes and 2 more join them making total
as 3 + 2 = 5 , therefore there are 5 aero planes now.

Question 5.
There are 4 Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 114
6 more Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 114 join them.
How many Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 114 are there now?
___4_____ + ___6_____ = ___10_____ Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 114
Answer:
4 + 6 = 10, there are 10 airplanes now.

Explanation:
Given there are 4 aero planes and 6 more join them making total as  4 + 6 = 10, therefore there are 10 aero planes now.

Lesson 4.8 Problem Solving: Addition within 20

Explore and Grow

Newton has 9 red crayons, 3 blue crayons, and 1 yellow crayon. How many crayons does he have in all?

________ crayons
Answer:
Newton has 9 + 3 + 1 = 13 crayons in all.

Explanation:
Given Newton has 9 red crayons, 3 blue crayons and 1 yellow crayon,
So the total number of crayons Newton has are 9 + 3 + 1 = 13 crayons,
therefore Newton has 13 crayons in all.

Show and Grow

Question 1.
You have 6 notebooks. You buy 5 more. How many notebooks do you have now?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 115
Circle what you know.
Underline what you need to find.
Solve:

_______ ○ _______ = _______                           _______ notebooks
Answer:
notebooks
I have 11 notebooks in all.

Explanation:
Given I have 6 note books and buy 5 more , So I know
I have 6 & 5 books, So I circled what I know, I need to find
how many total notebooks I do have now, so I underlined it,
making 6 + 5 = 11 notebooks. So I have  total 11 notebooks now.

Apply and Grow: Practice

Question 2.
There are 6 soccer balls, 10 basketballs, and 4 volleyballs. How many balls are there in all?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 116
Circle what you know.
Underline what you need to find.
Solve:
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 117
Answer:
balls
I have 20 balls in all.

Explanation:
Given there are  6 soccer balls, 10 basketballs, and 4 volleyballs, So I know 6 , 10 & 4 balls, So I circled what I know, I need to find
how many total balls there in all so I underlined it, making 6 + 10 + 4 = 20 balls, therefore there are total 20 balls in all.

Question 3.
You do 8 push-ups. Your friend does I fewer than you. How many push-ups do you and your friend do in all?

____15_____ push-ups
Answer:
In total, I and my friend does 15 push-ups in all.

Explanation:
Given I do 8 push-ups and friend does 1 fewer than me
means 8-1 = 7 push-ups now total number of push-ups are
8 + 7 = 15, So in total I and my friend does 15 push-ups in all.

Question 4.
YOU BE THE TEACHER
Newton has 9 magnets. Descartes has 8 more than Newton. Your friend uses a bar model to show how many magnets Descartes has. Is your friend correct? Show how you know.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 118Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 118
Answer:
No, your friend is not correct as in bar model it is showing Descartes has 8 magnets.

Descartes has 9 + 8 =  17 magnets.
Explanation :
Given Newton has 9 magnets and Descartes has 8 more than Newton means Descartes has 9 + 8 = 17 magnets. But the bar model is showing Descartes has only 8 magnets, I have shown in the figure Descartes have 9 + 8 = 17 Magnets.

Think and Grow: Modeling Real Life

You have 5 bracelets. You have 7 fewer than your friend. How many bracelets does your friend have?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 119
Circle what you know.
Underline what you need to find.
Solve:
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 120

________ + ________ = _________                                       ____12____ bracelets
Answer:

Friend have 12 bracelets.

Explanation:
Given I have 5 bracelets, I have 7 fewer than my friend means
friend have in total as 5 + 7 = 12 bracelets, So circled what I know
5, 7 and underlined the number how many bracelets does friend have,
So friend have 12 bracelets.

Show and Grow

Question 5.
Your friend finds 9 seashells. You find 6 more than your friend. How many seashells do you find?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 121
Circle what you know.
Underline what you need to find.
Solve:

____6_____ + ___9____ = __15_____                                          ___15_____ seashells
Answer:

I found 15 seashells.

Explanation:
Given friend finds 9 seashells and I found 6 more than my friend,
So the total number of shells I found are 9 + 6 = 15 seashells.
I Know friend has 9 and I have more 6 seashells,
So I circled it and underlined to find how many seashells I found.

Problem Solving: Addition within 20 Practice 4.8

Question 1.
You have 5 robots. Your friend gives you more. Now you have 14. How many robots did your friend give you?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 122
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 123
Answer:

Friend has 9 robots.

Explanation :
Given I have 5 robots and now I have 14 robots so I circled it,
I need to find how many friend gives I have underlined it.
Let us take friend gives X robots and we write equation as
5 + X = 14, therefore X = 14 – 5 = 9 means friend gave 9 robots.

Question 2.
There are some plates on a table. You add 8 more. Now there are 12. How many plates were on the table to start?

______4____ plates
Answer:
There were 4 plates on the table to start.

Explanation:
Let us take X as some plates on the table, I added 8 more and
there are 12 now means X + 8 = 12, Therefore X = 12 – 8 = 4,
So 4 plates are there on the table to start.

Question 3.
YOU BE THE TEACHER
Newton has 8 tickets. Descartes has 5 more than Newton. Your friend uses a number line to show how many tickets Descartes has. Is your friend correct? Show how you know.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 124
Answer:
Descartes has 13 Tickets,
Yes, Friend is correct.

Explanation:
Given Newton has 8 tickets and Descartes has 5 more than Newton means 8 + 5 = 13 tickets, Descartes has 13  tickets, Now my friend is showing this on number line he is starting from 8 because Newton has 8 and moving 5 units to right as Descartes has 5 more tickets till 13
which is the same as we did sum 8 +5 = 13, So yes, friend is correct.

Question 4.
Modeling Real Life You have 9 medals. You have 9 fewer than your friend. How many medals does your friend have?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 125
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 126

____18_______ medals
Answer:

Friend have 18 medals.

Explanation:
Given I have 9 medals and 9 fewer than friend, So friend have
9 + 9 what I have, So friend have 18 medals.

Review & Refresh

Question 5.
4 – 3 = ________
Answer:
4- 3 = 1

Explanation:
Given the equation as 4 – 3 = _____ means subtract 3 from 4 we get 1
making equation as 4- 3 = 1.

Question 6.
5 – 2 = ________
Answer:
5 – 2 = 3

Explanation:
Given the equation as 5 – 2 = _____ means subtract 2 from 5 we get 3
making equation as 5 – 2 = 3.

Question 7.
9 – 9 = _________
Answer:
9 – 9 = 0

Explanation:
Given the equation as 9 – 9 = _____ means subtract 9 from 9 we get 0
making equation as 9 – 9 = 0.

Question 8.
7 – 0 = _________
Answer:
7 – 0 = 7

Explanation:
Given the equation as 7 – 0 = _____ means subtract 0 from 7 we get 7
making equation as 7- 0 = 7.

Add Numbers within 20 Performance Task

Question 1.
You track the weather for a few weeks. Each week has 7 days.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 127

a. You track the weather every day for the first week. But you miss 1 day in the second week. How many days do you track the weather?

___13_______ days
Answer:
I have track of 13 days weather.

Explanation:
Given each week has 7 days, I have the track for every day for the first week,
Means the first week = 7 days, Now in the second week I missed 1 day means
7 – 1 = 6 days track in the second week, So the total number of days are
first week track + second week track = 7 + 6 = 13 , total I have track of 13 days weather.

b. You track the weather for 1 more week. How many days in all do you track the weather?

____20______ days
Answer:
I have track of 20 days weather in all.

Explanation :
Given that I have tracked for 1 more week, We got 13 days before
and now we have 1 more week means 7 days to 13 making total as
13 +  7 = 20 days, So now I have track of 20 days weather in all.

Question 2.
Your friend also tracks the weather. She records 9 sunny days and 5 cloudy days.
How many days does your friend track the weather?

____14______ days
Answer:
My friend has 14 days track record of the weather.

Explanation:
Given my friend records 9 sunny days and 5 cloudy days
track of the weather so in all my friend has track of
9 + 5 = 14 days. Therefore my friend has 14 days track record of the weather.

Question 3.
You record 10 rainy days in the first three weeks. Is the number of rainy days the same as the number of sunny days?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 128
Yes                                No
Show how you know:
Answer:
No, Number of rainy days are not the same as the number of sunny days.

Explanation:
Given I record 10 rainy days in first three weeks and number of
sunny days for three weeks are 4 + 3 + 4 = 11 days , now comparing
with rainy days as rainy days are less to number of sunny days we say
No, number of rainy days are not the same as the number of sunny days.

Add Numbers within 20 Chapter Practice

Add Doubles from 6 to 10 Homework & Practice 4.1

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 129
____10______ + ____10____ = ___20______
Answer:
10 + 10 = 20 dots

Explanation:
Given in block 1 we have 10 dots and in block 2 we have 10 dots,
So in total we have 10 + 10 = 20 dots.

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 130
Answer:

9 + 9 = 18 dots

Explanation:
Given in block 1 we have 9 dots and in block 2 we have 9 dots,
So in total we have 9 + 9 = 18 dots.

Use Doubles within 20 Homework & Practice 4.2

Use the doubles 6 + 6 to find each sum.

Question 3.
6 + 7 = ___13_____
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 131
Answer:
6 + 7  = 13

Explanation:
We Know the double  6 + 6 = 12 , The number 7 is one more than 6,
So we can write 7 as 6 + 1, The required sum is one more than double,
12 + 1 =13, Sum is 13, therefore 6 + 7 = 13, Double plus 1.

6 + 5 = _______
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 132
Answer:
6 + 5 = 11

Explanation:
We Know the double of 6 + 6 = 12 , The number 5 is one less than 6,
So we can write 5 as 6 – 1, The required sum is one less than double,
12 – 1 = 11, Sum is 11, therefore 6 + 5 = 11, Double minus 1.

Find the sum. Write the double you used.

Question 4.
7 + 8 = _______

____7_____ ○ ____8____ = ___15____
Answer:

7 + 8 = 15, 7 + 7 double is used

Explanation:
As the first addend is 7 and we know the double of 7 + 7 = 14 ,
The number 8 is one more than 7, So we can write 8 as 7 + 1,
The required sum is one more than double,
14 + 1 =15, Sum is 15, therefore 7 + 8 = 15, we used the double 7 +7.

Question 5.
10 + 9 = ___19______

_________ ○ ________ = _______
Answer:
10 + 9 = 19, we use double 10 + 10,

Explanation:
As the first addend is 10 and we know the double is 10 + 10,
10 + 10 = 20 , The number 9 is one less than 10,
So we can write 9 as 10 – 1, The required sum is one less than double,
20 – 1 = 19, Sum is 19, therefore 10 + 9 = 17, Double minus 1.
We have use double 10 + 10 .

Count on to Add within 20 Homework & Practice 4.3

Question 6.
11 + 6 = ___17____
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 133
Answer:

11 + 6 = 17

Explanation:
First we start from 11 and move 6 units we get 17 on the number line,
So sum is 11 +6 = 17 .

Add Three Numbers Homework & Practice 4.4

Question 7.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 134
Answer:

6 + 4 + 4 = 14

Explanation:
By using linking the cubes
a. First we add 4 + 4 = 8 and now we add 6 + 8 we get 14.
b. Here we add 6 + 4 = 10, now we add 10 + 4 =14,
c. Now here we add first 6 with 4 we get 10 to 10 we add 4
we will get 10 + 4 = 14, So we get sum as 14.

Question 8.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 135
Answer:

7 + 8 + 5 = 20

Explanation:
By using linking the cubes
a. First we add 7 + 8 = 15 and now we add 15+ 5 we get 20.
b. Here we add 8 + 5 = 13, now we add 13 + 7 =20,
c. Now here we add first 7 with 5 we get 12 to 12 we add 8
we will get 12 + 8 = 20, So we get sum as 20.

Question 9.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 136
Answer:

9 + 8 + 1 = 18

Explanation:
By using linking the cubes
a. First we add 9 + 8 = 17 and now we add 17 + 1 we get 18.
b. Here we add 8 + 1 =9, now we add 9 + 9 =18,
c. Now here we add first 9 with 1 we get 10 to 10 we add 8
we will get 10 + 8 = 18, So we get sum as 18.

Question 10.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 137
Answer:

2 + 3 + 5 = 10

Explanation:
By using linking the cubes
a. First we add 2 + 3 = 5 and now we add 5 + 5 we get 10.
b. Here we add 3 + 5 =8, now we add 8 + 2 =10,
c. Now here we add first 2 with 5 we get 7 to 7 we add 3
we will get 7 + 3 = 10, So we get sum as 10.

Add Three Numbers by Making a 10 Homework & Practice 4.5

Make a 10 to add.

Question 11.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 138
Answer:

Add 7,3 first to make 10 then add 9, sum = 7 + 9 + 3 = 19

Explanation:
To make 10 first we see adding 7 + 9 = 16 is not equal to 10,
next we add 9 + 3 = 12 not equal to 10, 7 + 3 = 10 which
is equal to 10 ,So to make 10 we add  7,3 first, then add 9,
So the sum is 7 + 9 + 3 = 19.

Question 12.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 139
Answer:

Add 9,1 first to make 10 then add 4, sum = 4 + 9 + 1 = 14

Explanation:
To make 10 first we see adding 4 + 9 = 13 is not equal to 10,
next we add 9 + 1 = 10 is equal to 10, 4 + 1 = 5 which
is not equal to 10 ,So to make 10 we add  9,1 first, then add 4,
So the sum is 4 + 9 + 1 = 14.

Question 13.
MP Number Sense
What do you know about the missing addends and the sum?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 140
Answer:
We know the sum of the missing addends is 10 and the total sum is 10 + 5 = 15.

Explanation:
From the picture it is showing that if we add the missing
addends we get sum as 10 and the total sum is 10 + 5 = 15.
So whenever we are adding two missing addends we should always check,
their sum should be always 10, Example if one number is 1
other number will be 10-1 = 9. So the missing numbers are (1,9).
It can be even (2,8), (3,7),(4,6),(5,5),(6,4),(7,3),(8,2) or (9,1).

Add 9 Homework & Practice 4.6

Question 14.
Make a 10 to add 9 + 4.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 141
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 142
Answer:

Sum as 9 + 1 + 3 = 13  or 9 + 4 = 13.

Explanation:
First we make sum 10 , For 9 to make 10 we need 10-9=1,
as we have to  add 1 to 9 now  we will minus 1 from 4 we get 4 – 1 = 3,
Now we write the total sum as 9 + 1 + 3 = 13  or 9 + 4 = 13.

Make a 10 to Add Homework & Practice 4.7

Question 15.
Make a 10 to add 8 + 5.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 143
Answer:

Sum as 8 + 2 + 3 = 13  or 8 + 5 = 13.

Explanation:
First we make sum 10 , For 8 to make 10 we need 10-8=2,
as we have to  add 2 to 8 now  we will minus 2 from 5 we get 5 – 2 = 3,
Now we write the total sum as 8 + 2 + 3 = 13  or 8 + 5 = 13.

Question 16.
7 + 7 = ?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 144
Answer:

Sum as 7 + 3 + 4  = 14 or 7 + 7 = 14.

Explanation:
First we make sum 10, For 7 to make 10 we need 10-7=3,
as we have to add 3 to 7 now we will minus 3 from 7 we get 7 – 3 = 4,
Now we write the total sum as 7 + 3 + 4  = 14 or 7 + 7 = 14.

Question 17.
9 + 8 = ?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 145
Answer:

Sum as 9 + 1 + 7  = 17 or 9 + 8 = 17.

Explanation:
First we make sum 10, For 9 to make 10 we need 10-9=1,
as we have to add 1 to 9 now we will minus 1 from 8 we get 8 – 1 = 7,
Now we write the total sum as 9 + 1 + 7  = 17 or 9 + 8 = 17.

Problem Solving: Addition within 20 Homework & Practice 4.8

Question 18.
Modeling Real Life
Your friend finds 7 insects. You find 9 more than your friend. How many insects do you find?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 146

_____16____ insects
Answer:
I found 16 insects.

Explanation:
Given a friend finds 7 insects, I find 9 more than my friend means
I have in total as 7 + 9 = 16 insects,
So I found 16 insects.

Conclusion:

Learn the Big Ideas Math Book 1st Grade Answer Key Chapter 4 Add Numbers within 20 provided and improve your math as well as problem solving skills. You can achieve greater heights and fall in love with Math with our Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20. Stay with us to get the solutions of all Big Ideas Math Grade 1 Chapters from 1 to 14.

Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles

Big Ideas Math Geometry Answers Chapter 6

Give your kid the right assistance he/she needs regarding the Big Ideas Math Geometry Chapter 6 Relationships within Triangles concepts. Practicing from the Big Ideas Math Geometry Answers Ch 6 Relationships within Triangles on a frequent basis help you to attempt the exam with confidence. You can avail these handy resources for BIM Book Geometry Chapter 6 Relationships within Triangles Solution Key and clear the exams with high scores. In fact, you can download the BIM Book Geometry Ch 6 Relationships within Triangles Answers for free of cost and prepare them offline too.

Big Ideas Math Book Geometry Answer Key Chapter 6 Relationships Within Triangles

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Relationships Within Triangles Maintaining Mathematical Proficiency

Write an equation of the line passing through point P that is perpendicular to the given line.

Question 1.
P(3, 1), y = \(\frac{1}{3}\)x – 5

Answer:
y = -3x + 10

Explanation:
The slope of the given line is \(\frac { 1 }{ 3 } \). Since the slope of the perpendicular line must be the negative reciprocal of the slope of the given line.
The slope of the perpendicular line = -3
Substitute the values in y = mx + c
1 = -3(3) + c
1 = -9 + c
c = 1 + 9
c = 10
use the slope intercept form of a linear equation again substitute m, c
y = -3x + 10

Question 2.
P(4, – 3), y = – x – 5

Answer:
y = \(\frac { 1 }{ 3 } \)x – \(\frac { 5 }{ 3 } \)

Explanation:
The slope of the given line is -3. Since the slope of the perpendicular line must be the negative reciprocal of the slope of the given line.
The slope of the perpendicular line = \(\frac { 1 }{ 3 } \)
Substitute the values in y = mx + c
-3 = –\(\frac { 1 }{ 3 } \)(4) + c
c = -3 + \(\frac { 4 }{ 3 } \) = \(\frac { -9 + 4 }{ 3 } \) = \(\frac { -5 }{ 3 } \)
use the slope intercept form of a linear equation again substitute m, c
y = \(\frac { 1 }{ 3 } \)x + \(\frac { -5 }{ 3 } \)
y = \(\frac { 1 }{ 3 } \)x – \(\frac { 5 }{ 3 } \)

Question 3.
P(- 1, – 2), y = – 4x + 13

Answer:
y = \(\frac { 1 }{ 4 } \)x – \(\frac { 7 }{ 4 } \)

Explanation:
The slope of the given line is -4. Since the slope of the perpendicular line must be the negative reciprocal of the slope of the given line.
The slope of the perpendicular line = \(\frac { 1 }{ 4 } \)
Substitute the values in y = mx + c
-2 = \(\frac { 1 }{ 4 } \)(-1) + c
c = -2 + \(\frac { 1 }{ 4 } \) = \(\frac { -8 + 1 }{ 4 } \) = \(\frac { -7 }{ 4 } \)
use the slope intercept form of a linear equation again substitute m, c
y = \(\frac { 1 }{ 4 } \)x + \(\frac { -7 }{ 4 } \)
y = \(\frac { 1 }{ 4 } \)x – \(\frac { 7 }{ 4 } \)

Write the sentence as an inequality.

Question 4.
A number w is at least – 3 and no more than 8.

Answer:
-3 ≤ w < 8

Explanation:
At least means ≥ and no more than means <
w ≥ -3 and w < 8
-3 ≤ w < 8

Question 5.
A number m is more than 0 and less than 11.

Answer:
0 < m < 11

Explanation:
more than means > and less than means <
m > 0 and m < 11
0 < m < 11

Question 6.
A number s is less than or equal to 5 or greater than 2.

Answer:
2 < s ≤ 5

Explanation:
less than or equal to means ≤ and greater than means >
s ≤ 5 or s > 2
2 < s ≤ 5

Question 7.
A number d is fewer than 12 or no less than – 7.

Answer:
-7 < d < 12

Explanation:
fewer than means < no less than means >
d < 12 or d > -7
-7 < d < 12

Question 8.
ABSTRACT REASONING
Is it possible for the solution of a compound inequality to be all real numbers? Explain your reasoning.

Answer:
Yes

Relationships Within Triangles Mathematical Practices

Monitoring Progress

Refer to the figures at the top of the page to describe each type of line, ray, or segment in a triangle.

Question 1.
perpendicular bisector

Answer:
The perpendicular bisector of a side of a triangle is a line perpendicular to the side and passes through its midpoint. Three perpendicular bisectors of sides meet at a point in a triangle.

Question 2.
angle bisector

Answer:
The angle bisector of an angle in a triangle is a straight line that divides the angle into two congruent angles.

Question 3.
median

Answer:
It is a line segment from a vertex to the midpoint of the opposite side.

Question 4.
altitude

Answer:
It is a segment from a vertex of the triangle, perpendicular to the side opposite that vertex of the triangle.

Question 5.
midsegment

Answer:
It is a segment that connects the mid points of two sides in a triangle.

6.1 Perpendicular and Angle Bisectors

Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 1

Exploration 1

Points on a Perpendicular Bisector

Work with a partner. Use dynamic geometry software.

a. Draw any segment and label it \(\overline{A B}\). ConStruct the perpendicular bisector of \(\overline{A B}\).
Answer:

b. Label a point C that is on the perpendicular bisector of \(\overline{A B}\) but is not on \(\overline{A B}\).
Answer:

c. Draw \(\overline{C A}\) and \(\overline{C B}\) and find their lengths. Then move point C to other locations on the perpendicular bisector and note the lengths of \(\overline{C A}\) and \(\overline{C B}\).
Answer:

d. Repeat parts (a) – (c) with other segments. Describe any relationships(s) you notice.
Answer:

Exploration 2

Points on an Angle Bisector

Work with a partner. Use dynamic geometry software.

a. Draw two rays \(\vec{A}\)B and \(\vec{A}\)C to form ∠BAC. Construct the bisector of ∠BAC.
Answer:

b. Label a point D on the bisector of ∠BAC.
Answer:

c. Construct and find the lengths of the perpendicular segments from D to the sides of ∠BAC. Move point D along the angle bisector and note how the lengths change.
Answer:

d. Repeat parts (a)-(c) with other angles. Describe an relationship(s) you notice.
USING TOOLS STRATEGICALLY
To be proficient in math, you need to visualize the results of varying assumptions, explore consequences, and compare predictions with data.
Answer:

Communicate Your Answer

Question 3.
What conjectures can you make about a point on the perpendicular bisector of a segment and a point on the bisector of an angle?
Answer:

Question 4.
In Exploration 2. what is the distance from point D to \(\vec{A}\)B when the distance from D to \(\vec{A}\)C is 5 units? Justify your answer.
Answer:

Lesson 6.1 Perpendicular and Angle Bisectors

Use the diagram and tile given information to find the indicated measure.

Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 2

Question 1.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 3 is the perpendicular bisector of \(\overline{W Y}\), and \(\overline{y Z}\) = 13.75. Find WZ.

Answer:
WZ = 13.75

Explanation:
Using the perpendicular bisector theorem YZ = WZ
13.75 = WZ

Question 2.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 3 is the perpendicular bisector of \(\overline{W Y}\), WZ = 4n – 13, and YZ = n + 17. Find YZ.

Answer:
YZ = 27

Explanation:
Using the perpendicular bisector theorem YZ = WZ
n + 17 = 4n – 13
4n – n = 17 + 13
3n = 30
n = 10
YZ = 10 + 17 = 27

Question 3.
Find WX when WZ = 20.5. WY = 14.8. and YZ = 20.5.

Answer:
WX = 7.4

Explanation:
Because WZ = ZY and ZX is the perpendicular bisector of WY by the converse of perpendicular bisector theorem. By the definition of segment sector, WY = 2WX
14.8 = 2WX
WX = 7.4

Use the diagram and the given information to find the indicated measure.

Question 4.
\(\vec{B}\)D bisects ∠ABC, and DC = 6.9, Find DA.

Answer:
DA = 6.9

Explanation:
By using the angle bisector theorem
DC = DA
DA = 6.9

Question 5.
\(\vec{B}\)D bisects ∠ABC, AD = 3z + 7, and CD = 2z + 11. Find CD.

Answer:
z = 4

Explanation:
By using the angle bisector theorem
DC = DA
2z + 11 = 3z + 7
3z – 2z = 11 – 7
z = 4

Question 6.
Find m∠ABC when AD = 3.2, CD = 3.2, and m∠DBC = 39°.

Answer:
m∠DBC = 78°

Explanation:
m∠DBC = 39 = m∠DBA
m∠ABC = m∠DBC + m∠DBA
= 39 + 39 = 78°

Question 7.
Do you have enough information to conclude that \(\vec{Q}\)S bisects ∠PQR? Explain.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 4
Answer:
Yes,
Because PS = SR.

Question 8.
Write an equation of the perpendicular bisector of the segment with endpoints (- 1, – 5) and (3, – 1).

Answer:
y + 5 = 0

Explanation:
Midpoint = (\(\frac { -1 + 3 }{ 2 } \), \(\frac { -5 – 1 }{ 2 } \))
= (\(\frac { 2 }{ 2 } \), \(\frac { -6 }{ 2 } \)) = (1, -3)
Slope of the line = \(\frac { -1 + 1 }{ 3 + 5 } \) = 0
Slope of the perpendicular line = 0
Equation of the perpendicular bisector is (y + 5) = 0(x + 1)
y + 5 = 0

Exercise 6.1 Perpendicular and Angle Bisectors

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
Point C is in the interior of ∠DEF. If ∠DEC and ∠CEF are congruent, then \(\vec{E}\)C is the ________ of ∠DEF.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.

Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 5

Is point B the same distance from both X and Z?

Answer:
Yes.

Is point B equidistant from X and Z?

Answer:
Yes

Is point B collinear with X and Z?

Answer:
No.

Is point B on the perpendicular bisector of \(\overline{X Z}\)?

Answer:
Yes

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, find the indicated measure. Explain your reasoning.

Question 3.
GH
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 3

Question 4.
QR
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 7

Answer:
QR = RS
So, QR = 1.3

Question 5.
AB
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 8
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 5

Question 6.
UW
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 9

Answer:
UW = 55

Explanation:
UV = WU
9x + 1 = 7x + 13
9x – 7x = 13 – 1
2x = 12
x = 6
UW = 7(6) + 13 = 42 + 13 = 55

In Exercises 7-10. tell whether the information in the diagram allows you to conclude that point P lies on the perpendicular bisector of \(\overline{L M}\). Explain your reasoning.

Question 7.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 10
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 7

Question 8.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 11

Answer:
PN ⊥ LM. So, LN – MN
Therefore, point P lies on the LM.

Question 9.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 12
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 9

Question 10.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 13

Answer:
Yes.

In Exercises 11-14. find the indicated measure. Explain your reasoning.

Question 11.
m∠ABD
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 14
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 11

Question 12.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 15

Answer:
PS = RS
PS = 12

Question 13.
m∠KJL
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 16
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 13

Question 14.
FG
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 17

Answer:
FG = 16

Explanation:
FG = GH
x + 11 = 3x + 1
3x – x = 11 – 1
2x = 10
x = 5
FG = 5 + 11 = 16

In Exercises 15 and 16, tell whether the information in the diagram allows you to conclude that \(\vec{E}\)H bisects ∠FEG. Explain your reasoning.

Question 15.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 18
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 15

Question 16.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 19

Answer:
Yes, because H is equidistant from EF and EG, EH bisects ∠FEG by the Angle Bisector Theorem.

In Exercises 17 and 18, tell whether the information in the diagram allows you to conclude that DB = DC. Explain your reasoning.

Question 17.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 20
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 17

Question 18.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 21

Answer:
Yes, because AD is perpendicular to BD and DC. using perpendicular bisector theorem DB = DC.

In Exercises 19-22, write an equation of the perpendicular bisector of the segment with the given endpoints.

Question 19.
M(1, 5), N(7, – 1)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 19

Question 20.
Q(- 2, 0), R(6, 12)

Answer:
An equation of the perpendicular bisector is y = –\(\frac { 3 }{ 7 } \)x + \(\frac { 36 }{ 7 } \)

Explanation:
Slope of QR = \(\frac { 12 + 2 }{ 6 – 0 } \) = \(\frac { 7 }{ 3 } \)
The slope of the perpendicular line m = –\(\frac { 3 }{ 7 } \)
Midpoint of QR = (\(\frac { -2 + 6 }{ 2 } \), \(\frac { 0 + 12 }{ 2 } \)
= (-2, 6)
y = mx + b
6 = –\(\frac { 3 }{ 7 } \)(-2) + b
6 = \(\frac { 6 }{ 7 } \) + b
b = \(\frac { 36 }{ 7 } \)
An equation of the perpendicular bisector is y = –\(\frac { 3 }{ 7 } \)x + \(\frac { 36 }{ 7 } \)

Question 21.
U(- 3, 4), V(9, 8)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 21

Question 22.
Y( 10, – 7), Z(- 4, 1)

Answer:
An equation of the perpendicular bisector is y = \(\frac { 7 }{ 4 } \)x – \(\frac { 33 }{ 4 } \)

Explanation:
Slope of YZ = \(\frac { 1 + 7 }{ -4 – 10 } \) = \(\frac { -4 }{ 7 } \)
The slope of the perpendicular line m = \(\frac { 7 }{ 4 } \)
Midpoint of YZ = (\(\frac { 10 – 4 }{ 2 } \), \(\frac { -7 + 1 }{ 2 } \))
= (3, -3)
y = mx + b
-3 = \(\frac { 7 }{ 4 } \)(3) + b
b = -3 – \(\frac { 21 }{ 4 } \) = \(\frac { -12 – 21 }{ 4 } \) = \(\frac { -33 }{ 4 } \)
An equation of the perpendicular bisector is y = \(\frac { 7 }{ 4 } \)x – \(\frac { 33 }{ 4 } \)

ERROR ANALYSIS
In Exercises 23 and 24, describe and correct the error in the student’s reasoning.

Question 23.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 22
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 23

Question 24.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 23

Answer:
CP is not perpendicular to BP. So, x = 5 is wrong.

Question 25.
MODELING MATHEMATICS
In the photo, the road is perpendicular to the support beam and \(\overline{A B} \cong \overline{C B}\). Which theorem allows you to conclude that \(\overline{A D} \cong \overline{C D}\)?
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 24
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 25

Question 26.
MODELING WITH MATHEMATICS
The diagram shows the position of the goalie and the puck during a hockey game. The goalie is at point G. and the puck is at point P.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 25
a. What should be the relationship between \(\vec{P}\)G and ∠APB to give the goalie equal distances to travel on each side of \(\vec{P}\)G?

Answer:
PG should be the perpendicular bisector of ∠APB.

b. How does m∠APB change as the puck gets closer to the goal? Does this change make it easier or more difficult for the goalie to defend the goal? Explain your reasoning.
Answer:

Question 27.
CONSTRUCTION
Use a compass and straightedge to construct a copy of \(\overline{X Y}\). Construct a perpendicular bisector and plot a point Z on the bisector so that the distance between point Z and \(\overline{X Y}\) is 3 centimeters. Measure \(\overline{X Z}\) and \(\overline{Y Z}\). Which theorem does this construction demonstrate?
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 26
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 27

Question 28.
WRITING
Explain how the Converse of the Perpendicular Bisector Theorem (Theorem 6.2) is related to the construction of a perpendicular bisector.

Answer:
The converse of the perpendicular bisector theorem states if a point is equidistant from both the endpoints of the line segments in the same plane, then that point is on the perpendicular bisector of the line segment.

Question 29.
REASONING
What is the value of x in the diagram?
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 27
(A) 13
(B) 18
(C) 33
(D) not enough information
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 29

Question 30.
REASONING
Which point lies on the perpendicular bisector of the segment with endpoints M(7, 5) and m(- 1, 5)?
(A) (2, 0)
(B) (3, 9)
(C) (4, 1)
(D) (1, 3)

Answer:
(B) (3, 9)

Explanation:
The distance from M to (3, 9) is = √(7 – 3)² + (5 – 9)² = √16 + 16 = √32
The distance from m to (3, 9) = √(-1 – 3)² + (5 – 9)² = √16 + 16 = √32

Question 31.
MAKING AN ARGUMENT
Your friend says it is impossible for an angle bisector of a triangle to be the same line as the perpendicular bisector of the opposite side. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 31

Question 32.
PROVING A THEOREM
Prove the Converse of the Perpendicular Bisector Theorem (Thm. 6.2). (Hint: Construct a line through point C perpendicular to \(\overline{A B}\) at point P.)
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 28
Given CA = CB
Prove Point C lies on the perpendicular bisector of \(\overline{A B}\).

Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 1
Compare △ACP, △CPB
AP = BP
CP = CP
∠APC = ∠BPC = 90°
So, △ACP ≅ △CPB using SAS congruence theorem
So, CA = CB

Question 33.
PROVING A THEOREM
Use a congruence theorem to prove each theorem.
a. Angle Bisector Theorem (Thin. 6.3)
b. Converse of the Angle Bisector Theorem (Thm. 6.4)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 33.1
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 33.2

Question 34.
HOW DO YOU SEE IT?
The figure shows a map of a city. The city is arranged so each block north to south is the same length and each block east to west is the same length.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 29
a. Which school is approximately equidistant from both hospitals? Explain your reasoning.
Answer:

b. Is the museum approximately equidistant from Wilson School and Roosevelt School? Explain your reasoning.
Answer:

Question 35.
MATHEMATICAL CONNECTIONS
Write an equation whose graph consists of all the points in the given quadrants that are equidistant from the x- and y-axes.
a. I and III
b. II and IV
c. I and II
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 35

Question 36.
THOUGHT PROVOKING
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry, all points are on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. In spherical geometry, is it possible for two lines to be perpendicular but not bisect each other? Explain your reasoning.
Answer:

Question 37.
PROOF
Use the information in the diagram to prove that \(\overline{A B} \cong \overline{C B}\) if and onI if points D, E, and B are collinear.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 30
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 37

Question 38.
PROOF
prove the statement in parts (a) – (c)
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 31
Given Plane P is a perpendicular bisector of \(\overline{X Z}\) at point Y.
Prove
a. \(\overline{X W} \cong \overline{Z W}\)
b. \(\overline{X V} \cong \overline{Z V}\)
c. ∠VXW ≅ ∠VZW

Answer:
a. By using the perpendicular bisector theorem \(\overline{X W} \cong \overline{Z W}\)
b. By using the perpendicular bisector theorem \(\overline{X V} \cong \overline{Z V}\)

Maintaining Mathematical Proficiency

Classify the triangle by its sides.

Question 39.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 32
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 39

Question 40.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 33

Answer:
The trinagle is scalene as all the sides have different lengths.

Question 41.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 34
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 41

Classify the triangles by its angles.

Question 42.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 35
Answer:
The triangle is scalene as all angles are different.

Question 43.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 36
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.1 Question 43

Question 44.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 37

Answer:
The trinagle is an obtuse triangle.

6.2 Bisectors of Triangles

Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 38

Exploration 1

Properties of the Perpendicular Bisectors of a Triangle

Work with a partner: Use dynamic geometry software. Draw any ∆ABC.
a. Construct the perpendicular bisectors of all three sides of ∆ABC. Then drag the vertices to change ∆ABC. ‘What do you notice about the perpendicular bisectors?
Answer:

b. Label a point D at the intersection of the perpendicular bisectors.
Answer:

c. Draw the circle with center D through vertex A of ∆ABC. Then drag the vertices to change ∆ABC. What do you notice?
Answer:

Exploration 2

Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 39

Properties of the Angle Bisectors of a Triangle

Work with a partner. Use dynamic geometry software. Draw any ∆ABC.

a. Construct the angle bisectors of all three angles of ∆ABC, Then drag the vertices to change ∆ABC. What do you notice about the angle bisectors?
Answer:

b. Label a point D at the intersection of the angle bisectors.
Answer:

c. Find the distance between D and \(\overline{A B}\). Draw the circle with center D and this distance as a radius. Then drag the vertices to change ∆ABC. What do you notice?
LOOKING FOR STRUCTURE
To be proficient in math, you need to see complicated things as single objects or as being composed of several objects.
Answer:

Communicate Your Answer

Question 3.
What conjectures can you make about the perpendicular bisectors and the angle bisectors of a triangle?
Answer:

Lesson 6.2 Bisectors of Triangles

Monitoring Progress

Question 1.
Three snack carts sell hot pretzels horn points A, B, and E. What is the location of the pretzel distributor if it is equidistant from the three carts? Sketch the triangle and show the location.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 40

Answer:
The location is the circumcenter
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 1

Find the coordinates of the circumcenter of the triangle with the given vertices.

Question 2.
R(- 2, 5), S(- 6, 5), T(- 2, – 1)

Answer:
The circumcenter is (4, 7.33)

Explanation:
Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let R(- 2, 5), S(- 6, 5), T(- 2, – 1) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then
PR = PS = PT
PR² = PS² = PT²
PR² = PS²
(x + 2)² + (y – 5)² = (x + 6)² + (y – 5)²
x² + 4x + 4 + y² – 10x + 25 = x² + 12x + 36 + y² -10y + 25
4x + 4 = 12x + 36
12x – 4x = 36 – 4
8x = 32
x = 4
PS² = PT²
(x + 6)² + (y – 5)² = (x + 2)² + (y + 1)²
x² + 12x + 36 + y² -10y + 25 = x² + 4x + 4 + y² + 2y + 1
12x – 4x – 10y – 2y + 61 = 5
8x – 12y = -56
8(4) – 12y = -56
32 – 12y = -56
-12y = -56 – 32
-12y = -88
y = 7.33
The circumcenter is (4, 7.33)

Question 3.
W(- 1, 4), X(1, 4), Y( 1, – 6)

Answer:
The circumcenter is (0.25, -1)

Explanation:
Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let W(- 1, 4), X(1, 4), Y( 1, – 6) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then
PW = PX = PY
PW² = PY² = PX²
PX² = PY²
(x – 1)² + (y – 4)² = (x – 1)² + (y + 6)²
x² -2x + 1 + y² – 8y + 16 = x² -2x + 1 + y² + 12y + 36
-8y – 12y = 36 – 16
-20y = 20
y = -1
PW² = PX²
(x + 1)² + (y – 4)² = (x – 1)² + (y – 4)²
x² + 2x + 4 + y² – 8y + 16 = x² – 2x + 1 + y² – 8y + 16
4x = 1
x = 0.25
The circumcenter is (0.25, -1)

Question 4.
In the figure shown, QM = 3x + 8 and QN = 7x + 2. Find QP.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 41

Answer:
QP = \(\frac { 23 }{ 2 } \)

Explanation:
QM = QN
3x + 8 = 7x + 2
7x – 3x = 8 – 2
4x = 6
x = \(\frac { 3 }{ 2 } \)
QP = QN
= 7(\(\frac { 3 }{ 2 } \)) + 2
= \(\frac { 23 }{ 2 } \)

Question 5.
Draw a sketch to show the location L of the lamppost in Example 4.

Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 2

Exercise 6.2 Bisectors of Triangles

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
When three or more lines, rays, or segments intersect in the same Point. they are called _____________ lines, rays, or segments.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 1

Question 2.
WHICH ONE DOESNT BELONG?
Which triangle does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 42

Answer:
The third triangle does not belong with the other three. Because the point P in the remaining triangles is the circumcenter. But P is not circumcenter in the third triangle.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, the perpendicular bisectors of ∆ABC intersect at point G and are shown in blue. Find the indicated measure.

Question 3.
Find BG
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 43
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 3

Question 4.
Find GA
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 44

Answer:
Because G is the circumcenter of ∆ABC, AG = BG = CG
AG = BG = 11
So, AG = 11

In Exercises 5 and 6, the angle bisectors of ∆XYZ intersect at point P and are shown in red. Find the indicated measure.

Question 5.
Find PB.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 45
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 5

Question 6.
Find HP.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 46

Answer:
Because P is the incenter of ∆XYZ, PH = PF = PK
Therefore, PK = 15
HP = 15

In Exercises 7-10. find the coordinates of the circumcenter of the triangle with the given vertice

Question 7.
A(2, 6), B(8, 6), C(8, 10)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 7

Question 8.
D(- 7, – 1), E(- 1, – 1), F(- 7, – 9)

Answer:
The circumcenter is (-4, -5)

Explanation:
Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let D(- 7, – 1), E(- 1, – 1), F(- 7, – 9) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then
PD = PE = PF
PD² = PE² = PF²
PD² = PE²
(x + 7)² + (y + 1)² = (x + 1)² + (y + 1)²
x² + 14x + 49 + y² + 2y +1 = x² + 2x + 1 + y² + 2y + 1
14x – 2x = 1 – 49
12x = -48
x = -4
PD² = PF²
(x + 7)² + (y + 1)² = (x + 7)² + (y + 9)²
x² + 14x + 49 + y² + 2y + 1 = x² + 14x + 49 + y² + 18y + 81
18y – 2y = 1 – 81
16y = -80
y = -5
The circumcenter is (-4, -5)

Question 9.
H(- 10, 7), J(- 6, 3), K(- 2, 3)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 9.1
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 9.2

Question 10.
L(3, – 6), M(5, – 3) , N (8, – 6)

Answer:
The circumcenter is (5.5, -5.5)

Explanation:
Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let L(3, – 6), M(5, – 3) , N (8, – 6) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then
PL = PM = PN
PL² = PM² = PN²
PL² = PN²
(x – 3)² + (y + 6)² = (x – 8)² + (y + 6)²
x² – 6x + 9 + y² + 12y + 36 = x² -16x + 64 + y² + 12y + 36
-16x + 6x = 9 – 64
-10x = -55
x = 5.5
PL² = PM²
(x – 3)² + (y + 6)² = (x – 5)² + (y + 3)²
x² – 6x + 9 + y² + 12y + 36 = x² – 10x + 25 + y² + 6y + 9
-6x + 10x + 45 = 6y – 12y + 34
4x = -6y -11
4(5.5) = -6y – 11
22 + 11 = -6y
33 = -6y
y = -5.5
The circumcenter is (5.5, -5.5)

In Exercises 11-14, N is the incenter of ∆ABC. Use the given information to find the indicated measure.

Question 11.
ND = 6x – 2
NE = 3x + 7
Find NF.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 47
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 11

Question 12.
NG = x + 3
NH = 2x – 3
Find NJ.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 48

Answer:
NJ = 9

Explanation:
NG = NH = NJ
x + 3 = 2x – 3
2x – x = 3 + 3
x = 6
By Incenter theorem, NG = NH = NJ
NJ = 6 + 3 = 9

Question 13.
NK = 2x – 2
NL = – x + 10
Find NM
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 49
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 13

Question 14.
NQ = 2x
NR = 3x – 2
Find NS.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 50

Answer:
NS = 4

Explanation:
NQ = NR
2x = 3x – 2
3x – 2x = 2
x = 2
NQ = 2 (2) = 4
By Incenter theorem NS = NR = NQ
So, NS = 4

Question 15.
P is the circumcenter of ∆XYZ. Use the given information to find PZ.
PX = 3x + 2
PY = 4x – 8
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 51
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 15

Question 16.
P is the circumcenter of ∆XYZ. Use the given information to find PY.
PX = 4x + 3
PZ = 6x – 11
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 52

Answer:
PY = 31

Explanation:
PX = PZ
4x + 3 = 6x – 11
6x – 4x = 3 + 11
2x = 14
x = 7
PZ = 6(7) – 11 = 42 – 11 = 31
By the incenter theorem, PX = PZ = PY
PY = 31

CONSTRUCTION
In Exercises 17-20. draw a triangle of the given type. Find the circumcenter. Then construct the circumscribed circle.

Question 17.
right
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 17

Question 18.
obtuse

Answer:

Question 19.
acute isosceles
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 19

Question 20.
equilateral

Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 3

CONSTRUCTION
In Exercises 21-24, copy the triangle with the given angle measures. Find the incenter. Then construct the inscribed circle.

Question 21.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 53
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 21

Question 22.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 54

Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 4

Question 23.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 55
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 23

Question 24.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 56

Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 5

ERROR ANALYSIS
In Exercises 25 and 26. describe and correct the error in identifying equal distances inside the triangle.

Question 25.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 57
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 25

Question 26.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 58

Answer:
Because point T is the intersection of the angle bisectors, it is the incenter. But, because TV and TF are not necessarily perpendicular to a side of the trinagle, there is not sufficient evidence to conclude that TV and TZ are congruent. Point T is equidistance from the sides of the triangle.

Question 27.
MODELING WITH MATHEMATICS
You and two friends plan to meet to walk your dogs together. You want the meeting place to be the same distance from each person’s house. Explain how you can use the diagram to locate the meeting place.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 59
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 27

Question 28.
MODELING WITH MATHEMATICS
You are placing a fountain in a triangular koi pond. YOU want the foutain to be the same distance from each edge of the Pond. Where should you place the fountain? Explain your reasoning. Use a sketch to support your answer.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 60

Answer:
A, B, C are edges of the pond.
Foutain to be the same distance from each edge of the Pond
Then, you need to find the circumcenter of the triangle.
Place the fountain at the circumcenter.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 6

CRITICAL THINKING
In Exercises 29-32, complete the statement with always, sometimes, or never. Explain your reasoning.

Question 29.
The circumenter of a scalene triangle is ______________ inside the triangle.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 29

Question 30.
If the perpendicular bisector of one side of a triangle intersects the opposite vertex. then the triangle is ______________ isosceles.

Answer:
If the perpendicular bisector of one side of a triangle intersects the opposite vertex. then the triangle is always isosceles.

Explanation:
The Vertex is equidistant to other vertices then.

Question 31.
The perpendicular bisectors of a triangle intersect at a point that is ______________ equidistant from the midpoints of the sides of the triangle.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 31

Question 32.
The angle bisectors of a triangle intersect at a point that is ______________ equidistant from the sides of the triangle.

Answer:
The angle bisectors of a triangle intersect at a point that is circumcenter equidistant from the sides of the triangle.

CRITICAL THINKING
In Exercises 33 and 34, find the coordinates of the circumcenter of the triangle with the given vertices.

Question 33.
A(2, 5), B(6, 6). C( 12. 3)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 33.1
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 33.2
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 33.3

Question 34.
D(- 9, – 5), E(- 5, – 9), F(- 2, – 2)

Answer:
The circumcenter is (4.45, 4.45)

Explanation:
Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let D(- 9, – 5), E(- 5, – 9), F(- 2, – 2) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then
PD = PE = PF
PD² = PE² = PF²
PD² = PE²
(x + 9)² + (y + 5)² = (x + 5)² + (y + 9)²
x² + 18x + 81 + y² + 10y + 25 = x² + 10x + 25 + y² + 18y + 81
18x – 10x = 18y – 10y
8x = 8y
x = y
PE² = PF²
(x + 5)² + (y + 9)² = (x + 2)² + (y + 2)²
x² + 10x + 25 + y² + 18y + 81 = x² + 4x + 4 + y² + 2y + 4
10x – 4x + 106 = 8 + 2y – 18y
6x + 106 = 8 – 16y
6x + 98 = -16x
22x = -98
x = 4.45
y = 4.45
The circumcenter is (4.45, 4.45)

MATHEMATICAL CONNECTIONS
In Exercises 35 and 36. find the a1ue of x that makes N the incenter of the triangle.

Question 35.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 61
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 35

Question 36.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 62

Answer:
N the incenter is \(\frac { 12 }{ 2 } \).

Explanation:
25² = 24² + (14x)²
625 = 576 + 196x²
49 = 196x²
x = \(\frac { 7 }{ 14 } \) = \(\frac { 1 }{ 2 } \)
The value of x will make N the incenter is \(\frac { 12 }{ 2 } \).

Question 37.
PROOF
Where is the circumcenter located in any right triangle? Write a coordinate proof of this result.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 37

Question 38.
PROVING A THEOREM
Write a proof of the Incenter Theorem (Theorem 6.6).
Given ∆ABC, \(\overline{A D}\) bisects∠CAB,
\(\overline{B D}\) bisects ∠CBA, \(\overline{D E}\) ⊥ \(\overline{A B}\), \(\overline{D F}\) ⊥ \(\overline{B C}\), and \(\overline{D G}\) ⊥ \(\overline{C A}\)
Prove The angle bisectors intersect at D, which is equidistant from \(\overline{A B}\), \(\overline{B C}\), and \(\overline{C A}\)
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 63

Answer:

Question 39.
WRITING
Explain the difference between the circumcenter and the incenter of a triangle.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 39

Question 40.
REASONING
Is the incenter of a triangle ever located outside the triangle? Explain your reasoning.

Answer:
No

Explanation:
Because the incenter is the center of an inscribed circle, it must always be inside the triangle.

Question 41.
MODELING WITH MATHEMATICS
You are installing a circular pool in the triangular courtyard shown. You want to have the largest pool possible on the site without extending into the walkway.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 64
a. Copy the triangle and show how to install the pool so that it just touches each edge. Then explain how you can he sure that you could not fit a larger pool on the site.
b. You want to have the largest pool possible while leaving at least I foot of space around the pool. Would the center of the pool be in the same position as in part (a)? Justify your answer.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 41

Question 42.
MODELING WITH MATHEMATICS
Archaeologists find three stones. They believe that the stones were once pail of a circle of stones with a community fire pit at its center. They mark the locations of stones A, B, and C on a graph. where distances are measured in feet.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 65
a. Explain how archaeologists can use a sketch to estimate the center of the circle of stones.

b. Copy the diagram and find the approximate coordinates of the point at which the archaeologists should look for the fire pit.

Answer:
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 6.2 3
To get the centre of the circle P, we need to find the circumcenter of the triangle ABC.
Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let A(2, 10), B(13, 6), C(6, 1) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then
PA = PB = PC
PA² = PB² = PC²
PA² = PB²
(x – 2)² + (y – 10)² = (x – 13)² + (y – 6)²
x² – 4x + 4 + y² – 20y + 100 = x² – 26x + 169 + y² – 12y + 36
-4x + 26x – 20y + 12y = 205 – 104
22x – 8y = 101 —- (i)
PB² = PC²
(x – 13)² + (y – 6)² = (x – 8)² + (y – 1)²
x² – 26x + 169 + y² – 12y + 36 = x² – 16x + 64 + y² – 2y + 1
-26x + 16x – 12y + 2y = 65 – 205
-10x – 10y = -140
x + y = 14
x = 14 – y — (ii)
Substitute (ii) in (i)
22(14 – y) – 8y = 101
308 – 22y – 8y = 101
308 – 101 = 30y
207 = 30y
y = 6.9
Substitute y = 6.9 in (ii)
x = 14 – 6.9
x = 7.1
The circumcenter is (7.1, 6.9)

Question 43.
REASONING
Point P is inside ∆ABC and is equidistant from points A and B. On which of the following segments must P be located?
(A) \(\overline{A B}\)
(B) the perpendicular bisector of \(\overline{A B}\)
(C) \(\overline{A C}\)
(D) the perpendicular bisector of \(\overline{A C}\)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 43

Question 44.
CRITICAL THINKING
A high school is being built for the four towns shown on the map. Each town agrees that the school should be an equal distance from each of the tourist towns. Is there a single point where they could agree to build the school? If so, find it. If not, explain why not. Justify your answer with a diagram.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 66
Answer:

Question 45.
MAKING AN ARGUMENT
Your friend says that the circumcenter of an equilateral triangle is also the incenter of the triangle. Is your friend correct? Explain in your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 45

Question 46.
HOW DO YOU SEE IT?
The arms of the windmill are the angle bisectors of the red triangle. What point of concurrency is the point that Connects the three arms?
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 67
Answer:

Question 47.
ABSTRACT REASONING
You are asked to draw a triangle and all its perpendicular bisectors and angle bisectors.
a. For which type of triangle would you need the fewest segments? What is the minimum number of segments you would need? Explain.
b. For which type of triangle would you need the most segments? What is the maximum number of segments you would need? Explain.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 47

Question 48.
THOUGHT PROVOKING
The diagram shows an official hockey rink used by the National Hockey League. Create a triangle using hockey players as vertices in which the center circle is inscribed in the triangle. The center dot should he the incenter of your triangle. Sketch a drawing of the locations of your hockey players. Then label the actual lengths of the sides and the angle measures in your triangle.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 68
Answer:

COMPARING METHODS
In Exercises 49 and 50. state whether you would use perpendicular bisectors or angle bisectors. Then solve the problem.

Question 49.
You need to cut the largest circle possible from an isosceles triangle made of paper whose sides are 8 inches, 12 inches, and 12 inches. Find the radius of the circle.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 49

Question 50.
On a map of a camp. You need to create a circular walking path that connects the pool at (10, 20), the nature center at (16, 2). and the tennis court at (2, 4). Find the coordinates of the center of the circle and the radius of the circle.

Answer:
The center of the circular path is located at (10, 10) and the radius of the circular path is 10 units.

Explanation:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 7

Let the centre of the circle be at O (x, y)
Slope of AB = \(\frac { 20 – 4 }{ 10 – 2 } \) = 2
The slope of XO must be \(\frac { -1}{ 2 } \) the negative reciprocal of the slope of AB as the 2 lines are perpendicular
Slope of XO = \(\frac { y – 12 }{ x – 6 } \) = \(\frac { -1 }{ 2 } \)
y – 12 = -0.5x + 3
0.5x + y = 12 + 3 = 15
x + 2y = 30
The slope of BC = \(\frac { 2 – 20 }{ 16 – 10 } \) = -3
The slope of XO must be \(\frac { 1 }{ 3 } \) = \(\frac { 11 – y }{ 13 – x } \)
33 – 3y = 13 – x
x – 3y = -33 + 13 = -20
Subtrcat two equations
x + 2y – x + 3y = 30 + 20
y = 10
x – 30 = -20
x = 10
r = √(10 – 2)² + (10 – 4)²
r = 10

Question 51.
CRITICAL THINKING
Point D is the incenter of ∆ABC. Write an expression for the length x in terms of the three side lengths AB, AC, and BC.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 69
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 51

Maintaining Mathematical Proficiency

The endpoints of \(\overline{A B}\) are given. Find the coordinates of the midpoint M. Then find AB.
Question 52.
A(- 3, 5), B(3, 5)

Answer:
AB = 6

Explanation:
Midpoint of AB = (\(\frac { -3 + 3 }{ 2 } \), \(\frac { 5 + 5 }{ 2 } \)) = (0, 5)
AB = √(3 + 3)² + (5 – 5)² = 6

Question 53.
A(2, – 1), B(10, 7)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 53

Question 54.
A(- 5, 1), B(4, – 5)

Answer:
AB = 10.81

Explanation:
Midpoint of AB = (\(\frac { -5 + 4 }{ 2 } \), \(\frac { 1 – 5 }{ 2 } \)) = (\(\frac { -1 }{ 2 } \), -2)
AB = √(4 + 5)² + (-5 – 1)² = √81 + 36 =  10.81

Question 55.
A(- 7, 5), B(5, 9)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 55

Write an equation of the line passing through point P that is perpendicular to the given line.
Graph the equations of the lines to check that they are perpendicular.
Question 56.
P(2, 8), y = 2x + 1

Answer:
y = \(\frac { -1 }{ 2 } \)x + 9

Explanation:
The slope of the given line m = 2
The slope of the perpendicular line M = \(\frac { -1 }{ 2 } \)
The perpendicular line passes through the given point P(2, 8) is
8 = \(\frac { -1 }{ 2 } \)(2) + b
b = 9
So, y = \(\frac { -1 }{ 2 } \)x + 9
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 6.2 4

Question 57.
P(6, -3), y = – 5
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 57

Question 58.
P(- 8, – 6), 2x + 3y = 18

Answer:
The perpendicular equation is y = \(\frac { 3 }{ 2 } \)x + 6

Explanation:
The given line is 2x + 3y = 18
y = \(\frac { -2 }{ 3 } \)x + 6
The slope of the given line m = \(\frac { -2 }{ 3 } \)
The slope of the perpendicular line is M = \(\frac { 3 }{ 2 } \)
The perpendicular line passing through the point P(-8, -6) is
-6 = \(\frac { 3 }{ 2 } \)(-8) + b
b = 6
The perpendicular equation is y = \(\frac { 3 }{ 2 } \)x + 6
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 6.2 5

Question 59.
P(- 4, 1), y + 3 = – 4(x + 3)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.2 Question 59

6.3 Medians and Altitudes of Triangles

Exploration 1

Finding Properties of the Medians of a Triangle

Work with a partner. Use dynamic geometry software. Draw any ∆ABC.

Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 70

a. Plot the midpoint of \(\overline{B C}\) and label it D, Draw \(\overline{A D}\), which is a median of ABC. Construct the medians to the other two sides of ∆ABC.
Answer:

b. What do you notice about the medians? Drag the vertices to change ∆ABC. Use your observations to write a conjecture about the medians of a triangle.
Answer:

c. In the figure above, point G divides each median into a shorter segment and a longer segment. Find the ratio of the length of each longer segment to the length of the whole median. Is this ratio always the same? Justify your answer.
Answer:

Exploration 2

Finding Properties of the Altitudes of a Triangle

Work with a partner. Use dynamic geometry software. Draw any ∆ABC.

Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 71

a. Construct the perpendicular segment from vertex A to \(\overline{B C}\). Label the endpoint D. \(\overline{A D}\) is an altitude of ∆ABC.
Answer:

b. Construct the altitudes to the other two sides of ∆ABC. What do you notice?
Answer:

c. Write a conjecture about the altitudes of a triangle. Test your conjecture by dragging the vertices to change ∆ABC.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.
Answer:

Communicate Your Answer

Question 3.
What conjectures can you make about the medians and altitudes of a triangle?
Answer:

Question 4.
The length of median \(\overline{R U}\) in ∆RST is 3 inches. The point 0f concurrency of the three medians of ∆RST divides \(\overline{R U}\) into two segments. What are the lengths of these two segments?
Answer:

Lesson 6.3 Medians and Altitudes of Triangles

Monitoring Progress

There are three paths through a triangular park. Each path goes from the midpoint of one edge to the opposite corner. The paths meet at point P.

Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 72

Question 1.
Find PS and PC when SC = 2100 feet.

Answer:
PS = 700, PC = 2100

Explanation:
PC = \(\frac { 2 }{ 3 } \)SC
SC = PS + PC
SC = PS + \(\frac { 2 }{ 3 } \)SC
PS = \(\frac { 1 }{ 3 } \)SC
So, PS = \(\frac { 1 }{ 3 } \)(2100) = 700
PC = \(\frac { 2 }{ 3 } \)(2100) = 1400

Question 2.
Find TC and BC when BT = 1000 feet.

Answer:
BC = 2000 ft, TC = 1000 ft

Explanation:
BT = TC
So, TC = 1000 ft
BC = BT + TC
BC = 1000 + 1000 = 2000 ft

Question 3.
Find PA and TA when PT = 800 feet.

Answer:
PA = 2400 ft, TA = 1600 ft

Explanation:
PT = \(\frac { 1 }{ 3 } \)PA
PA = 3PT
= 3 • 800 = 2400
TA = \(\frac { 2 }{ 3 } \)PA
TA = \(\frac { 2 }{ 3 } \)(2400)
= 1600

Find the coordinates of the centroid of the triangle with the given vertices.

Question 4.
F(2, 5), G(4, 9), H(6, 1)

Answer:
The centroid is (4, 5).

Explanation:
The centroid of the triangle FGH is O = (\(\frac { 2 + 4 + 6 }{ 3 } \), \(\frac { 5 + 9 + 1 }{ 3 } \))
= (\(\frac { 12 }{ 3 } \), \(\frac { 15 }{ 3 } \))
= (4, 5)

Question 5.
X(- 3, 3), Y(1, 5), Z(- 1, – 2)

Answer:
The centriod of given triangle is (-1, 2)

Explanation:
The centroid of the triangle XYZ is O = (\(\frac { -3 + 1 – 1 }{ 3 } \), \(\frac { 3 + 5 – 2 }{ 3 } \))
= (\(\frac { -3 }{ 3 } \), \(\frac { 6 }{ 3 } \))
= (-1, 2)

Tell whether the orthocenter of the triangle with the given vertices is inside, on, or outside the triangle. Then find the coordinates of the orthocenter.

Question 6.
A(0, 3), B(0, – 2), C(6, -3)

Answer:
The orthocenter lies outside the triangle.

Explanation:
The slope of the line BC = \(\frac { -3 + 2 }{ 6 – 0 } \) = \(\frac { -1 }{ 6 } \)
The slope of the perpendicular line = 6
The perpendicular line is (y – 3) = 6(x – 0)
y – 3 = 6x
y = 6x + 3
The slope of AC = \(\frac { -3 – 3 }{ 6 – 0 } \) = \(\frac { -6 }{ 6 } \) = -1
The slope of the perpendicular line = 1
The equation of perpendicular line (y + 2) = 1(x – 0)
y + 2 = x
Substitute y = 6x + 3 in above equation
6x + 3 + 2 = x
5x = -5
x = -1
Put x = -1 in y + 2 = x
y + 2 = -1
y = -3
The orthocenter is (-1, -3)
The orthocenter lies outside the triangle.

Question 7.
J(- 3, – 4), K(- 3, 4), L(5, 4)

Answer:
The orthocenter lies on the vertex of the triangle.

Explanation:
The slope of JL = \(\frac { 4 + 4 }{ 5 + 3 } \) = 2
The slope of perpendicular line is \(\frac { -1 }{ 2 } \)
The equation of perpendicular line is (y – 4) = \(\frac { -1 }{ 2 } \)(x + 3)
2y – 8 = -x – 3
x + 2y – 5 = 0
The slope of JK = \(\frac { 4 + 4 }{ -3 + 3 } \) = 0
The slope of the perpendicular line is 0
the equation of perpendicular line is (y – 4) = 0
y = 4
Substitute y = 4 in x + 2y – 5 = 0
x + 2(4) – 5 = 0
x + 3 = 0
x = -3
So, the ortho center is (-3, 4)
It lies on the vertex of the triangle.

Question 8.
WHAT IF?
In Example 4, you want to show that median \(\overline{B D}\) is also an angle bisector. How would your proof be different?
Answer:

Exercise 6.3 Medians and Altitudes of Triangles

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Name the four types of points of concurrency. Which lines intersect to form each of the points?
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 1

Question 2.
COMPLETE THE SENTENCE
The length of a segment from a vertex to the centroid is ______________ the length of the median from that vertex.

Answer:
The length of a segment from a vertex to the centroid is one-third of the length of the median from that vertex.

Monitoring progress and Modeling with Mathematics

In Exercises 3-6, point P is the centroid of ∆LMN. Find PN and QP.

Question 3.
QN = 9
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 73
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 3

Question 4.
QN = 21
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 74

Answer:
PN = 14 units, QP = 7 units

Explanation:
PN = \(\frac { 2 }{ 3 } \)QN
PN = \(\frac { 2 }{ 3 } \)(21)
PN = 14
QP = \(\frac { 1 }{ 3 } \)QN
= \(\frac { 1 }{ 3 } \)(21)
= 7

Question 5.
QN = 30
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 75
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 5

Question 6.
QN = 42
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 76

Answer:
PN = 28 units, QP = 14 units

Explanation:
PN = \(\frac { 2 }{ 3 } \)QN
PN = \(\frac { 2 }{ 3 } \)(42)
PN = 28
QP = \(\frac { 1 }{ 3 } \)QN
= \(\frac { 1 }{ 3 } \)(42)
= 14

In Exercises 7-10. point D is the centroid of ∆ ABC. Find CD and CE.

Question 7.
DE = 5
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 77
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 7

Question 8.
DE = 11
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 78

Answer:
CD = 22 units, CE = 33 units

Explanation:
DE = \(\frac { 1 }{ 3 } \)CE
11 = \(\frac { 1 }{ 3 } \) CE
CE = 33
CD = \(\frac { 2 }{ 3 } \) CE
CD = \(\frac { 2 }{ 3 } \)(33)
CD = 22

Question 9.
DE = 9
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 79
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 9

Question 10.
DE = 15
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 80

Answer:
CE = 45 units, CD = 30

Explanation:
DE = \(\frac { 1 }{ 3 } \)CE
15 = \(\frac { 1 }{ 3 } \) CE
CE = 45
CD = \(\frac { 2 }{ 3 } \) CE
CD = \(\frac { 2 }{ 3 } \)(45)
CD = 30

In Exercises 11-14. point G is the centroid of ∆ABC. BG = 6, AF = 12, and AE = 15. Find the length of the segment.

Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 81

Question 11.
\(\overline{F C}\)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 11

Question 12.
\(\overline{B F}\)

Answer:
\(\overline{B F}\) = 15

Explanation:
All the medians have same length.
so, AE = 15 = BF

Question 13.
\(\overline{A G}\)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 13

Question 14.
\(\overline{G E}\)

Answer:
\(\overline{G E}\) = 5 units

Explanation:
AE = AG + GE
15 = 10 + GE
GE = 5

In Exercises 15-18. find the coordinates of the centroid of the triangle with the given vertices.

Question 15.
A(2, 3), B(8, 1), C(5, 7)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 15.1
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 15.2
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 15.3

Question 16.
F(1, 5), G( – 2, 7), H(- 6, 3)

Answer:
The centroid is (\(\frac { -7 }{ 3 } \), 5)

Explanation:
The centroid of the trinagle = (\(\frac { 1 – 2 – 6 }{ 3 } \), \(\frac { 5 + 7 + 3 }{ 3 } \))
= (\(\frac { -7 }{ 3 } \), 5)

Question 17.
S(5, 5), T(11, – 3), U(- 1, I)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 17.1
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 17.2

Question 18.
X(1, 4), Y(7, 2), Z(2, 3)

Answer:
The centroid of the triangle is (\(\frac { 10 }{ 2 } \), 3)

Explanation:
The centroid of the trinagle = (\(\frac { 1 + 7 + 2 }{ 3 } \), \(\frac { 4 + 2 + 3 }{ 3 } \))
= (\(\frac { 10 }{ 2 } \), 3)

In Exercises 19-22. tell whether the orthocenter is inside, on, or outside the triangle. Then find the coordinates of the orthocenter.

Question 19.
L(0, 5), M(3, 1), N(8, 1)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 19.1
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 19.2

Question 20.
X(- 3, 2), Y(5, 2), Z(- 3, 6)

Answer:
The orthocenter lies on the vertex of the triangle.

Explanation:
The slope of YZ = \(\frac { 6 – 2 }{ -3 – 5 } \) = \(\frac { -1 }{ 2 } \)
The slope of the perpendicular line is 2
The equation of perpendicular line is (y – 2) = 2(x + 3)
y – 2 = 2x + 6
2x – y + 8 = 0
The slope of XZ = \(\frac { 6 – 2 }{ -3 + 3 } \) = 0
The equation of perpendicular line is (y – 2) = 0
y = 2
Substitute y = 2 in 2x – y + 8 = 0
2x – 2 + 8 = 0
2x + 6 = 0
x = -3
the orthocenter is (-3, 2)
The orthocenter lies on the vertex of the triangle.

Question 21.
A(- 4, 0), B(1, 0), C(- 1, 3)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 21.1
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 21.2

Question 22.
T(-2, 1), U( 2, 1), V(0, 4)

Answer:
The orthocenter lies inside the triangle ABC.

Explanation:
The slope of UV = \(\frac { 4 – 1 }{ 0 – 2 } \) = \(\frac { -3 }{ 2 } \)
The slope of the perpendicular line is \(\frac { 2 }{ 3 } \)
The equation of the perpendicular line is (y – 1) = \(\frac { 2 }{ 3 } \)(x + 2)
3(y – 1) = 2(x + 2)
3y – 3 = 2x + 2
2x – 3y + 5 = 0 — (i)
The slope of TV = \(\frac { 4 – 1 }{ 0 + 2 } \) = \(\frac { 3 }{ 2 } \)
The slope of the perpendicular line is \(\frac { -2 }{ 3 } \)
The equation of the perpendicular line is (y – 1) = \(\frac { -2 }{ 3 } \)(x – 2)
3(y – 1) = -2(x – 2)
3y – 3 = -2x + 4
2x + 3y – 7 = 0 —(ii)
Add two equations
2x – 3y + 5 + 2x + 3y – 7 = 0
4x – 2 = 0
x = 0.5
2x – 1.5 + 5 = 0
x = -1.75
So, the orthocenter is (0, 2.33)
The orthocenter lies inside the triangle ABC.

CONSTRUCTION
In Exercises 23-26, draw the indicated triangle and find its centroid and orthocenter.
Question 23.
isosceles right triangle
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 23

Question 24.
obtuse scalene triangle

Answer:
Construct the medians for all the sides of the trinagle by finding the midpoint. The point where all medians intersect is centroid. The orthocenter lies outside the triangle for an obtuse triangle.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 6.3 1

Question 25.
right scalene triangle
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 25

Question 26.
acute isosceles triangle

Answer:
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 6.3 2

ERROR ANALYSIS
In Exercises 27 and 28, describe and correct the error in finding DE. Point D is the centroid of ∆ABC.

Question 27.
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 82
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 27

Question 28.
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 83

Answer:
The length of DE should be 1/3 of length of AE because it is the shorter segment from the centroid
DE = \(\frac { 1 }{ 3 } \)AE
= \(\frac { 1 }{ 3 } \)(24)
= 8
DE = 8

PROOF
In Exercises 29 and 30, write a proof of the statement.

Question 29.
The angle bisector from the vertex angle to the base of an isosceles triangle is also a median.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 29.1
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 29.2

Question 30.
The altitude from the vertex angle to the base of an isosceles triangle is also a perpendicular bisector.
Answer:

CRITICAL THINKING
In Exercises 31-36, complete the statement with always, sometimes, or never. Explain your reasoning.

Question 31.
The centroid is _____________ on the triangle.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 31

Question 32.
The orthocenter is _____________ outside the triangle.

Answer:
The orthocenter is always outside the triangle opposite the longest leg, on the same side as the largest angle. The only time all three of these centers fall in the same spot is in the case of an equilateral triangle. In fact, in this case, the incenter falls in the same place as well.

Question 33.
A median is _____________ the same line segment as a perpendicular bisector.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 33

Question 34.
An altitude is ______________ the same line segment as an angle bisector.

Answer:

Question 35.
The centroid and orthocenter are _____________ the same point.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 35

Question 36.
The centroid is ______________ formed by the intersection oÍ the three medians.

Answer:
The centroid is a point formed by the intersection oÍ the three medians.

Question 37.
WRITING
Compare an altitude of a triangle with a perpendicular bisector of a triangle.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 37

Question 38.
WRITING
Compare a median. an altitude, and an angle bisector of a triangle.

Answer:
Median is a line segment joining a vertex of a triangle with the midoint of the opposite side. Angle bisector is a line segment joining a vertex of a trianglr with the opposite side such that the angle at the vertex split into two equal parts. While altitude is also a line segment joining a vertex of a triangle with opposite side such that the segment is perpendicular to the opposite side. In general, medians, angle bisectors and altitudes drawn from the same vertex of a triangle are different line segments. In an equilateral triagle, altitude, median, and angle bisector drawn from the same vertex overlap. In an isosceles trianglle, the altitude drawn to the base is the median and the angle bisector, the median drawn to the base is the altitude and the angle bisector, the bisector of the angle opposite to the base is the altitude and the median.

Question 39.
MODELING WITH MATHEMATICS
Find the area of the triangular part of the paper airplane wing that is outlined in red. Which special segment of the triangle did you use?
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 84
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 39

Question 40.
ANALYZING RELATIONSHIPS
Copy and complete the statement for ∆DEF with centroid K and medians
\(\overline{D H}\), \(\overline{E J}\), and \(\overline{F G}\).

a. EJ = ____ KJ

Answer:
EJ = 2KJ

Explanation:
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 6.3 3
Since KJ is the shorter distance between the two parts of the median, its length is 1/3 of the EJ.

b. DK = ____ KH

Answer:
DK = 2 KH
Since KH is the shorter distance between the two parts of the median, its length is 1/3 of the DH.

c. FG = ___ KF

Answer:
FG = \(\frac { 2 }{ 3 } \)KF
Since KF is the longest distance between two parts, its length is \(\frac { 2 }{ 3 } \) of FG

d. KG = ___ FG

Answer:
KG = 2FG
Since KG is the shorter distance between the two parts of the median, its length is 1/3 of the FG.

MATHEMATCAL CONNETIONS
In Exercises 41-44, point D is the centroid of ∆ABC. Use the given information to find the value of x.

Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 85

Question 41.
BD = 4x + 5 and BF = 9x
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 41

Question 42.
GD = 2x – 8 and GC = 3x + 3

Answer:
x = 9

Explanation:
GD = \(\frac { 1 }{ 3 } \)GC
2x – 8 = \(\frac { 1 }{ 3 } \)(3x + 3)
2x – 8 = x + 1
x = 9

Question 43.
AD = 5x and DE = 3x – 2
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 43

Question 44.
DF = 4x – 1 and BD = 6x + 4

Answer:
x = 3

Explanation:
BD = 2DF
6x + 4 = 2(4x – 1)
6x + 4 = 8x – 2
2x = 6
x = 3

Question 45.
MATHEMATICAL CONNECTIONS
Graph the lines on the same coordinate plane. Find the centroid of the triangle formed by their intersections.
y1 = 3x – 4
y2 = \(\frac{3}{4}\)x + 5
y2 = – \(\frac{3}{2}\)x – 4
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 45.1
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 45.2

Question 46.
CRITICAL THINKING
In what types of triangles can a vertex be one of the points of concurrency of the triangle? Explain your reasoning.

Answer:
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 6.3 4
In right angles, the orthocenter forms at the 90 degrees angle.

Question 47.
WRITING EQUATIONS
Use the numbers and symbols to write three different equations for PE.
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 86
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 87
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 47

Question 48.
HOW DO YOU SEE IT?
Use the figure.
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 88
a. What type of segment is \(\overline{K M}\)? Which point of concurrency lies on \(\overline{K M}\)?

Answer:
KM is the median drawn from K to JL.

b. What type of segments is \(\overline{K N}\)? Which point of concurrency lies on \(\overline{K N}\)?

Answer:
KN is the altitude drawn from K to JL.

c. Compare the areas of ∆JKM and ∆KLM. Do you think the areas of the triangles formed by the median of any triangle will always compare this way? Explain your reasoning.

Answer:

Question 49.
MAKING AN ARGUMENT
Your friend claims that it is possible for the circumcenter, incenter, centroid, and orthocenter to all be the same point. Do you agree? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 49

Question 50.
DRAWING CONCLUSIONS
The center of gravity of
a triangle, the point where a triangle can balance on the tip of a pencil, is one of the four points of concurrency. Draw and cut out a large scalene triangle on a piece of cardboard. Which of the four points of concurrency is the center of gravity? Explain.

Answer:
The median is the balancing line of the triangle
The centroid is the point of concurrency of three medians
The centroid is the center of gravity of the triangle.

Explanation:
The median is the balancing line of the triangle i.e median divides the triangle into two triangles of equal areas.
Assuming trinagle ABC
The area of ABC = 0.5bh
The median divides base into two equal parts, each forming a smaller triangle with the same height of the original one.
AD is the median of base BC
So, bD = CD = \(\frac { 1 }{ 2 } \) BC
The area of ABC = \(\frac { 1 }{ 2 } \) x BC x h
Area of ABD = \(\frac { 1 }{ 2 } \) x BD x h = \(\frac { 1 }{ 2 } \) x ABC area
Area of ACD = \(\frac { 1 }{ 2 } \) x CD x h = \(\frac { 1 }{ 2 } \) x ABC area
The centroid is the concurrency point of three medians.
So, the centroid is the gravity centre of the triangle.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 6.3 5

Question 51.
PROOF
Prose that a median of an equilateral triangle is also an angle bisector, perpendicular bisector, and altitude.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 51.1
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 51.2
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 51.3

Question 52.
THOUGHT PROVOKING
Construct an acute scalene triangle. Find the orthocenter, centroid, and circumcenter. What can you conclude about the three points of concurrency?

Answer:
The centroid, orthocenter and circumcenter are collinear.

Explanation:
Below is the scalene triangle with the centroid, orthocenter and circumcenter
Point G is the centroid, H is the orthocenter and J is the circumcenter.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 6.3 6
If we draw a line through these points, we can see all the points lie on the same line.
Therefore, we can conclude that the centroid, orthocenter and circumcenter are collinear.
Below mentioned is the construction part.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 6.3 6

Question 53.
CONSTRUCTION
Follow the steps to construct a nine-point circle. Why is it called a nine-point circle?
Step 1 Construct a large acute scalene triangle.
Step 2 Find the orthocenter and circumcenter of the triangle.
Step 3 Find the midpoint between the orthocenter and circumcenter.
Step 4 Find the midpoint between each vertex and the orthocenter.
Step 5 Construct a circle. Use the midpoint in Step 3 as the center of the circle, and the distance from the center to the midpoint of a side of the triangle as the radius.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 53

Question 54.
PROOF
Prove the statements in parts (a)-(c).
Given \(\overline{L P}\) and \(\overline{M Q}\) are medians of scalenc ∆LMN.
Point R is on \(\vec{L}\)P such that \(\overline{L P} \cong \overline{P R}\). Point S is on \(\vec{M}\)Q such that \(\overline{M Q} \cong \overline{Q S}\).
Prove
a. \(\overline{N S} \cong \overline{N R}\)
b. \(\overline{N S}\) and \(\overline{N R}\) are both parallel to \(\overline{L M}\).
c. R, N, and S are collinear.
Answer:

Maintaining Mathematical Proficiency

Determine whether \(\overline{A B}\) is parallel to \(\overline{C D}\).

Question 55.
A(5, 6), B (- 1, 3), C(- 4, 9), D(- 16, 3)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 55

Question 56.
A(- 3, 6), B(5, 4), C(- 14, – 10), D(- 2, – 7)

Answer:
AB is not parallel to CD.

Explanation:
Slope of AB = \(\frac { 4 – 6 }{ 5 + 3 } \) = \(\frac { -1 }{ 4 } \)
Slopeof CD = \(\frac { -7 + 10 }{ -2 + 14 } \) = \(\frac { 1 }{ 4 } \)
The slopes of AB and CD are not equal. So AB and CD are not parallel.

Question 57.
A (6, – 3), B(5, 2), C(- 4, – 4), D(- 5, 2)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.3 Question 57

Question 58.
A(- 5, 6), B(- 7, 2), C(7, 1), D(4, – 5)

Answer:
AB is parallel to CD.

Explanation:
Slope of AB = \(\frac { 2 – 6 }{ -7 + 5 } \) = 2
Slopeof CD = \(\frac { -5 – 1 }{ 4 – 7 } \) = 2
The slopes of AB and CD are equal. So, AB is parallel to CD.

6.1 and 6.3 Quiz

Find the indicated measure. Explain your reasoning.

Question 1.
UV
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 89

Answer:
UV = 15

Explanation:
SV = VU
2x + 11 = 8x – 1
8x – 2x = 11 + 1
6x = 12
x = 2
UV = 8(2) – 1 = 15

Question 2.
QP
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 90

Answer:
QP = 18

Explanation:
QP = QR
6x = 3x + 9
3x = 9
x = 3
QP = 6(3) = 18

Question 3.
m∠GJK
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 91

Answer:
m∠GJK = 59°

Explanation:
5x – 4 = 4x + 3
x = 7
∠JGK = 4(7) + 3 = 31
m∠GJK = 180 – (31 + 90) = 180 – 121 = 59

Find the coordinates of the circumcenter of the triangle with the given vertices.

Question 4.
A(- 4, 2), B(- 4, – 4), C(0, – 4)

Answer:
The circumcenter is (-2, -1)

Explanation:
Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let A(- 4, 2), B(- 4, – 4), C(0, – 4) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then
PA = PB = PC
PA² = PB² = PC²
PA² = PB²
(x + 4)² + (y – 2)² = (x + 4)² + (y + 4)²
x² + 8x + 16 + y² – 4y + 4 = x² + 8x + 16 + y² + 8y + 16
12y = -12
y = -1
PB² = PC²
(x + 4)² + (y + 4)² = (x – 0)² + (y + 4)²
x² + 8x + 16 + y² + 8y + 16 = x² + y² + 8y + 16
8x = -16
x = -2
The circumcenter is (-2, -1)

Question 5.
D(3, 5), E(7, 9), F(11, 5)

Answer:
The circumcenter is (7, 5)

Explanation:
Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let D(3, 5), E(7, 9), F(11, 5) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then
PD = PE = PF
PD² = PE² = PF²
PD² = PE²
(x – 3)² + (y – 5)² = (x – 7)² + (y – 9)²
x² – 6x + 9 + y² – 10y + 25 = x² – 14x + 49 + y² – 18y + 81
-6x + 14x – 10y + 18y = 130 – 34
8x + 8y = 96
x + y = 12 — (i)
PE² = PF²
(x – 7)² + (y – 9)² = (x – 11)² + (y – 5)²
x² – 14x + 49 + y² – 18y + 81 = x² – 22x + 121 + y² – 10y + 25
-14x + 22x – 18y + 10y = 146 – 130
8x – 8y = 16
x – y = 2 — (ii)
Add (i) & (ii)
x + y + x – y = 12 + 2
2x = 14
x = 7
Put x = 7 in (i)
7 + y = 12
y = 5
The circumcenter is (7, 5)

The incenter of ∆ABC is point N. Use the given information to find the indicated measure.

Question 6.
NQ = 2x + 1, NR = 4x – 9
Find NS.
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 92

Answer:
NS = 11

Explanation:
NQ = NR = NS
2x + 1 = 4x – 9
4x – 2x = 10
2x = 10
x = 5
NQ = 10 + 1 = 11
NS = 11

Question 7.
NU = – 3x + 6, NV = – 5x
Find NT.
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 93

Answer:
NT = 15

Explanation:
NU = NV = NT
-3x + 6 = -5x
-3x + 5x = -6
2x = -6
x = -3
NT = -5(-3) = 15

Question 8.
NZ = 4x – 10, NY = 3x – 1
Find NW.
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 94

Answer:
NW = 26

Explanation:
NZ = NY = NW
4x – 10 = 3x – 1
x = 9
NZ = 4(9) – 10 = 36 – 10 = 26
NW = 26

Find the coordinates of the centroid of the triangle wilt the given vertices.
Question 9.
J(- 1, 2), K(5, 6), L(5, – 2)

Answer:
Centroid of the triangle is (3, 2)

Explanation:
Centroid of the triangle = (\(\frac { -1 + 5 + 5 }{ 3 } \), \(\frac { 2 + 6 – 2 }{ 3 } \))
= (3, 2)

Question 10.
M(- 8, – 6), N(- 4, – 2), P(0, – 4)

Answer:
Centroid of the triangle is (-4, -4)

Explanation:
Centroid of the triangle = (\(\frac { -8 – 4 + 0 }{ 3 } \), \(\frac { -6 – 2 – 4 }{ 3 } \))
= (-4, -4)

Tell whether the orthocenter is inside, on, or outside the triangle. Then find its coordinates.

Question 11.
T(- 2, 5), U(0, 1), V(2, 5)

Answer:
The orthocenter lies inside the triangle TUV

Explanation:
The slope of TU = \(\frac { 1 – 5 }{ 0 + 2 } \) = -2
The slope of the perpendicular line is \(\frac { 1 }{ 2 } \)
The perpendicular line is y – 5 = \(\frac { 1 }{ 2 } \)(x – 2)
2y – 10 = x – 2
x – 2y + 8 = 0
The slope of UV = \(\frac { 5 – 1 }{ 2 – 0 } \) = 2
The slope of the perpendicular line is \(\frac { -1 }{ 2 } \)
The perpendicular line is y – 5 = \(\frac { -1 }{ 2 } \)(x + 2)
2y – 10 = -x – 2
x + 2y – 8 = 0
equate both equations
x – 2y + 8 = x + 2y – 8
-4y = -16
y = 4
x – 2(4) + 8 = 0
x = 0
So, the orthocenter is (0, 4)
The orthocenter lies inside the triangle TUV

Question 12.
X(- 1, – 4), Y(7, – 4), Z(7, 4)

Answer:
The orthocenter lies on the vertex.

Explanation:
The slope of XY = \(\frac { -4 + 4 }{ 7 + 1 } \) = 0
The slope of the perpendicular line is 0
The perpendicular line is y – 4 = 0
y = 4
The slope of XZ =\(\frac { 4 + 4 }{ 7 + 1} \) = 1
The slope of the perpendicular line is -1
The perpendicular line is y + 4 = -1(x – 7)
y + 4 = -x + 7
x + y – 3 = 0
x + 4 – 3 = 0
x = -1
So, the orthocenter is (-1, 4)
The orthocenter lies on the vertex.

Question 13.
A woodworker is culling the largest wheel possible from a triangular scrap of wood. The wheel just touches each side of the triangle, as shown.
Answer:

Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 95

a. Which point of concurrency is the center of the circle? What type
of segments are \(\overline{B G}\), \(\overline{C G}\), and \(\overline{A G}\)?
Answer:

b.
Which theorem can you use to prove that ∆BGF ≅ ∆BGE?
Answer:

c. Find the radius of the wheel to the nearest tenth of a centimeter. Justify your answer.

Answer:

Question 14.
The Deer County Parks Committee plans to build a park at point P, equidistant from the three largest cities labeled X, Y, and Z. The map shown was created b the committee.

Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 96

a. Which point of concurrency did the commIttee use as the location of the Park?
Answer:

b. Did the committee use the best point of concurrency for the location of the park? Ii not, which point would be better to use? Explain.
Answer:

6.4 The Triangle Midsegment Theorem

Exploration 1

Midsegments of a Triangle

Work with a partner. Use dynamic geometry software. Draw any ∆ABC.

Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 97

a. Plot midpoint D of \(\overline{A B}\) and midpoint E of \(\overline{B C}\). Draw \(\overline{D E}\), which is a midsegment of ∆ABC.
Answer:

b. Compare the slope and length of \(\overline{D E}\) with the slope and length of \(\overline{A C}\).
Answer:

c. Write a conjecture about the relationships between the midsegments and sides of a triangle. Test your conjecture by drawing the other midsegments of ∆ABC, dragging vertices to change ∆ABC. and noting whether the relationships hold.
Answer:

Exploration 2

Midsegments of a Triangle

Work with a partner. Use dynamic geometry software. Draw any ∆ABC.

Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 98

a. Draw all three midsegments of ∆ABC.
Answer:

b. Use the drawing to write a Conjecture about the triangle formed by the midsegments of the original triangle.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures.
Answer:

Communicate Your Answer

Question 3.
How are the midsegments of a triangle related to the sides of the triangle?
Answer:

Question 4.
In ∆RST. \(\overline{U V}\) is the rnidsegrnent connecting the midpoints of \(\overline{R S}\) and \(\overline{S T}\). Given
UV = 12, find RT.
Answer:

Lesson 6.4 The Triangle Midsegment Theorem

Monitoring progress

Use the graph of △ABC.

Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 99

Question 1.
In △ABC, show that midsegments \(\overline{D E}\) is parallel to \(\overline{A C}\) and that DE = \(\frac { 1 }{ 2 }\)AC.

Answer:
Find the coordinates of D and E by getting the midpoints of AB, BC
AB midpoint = D(\(\frac { 1 – 1 }{ 2 } \), \(\frac { 4 – 6 }{ 2 } \)) = D(0, -1)
Midpoint of BC = E(\(\frac { 5 – 1 }{ 2 } \), \(\frac { 4 + 0 }{ 2 } \)) = E(2, 2)
Slope of AC = \(\frac { 0 + 6 }{ 5 – 1 } \) =\(\frac { 3 }{ 2 } \)
Slope of DE = \(\frac { 2 + 1 }{ 2 – 0 } \) = \(\frac { 3 }{ 2 } \)
The slopes are same, so the lines are not parallel.
DE = √(2 – 0)² + (2 + 1)² = √13
AC = √(5 – 1)² + (0 + 6)² = √16 + 36 = √52
So, DE = \(\frac { 1 }{ 2 }\)AC

Question 2.
Find the coordinates of the endpoints of midsegments \(\overline{E F}\), Which opposite \(\overline{A B}\). show that \(\overline{E F}\) is parallel to \(\overline{A B}\) and that EF = \(\frac { 1 }{ 2 }\)AB.

Answer:
The coordinates of E = (\(\frac { -1 + 5 }{ 2 } \), \(\frac { 0 + 4 }{ 2 } \)) = (2, 2)
The coordinates of F = (\(\frac { 1 + 5 }{ 2 } \), \(\frac { 0 + 6 }{ 2 } \)) = (3, 3)
The slope of AB = \(\frac { 4 + 6 }{ -1 – 1 } \) = -5
The slope of EF = \(\frac { 3 – 2 }{ 3 – 2 } \) = 1
The slopes are different. So the lines are not parallel
EF = √(3 – 2)² + (3 – 2)² = √2
AB = √(-1 – 1)² + (4 + 6)² = √4 + 100 = √104

Question 3.
In Example 2, find the coordinates of F, the midpoint of \(\overline{O C}\). Show that \(\overline{F E}\) || \(\overline{O B}\) and FE = \(\frac { 1 }{ 2 }\)OB.

Answer:
The coordinates of F = (\(\frac { 0 + 2p }{ 2 } \), \(\frac { 0 + 0 }{ 2 } \))
= (p, 0)
The slope of OB = \(\frac { 2r – 0 }{ 2q – 0 } \) = \(\frac { r }{ q } \)
The slope of FE = \(\frac { r – 0 }{ q + p – p }\) = \(\frac { r }{ q }\)
The slopes are equal. So, the lines are parallel.
FE = √(q + p – p)² + (r – 0)² = √q² + r²
OB = √(2q – 0)² + (2r – 0)² = √4q² + 4r²  = 2√q² + r²
OB = 2(FE)
FE = \(\frac { 1 }{ 2 }\)OB.

Question 4.
Copy the diagram in Example 3. Draw and name the third midsegment.
Then find the length of \(\overline{V S}\) when the length of the third midsegment is 81 inches.

Answer:
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 6.4 1
The length of third midsegment = 81 in
The length of VS = \(\frac { 81 }{ 2 } \) = 40.5

Question 5.
In Example 4. if F is the midpoint of \(\overline{C B}\), what do you know about \(\overline{D F}\)?

Answer:
DF is parallel to AE.

Question 6.
WHAT IF?
In Example 5, you jog down Peach Street to Plum Street, over Plum Street to Cherry Street. up Cherry Street to Pear Street. over Pear Street to Peach Street. and then back home up Peach Street. Do you jog more miles in Example 5? Explain.

Answer:
The distance you jog = 2.25 + 1.4 + 1.3 + .65 = 5.6

Exercise 6.4 The Triangle Midsegment Theorem

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
The ___________ of a triangle is a segment that connects the midpoints of two sides of the triangle.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 1

Question 2.
COMPLETE THE SENTENCE
If \(\overline{D E}\) is the midsegment opposile \(\overline{A C}\) in ∆ABC, then \(\overline{D E}\) || \(\overline{A C}\) and DE = ________ AC by the Triangle Midsegrnent Theorem

Answer:
If \(\overline{D E}\) is the midsegment opposile \(\overline{A C}\) in ∆ABC, then \(\overline{D E}\) || \(\overline{A C}\) and DE = \(\frac { 1 }{ 2 } \) AC by the Triangle Midsegrnent Theorem (Theorem 6.8).

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, use the graph of ∆ABC with midsegments \(\overline{D E}\), \(\overline{E F}\), and \(\overline{D F}\).

Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 100

Question 3.
Find the coordinates of points D, E, and F.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 3

Question 4.
Show that \(\overline{D E}\) is parallel to \(\overline{C B}\) and that DE = \(\frac{1}{2}\)CB.

Answer:
The slope of DE = \(\frac { 0 + 2 }{ -2 + 4 } \) = 1
The slope of CB = \(\frac { -2 + 6 }{ 1 + 3 } \)  = 1
The slopes are same. So \(\overline{D E}\) is parallel to \(\overline{C B}\)
DE = √(-2 + 4)² + (0 + 2)² = √4 + 4 = √8
CB = √(1 + 3)² + (-2 + 6)² = √16 + 16 = √32
So, DE = \(\frac{1}{2}\)CB

Question 5.
Show that \(\overline{E F}\) is parallel to \(\overline{A C}\) and that EF = \(\frac{1}{2}\)AC.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 5

Question 6.
Show that \(\overline{D F}\) is parallel to \(\overline{A B}\) and that DF = \(\frac{1}{2}\)AB.

Answer:
The slope of DF = \(\frac { -4 + 2 }{ -1 + 4 } \) = \(\frac { -2 }{ 3 } \)
The slope of AB = \(\frac { -2 – 2 }{ 1 + 5 } \) = \(\frac { -4 }{ 6 } \) = \(\frac { -2 }{ 3 } \)
The slopes are equal. So, the lines are parallel.
DF = √(-1 + 4)² + (-4 + 2)² = √9 + 4 = √13
AB = √(1 + 5)² + (-2 – 2)² = √36 + 16 = √52
So, DF = \(\frac{1}{2}\)AB.

In Exercises 7-10, \(\overline{D E}\) is a midsegment of ∆ABC Find the value of x.

Question 7.
Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 101
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 7

Question 8.
Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 102

Answer:
x = 10

Explanation:
DE = \(\frac { 1 }{ 2 } \)AB
5 = \(\frac { 1 }{ 2 } \)x
x = 10

Question 9.
Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 103
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 9

Question 10.
Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 104

Answer:
x = 8

Explanation:
BE = EC
x = 8

In Exercise 11-16, \(\overline{X J} \cong \overline{J Y}\), \(\overline{Y L} \cong \overline{L Z}\), and \(\overline{X K} \cong \overline{K Z}\). Copy and complete the statement.

Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 105

Question 11.
\(\overline{J K}\) || __________
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 11

Question 12.
\(\overline{X Y}\) || __________

Answer:
\(\overline{X Y}\) || KL

Question 13.
\(\overline{J L}\) || __________
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 13

Question 14.
\(\overline{J L}\) ≅ __________ ≅ __________

Answer:
\(\overline{J L}\) ≅ JK ≅ KL

Question 15.
\(\overline{J Y}\) ≅ __________ ≅ __________
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 15

Question 16.
\(\overline{J K}\) ≅ __________ ≅ __________

Answer:
\(\overline{J K}\) ≅ JL ≅ KL

MATHEMATICAL CONNECTIONS
In Exercises 17-19. use ∆GHJ, where A, B, and C are midpoints of the sides.
Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 106

Question 17.
When AB = 3x + 8 and GJ = 2x + 24, what is AB?
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 17

Question 18.
When AC = 3y – 5 and HJ = 4y + 2, what is HB?

Answer:
HB = 13

Explanation:
AC = 0.5(HJ)
3y – 5 = 0.5(4y + 2)
3y – 5 = 2y + 1
3y – 2y = 1 + 5
y = 6
HB = 0.5(HJ)
= 0.5(4(6) + 2) = 0.5(26) = 13

Question 19.
When GH = 7 – 1 and CB = 4z – 3. what is GA?
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 19

Question 20.
ERROR ANALYSIS
Describe and correct the error.
Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 107

Answer:
BC = \(\frac { 1 }{ 2 } \)DE
10 = \(\frac { 12 }{ 2 } \)(5)
According to triangle midsegment theorem, AD = DB and AE = EC.

Question 21.
MODELING WITH MATHEMATICS
The distance between consecutive bases on a baseball held is 90 feet. A second baseman stands halfway between first base and second base, a shortstop stands hallway between second base and third base, and a pitcher stands halfway between first base and third base. Find the distance between the shortstop and the pitcher.
Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 108
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 21

Question 22.
PROVING A THEOREM
Use the figure from Example 2 to prose the Triangle Midsegment Theorem (Theorem 6.8) for midsegment \(\overline{D F}\), where F is the midpoint of \(\overline{O C}\).
Answer:

Question 23.
CRITICAL THINKING
\(\overline{X Y}\) is a midsegment of ∆LMN. Suppose \(\overline{D E}\) is called a “quarter segment” of ∆LMN. What do you think an “eighth segment” would be? Make conjectures about the properties of a quarter segment and an eighth segment. Use variable coordinates to verify your conjectures.
Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 109
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 23.1
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 23.2

Question 24.
THOUGHT PROVOKING
Find a real-life object that uses midsegments as part of its structure. Print a photograph of the object and identify the midsegments of one of the triangles in the structure.
Answer:

Question 25.
ABSTRACT REASONING
To create the design shown. shade the triangle formed by the three midsegments of the triangle. Then repeat the process for each unshaded triangle.
Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 110
a. What is the perimeter of the shaded triangle in Stage 1?
b. What is the total perimeter of all the shaded triangles in Stage 2?
c. What is the total perimeter of all the shaded triangles in Stage 3?
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 25.1
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 25.2

Question 26.
HOW DO YOU SEE IT?
Explain how you know that the yellow triangle is the midsegment triangle of the red triangle in the pattern of floor tiles shown.
Big Ideas Math Geometry Solutions Chapter 6 Relationships Within Triangles 111

Answer:
Because the yellow triangle vertices are located at the midpoints of the red triangle sides.

Question 27.
ATTENDING TO PRECISION
The points P(2, 1), Q(4, 5), and R(7, 4) are the midpoints of the sides of a triangle. Graph the three midsegments. Then show how to use your graph and the properties of midsemeriLs to draw the original triangle. Give the coordinates of each vertex.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 27

Maintaining Mathematical Proficiency

Find a counter example to show that the conjecture is false.

Question 28.
The difference of two numbers is always less than the greater number.

Answer:
When we subtract -3 from 4 i.e 4 -(-3) = 7. The number is greater than the individual integers. So, the subtraction of a negative number from a positive number gives an answer greater than both numbers.

Question 29.
An isosceles triangle is always equilateral.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.4 Question 29

6.5 Indirect Proof and Inequalities in One Triangle

Exploration 1

Comparing Angle Measures and Side Lengths

Work with a partner: Use dynamic geometry software. Draw any scalene ∆ABC

a. Find the side lengths and angle measures of the triangle.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 112
Answer:

b. Order the side lengths. Order the angle measures. What do you observe?
Answer:

c. Drug the vertices of ∆ABC to form new triangles. Record the side lengths and angle measures in a table. Write a conjecture about your findings.
Answer:

Exploration 2

A Relationship of the Side Lengths of a Triangle

Work with a partner. Use dynamic geometry software. Draw any ∆ABC.

a. Find the side lengths of the triangle.
Answer:

b. Compare each side length with the sum of the other two side lengths.
ATTENDING TO PRECISION
To be proficient in math, you need to express numerical answers with a degree of precision appropriate for the content.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 113
Answer:

c. Drag the vertices of ∆ABC to form new triangles and repeat parts (a) and (b). Organize your results in a table. Write a conjecture about your findings.
Answer:

Communicate Your Answer

Question 3.
How are the sides related to the angles of a triangle? How are any two sides of a triangle related to the third side?
Answer:

Question 4.
Is it possible for a triangle to have side lengths of 3, 4, and 10? Explain.
Answer:

Lesson 6.5 Indirect Proof and Inequalities in One Triangle

Monitoring Progress

Question 1.
Write in indirect proof that a scalene triangle cannot have two congruent angles.

Answer:

Question 2.
List the angle of ∆PQR in order from smallest to largest.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 114

Answer:
The sides from the largest to smallest are PQ, QR, PR
The angles opposite to sides are ∠R, ∠P, ∠Q
The angles from smallest to the largest are ∠Q, ∠P, ∠R

Question 3.
List the sides of ∆RST in order from shortest to longest.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 115

Answer:
The angles from shortest to longest are ∠R, ∠T, ∠S

Question 4.
A triangle has one side of length 12 inches and another side of length 20 inches. Describe the possible lengths of the third side.

Answer:
The length of the third side must be greater than 8 and lesser than 32.

Explanation:
Let x represent the length of the third side. Draw diagrams to help visualize the small and large values of x. Then use the Triangle Inequality Theorem to write and solve inequalities.
The smallest value of x is x + 12 > 20. So, x > 8
The largest value of x is 12 + 20 > x
32 > x
So, 32 > x or 32 < x
The length of the third side must be greater than 8 and lesser than 32.

Decide Whether it is possible to construct a triangle with the given side lengths. Explain your reasoning.

Question 5.
4 ft, 9 ft, 10 ft

Answer:
It is not possible to construct a triangle.

Explanation:
According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle is greater than the length of the third side.
So, 9 + 4 = 13
13 is not greater than 10.
So, it is not possible to construct a triangle.

Question 6.
8 ft, 9 ft, 18 ft

Answer:
It is possible to construct a triangle.

Explanation:
According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle is greater than the length of the third side.
8 + 9 = 17 > 18
So, it is possible to construct a triangle.

Question 7.
5 cm, 7 cm, 12 cm

Answer:
It is not possible to construct a triangle.

Explanation:
According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle is greater than the length of the third side.
7 + 5 = 12
12 is not greater than 12
so, it is not possible to construct a triangle.

Exercise 6.5 Indirect Proof and Inequalities in One Triangle

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Why is an indirect proof also called a proof by contradiction?
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 1

Question 2.
WRITING
How can you tell which side of a triangle is the longest Irom the angle measures of the
triangle? How can you tell which side is the shortest?

Answer:
The angle which is opposite to the longest side is the longest angle.

Monitoring progress and Modeling with Mathematics

In Exercises 3-6, write the first step in an indirect proof of the statement.

Question 3.
If WV + VU ≠ 12 inches and VU = 5 inches, then WV ≠ 7 inches.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 3

Question 4.
If x and y are odd integers. then xy is odd.

Answer:
Assume x = 1, y = 3
xy = 3

Question 5.
In ∆ABC. if m∠A = 100°, then ∠B is not a right angle.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 5

Question 6.
In ∆JKL, if M is the midpoint of \(\overline{K L}\), then \(\overline{J M}\) is a median.

Answer:
Assume temporarily JM is a median.

In Exercises 7 and 8, determine which two statements contradict each other. Explain your reasoning.

Question 7.
(A) ∆LMN is a right triangle.
(B) ∠L ≅∠V
(C) ∆LMN is equilateral.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 7

Question 8.
(A) Both ∠X and ∠Y have measures greater than 20°.
(B) Both ∠X and ∠Y have measures less than 30°.
(C) m∠X + m∠Y = 62°
Answer:
B and C. If ∠X and ∠Y are less than 30, then  m∠X + m∠Y is not equal to 62.

In Exercises 9 and 10, use a ruler and protractor to draw the given type of triangle. Mark the largest angle and longest side in red and the smallest angle and shortest side in blue. What do you notice?

Question 9.
acute scalene
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 9

Question 10.
right scalene

Answer:
To construct a scalene triangle draw a segment and label it AB. Assuring that three sides have different lengths. Draw an arc with center A and AC as radius draw an arc and draw another arc taking BC as radius and B as centre. Label the point of intersection of two arcs as C.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 6.5 1
The largest side is BC. So, the largest angle is ∠A. The smallest side is AB and the smallest angle is ∠C.

In Exercises 11 and 12, list the angles of the given triangle from smallest to largest.

Question 11.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 116
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 11

Question 12.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 117
Answer:
The sides of △JKL from smallest to largest are KL, JL, JK. So by the triangle, longer side theorem, the angles from smallest to largest are ∠J, ∠K, ∠L.

In Exercises 13-16, list the sides of the given triangle from shortest to longest.

Question 13.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 118
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 13

Question 14.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 119
Answer:
The angles of the triangle from smallest to largest are ∠Z, ∠X, ∠Y. So by the trinagle larger angle theorem, the sides from shortest to largest are XY, YZ, and XZ.

Question 15.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 120
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 15

Question 16.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 121
Answer:
∠D = 180 – (90 + 33) = 180 – 57
The angles from smallest to largest are ∠G, ∠D, ∠F. So by the triangle larger angle theorem, the sides from shortest to longest are FD, FG, and DG.

In Exercises 17-20, describe the possible lengths of the third side of the triangle given the lengths of the other to sides.

Question 17.
5 inches, 12 inches
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 17

Question 18.
12 feet, 18 feet

Answer:
The possible lengths of the third side are greater than 6 and less than 30 ft

Explanation:
x + 12 > 18
x > 6
12 + 18 > x
30 > x or x < 30
The possible lengths of the third side are greater than 6 and less than 30 ft

Question 19.
2 feet, 40 inches
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 19

Question 20.
25 meters, 25 meters

Answer:
The possible lengths of the third side are greater than 0 and less than 50 m.

Explanation:
x + 25 > 25
x > 0
25 + 25 > x
50 > x or x < 50
The possible lengths of the third side are greater than 0 and less than 50 m.

In Exercises 21-23, is it possible to construct a triangle with the given side lengths? If not, explain why not.

Question 21.
6, 7, 11
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 21

Question 22.
3, 6, 9

Answer:
No

Explanation:
3 + 6 = 9 → 9 = 9
6 + 9 = 15 → 15 < 9
9 + 3 = 12 → 12 < 9
No, the sum of any two side lengths of a triangle is less than the length of the third side.

Question 23.
28, 17, 46
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 23

Question 24.
35, 120, 125

Answer:
35 + 120 = 155 → 155 < 125
So, No.

Question 25.
ERROR ANALYSIS
Describe and correct the error in writing the first step of an indirect proof.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 122
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 25

Question 26.
ERROR ANALYSIS
Describe and correct the error in labeling the side lengths 1, 2, and √3 on the triangle.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 123

Answer:
Number 1 is opposite to 90 degrees and 2 is opposite to 60 degrees and the remaining is √3 on the triangle.

Question 27.
REASONING
You are a lawyer representing a client who has been accused of a crime. The crime took place in Los Angeles, California. Security footage shows your client in New York at the time of the crime. Explain how to use indirect reasoning to prove your client is innocent.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 27

Question 28.
REASONING
Your class has fewer than 30 students. The teacher divides your class into two groups. The first group has 15 students. Use indirect reasoning to show that the second group must have fewer than 15 students.

Answer:
The assumption that N2 ≥ 15 is false.

Explanation:
The number of students in a class N < 30
The number of students in the first group N1 = 15
The number of students in the second group N2
N = N1 + N2
Assume temporarily that the number of student in the second group has 15 or more students
N2 ≥ 15
N1 + N2 ≥ 15 + 15
30 ≥ 30
So, the assumption that N2 ≥ 15 is false.

Question 29.
PROBLEM SOLVING
Which statement about ∆TUV is false?
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 124
(A) UV > TU
(B) UV + TV > TU
(C) UV < TV
(D) ∆TUV is isosceles.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 29

Question 30.
PROBLEM SOLVING
In ∆RST. which is a possible side length for ST? Select all that apply.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 125
(A) 7
(B) 8
(C) 9
(D) 10
Answer:
A.

Explanation:
∠R = 180 – (65 + 56) = 180 – 121 = 59, which indicates that ∆RST i isosceles. By the trinagle longer side theorem, ST is 7

Question 31.
PROOF
Write an indirect proof that an odd number is not divisible by 4.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 31

Question 32.
PROOF
Write an indirect proof of the statement
“In ∆QRS, if m∠Q + m∠R = 90°, then m∠S = 90°.”
Answer:

Question 33.
WRITING
Explain why the hypotenuse of a right triangle must always be longer than either leg.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 33

Question 34.
CRITICAL THINKING
Is it possible to decide if three side lengths form a triangle without checking all three inequalities shown in the Triangle Inequality Theorem (Theorem 6. 11)? Explain your reasoning.
Answer:

Question 35.
MODELING WITH MATHEMATICS
You can estimate the width of the river from point A to the tree at point B by measuring the angle to the tree at several locations along the riverbank. The diagram shows the results for locations C and D.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 126
a. Using ∆BCA and ∆BDA, determine the possible widths of the river. Explain your reasoning.
b. What could you do if you wanted a closer estimate?
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 35

Question 36.
MODELING WITH MATHEMATICS
You travel from Fort Peck Lake to Glacier National Park and from Glacier National Park to Granite Peak.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 127
a. Write two inequalities to represent the possible distances from Granite Peak back to Fort Peck Lake.
Answer:

b. How is your answer to part (a) affected if you know that m∠2 < m∠1 and m∠2 < m∠3?
Answer:

Question 37.
REASONING
In the figure. \(\overline{X Y}\) bisects ∠WYZ. List all six angles of ∆XYZ and ∆WXY in order from smallest to largest. Explain ‘our reasoning.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 128
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 37

Question 38.
MATHEMATICAL CONNECTIONS
In ∆DEF, m∠D = (x + 25)°. m∠E = (2x – 4)°, and in m∠F = 63°. List the side lengths and angle measures of the triangle in order from least to greatest.

Answer:

Explanation:
sum of angles is 180 degrees
x + 25 + 2x – 4 + 63 = 180
3x + 84 = 180
3x = 96
x = 32
m∠D = 32 + 25 = 57
m∠E = 2(32) – 4 = 60
m∠F = 63
Thwe angles from least to greatest are m∠D, m∠E, m∠F and the side lengths are EF, DF, DE.

Question 39.
ANALYZING RELATIONSHIPS
Another triangle inequality relationship is given by the Exterior Angle Inequality Theorem. It states:

The measures of an exterior angle of a triangle is greater than the measure of either of the nonadjacent interior angles.

Explain how you know that m∠1 > m∠A and m∠1 > m∠B in ∆ABC with exterior angle ∠1.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 129
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 39

MATHEMATICAL CONNECTIONS
In Exercises 40 and 41, describe the possible values of x.

Question 40.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 130
Answer:
The possible values of x are x > \(\frac { 5 }{ 3 } \) and x < 15

Explanation:
JK + JL > KL
x + 11 + 5x – 9 > 2x +10
6x + 2 > 2x + 10
4x > 8
x > 2
JK + KL > JL
x + 11 + 2x + 10 > 5x – 9
3x + 21 > 5x – 9
30 > 2x
15 > x
JL + KL > JK
5x – 9 + 2x + 10 > x + 11
7x + 1 > x + 11
6x > 10
x > \(\frac { 5 }{ 3 } \)
The possible values of x are x > \(\frac { 5 }{ 3 } \) and x < 15

Question 41.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 131
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 41

Question 42.
HOW DO YOU SEE IT?
Your house is on the corner of Hill Street and Eighth Street. The library is on the corner of View Street and Seventh Street. What is the shortest route to et from your house to the library? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 132
Answer:
Through Washington ave is the shortest route to reach from house to the library.

Question 43.
PROVING A THEOREM
Use the diagram to prove the Triangle Longer Side Theorem (Theorem 6.9).
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 133
Given BC > AB, BD = BA
Prove m∠BAC > m∠C
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 43

Question 44.
USING STRUCTURE
The length of the base of an isosceles triangle is l. Describe the possible lengths for each leg. Explain our reasoning.
Answer:
If other lengths are x and x
Then, x + x > l
2x > l
x > l/2
x + l > x
The possible lengths of each leg are l are l/2.

Question 45.
MAKING AN ARGUMENT
Your classmate claims to have drawn a triangle with one side length of 13 inches and a perimeter of 2 feet. Is this possible? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 45

Question 46.
THOUGHT PROVOKING
Cut two pieces of string that are each 24 centimeters long. Construct an isosceles triangle out of one string and a scalene triangle out of the other. Measure and record the side lengths. Then classify each triangle by its angles.
Answer:

Question 47.
PROVING A THEOREM
Prove the Triangle Inequality Theorem (Theorem 6. 11).
Given ∆ABC
Prove AB + BC > AC, AC + BC > AB, and AB + AC > BC
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 47.1
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 47.2
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 47.3

Question 48.
ATTENDING TO PRECISION
The perimeter of ∆HGF must be between what two integers? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 134
Answer:

Question 49.
PROOF
Write an indirect proof that a perpendicular segment is the shortest segment from a point to a plane.
Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 135
Given \(\overline{P C}\) ⊥ palne M
Prove \(\overline{P C}\) is the shortest from P to plane M.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 49

Maintaining Mathematical Proficiency

Name the indicated angle between the pair of sides given.

Big Ideas Math Answer Key Geometry Chapter 6 Relationships Within Triangles 136

Question 50.
\(\overline{A E}\) and \(\overline{B E}\)
Answer:
The included angle is \(\overline{A E}\) and \(\overline{B E}\) is ∠AEB

Question 51.
\(\overline{A C}\) and \(\overline{D C}\)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 51

Question 52.
\(\overline{A D}\) and \(\overline{D C}\)
Answer:
The included angle is \(\overline{A D}\) and \(\overline{D C}\) is ∠ADC

Question 53.
\(\overline{C E}\) and \(\overline{B E}\)
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.5 Question 53

6.6 Inequalities in Two Triangles

Exploration 1

Comparing Measures in Triangles

Work with a partner. Use dynamic geometry software.

a. Draw ∆ABC, as shown below.
Answer:

b. Draw the circle with center C(3, 3) through the point A(1, 3).
Answer:

c. Draw ∆DBC so that D is a point on the circle.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 137
Answer:

d. Which two sides of ∆ABC are congruent to two sides of ∆DBC? Justify your answer.
Answer:

e. Compare the lengths of \(\overline{A B}\) and \(\overline{D B}\). Then compare the measures of ∠ACB and ∠DCB. Are the results what you expected? Explain.
Answer:

f. Drag point D to several locations on the circle. At each location, repeat part (e). Copy and record your results in the table below.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 138
Answer:

g. Look for a pattern of the measures in our table. Then write a conjecture that summarizes your observations.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures.
Answer:

Communicate Your Answer

Question 2.
If two sides of one triangle are congruent to two sides of another triangle, what can you say about the third sides of the triangles?
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 139
Answer:

Question 3.
Explain how you can use the hinge shown at the left to model the concept described in Question 2.
Answer:

Lesson 6.6 Inequalities in Two Triangles

Monitoring Progress

Use the diagram

Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 140

Question 1.
If PR = PS and m∠QPR > m∠QPS, which is longer, \(\overline{S Q}\) or \(\overline{R Q}\)?

Answer:
Given that,
PR = PS and PQ ≅ PQ by the reflexive property of congruence.
m∠QPR > m∠QPS.
So, two sides of △PSQ are congruent to two sides of △PRQ and included angle in △PRQ is larger.
By the hinge theorem RQ > SQ.

Question 2.
If PR = PS and RQ < SQ, which is larger, ∠RPQ or ∠SPQ?

Answer:
Given that,
PR = PS, RQ < SQ
PQ ≅ PQ by the reflexive property of congruence theorem.
So, two sides of △PSQ are congruent to two sides of △PRQ and third side of △PSQ is longer.
By the converse of the hinge theorem, ∠SPQ > ∠RPQ.

Question 3.
Write a temporary assumption you can make to prove the Hinge Theorem indirectly. What two cases does that assumption lead to?

Answer:
Hinge theorem states that if two sides of the triangle are congruent to two sides of another triangle and the included angle of the first is larger than the included angle of the second, then the third side of the first triangle is longer than the third side of the second triangle.

Question 4.
WHAT IF?
In Example 5, Group C leaves camp and travels 2 miles due north. then turns 40° towards east and travels 1.2 miles. Compare the distances from camp for all three groups.

Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 1
group A = 180 – 45 = 135
group B = 180 – 30 = 150
Group C = 180 – 40 = 140
Because 150 > 140 > 135, the distance Group B from the camp is greater than the distance Group C, A. by the hinge theorem.
The distances are Group B > group C > group A.

Exercise 6.6 Inequalities in Two Triangles

Vocabulary and Core Concept Check

Question 1.
WRITING
Explain why Theorem 6.12 is named the “Hinge Theorem.”
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 Question 1

Question 2.
COMPLETE THE SENTENCE
In ∆ABC and ∆DEF, \(\overline{A B} \cong \overline{D E}\), \(\overline{B C} \cong \overline{E F}\). and AC < DF. So m∠_______ > m ∠ _______ by the Converse of the Hinge Theorem

Answer:
In ∆ABC and ∆DEF, \(\overline{A B} \cong \overline{D E}\), \(\overline{B C} \cong \overline{E F}\). and AC < DF. So m∠E > m ∠B by the Converse of the Hinge Theorem

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, copy and complete the statement with <, >, or = Explain your reasoning.

Question 3.
m∠1 ________ m∠2
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 141
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 Question 3

Question 4.
m∠1 ________ m∠2
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 142

Answer:
m∠1 < m∠2 by the converse of the hinge theorem.

Question 5.
m∠1 ________ m∠2
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 143
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 Question 5

Question 6.
m∠1 ________ m∠2
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 144

Answer:
m∠1 > m∠2 by the converse of the hinge theorem.

In Exercises 7-10. copy and complete the statement with <, >, or =. Explain your reasoning.

Question 7.
AD ________ CD
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 145
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 Question 7

Question 8.
MN ________ LK
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 146

Answer:
MN < KL by the converse of the hinge theorem.

Question 9.
TR ________ UR
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 147
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 Question 9

Question 10.
AC ________ DC
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 148

Answer:
AC > Dc by the converse of the hinge theorem.

PROOF
In Exercises 11 and 12, write a proof.

Question 11.
Given \(\overline{X Y} \cong \overline{Y Z}\), m∠WYZ > m∠WYX
Prove WZ > WX
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 149
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 Question 11

Question 12.
Given \(\overline{B C} \cong \overline{D A}\), DC < AB
Prove m∠BCA > m∠DAC
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 150

Answer:
Given that, AD ≅ BC
AC ≅ AC by the reflexive property of congruence theorem
DC < AB is given
So, m∠BCA > m∠DAC by the hinge theorem.

In Exercises 13 and 14, you and your friend leave on different flights from the same airport. Determine which flight is farther from the airport. Explain your reasoning.

Question 13.
Your flight: Flies 100 miles due west, then turns 20° toward north and flies 50 miles.
Friend’s flight: Flies 100 miles due north, then turns 30° toward east and flies 50 miles.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 Question 13

Question 14.
Your flight: Flies 21omiles due south, then turns 70° toward west and flies 80 miles.
Friend’s flight: Flies 80 miles due north, then turns 50° toward east and flies 210 miles.

Answer:
My friend flight is farther from the airport.

Explanation:
The figure below represents the situation.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 6.6 1
I flew 210 miles north to point B then runed 70° to the west and flew 80 miles to point C.
My friend flew 80 miles north to point D then turned 50° east and flew 210 miles to point F.
We should take into consideration that 70° to west means 110° and 50° to east means 130°
AB = DF
BC = AD
m∠ABC < m∠ADF
So, AC < AF
My friend flight is farther from the airport.

Question 15.
ERROR ANALYSIS
Describe and correct the error in using the Hinge Theorem (Theorem 6.12).
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 151
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 Question 15

Question 16.
REPEATED REASONING
Which is possible measure for ∠JKM? Select all that apply.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 152
(A) 15°
(B) 22°
(C) 25°
(D) 35°

Answer:
(D) 35°

Explanation:
KL ≅ KJ is given
KM ≅ KM by the reflexive property of congruence
LM > MJ
So, m∠LKM > m∠MKJ
25 > 35

Question 17.
DRAWING CONCLUSIONS
The path from E to F is longer than the path from E to D. The path from G to D is the same length as the path from G to F. What can you conclude about the angles of the paths? Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 153
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 Question 17

Question 18.
ABSTRACT REASONING
In ∆EFG, the bisector of ∠F intersects the bisector of ∠G at point H. Explain why \(\overline{F G}\) must be longer than \(\overline{F H}\) or \(\overline{H G}\).

Answer:
The sum of interior angles of a triangle is 180 degrees
The sum of angles F and G cannot be larger than 180 degrees
∠HFG is half of the measurement of ∠F. ∠HGF is half of the measurement of ∠G. thus, ∠HFG + ∠HGF < 90°
∠FHG = 180° – ∠HGF – ∠HFG > 90°

Question 19.
ABSTRACT REASONING
\(\overline{N R}\) is a median of ∆NPQ, and NQ > NP Explain why ∠NRQ is obtuse.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 Question 19

MATHEMATICAL CONNECTIONS
In Exercises 20 and 21, write and solve an inequality for the possible values of x.

Question 20.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 154
Answer:
3x + 2 = x + 3
3x – x = 3 – 2
2x = 1
x = \(\frac { 1 }{ 2 } \)

Question 21.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 155
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 Question 21

Question 22.
HOW DO YOU SEE IT?
In the diagram, triangles are formed by the locations of the players on the basketball court. The dashed lines represent the possible paths of the basketball as the players pass. How does m∠ACB compare with m∠ACD?
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 156

Answer:
BC ≅ CD
AC ≅ AC by the reflexive property of congruence theorem
AB < AD
So, m∠ACB < m∠ACD by the converse of the hinge theorem.

Question 23.
CRITICAL THINKING
In ∆ABC, the altitudes from B and C meet at point D, and m∠BDC. What is true about ∆ABC? Justify your answer.
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 Question 23

Question 24.
THOUGHT PROVOKING
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry, all points are on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. In spherical geometry, state an inequality involving the sum of the angles of a triangle. Find a formula for the area of a triangle in spherical geometry.

Answer:

Maintaining Mathematical proficiency

Find the value of x.

Question 25.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 157
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 Question 25

Question 26.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 158

Answer:
As two sides are congruent, two angles opposite to the congruent sides are also congruent by the converse of the isosceles trinagle theorem.
36 + x + x = 180
36 + 2x = 180
2x = 144
x = 72

Question 27.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 159
Answer:
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 6.6 Question 27

Question 28.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 160
Answer:
The sum of interior angles in  a triangle are 180 degrees
44 + 64 + y = 180
y = 180 – 108
y = 72
x + y = 180
x + 72 = 180
x = 108

Relationships Within Triangles Chapter Review

6.1 Perpendicular and Angle Bisectors

Find the indicated measure. Explain your reasoning.

Question 1.
DC
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 161

Answer:
DC = 20

Explanation:
AD = CD
20 = 7x – 15
7x = 35
x = 5
DC = 7(5) – 15 = 20

Question 2.
RS
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 162

Answer:
RS = 23

Explanation:
PS = SR
6x + 5 = 9x – 4
9x – 6x = 9
3x = 9
x = 3
RS = 9(3) – 4 = 23

Question 3.
m∠JFH
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 163

Answer:
m∠JFH = 47°

Explanation:
m∠JFH = m∠JFG
So, m∠JFH = 47°

6.2 Bisectors of Triangles

Find the coordinates of the circumcenter of the triangle with the given vertices.

Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 164

Question 4.
T(- 6, – 5), U(0, – 1), V(0, – 5)

Answer:
The circumcenter is (-3, -3)

Explanation:
Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let T(- 6, – 5), U(0, – 1), V(0, – 5) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then
PT = PU = PV
PT² = PU² = PV²
PT² = PU²
(x + 6)² + (y + 5)² = (x – 0)² + (y + 1)²
x² + 12x + 36 + y² + 10y + 25 = x² + y² + 2y + 1
12x + 8y + 61 = 1
12x + 8y + 60 = 0
3x + 2y + 15 = 0
PU² = PV²
(x – 0)² + (y + 1)² = (x – 0)² + (y + 5)²
x² + y² + 2y + 1 = x² + y² + 10y + 25
8y + 24 = 0
8y = -24
y = -3
Substitute y = -3 in 3x + 2y + 15 = 0
3x + 2(-3) + 15 = 0
3x – 6 + 15 = 0
3x = -9
x = -3
The circumcenter is (-3, -3)

Question 5.
X(- 2, 1), Y(2, – 3), Z(6, – 3)

Answer:
The circumcenter is (4, 3)

Explanation:
Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let X(- 2, 1), Y(2, – 3), Z(6, – 3) be the vertices of the given triangle and let P(x, y) be the circumcentre of this triangle. Then
PX = PY = PZ
PX² = PY² = PZ²
PX² = PY²
(x + 2)² + (y – 1)² = (x – 2)² + (y + 3)²
x² + 4x + 4 + y² – 2y + 1 = x² – 4x + 4 + y² + 6y + 9
8x – 8y + 5 = 13
8x – 8y – 8 = 0
x – y = 1
PY² = PZ²
(x – 2)² + (y + 3)² = (x – 6)² + (y + 3)²
x² – 4x + 4 + y² + 6y + 9 = x² – 12x + 36 + y² + 6y + 9
8x = 32
x = 4
Substitue x = 4 in x – y = 1
4 – y = 1
-y = -3
y = 3
The circumcenter is (4, 3)

Question 6.
Point D is the incenter of ∆LMN. Find the value of x.

Answer:
x = 13

6.3 Medians and Altitudes of Triangles

Find the coordinates of the centroid of the triangle with the given vertices.

Question 7.
A(- 10, 3), B(- 4, 5), C(- 4, 1)

Answer:
The centroid is (-6, 3)

Explanation:
The centroid = (\(\frac { -10 – 4 – 4 }{ 3 } \), \(\frac { 3 + 5 + 1 }{ 3 } \))
= (-6, 3)

Question 8.
D(2, – 8), E(2, – 2), F(8, – 2)

Answer:
The centroid is (4, -4)

Explanation:
The centroid = (\(\frac { 2 + 2 + 8 }{ 3 } \), \(\frac { -8 – 2 – 2 }{ 3 } \))
= (4, -4)

Tell whether the orthocenter of the triangle with the given vertices is inside, on, or outside the triangle. Then find the coordinates of the orthocenter.

Question 9.
G(1, 6), H(5, 6), J(3, 1)

Answer:
The orthocenter is (3, 5.2)
The orthocenter lies inside the triangle.

Explanation:
The slope of the line HJ = \(\frac { 1 – 6 }{ 3 – 5} \) = \(\frac { 5 }{ 2 } \)
The slope of the perpendicular line = \(\frac { -2 }{ 5 } \)
The perpendicular line is (y – 6) = \(\frac { -2 }{ 5 } \)(x – 1)
5(y – 6) = -2(x – 1)
5y – 30 = -2x + 2
2x + 5y – 32 = 0 — (i)
The slope of GJ = \(\frac { 1 – 6 }{ 3 – 1 } \) = \(\frac { -5 }{ 2 } \)
The slope of the perpendicular line = \(\frac { 2 }{ 5 } \)
The equation of perpendicular line (y – 6) = \(\frac { 2 }{ 5 } \)(x – 5)
5(y – 6) = 2(x – 5)
5y – 30 = 2x – 10
2x – 5y + 20 = 0 — (ii)
Equate both equations
2x + 5y – 32 = 2x – 5y + 20
10y = 52
y = 5.2
Substitute y = 5.2 in (i)
2x + 5(5.2) – 32 = 0
2x + 26 – 32 = 0
2x = 6
x = 3
The orthocenter is (3, 5.2)
The orthocenter lies inside the triangle.

Question 10.
K(-8, 5), L(- 6, 3), M(0, 5)

Answer:
The othrocenter is (-6, -1)
The orthocenter lies outside of the triangle

Explanation:
The slope of LM = \(\frac { 5 – 3}{ 0 + 6} \) = \(\frac { 1 }{ 3 } \)
The slope of the perpendicular line = -3
The perpendicular line is (y – 5) = -3(x + 8)
y – 5 = -3x – 24
3x + y + 19 = 0 —- (ii)
The slope of KL = \(\frac { 3 – 5 }{ -6 + 8 } \) = -1
The slope of the perpendicular line = \(\frac { 1 }{ 2 } \)
The equation of perpendicular line (y – 5) = \(\frac { 1 }{ 2 } \)(x – 0)
2y – 10 = x —- (ii)
Substitute (ii) in (i)
3(2y – 10) + y + 19 = 0
6y – 30 + y + 19 = 0
7y – 11 = 0
y = \(\frac { 11 }{ 7 } \)
x = -6
The othrocenter is (-6, -1)
The orthocenter lies outside of the triangle

6.4 The Triangle Midsegment Theorem

Find the coordinates of the vertices of the midsegment triangle for the triangle with the given vertices.

Question 11.
A(- 6, 8), B(- 6, 4), C(0, 4)

Answer:
The midsegment of AB = (-6, 6)
The midsegment of BC = (-3, 4)
The midsegment of AC = (-3, 6)

Explanation:
The midsegment of AB = (\(\frac { -6 – 6 }{ 2 } \), \(\frac { 8 + 4 }{ 2 } \))
= (-6, 6)
The midsegment of BC = (\(\frac { -6 + 0 }{ 2 } \), \(\frac { 4 + 4 }{ 2 } \))
= (-3, 4)
The midsegment of AC = (\(\frac { -6 + 0 }{ 2 } \), \(\frac { 8 + 4 }{ 2 } \))
= (-3, 6)

Question 12.
D(- 3, 1), E(3, 5), F(1, – 5)

Answer:
The midsegment of DE = (0, 3)
The midsegment of EF = (2, 0)
The midsegment of DF = (-1, -2)

Explanation:
The midsegment of DE = (\(\frac { -3 + 3 }{ 2 } \), \(\frac { 1 + 5 }{ 2 } \))
= (0, 3)
The midsegment of EF = (\(\frac { 3 + 1 }{ 2 } \), \(\frac { 5 – 5 }{ 2 } \))
= (2, 0)
The midsegment of DF = (\(\frac { -3 + 1 }{ 2 } \), \(\frac { 1 – 5 }{ 2 } \))
= (-1, -2)

6.5 Indirect Proof and Inequalities in One Triangle

Describe the possible lengths of the third side of the triangle given the lengths of the other two sides.

Question 13.
4 inches, 8 inches

Answer:
4 < x < 12

Explanation:
4 + 8 > x
12 > x
4 + x > 8
x > 4
8 + x > 4
x > -4
4 < x < 12

Question 14.
6 meters, 9 meters

Answer:
3 < x < 15

Explanation:
6 + 9 > x
15 > x
6 + x > 9
x > 3
9 + x > 6
x > -3
3 < x < 15

Question 15.
11 feet, 18 feet

Answer:
7 < x < 29

Explanation:
11 + 18 > x
29 > x
11 + x > 18
x > 7
18 + x > 11
x > -7
7 < x < 29

Question 16.
Write an indirect proof 0f the statement “In ∆XYZ, if XY = 4 and XZ = 8. then YZ > 4.”

Answer:
4 + 8 > x
12 > x
4 + x > 8
x > 4
8 + x > 4
x > -4
So, 12 > x > 4
Therefore, YZ > 4

6.6 Inequalities in Two Triangles

Use the diagram.

Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 165

Question 17.
If RQ = RS and m∠QRT > m∠SRT, then how does \(\overline{Q T}\) Compare to \(\overline{S T}\)?

Answer:
TQ > ST

Explanation:
Given that,
RQ = RS and m∠QRT > m∠SRT
RT = RT by the reflexive property of congruence theorem
TQ > ST by the converse of the hinge theorem.

Question 18.
If RQ = RS and QT > ST, then how does ∠QRT compare to ∠SRT?

Answer:
Given that,
RQ = RS and QT > ST
RT = RT by the reflexive property of congruence theorem
So, m∠QRT > m∠SRT by the converse of the hinge theorem.

Relationships Within Triangles Chapter Test

In Exercise 1 and 2, \(\overline{M N}\) is a midsegment of ∆JKL. Find the value of x.

Question 1.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 166

Answer:
x = 6

Explanation:
KL = 0.5(MN)
12 = 0.5x
x = 6

Question 2.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 167

Answer:
x = 9

Explanation:
JM = ML
x = 9

Find the indicated measure. Identify the theorem you use.

Question 3.
ST
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 168

Answer:
ST = 17

Explanation:
RS = ST by the perpendicular bisector theorem
3x + 8 = 7x – 4
7x – 3x = 12
4x = 12
x = 3
ST = 7(3) – 4 = 21 – 4 = 17

Question 4.
WY
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 169

Answer:
WY = 32

Explanation:
6x + 2 = 9x – 13 by the angle bisector theorem
3x = 15
x = 5
WY = 6(5) + 2 = 32

Question 5.
BW
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 170

Answer:
BW = 20

Explanation:
WC = BW = AW by the incenter theorem
BW = 20

Copy and complete the statement with <, >, or =.

Question 6.
AB _____ CB

Answer:
AB > CB

Question 7.
m∠1 _____ m∠2

Answer:
m∠1 < m∠2 by the converse of the hinge theorem

Question 8.
m∠MNP ________ m∠NPM

Answer:
m∠MNP < m∠NPM

Question 9.
Find the coordinates of the circumcenter, orthocenter, and centroid of the triangle with vertices A(0, – 2), B(4, – 2), and C(0, 6).

Answer:
Circumcenter is (1.5, -0.5)
orthoceter is (0, -2)
centroid is (1.3, 0.7)

Explanation:
Graph the triangle
Draw the bisector of every angle and their intersection to get the circumcenter
Draw the altitudes and their intersections to get orthocenter
Find the midpoint of all sides and then connect it with the opposite vertex. The intersection of these medians is the centroid.
Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles 1
Circumcenter is (1.5, -0.5)
orthoceter is (0, -2)
centroid is (1.3, 0.7)

Question 10.
Write an indirect proof of the Corollary to the Base Angles Theorem (Corollary 5.2): If ∆PQR is equilateral, then it is equiangular.

Answer:
Assume for contradiction that there is an equilateral triangle namely △ABC. So, AB = BC = AC. According to the base angles theorem, if two sides of a triangle are congruent, then the angles opposite them are congruent.
Since AB = AC, the base angle theorem tells that m∠B = m∠C
Since AB = BC, the base angle theorem tells that m∠A = m∠C
By the transitive property of equality m∠A = m∠B = m∠C, which contradicts the assumption that triangle ABC is equiangular. Thus, the corollary is true.

Question 11.
∆DEF is a right triangle with area A. Use the area for ∆DEF to write an expression for the area of ∆GEH. Justify your answer.
Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 171

Answer:
The area of ∆DEF = A
As HG is the midsegment of ∆DEF.
Area of ∆GEH = A/2

Question 12.
Two hikers start at a visitor center. The first hikes 4 miles hikes due west, then turns 40° toward south and hikes 1.8 miles. The second hikes 4 miles due east, then turns 52° toward north and hikes 1 .8 miles. Which hiker is farther from the visitor center? Explain how you know.

Answer:
The second hiker is farther away from the visitor center as all things being equal cos 128 < cos 140 from distance² = 4² + 1.8² — 2 x 4 x 1.8 cos (angle turned)

In Exercises 13-15, use the map.

Big Ideas Math Answers Geometry Chapter 6 Relationships Within Triangles 172

Question 13.
Describe the possible lengths of Pine Avenue.

Answer:
The possible lengths of pine avenue are between 2 and 16.

Explanation:
7 + 9 > x
16 > x
7 + x > 9
x > 2
9 + x > 7
x > -2
2 < x < 16
The possible lengths of pine avenue are between 2 and 16.

Question 14.
You ride your bike along a trail that represents the shortest distance from the beach to Main Street. You end up exactly halfway between your house and the movie theatre. How long is Pine Avenue? Explain.

Answer:
halfway between your house and the movie theatre = 7/2 = 3.5
Pine Avenue is 3.5 mi

Question 15.
A market is the same distance from your house, the movie theater, and the beach. Copy the map and locate the market.

Answer:
Centroid is the market.

Relationships Within Triangles Cumulative Assessment

Question 1.
Which definitions(s) and/or theorem(s) do you need to use to prove the Converse of the Perpendicular Bisector Theorem (Theorem 6.2)? Select all that apply.
Given CA = CB
Prove Point C lies on the perpendicular bisector of AB.
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 173

Definition of perpendicular bisectorDefinition of angle bisector
Definition of segment congruenceDefinition of angle congruence
Base Angles Theorem (Theorem 5.6)Converse of the Base Angles Theorem (Theorem 5.7)
ASA Congruence Theorem (Theorem 5.10)AAS Congruence Theorem (Theorem 5.11)

Answer:
Definition of perpendicular bisector is used to prove the Converse of the Perpendicular Bisector Theorem.

Question 2.
Use the given information to write a two-column proof.
Given \(\overline{Y G}\) is the perpendicular bisector of \(\overline{D F}\).
Prove ∆DEY ≅ ∆FEY
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 174
Answer:

Question 3.
What are the coordinates of the centroid of ∆LMN?
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 175
(A) (2, 5)
(B) (3, 5)
(C) (4, 5)
(D) (5, 5)

Answer:
(B) (3, 5)

Explanation:
The coordinates of L(3, 8), M (1, 5), N(5, 2)
Centroid = (\(\frac { 3 + 1 + 5 }{ 3 } \), \(\frac { 8 + 5 + 2 }{ 3 } \))
= (3, 5)

Question 4.
Use the steps in the construction to explain how you know that the circle is circumscribed about ∆ABC.
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 176

Answer:
Draw a perpendicular bisector for AB
Draw a perpendicular bisector for BC
Extend those bisectors to meet at one point.
By taking the point of intersection of perpendicular bisectors as radius, draw a circle that connects three vertices of the triangle.

Question 5.
Enter the missing reasons in the proof of the Base Angles Theorem (Theorem 5.6).
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 177
Given \(\overline{A B} \cong \overline{\Lambda C}\)
Prove ∠B = ∠C

StatementsReasons
1. Draw \(\overline{A D}\), the angle bisector of ∠CAB1. Construction of angle bisector
2. ∠CAD ≅ ∠BAD2. ________________________
3. \( \overline{\Lambda B} \cong \overline{A C} \)3. ________________________
4. \( \overline{D A} \cong \overline{D A} \)4. ________________________
5. ∆ADB ≅ ∆ADC5. ________________________
6. ∠B ≅ ∠C6. ________________________

Answer:

StatementsReasons
1. Draw \(\overline{A D}\), the angle bisector of ∠CAB1. Construction of angle bisector
2. ∠CAD ≅ ∠BAD2. Angle Bisector Congruence Theorem
3. \( \overline{\Lambda B} \cong \overline{A C} \)3. Given
4. \( \overline{D A} \cong \overline{D A} \)4. Reflexive property of congruence theorem
5. ∆ADB ≅ ∆ADC5. SAS congruence theorem
6. ∠B ≅ ∠C6. Angle congruence theorem

Question 6.
Use the graph of ∆QRS.
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 178
a. Find the coordinates of the vertices of the midsegment triangle. Label the vertices T, U, and V.

Answer:
T(0, 7), U(2, 4), V(-1, 5)

Explanation:
The coordinates of Q(-3, 8), R(3, 6), S(1, 2)
The midpoint of QR = T = (\(\frac { -3 +3 }{ 2 } \), \(\frac { 8 + 6 }{ 2 } \)) = (0, 7)
Midpoint of RS = U = (\(\frac { 3 + 1 }{ 2 } \), \(\frac { 6 + 2 }{ 2 } \)) = (2, 4)
Midpoint of SR = V = (\(\frac { 1 – 3 }{ 2 } \), \(\frac { 2 + 8 }{ 2 } \)) = (-1, 5)

b. Show that each midsegment joining the midpoints of two sides is parallel to the third side and is equal to half the length of the third side.
Answer:
The slope of RS = \(\frac { 2 – 6 }{ 1 – 3 } \) = 2
The slope of TV = \(\frac { 5 – 7 }{ -1 – 0 } \) = 2
As the slopes are same. The lines are parallel.
TV = √(-1 – 0)² + (5 – 7)² = √1 + 4 = √5
RS = √(1 – 3)² + (2 – 6)² = √4 + 16 = √20
TV = 0.5(RS)

Question 7.
A triangle has vertices X(- 2, 2), Y(1, 4), and Z(2, – 2). Your friend claims that a translation of (x, y) → (x + 2, y – 3) and a translation by a scale factor of 3 will produce a similarity transformation. Do you support our friend’s claim? Explain our reasoning.

Answer:

Question 8.
The graph shows a dilation of quadrilateral ABCD b a scale factor of 2. Show that the line containing points B and D is parallel to the line Containing points B’ and D’.
Big Ideas Math Geometry Answer Key Chapter 6 Relationships Within Triangles 179
Answer:
The coordinates of B(4, 1), D(1, -1), B'(8, 2), D'(2, -2)
The slope of BD = \(\frac { -1 – 1 }{ 1 – 4 } \) = \(\frac { 2 }{ 3 } \)
The slope of B’D’ = \(\frac { -2 – 2 }{ 2 – 8 } \) = \(\frac { -4 }{ -6 } \) = \(\frac { 2 }{ 3 } \)
The slopes are equal. So, BD is parallel to B’D’.

Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays

Big Ideas Math Answers Grade 8 Chapter 6

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Big Ideas Math Book 8th Grade Answer Key Chapter 6 Data Analysis and Displays

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Performance Task

Lesson: 1 Scatter Plots

Lesson: 2 Lines of Fit

Lesson: 3 Two-Way Tables

Lesson: 4 Choosing a Data Display

Chapter: 6 – Data Analysis and Displays 

Data Analysis and Displays STEAM Video/Performance Task

STEAM Video

Fuel Economy
The fuel economy of a vehicle is a measure of the effciency of the vehicle’s engine. What are the benefits of using a car with high fuel economy?
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 1
Watch the STEAM Video “Fuel Economy.” Then answer the following questions.
1. Tory says that the footprint of a vehicle is the area of the rectangle formed by the wheel base and the track width. What is the footprint of a car with a wheel base of 106 inches and a track width of 61 inches?
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 2
2. The graph shows the relationship between the fuel economy and the footprint for four vehicles.
a. What happens to the fuel economy as the footprint increases?
b. Plot the point (50, 40) on the graph. What does this point represent? Does the point fit in with the other points? Explain.

Answer:
1.The footprint of a car = 6,466 sq inches.

Explanation:
In the above-given question,
Tory says that the footprint of a vehicle is the area of the rectangle formed by the wheelbase and the track width.
area of rectangle = length  x width
Given that the footprint of a car = 106 inches.
width with 61 inches.
area = 106 x 61
footprint = 6,466 sq inches.

Answer:
2. a.The fuel economy increases when the footprint increases.

Explanation:
In the above-shown video,
tory says that whenever the footprint increases the fuel economy also increases.
whenever the footprint decreases the fuel economy decreases.

Answer:
2.b.The point (50, 40) represents the outlier.

Explanation:
In the above-given graph,
the point (50, 40) lies in the graph.
it represents the outlier of the graph.
Big-Ideas-Math-Solutions-Grade-8-Chapter-6-Data Analysis and Displays-6-1

Performance Task

Cost vs. Fuel Economy
After completing this chapter, you will be able to use the STEAM concepts you learned to answer the questions in the Video Performance Task. You will be given fuel economies and purchase prices of hybrid and non hybrid car models.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 3
You will be asked to create graphs to compare car models. Why might you want to know the relationship between the fuel economy and the purchase price of a vehicle?

Answer:
The relationship between the fuel economy and the purchase price of a vehicle is proportional.

Explanation:
In the above-given figure,
Given that the city fuel Economy and the purchase price of the cars.
for car A (21.8, 24)
for car B(22.4, 22)
for car C(40.1, 18)
if the fuel economy increases the purchase price also increases.
whenever the economy decreases the purchase price also decreases.

Data Analysis and Displays Getting Ready for Chapter 6

Chapter Exploration
1. Work with a partner. The table shows the number of absences and the final grade for each student in a sample.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 4
a.Write the ordered pairs from the table. Then plot them in a coordinate plane.
b. Describe the relationship between absences and final grade.
c. MODELING A student has been absent6 days. Use the data to predict the student’s final grade. Explain how you found your answer.

Answer:
a. (0, 95), (3, 88), (2, 90), (5, 83), (7, 79), (9, 70), (4, 85), (1, 94), (10, 65), (8, 75).
b. the relationships between the absences and the final grade is decreasing when the absences increases.
c. The student’s final grade is 80.

Explanation:
a. From the above-given figure,
The ordered pairs are:
(0, 95), (3, 88), (2, 90), (5, 83), (7, 79), (9, 70), (4, 85), (1, 94), (10, 65), (8, 75).
Big-Ideas-Math-Solutions-Grade-8-Chapter-6-Data Analysis and Displays-6-2

B. whenever the final grade is decreasing the absences also decrease.
whenever the final grade increases the absence also increases.
c. Given that the student has been absent for 6 days.
The student’s final grade is 80.

2. Work with a partner. Match the data sets with the most appropriate scatter plot. Explain your reasoning.
a. month of birth and birth weight for infants at a day care
b. quiz score and test score of each student in a class
c. age and value of laptop computers
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 5

Vocabulary
The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
scatter plot
two-way table
line of fit
joint frequency

Answer:
Scatter plot = A scatter plot uses dots to represent values for two different numeric variables. The position of each dot on the horizontal and vertical axis indicates values for an individual data point.
Two-way table = A two-way table is a way to display frequencies or relative frequencies for two categorical variables.
Line of fit = Line of fit refers to a line through a scatter plot of data points that best expresses the relationship between those points.
Joint frequency = Joint frequency is joining one variable from the row and one variable from the column.

Explanation:
Scatter plot = A scatter plot uses dots to represent values for two different numeric variables. The position of each dot on the horizontal and vertical axis indicates values for an individual data point.
Two-way table = A two-way table is a way to display frequencies or relative frequencies for two categorical variables.
Line of fit = Line of fit refers to a line through a scatter plot of data points that best expresses the relationship between those points.
Joint frequency = Joint frequency is joining one variable from the row and one variable from the column.

Lesson 6.1 Scatter Plots

EXPLORATION 1

Work with a partner. The weights and circumferences of several sports balls are shown.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 1
a. Represent the data in the coordinate plane. Explain your method.
b. Is there a relationship between the size and the weight of a sports ball? Explain your reasoning.
c. Is it reasonable to use the graph to predict the weights of the sports balls below? Explain your reasoning.
Kickball : circumference = 26 in.
Bowling ball : circumference = 27 in.
Answer:
a.(21, 30), (5, 9), (1.6, 5.3), (16, 28), (2, 8), (1.4, 7), (7, 12), (10, 26).

Explanation:
Big-Ideas-Math-Solutions-Grade-8-Chapter-6-Data Analysis and Displays-6.1-1

Answer:
b. The weight is measured in inches and size is measured in ounces.

Explanation:
In the above-given figure,
the size and the weight of the balls are given.
size and weight of basketball = (21, 30).
size and weight of baseball = (5, 9).
size and weight of golfball = (1.6, 5.3).
size and weight of soccerball = (16, 28).
size and weight of tennis = (2, 8).
size and weight of racquetball = (1.4, 7).
size and weight of softball = (7, 12).
size and weight of volleyball = (10, 26)

Answer:
c. No, it is not reasonable to use the graph.

Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 2

Try It

Question 1.
Make a scatter plot of the data. Identify any outliers, gaps, or clusters.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 3

Answer:
outliers = (120, 70)
gaps =(10, 62) to (45, 85)
clusters =(80, 95), (90, 97), (80, 91)

Explanation:
outliers =(120, 70)
gaps = (10, 62) to (45, 85)
clusters = (80, 95), (90, 97), (80, 91)

Big-Ideas-Math-Solutions-Grade-8-Chapter-6-Data Analysis and Displays-6.1-2

Question 2.
Describe the relationship between the data in Example 1.

Answer:
Linear relationship.

Explanation:
In the above-given graph,
the relationship used is a linear relationship.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 3.
SCATTER PLOT
Make a scatter plot of the data. Identify any outliers, gaps, or clusters. Then describe the relationship between the data.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 4
Answer:
outliers = (3,24)
clusters = 22 to 36
gaps = (4, 27), (8, 36)

Explanation:
outliers = (3,24)
clusters = 22 to 36
gaps = (4, 27), (8, 36)

Big-Ideas-Math-Solutions-Grade-8-Chapter-6-Data Analysis and Displays-6.1-3

Question 4.
WHICH ONE DOESN’T BELONG?
Using the scatter plot, which point does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 5

Answer:
The point (3.5, 3) does not belong with the other three.

Explanation:
In the above-given figure
The points (1,8),  (3, 6.5), and (8, 2) lies in the coordinate plane.
the point (3.5, 3) does not belong with the other three.
the point (3.5, 3) is an outlier.
Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
The table shows the high school and college grade point averages (GPAs) of 10 students. What college GPA do you expect for a high school student with a GPA of 2.7?
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 6

Answer:
The college GPA I expect for a high school student with a GPA of 2.7 is 2.45.

Explanation:
In the above-given points,
given that the college GPA for high school students.
college GPA for 2.4 = high school students of 2.6
so I am expecting the 2.45 for 2.7.

Question 6.
The scatter plot shows the ages of 12 people and the numbers of pets each person owns. Identify any outliers, gaps, or clusters. Then describe the relationship between the data.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 7

Answer:
outliers = (40, 6)
clusters = (20, 2) to (70, 1)
gaps = (0, 30), (1, 35), (2, 50) and so on.

Explanation:
Given that,
the person’s age (years) in the x-axis.
a number of pets owned in the y-axis.
outliers = (40, 6)
clusters = (20, 2) to (70, 1)
gaps = (0, 30), (1, 35), (2, 50) and so on.

Scatter Plots Homework & Practice 6.1

Review & Refresh

Solve the system. Check your solution.
Question 1.
y = – 5x + 1
y = – 5x – 2

Answer:
There is no solution for the given equation.

Explanation:
Given that y = – 5x + 1
y = – 5x – 2
so there is no solution for the given equation.

Question 2.
2x + 2y = 9
x = 4.5 – y

Answer:
9 = 9

Explanation:
Given that,
2x + 2y = 9
x = 4.5 – y
2(4.5 – y) + 2y = 9
9 – 2y + 2y = 9
-2y and + 2y get cancelled on both sides.
9 = 9

Question 3.
y = – x
6x + y = 4

Answer:
x = (4/5 , -4/5)

Explanation:
Given that y = -x
6x + y = 4
6x + (-x) = 4
6x – x = 4
5x = 4
x = (4/5)

Question 4.
When graphing a proportional relationship represented by y = mx, which point is not on the graph?
A. (0, 0)
B. (0, m)
C. (1, m)
D. (2, 2m)

Answer:
Point A is not on the graph.

Explanation:
In the above question,
given that the points are:
(0, 0)
(0, m)
(1, m)
(2, 2m)
the point (0, 0) is not in the graph.

Concepts, Skills, &Problem Solving

USING A SCATTER PLOT The table shows the average prices (in dollars) of jeans sold at different stores and the numbers of pairs of jeans sold at each store in one month. (See Exploration 1, p. 237.)

Question 5.
Represent the data in a coordinate plane.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 8

Answer:
The points are (22, 152), (40, 94), (28, 134), (35, 110), and (46, 81)

Explanation:
In the above-given figure,
The points are (22, 152), (40, 94), (28, 134), (35, 110), and (46, 81)
Big-Ideas-Math-Solutions-Grade-8-Chapter-6-Data Analysis and Displays-6.1-4

Question 6.
Is there a relationship between the average price and the number sold? Explain your reasoning.

Answer:
The linear relationship.

Explanation:
In the above-given figure,
the relationship given is linear relationship.

MAKING A SCATTER PLOT Make a scatter plot of the data. Identify any outliers, gaps, or clusters.
Question 7.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 9

Answer:
Outliers = (102, 63)
gaps = x from 40 to 44
clusters = 82 to 89

Explanation:
outliers = (102, 63)
gaps = x from 40 to 44
clusters = 82 to 89
Big-Ideas-Math-Solutions-Grade-8-Chapter-6-Data Analysis and Displays-6.1-5

Question 8.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 10

Answer:
Outliers = (0, 5.5)
gaps = x from 4.5 to 5.5
clusters = 1.5 to 2.5

Explanation:
outliers = (0, 5.5)
gaps = x from 4.5 to 5.5
clusters = 1.5 to 2.5
Big-Ideas-Math-Solutions-Grade-8-Chapter-6-Data Analysis and Displays-6.1-6

IDENTIFYING RELATIONSHIPS Describe the relationship between the data. Identify any outliers, gaps, or clusters.
Question 9.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 11

Answer:
Outliers = (15, 10)
gaps = from x = 15 to x = 25
clusters = 0
Negative linear relationship.

Explanation:
Outliers = (15, 10)
gaps = from x = 15 to x = 25
clusters = 0
There are no clusters.

Question 10.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 12

Answer:
There are no clusters.
gaps = from x = 4 to x = 36
outliers.

Explanation:
In the above-given figure,
there are no clusters.
gaps = from x = 4 to x = 36
no outliers.

Question 11.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 13

Answer:
There is no relationship.
there are no clusters.
no gaps.
no outliers.

Explanation:
In the above-given graph,
there are no clusters.
no gaps.
no clusters.
there is no relationship.

Question 12.
CRITICAL THINKING
The table shows the average price per pound for honey at a store from 2014 to 2017. Describe the relationship between the data.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 14

Answer:
The relationship is a positive linear relationship.

Explanation:
In the above-figure,
given points are:
(2014, $4.65), (2015, $5.90), (2016, $6.50), and (2017, $7.70)
so the above given is a positive linear relationship.

Question 13.
MODELING REAL LIFE
The scatter plot shows the amount of rainfall and the amount of corn produced by a farm over the last 10 years. Describe the relationship between the amount of rainfall and the amount of corn produced.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 15

Answer:
The relationship is a positive linear relationship.

Explanation:
In  the above-given figure,
outliers = (49, 80)
clusters = from x = 190 to 220.

Question 14.
OPEN-ENDED
Describe a set of real-life data that has a negative linear relationship.
Answer:

Question 15.
MODELING REAL LIFE
The scatter plot shows the total earnings (wages and tips) of a food server during one day.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 16
a. About how many hours must the server work to earn $70?
b. About how much does the server earn for 5 hours of work?
c. Describe the relationship shown by the data.

Answer:
a. 3.5 h
b. 85 $
c. positive linear relationship.

Explanation:
In the above-given graph,
given that,
a. the hours must server work to earn $70 = 3.5 h
b. The server earns for 5 hours of work = $ 85.
c. the relationship is shown by the data = positive linear relationship.

Question 16.
PROBLEM SOLVING
The table shows the memory capacities (in gigabytes) and prices (in dollars) of tablet computers. (a) Make a scatter plot of the data. Then describe the relationship between the data. (b) Identify any outliers, gaps, or clusters. Explain why they might exist.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 17

Answer:
Outliers =(16, 50)
gaps = 128 on x.
clusters = 64, 32, 64

Explanation:
Outliers =(16, 50)
gaps =128 on x.
clusters = 64, 32, 64.

Big-Ideas-Math-Solutions-Grade-8-Chapter-6-Data Analysis and Displays-6.1-7

Question 17.
PATTERNS
The scatter plot shows the numbers of drifting scooters sold by a company.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays 6.1 18
a. In what year were1000 scooters sold?
b. About how many scooters were sold in 2015?
c. Describe the relationship shown by the data.
d. Assuming this trend continues, in what year are about 500 drifting scooters sold?

Answer:
a. 2014
b. about 950 scooters.
c. negative linear relationship.
d. 2019.

Explanation:
In the above-given figure,
Given that the number of vehicles sold in the year.
a. 2014
b. about 950 scooters.
c. negative linear relationship.
d. 2019

Question 18.
DIG DEEPER!
Sales of sunglasses and beach towels at a store show a positive linear relationship in the summer. Does this mean that the sales of one item cause the sales of the other item to increase? Explain.

Answer:
Yes.

Explanation:
In the above-figure,
given that the sales of the sunglasses and beach towels at a store show a positive linear relationship.
yes the sales of one item cause the sales of the other item to increase.

Lesson 6.2 Lines of Fit

EXPLORATION 1

Representing Data by a Linear Equation
Work with a partner. You have been working on a science project for 8 months. Each month, you measured the length of a baby alligator.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 1
a. Use a scatter plot to draw a line that you think best describes the relationship between the data.
b. Write an equation for your line in part(a).
c. MODELING Use your equation in part(b) to predict the length of the baby alligator next September.

Answer:
a. The relation is a linear relationship.

Explanation:
Big-Ideas-Math-Solutions-Grade-8-Chapter-6-Data Analysis and Displays-6.1-8

Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 2

Try It

Question 1.
The table shows the numbers of people who attend a festival over an eight-year period. (a) Make a scatter plot of the data and draw a line of fit. (b) Write an equation of the line of fit. (c) Interpret the slope and the y-intercept of the line of fit.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 3

Answer:
The order pairs (1, 420), (2, 500), (3, 650), (4, 900), (5, 1100), (6, 1500), (7, 1750), (8, 2400)

Explanation:
Big-Ideas-Math-Solutions-Grade-8-Chapter-6-Data Analysis and Displays-6.2-2

Question 2.
Find an equation of the line of best fit for the data in Example 1. Identify and interpret the correlation coefficient.
Answer:

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 3.
FINDING A LINE OF FIT
The table shows the numbers of days spent training and the race times for several people in a race.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 4
a. Make a scatter plot of the data and draw a line of fit.
b. Write an equation of the line of fit.
c. Interpret the slope and the y-intercept of the line of fit.

Answer:

Question 4.
IDENTIFYING RELATIONSHIPS
Find an equation of the line of best fit for the data at the left. Identify and interpret the correlation coefficient
Answer:

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
The ordered pairs show amounts y (in inches) of rainfall equivalent x to inches of snow. About how many inches of rainfall are equivalent to 6 inches of snow? Justify your answer.
(16, 1.5) (12, 1.3) (18, 1.8) (15, 1.5) (20, 2.1) (23, 2.4)
Answer:

Question 6.
The table shows the heights (in feet) of a high jump bar and the number of people who successfully complete each jump. Identify and interpret the correlation coefficient.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 5
Answer:

Lines of Fit Homework & Practice 6.2

Review & Refresh

Describe the relationship between the data. Identify any outliers, gaps, or clusters.
Question 1.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 6

Answer:
Negative linear relationship.
outliers = (6, 10)
clusters = 0
gaps = 0

Explanation:
In the above-given figure,
The relationship is negative linear relationship.
outliers = (6, 10)
cluster = 0
gaps = 0
there are no clusters and no gaps.

Question 2.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 7

Answer:

Question 3.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 8

Answer:

positive linear relationships.
outliers = 0
gaps = 0
clusters = x = 11 to x = 15

Explanation:
In the above-given figure,
given that
positive linear relationship.
outliers = 0
gaps = 0
clusters = x = 11 to x = 15

Write the fraction as a decimal and a percent.
Question 4.
\(\frac{29}{100}\)

Answer:
Decimal = 0.29
percent = 29 %

Explanation:
Given that
(29/100)
0.29
percent = 29%
decimal = 0.29

Question 5.
\(\frac{7}{25}\)

Answer:
Decimal = 0.28
percent = 28%

Explanation:
Given that
(7/25) = 0.28
decimal = 0.28
percent = 28

Question 6.
\(\frac{35}{50}\)

Answer:
Decimal = 0.7
percent = 0.007

Explanation:
Given that
(35/50) = 0.7
decimal = 0.7
percent = 0.007

Concepts, Skills, &Problem Solving
REPRESENTING DATA BY A LINEAR EQUATION Use a scatter plot to draw a line that you think best describes the relationship between the data. (See Exploration 1, p. 243.)
Question 7.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 9

Answer:
The points are (0,0), (1, 0.8), (2, 1.50), (3, 2.20), (4, 3.0), (5, 3.75)

Explanation:
In the above-given figure,
Given that :
the points are (0, 0), (1, 0.8), (2, 1.50), (3, 2.20), (4, 3.0), (5, 3.75)
The blue berries are in the x-axis.
weight is measured in pounds.
weight is shown in the y-axis.
Big-Ideas-Math-Solutions-Grade-8-Chapter-6-Data Analysis and Displays-6.2-3
Question 8.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 10

Answer:
The given points are (0,91), (2, 82), (4, 74), (6, 65), (8, 55), (10, 43).

Explanation:
In the above-given figure,
Given that :
the points are (0, 91, (2, 82), (4, 74), (6, 65), (8, 55), (10, 43)
The Age is given on the x-axis.
value is measured in dollars.
value is given in the y-axis.
Big-Ideas-Math-Solutions-Grade-8-Chapter-6-Data Analysis and Displays-6.2-4
Question 9.
FINDING A LINE OF FIT
The table shows the daily high temperatures (°F)and the numbers of hot chocolates sold at a coffee shop for eight randomly selected days.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 11
a. Make a scatter plot of the data and draw a line of fit.
b.Write an equation of the line of fit.
c. Interpret the slope and the y-intercept of the line of fit.

Answer:
a.The given points are (30, 45), (36, 43), (44, 36), (51, 35), (60, 30), (68, 27), (75, 23), (82, 17).
b. y = -0.5x + 60
c. you could expect that 60 hot chocolates are sold when the temperature is 0 degree f, and the sales decrease by 1 hot chocolate for every 2 degrees f increase in temperature.

Explanation:
a.The given points are (30, 45), (36, 43), (44, 36), (51, 35), (60, 30), (68, 27), (75, 23), (82, 17).
b. y = -0.5x + 60
c. you could expect that 60 hot chocolates are sold when the temperature is 0 degree f, and the sales decrease by 1 hot chocolate for every 2 degrees f increase in temperature.

Big-Ideas-Math-Solutions-Grade-8-Chapter-6-Data Analysis and Displays-6.2-5

Question 10.
NUMBER SENSE
Which correlation coefficient indicates a stronger relationship: – 0.98 or 0.91? Explain.

Answer:
0.91 indicates a stronger correlation coefficient.

Explanation:
In the above-given question,
-0.98 is a negative value and 0.91 is a positive value.
So 0.91 indicates a stronger correlation coefficient.

Question 11.
IDENTIFYING RELATIONSHIPS
The table shows the admission costs (in dollars) and the average number of daily visitors at an amusement park each year for the past 8 years. Find an equation of the line of best fit. Identify and interpret the correlation coefficient.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 12

Answer:
The equation for the line of best fit is Y = -4.9x + 1042
about -0.969.
strong negative correlation.

Explanation:
In the above-given figure,
The given points are (20, 940), (21, 935), (22, 940), (24, 925), (25, 920), (27, 905), (28, 910), and (30, 890)
The equation for the line of best fit is y = -4.9x + 1042.
about -0.969.
strong negative correlation.

Question 12.
REASONING
The table shows the weights(in pounds) and the prescribed dosages (in milligrams) of medicine for six patients.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 13
a. Find an equation of the line of best fit. Identify and interpret the correlation coefficient.
b. Interpret the slope of the line of best fit.
c. A patient who weighs 140 pounds is prescribed 135 milligrams of medicine. How does this affect the line of best fit?
Answer:

Question 13.
MODELING REAL LIFE
The table shows the populations (in millions) and the numbers of electoral votes assigned for eight states in the 2016 presidential election.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 14
a. Find an equation of the line of best fit. Identify and interpret the correlation coefficient.
b. Interpret the slope of the line of best fit.
c. Interpret the y-intercept of the line of best fit.
d. RESEARCH Research the Electoral College to explain the meaning of your answer in part(c).

Answer:
a. y = 1.3 x + 2; about 0.9995; strong positive correlation.
b. The number of electoral votes increases by 1.3 for every increase of 1  million people in the state.
c. A state with a population of 0 has 2 electoral votes.
d. The number of electoral votes a state has is based on the number of members that the state has in congress. Each state has 2 senators, plus a number of members of the House of Representatives based on its population. so, the y-intercept is 2 because a hypothetical state with no population would still have 2 senators.

Explanation:
a. y = 1.3 x + 2; about 0.9995; strong positive correlation.
b. The number of electoral votes increases by 1.3 for every increase of 1  million people in the state.
c. A state with a population of 0 has 2 electoral votes.
d. The number of electoral votes a state has is based on the number of members that the state has in congress. Each state has 2 senators, plus a number of members of the House of Representatives based on its population. so, the y-intercept is 2 because a hypothetical state with no population would still have 2 senators.

Question 14.
MODELING REAL LIFE
The table shows the numbers (in millions) of active accounts for two social media websites over the past five years. Assuming this trend continues, how many active accounts will Website B have when Website A has 280 million active accounts? Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 15

Answer:

Question 15.
DIG DEEPER!
The table shows the heights y(in feet) of a baseball x seconds after it was hit.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays 6.2 16
a. Predict the height after 5 seconds.
b. The actual height after 5 seconds is about 3 feet. Why might this be different from your prediction?

Answer:
a. 251 ft.
b. The height of the baseball is not linear.

Explanation:
a. The height after 5 seconds is 251 feet.
Given that the seconds on the x-axis and height on the y-axis.
the points are (0, 3), (0.5, 39), (1, 67), (1.5, 87), and (2, 99).
b. The actual height after 5 seconds is about 3 feet.

Lesson 6.3 Two-Way Tables

EXPLORATION 1

Analyzing Data
Work with a partner. You are the manager of a sports shop. The table shows the numbers of soccer T-shirts that your shop has left in stock at the end of a soccer season.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 1
a. Complete the table.
b. Are there any black-and-gold XL T-shirts in stock? Justify your answer.
c. The numbers of T-shirts you ordered at the beginning of the soccer season are shown below. Complete the table.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 2
d. REASONING How would you alter the numbers of T-shirts you order for the next soccer season?
Answer:

Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 3

Try It

Question 1.
How many students in the survey above studied for the test and failed?
Answer:

Question 2.
You randomly survey students in a cafeteria about their plans for a football game and a school dance. The two-way table shows the results. Find and interpret the marginal frequencies for the survey.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 4
Answer:

Question 3.
You randomly survey students about whether they buy a school lunch or pack a lunch. The results are shown. Make a two-way table that includes the marginal frequencies.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 5
Answer:

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 4.
READING A TWO-WAY TABLE
The results of a music survey are shown in the two-way table. How many students dislike both country and jazz? How many students like country but dislike jazz?
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 6
Answer:

Question 5.
MAKING A TWO-WAY TABLE
You randomly survey students about their preference for a class field trip. The results are shown in the tally sheets. Make a two-way table that includes the marginal frequencies.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 7
Answer:

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 6.
The results of a voting survey are shown in the two-way table. For each age group, what percent of voters prefer Candidate A? Candidate B? Determine whether there is a relationship between age and candidate preference.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 8
Answer:

Question 7.
You randomly survey 40 students about whether they play an instrument. You find that8 males play an instrument and 13 females do not play an instrument. A total of 17 students in the survey play an instrument. Make a two-way table that includes the marginal frequencies.
Answer:

Question 8.
Collect data from each student in your math class about whether they like math and whether they like science. Is there a relationship between liking math and liking science? Justify your answer.
Answer:

Two-Way Tables Homework & Practice 6.3

Review & Refresh

Find an equation of the line of best fit for the data.
Question 1.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 9

Answer:
The line y = 12.6x + 75.8 best fit for the data.

Explanation:
In the above-given figure,
Given that the points are (0,75), (1, 91), (2, 101), (3, 109) and (4, 129).
The line y = 12.6x + 75.8 is the best fit for the data.

Question 2.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 10
Answer:

The vertices of a triangle are A (1, 2), B (3, 1), and C (1, – 1). Draw the figure and its image after the translation.
Question 3.
4 units left
Answer:

Question 4.
2 units down
Answer:

Question 5.
(x – 2, y + 3)
Answer:

Concepts, Skills, &Problem Solving

ANALYZING DATA In Exploration 1, determine how many of the indicated T-shirt are in stock at the end of the soccer season. (See Exploration 1, p. 249.)
Question 6.
black-and-white M

Answer:
4 T-shirts are in stock at the end of the soccer season.

Explanation:
In the above-given Exploration 1,
Given that The T-shirts are in stock.
4 T-shirts are in stock at the end of the soccer season.

Question 7.
blue-and-gold XXL

Answer:
0 shirts.

Explanation:
In the above-given Exploration 1,
Given that The T-shirts are in stock.
0 T-shirts are in stock at the end of the soccer season.

Question 8.
blue-and-white L

Answer:
1 T-shirt.

Explanation:
In the above-given Exploration 1,
Given that the T-shirts are in stock.
1 T-shirt is in stock at the end of the soccer season.

READING A TWO-WAY TABLE You randomly survey students about participating in a yearly fundraiser. The two-way table shows the results.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 11
Question 9.
How many female students participateFundraiserin the fundraiser?

Answer:
51 students participate.

Explanation:
In the above-given table,
Given that male and female students are participated in the fundraiser.
so 51 female students participate.

Question 10.
How many male students do not participate in the fundraiser?

Answer:
30 male students do not participate.

Explanation:
In the above-given table,
Given that male and female students are participated in the fundraiser.
so 30 male students do not participate.

FINDING MARGINAL FREQUENCIES Find and interpret the marginal frequencies.
Question 11.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 12

Answer:
71 students are juniors.
75 students are seniors.
93 students will attend the school play.
53 students will not attend the school play.
146 students were surveyed.

Explanation:
In the above-given table,
Given that students of the class participate in the school play.
71 students are juniors.
75 students are seniors.
93 students will attend the school play.
53 students will not attend the school play.
146 students were surveyed.

Question 12.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 13

Answer:
The data plan of 78 people is limited for the cell phone company A.
The data plan of  94 people is limited for the cell phone company B.
The data plan of 175 people is unlimited for the cell phone company A.
The data plan of 135 people is unlimited for the cell phone company B.
482 people were surveyed.

Explanation:
In the above-given table,
The data plan of the cell phone company are given.
The data plan of 78 people is limited for the cell phone company A.
The data plan of  94 people is limited for the cell phone company B.
The data plan of 175 people is unlimited for the cell phone company A.
The data plan of 135 people is unlimited for the cell phone company B.
482 people were surveyed.

Question 13.
MAKING A TWO-WAY TABLE
A researcher randomly surveys people with a medical condition about whether they received a treatment and whether their condition improved. The results are shown. Make a two-way table that includes the marginal frequencies.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 14

Answer:
The people who improved with treatment = 34.
The people who did not improve with treatment = 10
The people who improved with no treatment = 12.
The people who did not improve with no treatment = 29
Totally are about 85 people.

Explanation:
The people who improved with treatment = 34.
The people who did not improve with treatment = 10
The people who improved with no treatment = 12.
The people who did not improve with no treatment = 29
Totally are about 85 people.

Question 14.
MODELING REAL LIFE
You randomly survey students in your school about the color of their eyes. The results are shown in the tables.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 15
a. Make a two-way table.
b. Find and interpret the marginal frequencies for the survey.
c. For each eye color, what percent of the students in the survey are male? female? Organize the results in a two-way table.
Answer:

Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 16
Question 15.
REASONING
Use the information from Exercise 14. For each gender, what percent of the students in the survey have green eyes? blue eyes? brown eyes? Organize the results in a two-way table.
Answer:

Question 16.
CRITICAL THINKING
What percent of students in the survey in Exercise 14 are either female or have green eyes? What percent of students in the survey are males who do not have green eyes? Find and explain the sum of these two percents.
Answer:

Question 17.
MODELING REAL LIFE
You randomly survey people in your neighborhood about whether they have at least $1000 in savings. The results are shown in the tally sheets. For each age group, what percent of the people have at least $1000 in savings? do not have at least $1000 in savings? Determine whether there is a relationship between age and having at least $1000 in savings.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 17
Answer:

Question 18.
DIG DEEPER!
The three-dimensional bar graph shows information about the numbers of hours students at a high school work at part-time jobs during the school year.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays 6.3 18
a. Make a two-way table that represents the data. Use estimation to find the entries in your table.
b. A newspaper article claims that more males than females drop out of high school to work full-time. Do the data support this claim? Explain your reasoning.
Answer:

Lesson 6.4 Choosing a Data Display

EXPLORATION 1

Displaying Data
Work with a partner. Analyze and display each data set in a way that best describes the data. Explain your choice of display.

a. NEW ENGLAND ROADKILL A group of schools in New England participated in a two-month study. They reported 3962 dead animals.
Birds: 307
Mammals: 2746
AmphibiAnswer: 145
Reptiles: 75
Unknown: 689
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 1
b. BLACK BEAR ROADKILL The data below show the numbers of black bears killed on a state’s roads each year for 20 years.
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 2
c. RACCOON ROADKILL A one-week study along a four-mile section of road found the following weights (in pounds) of raccoons that had been killed by vehicles.
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 3
d. What can be done to minimize the number of animals killed by vehicles?
Answer:

Try It

Choose an appropriate data display for the situation. Explain your reasoning.
Question 1.
the population of the United States divided into age groups
Answer:

Question 2.
the number of students in your school who play basketball, football, soccer, or lacrosse
Answer:

Tell whether the data display is appropriate for representing the data in Example 2. Explain your reasoning.
Question 3.
dot plot
Answer:

Question 4.
circle graph
Answer:

Question 5.
stem-and-leaf plot
Answer:

Question 6.
Which bar graph is misleading? Explain.
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 4
Answer:

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

CHOOSING A DATA DISPLAY Choose an appropriate data display for the situation. Explain your reasoning.
Question 7.
the percent of band students playing each instrument
Answer:

Question 8.
a comparison of the amount of time spent using a tablet computer and the remaining battery life
Answer:

Question 9.
IDENTIFYING A MISLEADING DISPLAY
Is the box-and-whisker plot misleading? Explain.
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 5
Answer:

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 10.
An employee at an animal shelter creates the histogram shown. A visitor concludes that the number of 7-year-old to 9-year-old dogs is triple the number of 1-year-old to 3-year-old dogs. Determine whether this conclusion is accurate. Explain.
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 6
Answer:

Question 11.
DIG DEEPER!
A business manager creates the line graph shown. (a) How do the data appear to change over time? Explain why this conclusion may not be accurate. (b) Why might the business manager want to use this line graph?
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 7
Answer:

Choosing a Data Display Homework & Practice 6.4

Review & Refresh

You randomly survey students about whether they recycle. The two-way table shows the results.
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 8
Question 1.
How many male students recycle? How many female students do not recycle?
Answer:

Question 2.
Find and interpret the marginal frequencies.
Answer:

Find the slope and the y-intercept of the graph of the linear equation.
Question 3.
y = 4x + 10
Answer:

Question 4.
y = – 3.5x – 2
Answer:

Question 5.
y – 8 = – x
Answer:

Concepts, Skills, & Problem Solving

Question 6.
DISPLAYING DATA
Analyze and display the data in a way that best describes the data. Explain your choice of display. (See Exploration 1, p. 255.)
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 9

Answer:

CHOOSING A DATA DISPLAY Choose an appropriate data display for the situation. Explain your reasoning.
Question 7.
a student’s test scores and how the scores are spread out

Answer:
stem and leaf plot shows how data is distributed.

Question 8.
the prices of different televisions and the numbers of televisions sold
Answer:

Question 9.
the outcome of rolling a number cube
Answer:

Question 10.
the distance a person drives each month
Answer:

Question 11.
IDENTIFYING AN APPROPRIATE DISPLAY
A survey asked 800 students to choose their favorite school subject. The results are shown in the table. Tell whether each data display is appropriate for representing the portion of students who prefer math. Explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 10
Answer:

Question 12.
IDENTIFYING AN APPROPRIATE DISPLAY
The table shows how many hours you worked as a lifeguard from May to August. Tell whether each data display is appropriate for representing how the number of hours worked changed during the 4 months. Explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 11
Answer:

Question 13.
WRITING
When should you use a histogram instead of a bar graph to display data? Use an example to support your answer.
Answer:

IDENTIFYING MISLEADING DISPLAYS Which data display is misleading? Explain.
Question 14.
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 12
Answer:

Question 15.
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 13
Answer:

Question 16.
REASONING
What type of data display is appropriate for showing the mode of a data set?
Answer:

Question 17.
CRITICAL THINKING
The director of a music festival creates the data display shown. A customer concludes that the ticket price for Group C is more than double the ticket price for Group A. Determine whether this conclusion is accurate. Explain.
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 14
Answer:

Question 18.
PATTERNS
A scientist gathers data about a decaying chemical compound and creates the scatter plot shown.
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 15
a.The scientist concludes that there is a negative linear relationship between the data. Determine whether this conclusion is accurate. Explain.
b. Estimate the amount of the compound remaining after 1 hour, 3 hours, 5 hours, and 7 hours.
Answer:

Question 19.
REASONING
A survey asks 100 students to choose their favorite sports. The results are shown in the circle graph.
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 16
a. Explain why the graph is misleading.
b. What type of data display is more appropriate for the data? Explain.
Answer:

Question 20.
STRUCTURE
With the help of computers, mathematicians have computed and analyzed trillions of digits of the irrational number π. One of the things they analyze is the frequency of each of the numbers 0 through 9. The table shows the frequency of each number in the first 100,000 digits of π.
a. Display the data in a bar graph.
b. Display the data in a circle graph.
c. Which data display is more appropriate? Explain.
d. Describe the distribution.
Big Ideas Math Solutions Grade 8 Chapter 6 Data Analysis and Displays 6.4 17
Answer:

Data Analysis and Displays Connecting Concepts

Using the Problem-Solving Plan
Question 1.
You randomly survey middle school students about whether they prefer action, comedy, or animation movies. The two-way table shows the results. Estimate the probability that a randomly selected middle school student prefers action movies.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cc 1
Understand the problem.
You know the results of a survey about movie preference. You are asked to estimate the probability that a randomly selected middle school student prefers action movies.

Make a plan.
Find the marginal frequencies for the data. Then use the marginal frequencies to find the probability that a randomly selected middle school student prefers action movies.

Solve and check.
Use the plan to solve the problem. Then check your solution.
Answer:

Question 2.
An equation of the line of best fit for a data set is y = – 0.68x + 2.35. Describe what happens to the slope and the y-intercept of the line when each y-value in the data set increases by 7.
Answer:

Question 3.
On a school field trip, there must be 1 adult chaperone for every 16 students. There are 8 adults who are willing to be a chaperone for the trip, but only the number of chaperones that are necessary will attend. Ina class of 124 students, 80 attend the trip. Make a two-way table that represents the data.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cc 2
Answer:

Performance Task

Cost vs. Fuel Economy
At the beginning of this chapter, you watched a STEAM Video called “Fuel Economy.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cc 3

Data Analysis and Displays Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 1

Graphic Organizers

You can use Information Frame an to help organize and remember a concept. Here is an example of an Information Frame for scatter plots.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 2

Choose and complete a graphic organizer to help you study the concept.
1. lines of fit
2. two-way tables
3. data displays
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 3

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 4

6.1 Scatter Plots   (pp. 237–242)
Learning Target: Use scatter plots to describe patterns and relationships between two quantities.

Question 1.
Make a scatter plot of the data. Identify any outliers, gaps, or clusters.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 5
Answer:

Describe the relationship between the data. Identify any outliers, gaps, or clusters.
Question 2.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 6
Answer:

Question 3.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 7
Answer:

Question 4.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 8
Answer:

Question 5.
Your school is ordering custom T-shirts. The scatter plot shows the numbers of T-shirts ordered and the cost per shirt. Describe the relationship between the numbers of T-shirts ordered and the cost per T-shirt.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 9.1
Answer:

Question 6.
Describe a set of real-life data that has each relationship.
a. positive linear relationship
b. no relationship
Answer:

Question 7.
The table shows the numbers of hours a waitress works and the amounts she earns in tips. How many hours do you expect the waitress to work when she earns $42 in tips?
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 9
Answer:

6.2 Lines of Fit   (pp. 243–248)
Learning Target: Use lines of fit to model data.

Question 8.
The table shows the numbers of students at a middle school over a 10-year period.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 10
a. Make a scatter plot of the data and draw a line of fit.
b. Write an equation of the line of fit.
c. Interpret the slope and the y-intercept of the line of fit.
d. Predict the number of students in year 11.
Answer:

Question 9.
Find an equation of the line of best fit for the data in Exercise 8. Identify and interpret the correlation coefficient.
Answer:

Question 10.
The table shows the revenue (in millions of dollars) for a company over an eight-year period. Assuming this trend continues, how much revenue will there be in year 9?
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 11.1
Answer:

6.3 TwoWay Tables   (pp. 249–254)
Learning Target: Use two-way tables to represent data. You randomly survey students about participating in the science fair. The two-way table shows the results.

Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 11
Question 11.
How many male students participate in the science fair?
Answer:

Question 12.
How many female students do not participate in the science fair?
Answer:

Question 13.
You randomly survey students in your school about whether they liked a recent school play. The two-way table shows the results. Find and interpret the marginal frequencies.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 13
Answer:

Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 14
You randomly survey people at a mall about whether they like the new food court. The results are shown.
Question 14.
Make a two-way table that includes the marginal frequencies.
Answer:

Question 15.
For each group, what percent of the people surveyed like the food court? dislike the food court? Organize your results in a two-way table.
Answer:

Question 16.
Does your table in Exercise 15 show a relationship between age and whether people like the food court?
Answer:

6.4 Choosing a Data Display (pp. 255–262)

Learning Target: Use appropriate data displays to represent situations.

Choose an appropriate data display for the situation. Explain your reasoning.
Question 17.
the numbers of pairs of shoes sold by a store each week
Answer:

Question 18.
the percent of votes that each candidate received in an election.
Answer:

Question 19.
Bird banding is attaching a tag to a bird’s wing or leg to track the movement of the bird. This provides information about the bird’s migration patterns and feeding behaviors. The table shows the numbers of robins banded in Pennsylvania over 5 years. Tell whether each data display is appropriate for representing how the number of bandings changed during the 5 years. Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 6 Data Analysis and Displays cr 19
Answer:

Question 20.
Give an example of a bar graph that is misleading. Explain your reasoning.
Answer:

Question 21.
Give an example of a situation where a dot plot is an appropriate data display. Explain your reasoning.
Answer:

Data Analysis and Displays Practice Test

Question 1.
The graph shows the population (in millions) of the United States from 1960 to 2010.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays pt 1
a. In what year was the population of the United States about 180 million?
b. What was the approximate population of the United States in 1990?
c. Describe the relationship shown by the data.
Answer:

Question 2.
The table shows the weight of a baby over several months.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays pt 2
a. Make a scatter plot of the data and draw a line of fit.
b. Write an equation of the line of fit.
c. Interpret the slope and the y-intercept of the line of fit.
Answer:

Question 3.
You randomly survey students at your school about what type of books they like to read. The two-way table shows your results. Find and interpret the marginal frequencies.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays pt 3
Answer:

Choose an appropriate data display for the situation. Explain your reasoning.
Question 4.
magazine sales grouped by price range
Answer:

Question 5.
the distance a person hikes each week
Answer:

Question 6.
The table shows the numbers of AP exams (in thousands) taken from 2012 to 2016, where x = 12 represents the year 2012. Find an equation of the line of best fit. Identify and interpret the correlation coefficient.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays pt 6
Answer:

Question 7.
You randomly survey shoppers at a supermarket about whether they use reusable bags. Of 60 male shoppers,15 use reusable bags. Of 110 female shoppers,60 use reusable bags. Organize your results in a two-way table. Include the marginal frequencies. Estimate the probability that a randomly selected male shopper uses reusable bags.
Big Ideas Math Answers 8th Grade Chapter 6 Data Analysis and Displays pt 7
Answer:

Data Analysis and Displays Cumulative Practice

Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays cp 1
Question 1.
What is the solution of the system of linear equations?
y = 2x – 1
y = 3x + 5
A. ( 13, 6)
B. (- 6, – 13)
C. (- 13, 6)
D. (- 6, 13)
Answer:

Question 2.
The diagram shows parallel lines cut by a transversal. Which angle is the corresponding angle for ∠6 ?
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays cp 2
F. ∠2
G. ∠3
H. ∠4
I. ∠8
Answer:

Question 3.
You randomly survey students in your school. You ask whether they have jobs. You display your results in the two-way table. How many male students do not have a job?
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays cp 3
Answer:

Question 4.
Which scatter plot shows a negative relationship between x and y?
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays cp 4
Answer:

Question 5.
A system of two linear equations has no solution. What can you conclude about the graphs of the two equations?
F. The lines have the same slope and the same y-intercept.
G. The lines have the same slope and different y-intercepts.
H. The lines have different slopes and the same y-intercept.
I. The lines have different slopes and different y-intercepts.
Answer:

Question 6.
What is the solution of the equation?
0.22(x + 6) = 0.2x + 1.8
A. x = 2.4
B. x = 15.6
C. x = 24
D. x = 156
Answer:

Question 7.
A person who is 5\(\frac{1}{2}\) feet tall casts a 3\(\frac{1}{2}\) -foot-long shadow. A nearby flagpole casts a 28-foot-long shadow. What is the height (in feet) of the flag pole?
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays cp 7
Answer:

Question 8.
A store records total sales (in dollars) each month for three years. Which type of graph can best show how sales increase over this time period?
F. circle graph
G. line graph
H. histogram
I. stem-and-leaf plot
Answer:

Question 9.
Trapezoid KLMN is graphed in the coordinate plane shown.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays cp 9
Rotate Trapezoid 90° clockwise about the origin. What are the M’, coordinates of point, the image of point M after the rotation?
A. (- 3, – 2)
B. (- 2, – 3)
C. (- 2, 3)
D. (3, 2)
Answer:

Question 10.
The table shows the numbers of hours students spent watching television from Monday through Friday for one week and their scores on a test that Friday.
Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays cp 10
Part A Make a scatter plot of the data.
Part B Describe the relationship between the hours of television watched and the test scores.
Part C Explain how to justify your answer in PartB using the linear regression feature of a graphing calculator.
Answer:

Conclusion:

Get the free access to Download Big Ideas Math Answers Grade 8 Chapter 6 Data Analysis and Displays from here. All the solutions are prepared in a simple manner. Test yourself by answering the questions given at the end of the chapter. Keep in touch with us to get the Solutions of all Big Ideas Math Grade 8 Chapters.

Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations

Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations

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Kids will discover more preparation sections in this Big Ideas Math book Ch 10 Answer Key of Algebra 1 such as Exercises 10.1 to 10.4 Questions, Chapter Test, Review Tests, Cumulative Practice, Quizzes, etc. Therefore, go through this article and gain more subject knowledge and also enhance your math skills by concerning the Chapter 10 Radical Functions and Equations BIM Algebra 1 Answer Key.

Topic-wise Big Ideas Math Book Algebra 1 Solution Key Chapter 10 Radical Functions and Equations

The solutions of BIM Algebra 1 Chapter 10 Radical Functions and Equations are explained by the expert professionals as per the common core standards. So, students can trust this ch 10 Radical Functions and Equations Big Ideas math book Answers and make use of them at their preparation. It helps to improve your math knowledge and also clarify your queries on algebra 1 ch 1o Radical Functions and Equations complex topics. Download the Chapter 10 Big Ideas Math Algebra 1 Radical Functions and Equations Answer key from the links without a single penny and ace up your preparation.

Radical Functions and Equations Maintaining Mathematical Proficiency

Evaluate the expression.
Question 1.
7\(\sqrt{25}\) + 10
Answer:

Question 2.
-8 – \(\sqrt{\frac{64}{16}}\)
Answer:

Question 3.
\(5\left(\frac{\sqrt{81}}{3}-7\right)\)
Answer:

Question 4.
-2(3\(\sqrt{9}\) + 13)
Answer:

Graph f and g. Describe the transformations from the graph of f to the graph of g.
Question 5.
f(x) = x; g(x) = 2x – 2
Answer:

Question 6.
f(x) = x; g(x) = \(\frac{1}{3}\)x + 5
Answer:

Question 7.
f(x) = x; g(x) = -x + 3
Answer:

Question 8.
ABSTRACT REASONING
Let a and b represent constants, where b ≥ 0. Describe the transformations from the graph of m(x) = ax + b to the graph of n(x) = -2ax – 4b.
Answer:

Radical Functions and Equations Mathematical Practices

Mathematically proficient students distinguish correct reasoning from flawed reasoning.

Monitoring Progress

Question 1.
Which of the following square roots are rational numbers? Explain your reasoning.
\(\sqrt{0}, \sqrt{1}, \sqrt{3}, \sqrt{4}, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{9}\)
Answer:

Question 2.
The sequence of steps shown appears to prove that 1 = 0. What is wrong with this argument?
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 2
Answer:

Lesson 10.1 Graphing Square Root Functions

Essential Question What are some of the characteristics of the graph of a square root function?

EXPLORATION 1

Graphing Square Root Functions
Work with a partner.

  • Make a table of values for each function.
  • Use the table to sketch the graph of each function.
  • Describe the domain of each function.
  • Describe the range of each function.

Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 1
Answer:

EXPLORATION 2

Writing Square Root Functions
Work with a partner. Write a square root function, y = f (x), that has the given values. Then use the function to complete the table.
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 2
Answer:

Communicate Your Answer

Question 3.
What are some of the characteristics of the graph of a square root function?
Answer:

Question 4.
Graph each function. Then compare the graph to the graph of f(x) = \(\sqrt{x}\).
a. g(x) = \(\sqrt{x-1}\)
b. g(x) = \(\sqrt{x-1}\)
c. g(x) = 2\(\sqrt{x}\)
d. g(x) = -2 \(\sqrt{x}\)
Answer:

Monitoring Progress

Describe the domain of the function.
Question 1.
f(x) = 10 \(\sqrt{x}\)
Answer:

Question 2.
y = \(\sqrt{2x}\) + 7
Answer:

Question 3.
h(x) = \(\sqrt{-x+1}\)
Answer:

Graph the function. Describe the range.
Question 4.
g(x) = \(\sqrt{x}\) – 4
Answer:

Question 5.
y = \(\sqrt{2x}\) + 5
Answer:

Question 6.
n(x) = 5\(\sqrt{x}\)
Answer:

Graph the function. Compare the graph to the graph of f(x) = \(\sqrt{x}\) .
Question 7.
h(x) = \(\sqrt{\frac{1}{4} x}\)
Answer:

Question 8.
g(x) = \(\sqrt{x}\) – 6
Answer:

Question 9.
m(x) = -3\(\sqrt{x}\)
Answer:

Question 10.
Let g(x) = \(\frac{1}{2} \sqrt{x+4}+1\). Describe the transformations from the graph of f(x) = \(\sqrt{x}\) to the graph of g. Then graph g.
Answer:

Question 11.
In Example 5, compare the velocities by finding and interpreting their average rates of change over the interval d = 30 to d = 40.
Answer:

Question 12.
WHAT IF?
At what depth does the velocity of the tsunami exceed 100 meters per second?
Answer:

Graphing Square Root Functions 10.1 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
A ________ is a function that contains a radical expression with the independent variable in the radicand.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 1

Question 2.
VOCABULARY
Is y = 2x\(\sqrt{5}\) a square root function? Explain.
Answer:

Question 3.
WRITING
How do you describe the domain of a square root function?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 3

Question 4.
REASONING
Is the graph of g(x) = 1.25\(\sqrt{x}\) a vertical stretch or a vertical shrink of the graph of f(x) = \(\sqrt{x}\)? Explain.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–14, describe the domain of the function.
Question 5.
y = 8\(\sqrt{x}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 5

Question 6.
y = \(\sqrt{4x}\)
Answer:

Question 7.
y = 4 + \(\sqrt{-x}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 7

Question 8.
y = \(\sqrt{-\frac{1}{2^{x}}}\) + 1
Answer:

Question 9.
h(x) = \(\sqrt{x-4}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 9

Question 10.
p(x) = \(\sqrt{x+7}\)
Answer:

Question 11.
f(x) = \(\sqrt{-x+8}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 11

Question 12.
g(x) = \(\sqrt{-x-1}\)
Answer:

Question 13.
m(x) = 2\(\sqrt{x+4}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 13

Question 14.
n(x) = \(\frac{1}{2} \sqrt{-x}-2\)
Answer:

In Exercises 15–18, match the function with its graph. Describe the range.
Question 15.
y = \(\sqrt{x-3}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 15

Question 16.
y = 3\(\sqrt{x}\)
Answer:

Question 17.
y = \(\sqrt{x}\) – 3
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 17

Question 18.
y = \(\sqrt{-x+3}\)
Answer:

Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 3

In Exercises 19–26, graph the function. Describe the range.
Question 19.
y = \(\sqrt{3x}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 19

Question 20.
y = 4\(\sqrt{-x}\)
Answer:

Question 21.
y = \(\sqrt{x}\) + 5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 21

Question 22.
y = -2 + \(\sqrt{x}\)
Answer:

Question 23.
f(x) = – \(\sqrt{x-3}\)
Answer:

Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 23

Question 24.
g(x) = \(\sqrt{x+4}\)
Answer:

Question 25.
h(x) = \(\sqrt{x+2}\) – 2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 25

Question 26.
f(x) = –\(\sqrt{x-1}\) + 3
Answer:

In Exercises 27–34, graph the function. Compare the graph to the graph of f (x) = \(\sqrt{x}\).
Question 27.
g(x) = \(\frac{1}{4} \sqrt{x}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 27

Question 28.
r(x) = \(\sqrt{2x}\)
Answer:

Question 29.
h(x) = \(\sqrt{x+3}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 29

Question 30.
q(x) = \(\sqrt{x}\) + 8
Answer:

Question 31.
p(x) = \(\sqrt{-\frac{1}{3} x}\)x
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 31

Question 32.
g(x) = -5\(\sqrt{x}\)
Answer:

Question 33.
m(x) = –\(\sqrt{x}\) – 6
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 33

Question 34.
n(x) = –\(\sqrt{x}\) – 4
Answer:

Question 35.
ERROR ANALYSIS
Describe and correct the error in graphing the function y = \(\sqrt{x}\) + 1 .
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 4
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 35

Question 36.
ERROR ANALYSIS
Describe and correct the error in comparing the graph of g(x) = \(-\frac{1}{4} \sqrt{x}\) to the graph of f (x) = \(\sqrt{x}\).
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 5
Answer:

In Exercises 37–44, describe the transformations from the graph of f (x) = \(\sqrt{x}\) to the graph of h. Then graph h.
Question 37.
h(x) = 4\(\sqrt{x+2}\) – 1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 37.1
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 37.2

Question 38.
h(x) = \(\frac{1}{2} \sqrt{x-6}\)+ 3
Answer:

Question 39.
h(x) = 2\(\sqrt{-x}\) – 6
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 39.1
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 39.2

Question 40.
h(x) = –\(\sqrt{x-3}\) – 2
Answer:

Question 41.
h(x) = \(\frac{1}{3} \sqrt{x+3}\) + 3
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 41.1
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 41.2

Question 42.
h(x) = 2\(\sqrt{x-1}\) + 4
Answer:

Question 43.
h(x) = -2\(\sqrt{x-1}\) + 5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 43.1
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 43.2

Question 44.
h(x) = -5\(\sqrt{x+2}\) – 1
Answer:

Question 45.
COMPARING FUNCTIONS
The model S(d ) = \(\sqrt{30df}\) represents the speed S (in miles per hour) of a van before it skids to a stop, where f is the drag factor of the road surface and d is the length (in feet) of the skid marks. The drag factor of Road Surface A is 0.75. The graph shows the speed of the van on Road Surface B. Compare the speeds by finding and interpreting their average rates of change over the interval d = 0 to d = 15.
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 6
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 45

Question 46.
COMPARING FUNCTIONS
The velocity v (in meters per second) of an object in motion is given by v(E ) = \(\sqrt{\frac{2 E}{m}}\), where E is the kinetic energy of the object (in joules) and m is the mass of the object (in kilograms). The mass of Object A is 4 kilograms. The graph shows the velocity of Object B. Compare the velocities of the objects by finding and interpreting the average rates of change over the interval E = 0 to E = 6.
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 7
Answer:

Question 47.
OPEN-ENDED
Consider the graph of y = \(\sqrt{x}\).
a. Write a function that is a vertical translation of the graph of y = \(\sqrt{x}\).
b. Write a function that is a reflection of the graph of y = \(\sqrt{x}\).
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 47

Question 48.
REASONING
Can the domain of a square root function include negative numbers? Can the range include negative numbers? Explain your reasoning.
Answer:

Question 49.
PROBLEM SOLVING
The nozzle pressure of a fire hose allows firefighters to control the amount of water they spray on a fire. The flow rate f(in gallons per minute) can be modeled by the function f = 12\(\sqrt{p}\), where p is the nozzle pressure (in pounds per square inch).
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 8
a. Use a graphing calculator to graph the function. At what pressure does the flow rate exceed 300 gallons per minute?
b. What happens to the average rate of change of the flow rate as the pressure increases?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 49

Question 50
PROBLEM SOLVING
The speed s (in meters per second) of a long jumper before jumping can be modeled by the function s = 10.9\(\sqrt{h}\), where h is the maximum height (in meters from the ground) of the jumper.
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 9
a. Use a graphing calculator to graph the function. A jumper is running 9.2 meters per second. Estimate the maximum height of the jumper.
b. Suppose the runway and pit are raised on a platform slightly higher than the ground. How would the graph of the function be transformed?
Answer:

Question 51.
MATHEMATICAL CONNECTIONS
The radius r of a circle is given by r = \(\sqrt{\frac{A}{\pi}}\), where A is the area of the circle.
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 10
a. Describe the domain of the function. Use a graphing calculator to graph the function.
b. Use the trace feature to approximate the area of a circle with a radius of 5.4 inches.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 51

Question 52.
REASONING
Consider the function f(x) = 8a\(\sqrt{x}\).
a. For what value of a will the graph of f be identical to the graph of the parent square root function?
b. For what values of a will the graph of f be a vertical stretch of the graph of the parent square root function?
c. For what values of a will the graph of f be a vertical shrink and a reflection of the graph of the parent square root function?
Answer:

Question 53.
REASONING
The graph represents the function f(x) = \(\sqrt{x}\).
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 11
a. What is the minimum value of the function?
b. Does the function have a maximum value? Explain.
c. Write a square root function that has a maximum value. Does the function have a minimum value? Explain.
d. Write a square root function that has a minimum value of -4.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 53

Question 54.
HOW DO YOU SEE IT?
Match each function with its graph. Explain your reasoning.
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 12
Answer:

Question 55.
REASONING
Without graphing, determine which function’s graph rises more steeply, f(x) = 5\(\sqrt{x}\) or g(x) = \(\sqrt{5x}\). Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 55

Question 56.
THOUGHT PROVOKING
Use a graphical approach to find the solutions of x – 1 = \(\sqrt{5x-9}\). Show your work. Verify your solutions algebraically.
Answer:

Question 57.
OPEN-ENDED
Write a radical function that has a domain of all real numbers greater than or equal to -5 and a range of all real numbers less than or equal to 3.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 57

Maintaining Mathematical Proficiency

Evaluate the expression.(Section 6.2)
Question 58.
\(\sqrt [ 3]{ 343 }\)
Answer:

Question 59.
\(\sqrt [ 3]{ -64 }\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 59

Question 60.
\(-\sqrt[3]{-\frac{1}{27}}\)
Answer:

Factor the polynomial.(Section 7.5)
Question 61.
x2 + 7x + 6
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 61

Question 62.
d2 – 11d + 28
Answer:

Question 63.
y2 – 3y – 40
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 10 Radical Functions and Equations 10.1 a 63

Lesson 10.2 Graphing Cube Roots Functions

Essential Question What are some of the characteristics of the graph of a cube root function?

EXPLORATION 1

Graphing Cube Root Functions
Work with a partner.

  •  Make a table of values for each function. Use positive and negative values of x.
  • Use the table to sketch the graph of each function.• Describe the domain of each function.
  • Describe the range of each function.

Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 1
Answer:

EXPLORATION 2

Writing Cube Root Functions
Work with a partner. Write a cube root function, y = f(x), that has the given values. Then use the function to complete the table.
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 2
Answer:

Communicate Your Answer

Question 3.
What are some of the characteristics of the graph of a cube root function?
Answer:

Question 4.
Graph each function. Then compare the graph to the graph of f (x) = \(\sqrt [ 3]{ x }\) .
a. g(x) = \(\sqrt [ 3]{ x-1 }\)
b. g(x) = \(\sqrt [ 3]{ x-1 }\)
c. g(x) = 2 \(\sqrt [ 3]{ x }\)
d. g(x) = -2\(\sqrt [ 3]{ x }\)
Answer:

Monitoring Progress

Graph the function. Compare the graph to the graph of f(x) = \(\sqrt [ 3]{ x }\).
Question 1.
h(x) = \(\sqrt [ 3]{ x }\) + 3
Answer:

Question 2.
m(x) = \(\sqrt [ 3]{ x }\) – 5
Answer:

Question 3.
g(x) = 4\(\sqrt [ 3]{ x }\)
Answer:

Graph the function. Compare the graph to the graph of f(x) = \(\sqrt [ 3]{ x }\).
Question 4.
g(x) =\(\sqrt [3]{ 0.5x+5 }\) + 5
Answer:

Question 5.
h(x) = 4\(\sqrt [3]{ x }\) – 1
Answer:

Question 6.
n(x) = \(\sqrt [ 3]{ 4-x }\)
Answer:

Question 7.
Let g(x) = \(-\frac{1}{2} \sqrt[3]{x+2}\) – 4. Describe the transformations from the graph of f (x) = \(\sqrt [ 3]{ x }\) to the graph of g. Then graph g.
Answer:

Question 8.
In Example 4, compare the average rates of change over the interval x = 2 to x = 10.
Answer:

Question 9.
WHAT IF?
Estimate the age of an elephant whose shoulder height is 175 centimeters.
Answer:

Graphing Cube Roots Functions 10.2 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The __________ of the radical in a cube root function is 3.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 1

Question 2.
WRITING
Describe the domain and range of the function f(x) = \(\sqrt [3]{ x-4 }\) + 1.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, match the function with its graph.
Question 3.
y = 3\(\sqrt [3]{ x+2 }\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 3

Question 4.
y = 3\(\sqrt [3]{ x-2 }\)
Answer:

Question 5.
y = 3\(\sqrt [3]{ x+2 }\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 5

Question 6.
y = \(\sqrt [3]{ x }\) – 2
Answer:

Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 3

In Exercises 7–12, graph the function. Compare the graph to the graph of f(x) = \(\sqrt [ 3]{ x }\).
Question 7.
h(x) = \(\sqrt [3]{ x-4 }\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 7

Question 8.
g(x) = \(\sqrt [3]{ x+1 }\)
Answer:

Question 9.
m(x) = \(\sqrt [3]{ x+5 }\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 9

Question 10.
q(x) = \(\sqrt [3]{ x }\) – 3
Answer:

Question 11.
p(x) = 6\(\sqrt [3]{ x }\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 11

Question 12.
j(x) = \(\sqrt[3]{\frac{1}{2} x}\)
Answer:

In Exercises 13–16, compare the graphs. Find the value of h, k, or a.
Question 13.
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 13

Question 14.
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 5
Answer:

Question 15.
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 15

Question 16.
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 7
Answer:

In Exercises 17–26, graph the function. Compare the graph to the graph of f(x) = \(\sqrt [3]{ x }\).
Question 17.
r(x) = – \(\sqrt [3]{ x-2 }\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 17

Question 18.
h(x) = – 3\(\sqrt [3]{ x+3 }\)
Answer:

Question 19.
k(x) = 5\(\sqrt [3]{ x+1 }\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 19

Question 20.
j(x) = 0.5\(\sqrt [3]{ x-4 }\)
Answer:

Question 21.
g(x) = 4\(\sqrt [3]{ x }\) – 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 21

Question 22.
m(x) = 3\(\sqrt [3]{ x }\) + 7
Answer:

Question 23.
n(x) = \(\sqrt [3]{ -8x }\) – 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 23

Question 24.
v(x) =\(\sqrt [3]{ 5x }\) + 2
Answer:

Question 25.
q(x) = \(\sqrt[3]{2(x+3)}\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 25

Question 26.
p(x) = \(\sqrt[3]{3(1-x)}\)
Answer:

In Exercises 27–32, describe the transformations from the graph of f(x) = \(\sqrt [3]{ x }\) to the graph of the given function. Then graph the given function.
Question 27.
g(x) = \(\sqrt [3]{ x-4 }\) + 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 27.1
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 27.2

Question 28.
n(x) = \(\sqrt [3]{ x+1 }\) – 3
Answer:

Question 29.
j(x) = -5\(\sqrt [3]{ x+3 }\) + 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 29.1
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 29.2

Question 30.
k(x) = 6\(\sqrt [3]{ x-9 }\) – 5
Answer:

Question 31.
v(x) = \(\frac{1}{3} \sqrt[3]{x-1}\) + 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 31.1
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 31.2

Question 32.
h(x) = -\frac{3}{2} \sqrt[3]{x+4} – 3
Answer:

Question 33.
ERROR ANALYSIS
Describe and correct the error in graphing the function f(x) = \(\sqrt [3]{ x-3 }\).
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 33

Question 34.
ERROR ANALYSIS
Describe and correct the error in graphing the function h(x) = \(\sqrt [3]{ x }\) + 1.
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 9
Answer:

Question 35.
COMPARING FUNCTIONS
The graph of cube root function q is shown. Compare the average rate of change of q to the average rate of change of f(x) = 3\(\sqrt [3]{ x }\) over the interval x = 0 to x = 6.
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 35

Question 36.
COMPARING FUNCTIONS
The graphs of two cube root functions are shown. Compare the average rates of change of the two functions over the interval x = -2 to x = 2.
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 11
Answer:

Question 37.
MODELING WITH MATHEMATICS
For a drag race car that weighs 1600 kilograms, the velocity v (in kilometers per hour) reached by the end of a drag race can be modeled by the function v = 23.8\(\sqrt [3]{ p }\), where p is the car’s power (in horsepower). Use a graphing calculator to graph the function. Estimate the power of a 1600-kilogram car that reaches a velocity of 220 kilometers per hour.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 37

Question 38.
MODELING WITH MATHEMATICS
The radius r of a sphere is given by the function r = \(\sqrt[3]{\frac{3}{4 \pi}} V\), where V is the volume of the sphere. Use a graphing calculator to graph the function. Estimate the volume of a spherical beach ball with a radius of 13 inches.
Answer:

Question 39.
MAKING AN ARGUMENT
Your friend says that all cube root functions are odd functions. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 39

Question 40.
HOW DO YOU SEE IT?
The graph represents the cube root function f(x) = \(\sqrt [3]{ x }\).
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 12
a. On what interval is f negative? positive?
b. On what interval, if any, is f decreasing? increasing?
c. Does f have a maximum or minimum value? Explain.
d. Find the average rate of change of f over the interval x = -1 to x = 1.
Answer:

Question 41.
PROBLEM SOLVING
Write a cube root function that passes through the point (3, 4) and has an average rate of change of -1 over the interval x = -5 to x = 2.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 41.1
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 41.2

Question 42.
THOUGHT PROVOKING
Write the cube root function represented by the graph. Use a graphing calculator to check your answer.
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 13
Answer:

Maintaining Mathematical Proficiency

Factor the polynomial.(Section 7.6)
Question 43.
3x2 + 12x – 36
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 43

Question 44.
2x2 – 11x + 9
Answer:

Question 45.
4x2 + 7x – 15
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 45.1
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 45.2

Solve the equation using square roots.(Section 9.3)
Question 46.
x2 – 36 = 0
Answer:

Question 47.
5x2 + 20 = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 47

Question 48.
(x + 4)2 – 81
Answer:

Question 49.
25(x – 2)2 = 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 a 49

Radical Functions and Equations Study Skills: Making Note Cards

10.1–10.2 What Did YouLearn?

Core Vocabulary
square root function, p. 544
radical function, p. 545
cube root function, p. 552

Core Concepts
Lesson 10.1
Square Root Functions, p. 544
Transformations of Square Root Functions, p. 545
Comparing Square Root Functions Using Average Rates of Change, p. 546

Lesson 10.2
Cube Root Functions, p. 552
Comparing Cube Root Functions
Using Average Rates of Change, p. 554

Mathematical Practices

Question 1.
In Exercise 45 on page 549, what information are you given? What relationships are present? What is your goal?
Answer:

Question 2.
What units of measure did you use in your answer to Exercise 38 on page 556? Explain your reasoning.
Answer:

Study Skills: Making Note Cards

Invest in three different colors of note cards. Use one color for each of the following: vocabulary words, rules, and calculator keystrokes.

  • Using the first color of note cards, write a vocabulary word on one side of a card. On the other side, write the definition and an example. If possible, put the definition in your own words.
  • Using the second color of note cards, write a rule on one side of a card. On the other side, write an explanation and an example.
  • Using the third color of note cards, write a calculation on one side of a card. On the other side, write the keystrokes required to perform the calculation.
    Use the note cards as references while completing your homework. Quiz yourself once a day.
    Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations 10.2 16

Radical Functions and Equations 10.1–10.2 Quiz

Describe the domain of the function.(Lesson 10.1)
Question 1.
y = \(\sqrt{x-3}\)
Answer:

Question 2.
f(x) = 15\(\sqrt{x}\)
Answer:

Question 3.
y = \(\sqrt{3-x}\)
Answer:

Graph the function. Describe the range. Compare the graph to the graph of f(x) = \(\sqrt{x}\). (Lesson 10.1)
Question 4.
g(x) = \(\sqrt{x}\) + 5
Answer:

Question 5.
n(x) = \(\sqrt{x-4}\)
Answer:

Question 6.
r(x) = –\(\sqrt{x-1}\) + 1
Answer:

Graph the function. Compare the graph to the graph of f(x) = \(\sqrt [3]{ x }\). (Lesson 10.2)
Question 7.
b(x) = \(\sqrt [3]{ x+2 }\)
Answer:

Question 8.
h(x) = -3\(\sqrt [3]{ x-6 }\)
Answer:

Question 9.
q(x) = \(\sqrt [3]{ -4-x }\)
Answer:

Compare the graphs. Find the value of h, k, or a. (Lesson 10.1 and Lesson 10.2)
Question 10.
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations q 10
Answer:

Question 11.
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations q 11
Answer:

Question 12.
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations q 12
Answer:

Describe the transformations from the graph of f to the graph of h. Then graph h. (Section 10.1 and Section 10.2)
Question 13.
f(x) = \(\sqrt{x}\); h(x) = -3 \(\sqrt{x+2}\) + 6
Answer:

Question 14.
f(x) = \(\sqrt[3]{x}\); h(x) = \(\frac{1}{2} \sqrt[3]{x}-3\)
Answer:

Question 15.
The time t (in seconds) it takes a dropped object to fall h feet is given by t = \(\frac{1}{4} \sqrt{h}\). (Section 10.1)
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations q 15
a. Use a graphing calculator to graph the function. Describe the domain and range.
b. It takes about 7.4 seconds for a stone dropped from the New River Gorge Bridge in West Virginia to reach the water below. About how high is the bridge above the New River?
Answer:

Question 16.
The radius r of a sphere is given by the function r = \(\sqrt[3]{\frac{3}{4 \pi} V}\), where V is the volume of the sphere. Spaceship Earth is a spherical structure at Walt Disney World that has an inner radius of about 25 meters. Use a graphing calculator to graph the function. Estimate the volume of Spaceship Earth. (Section 10.2)
Answer:

Question 17.
The graph of square root function g is shown. Compare the average rate of change of g to the average rate of change of h(x) = \(\sqrt[3]{\frac{3}{2} x}\)x over the interval x = 0 to x = 3.
Big Ideas Math Algebra 1 Answers Chapter 10 Radical Functions and Equations q 17
Answer:

Lesson 10.3 Solving Radical Equations

Essential Question How can you solve an equation that contains square roots?

EXPLORATION 1

Analyzing a Free-Falling Object
Work with a partner. The table shows the time t (in seconds) that it takes a free-falling object (with no air resistance) to fall d feet.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 1
a. Use the data in the table to sketch the graph of t as a function of d. Use the coordinate plane below.
b. Use your graph to estimate the time it takes the object to fall 240 feet.
c. The relationship between d and t is given by the function t = \(\sqrt{\frac{d}{16}}\).
Use this function to check your estimate in part (b).
d. It takes 5 seconds for the object to hit the ground. How far did it fall? Explain your reasoning.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 2

EXPLORATION 2

Solving a Square Root Equation
Work with a partner. The speed s (in feet per second) of the free-falling object in Exploration 1 is given by the functions
s = \(\sqrt{64d}\).
Find the distance the object has fallen when it reaches each speed.
a. s = 8 ft/sec
b. s = 16 ft/sec
c. s = 24 ft/sec

Communicate Your Answer

Question 3.
How can you solve an equation that contains square roots?
Answer:

Question 4.
Use your answer to Question 3 to solve each equation.
a. 5 = \(\sqrt{x}\) + 20
b. 4 = \(\sqrt{x-18}\)
c. \(\sqrt{x}\) + 2 = 3
d. -3 = -2\(\sqrt{x}\)
Answer:

Monitoring Progress

Solve the equation. Check your solution.
Question 1.
\(\sqrt{x}\) = 6
Answer:

Question 2.
\(\sqrt{x}\) – 7 = 3
Answer:

Question 3.
\(\sqrt{y}\) + 15 = 22
Answer:

Question 4.
1 – \(\sqrt{c}\) = -2
Answer:

Solve the equation. Check your solution.
Question 5.
\(\sqrt{x+4}\) + 7 = 11
Answer:

Question 6.
15 = 6 + \(\sqrt{3w-9}\)
Answer:

Question 7.
\(\sqrt{3x+1}\) = \(\sqrt{4x-7}\)
Answer:

Question 8.
\(\sqrt{n}\) = \(\sqrt{5n-1}\)
Answer:

Question 9.
\(\sqrt [3]{ y }\) = 4 = 1
Answer:

Question 10.
\(\sqrt [3]{ 3c+7 }\) = 10
Answer:

Solve the equation. Check your solution(s).
Question 11.
\(\sqrt{4-3x}\) = x
Answer:

Question 12.
\(\sqrt{3m}\) + 10 = 1
Answer:

Question 13.
p + 1 = \(\sqrt{7p+15}\)
Answer:

Question 14.
What is the length of a pendulum that has a period of 2.5 seconds?
Answer:

Solving Radical Equations 10.3 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Why should you check every solution of a radical equation?
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which equation does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 3
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, solve the equation. Check your solution.
Question 3.
\(\sqrt{x}\) = 9
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 3

Question 4.
\(\sqrt{x}\) = 4
Answer:

Question 5.
7 = \(\sqrt{x}\) – 5
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 5

Question 6.
\(\sqrt{p}\) – 7 = -1
Answer:

Question 7.
\(\sqrt{c}\) + 12 = 23
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 7

Question 8.
\(\sqrt{x}\) + 6 = 8
Answer:

Question 9.
4 – \(\sqrt{x}\) = 2
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 9

Question 10.
-8 = 7 = \(\sqrt{r}\)
Answer:

Question 11.
3\(\sqrt{y}\) – 18 = -3
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 11

Question 12.
2\(\sqrt{q}\) + 5 = 11
Answer:

In Exercises 13–20, solve the equation. Check your solution.
Question 13.
\(\sqrt{a-3}\) + 5 = 9
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 13

Question 14.
\(\sqrt{b+7}\) – 5 = -2
Answer:

Question 15.
2 \(\sqrt{x+4}\) = 16
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 15

Question 16.
5\(\sqrt{y-2}\) = 10
Answer:

Question 17.
-1 = \(\sqrt{5r+1}\) – 7
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 17

Question 18.
2 = \(\sqrt{4s-4}\) – 4
Answer:

Question 19.
7 + 3\(\sqrt{3p-9}\) = 25
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 19

Question 20.
19 – 4\(\sqrt{3c-11}\) = 11
Answer:

Question 21.
MODELING WITH MATHEMATICS
The Cave of Swallows is a natural open-air pit cave in the state of San Luis Potosí, Mexico. The 1220-foot- deep cave was a popular destination for BASE jumpers. The function t = \(\frac{1}{4} \sqrt{d}\) represents the time t (in seconds) that it takes a BASE jumper to fall d feet. How far does a BASE jumper fall in 3 seconds?
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 4.1
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 21

Question 22.
MODELING WITH MATHEMATICS
The edge length s of a cube with a surface area of A is given by s = \(\sqrt{\frac{A}{6}}\). What is the surface area of a cube with an edge length of 4 inches?
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 4
Answer:

In Exercises 23–26, use the graph to solve the equation.
Question 23.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 5
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 23

Question 24.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 6
Answer:

Question 25.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 7
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 25

Question 26.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 8
Answer:

In Exercises 27–34, solve the equation. Check your solution. (See Example 3.)
Question 27.
\(\sqrt{2x-9}\) = \(\sqrt{x}\)
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 27

Question 28.
\(\sqrt{y+1}\) = \(\sqrt{4y-8}\)
Answer:

Question 29.
\(\sqrt{3g+1}\) = \(\sqrt{7g-19}\)
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 29

Question 30.
\(\sqrt{8h-7}\) = \(\sqrt{6h+7}\)
Answer:

Question 31.
\(\sqrt{\frac{p}{2}-2}\) = \(\sqrt{p-8}\)
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 31

Question 32.
\(\sqrt{2v-5}\) = \(\sqrt{\frac{v}{3}+5}\)
Answer:

Question 33.
\(\sqrt{2c+1}\) = \(\sqrt{4c}\) = 0
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 33

Question 34.
\(\sqrt{5r}\) – \(\sqrt{8r-2}\) = 0
Answer:

MATHEMATICAL CONNECTIONS In Exercises 35 and 36, find the value of x.
Question 35.
Perimeter = 22 cm
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 9
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 35

Question 36.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 10
Answer:

In Exercises 37–44, solve the equation. Check your solution.
Question 37.
\(\sqrt [3]{ x }\) = 4
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 37

Question 38.
\(\sqrt [3]{ y }\) = 2
Answer:

Question 39.
6 = 3\(\sqrt [3]{ 8g }\)
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 39

Question 40.
\(\sqrt [3]{ r+19 }\) = 3
Answer:

Question 41.
\(\sqrt [3]{ 2x+9 }\) = -3
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 41

Question 42.
-5 = \(\sqrt [3]{ 10x+15 }\)
Answer:

Question 43.
\(\sqrt [3]{ y+6 }\) = \(\sqrt [3]{ 5y-2 }\)
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 43

Question 44.
\(\sqrt [3]{ 7j-2 }\) = \(\sqrt [3]{ j+4 }\)
Answer:

In Exercises 45–48, determine which solution, if any, is an extraneous solution.
Question 45.
\(\sqrt{6x-5}\) = x; x = 5, x = 1
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 45

Question 46.
\(\sqrt{2y+3}\) = y; y = -1, y = 3
Answer:

Question 47.
\(\sqrt{12p+16}\) = -2p; p = -1, p = 4
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 47

Question 48.
-3g = \(\sqrt{-18-27}\); g = -2, g = -1
Answer:

In Exercises 49–58, solve the equation. Check your solution(s).
Question 49.
y = \(\sqrt{5y-4}\)
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 49

Question 50.
\(\sqrt{-14x-9x}\) = x
Answer:

Question 51.
\(\sqrt{1-3a}\) = 2a
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 51.1
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 51.2

Question 52.
2q = \(\sqrt{10q-6}\)
Answer:

Question 53.
9 + \(\sqrt{5p}\) = 4
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 53

Question 54.
\(\sqrt{3n}\) – 11 = -5
Answer:

Question 55.
\(\sqrt{2m+2}\) – 3 = 1
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 55

Question 56.
15 + \(\sqrt{4b-8}\) = 13
Answer:

Question 57.
r + 4 = \(\sqrt{-4r-19}\)
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 57.1
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 57.2

Question 58.
\(\sqrt{3-s}\) = s – 1
Answer:

ERROR ANALYSIS In Exercises 59 and 60, describe and correct the error in solving the equation.
Question 59.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 11
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 59

Question 60.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 12
Answer:

Question 61.
REASONING
Explain how to use mental math to solve \(\sqrt{2x}\) + 5 = 1.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 61

Question 62.
WRITING
Explain how you would solve \(\sqrt [4]{ m+4 }\) – \(\sqrt [4]{ 3m }\) = 0.
Answer:

Question 63.
MODELING WITH MATHEMATICS
The formula V = \(\sqrt{PR}\) relates the voltage V (in volts), power P (in watts), and resistance R (in ohms) of an electrical circuit. The hair dryer shown is on a 120-volt circuit. Is the resistance of the hair dryer half as much as the resistance of the same hair dryer on a 240-volt circuit? Explain your reasoning.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 13
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 63

Question 64.
MODELING WITH MATHEMATICS
The time t (in seconds) it takes a trapeze artist to swing back and forth is represented by the function t = 2π \(\sqrt{\frac{r}{32}}\), where r is the rope length (in feet). It takes the trapeze artist 6 seconds to swing back and forth. Is this rope \(\frac{3}{2}\) as long as the rope used when it takes the trapeze artist 4 seconds to swing back and forth? Explain your reasoning.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 14
Answer:

REASONING In Exercises 65–68, determine whether the statement is true or false. If it is false, explain why.
Question 65.
If \(\sqrt{a}\) = b, then (\(\sqrt{a}\))2 = b2.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 65

Question 66.
If \(\sqrt{a}\) = \(\sqrt{b}\), then a = b.
Answer:

Question 67.
If a2 = b2, then a = b.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 67

Question 68.
If a2 = \(\sqrt{b}\), then a4 = (\(\sqrt{b}\))2
Answer:

Question 69.
COMPARING METHODS
Consider the equation x + 2 = \(\sqrt{2x-3}\).
a. Solve the equation by graphing. Describe the process.
b. Solve the equation algebraically. Describe the process.
c. Which method do you prefer? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 69.1
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 69.2

Question 70.
HOW DO YOU SEE IT?
The graph shows two radical functions.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 15
a. Write an equation whose solution is the x-coordinate of the point of intersection of the graphs.
b. Use the graph to solve the equation.
Answer:

Question 71.
MATHEMATICAL CONNECTIONS
The slant height s of a cone with a radius of r and a height of h is given by s = \(\sqrt{r^{2}+h^{2}}\). The slant heights of the two cones are equal. Find the radius of each cone.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 16
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 71

Question 72.
CRITICAL THINKING
How is squaring \(\sqrt{x+2}\) different from squaring \(\sqrt{x}\) + 2?
Answer:

USING STRUCTURE In Exercises 73–78, solve the equation. Check your solution.
Question 73.
\(\sqrt{m+15}\) = \(\sqrt{m}\) + \(\sqrt{5}\)
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 73.1
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 73.2

Question 74.
2 = \(\sqrt{x+1}\) = \(\sqrt{x+2}\)
Answer:

Question 75.
\(\sqrt{5y+9}\) + \(\sqrt{5y}\) = 9
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 75.1
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 75.2

Question 76.
\(\sqrt{2c-8}\) – \(\sqrt{2c}\) – 4 = 0
Answer:

Question 77.
2\(\sqrt{1+4h}\) – 4\(\sqrt{h}\) – 2 = 0
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 77.1
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 77.2

Question 78.
\(\sqrt{20-4z}\) + 2\(\sqrt{-z}\) = 10
Answer:

Question 79.
OPEN-ENDED
Write a radical equation that has a solution of x = 5.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 79

Question 80.
OPEN-ENDED
Write a radical equation that has x = 3 and x = 4 as solutions.
Answer:

Question 81.
MAKING AN ARGUMENT
Your friend says the equation \(\sqrt{(2 x+5)^{2}}\) = 2x + 5 is always true, because after simplifying the left side of the equation, the result is an equation with infinitely many solutions. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 81

Question 82.
THOUGHT PROVOKING
Solve the equation \(\sqrt [3]{ x+1 }\) = \(\sqrt{x-3}\). Show your work and explain your steps.
Answer:

Question 83.
MODELING WITH MATHEMATICS
The frequency f (in cycles per second) of a string of an electric guitar is given by the equation f = \(\frac{1}{2 \ell} \sqrt{\frac{T}{m}}\), where ℓ is the length of the string (in meters), T is the string’s tension (in newtons), and m is the string’s mass per unit length (in kilograms per meter). The high E string of an electric guitar is 0.64 meter long with a mass per unit length of 0.000401 kilogram per meter.
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 17
a. How much tension is required to produce a frequency of about 330 cycles per second?
b. Would you need more or less tension to create the same frequency on a string with greater mass per unit length? Explain.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 83.1
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 83.2

Maintaining Mathematical Proficiency

Find the product.(Section 7.2)
Question 84.
(x + 8)(x – 2)
Answer:

Question 85.
(3p – 1)(4p + 5)
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 85

Question 86.
(s + 2)(s2 + 3s – 4)
Answer:

Graph the function. Compare the graph to the graph of f(x) = x2.(Section 8.1)
Question 87.
r(x) = 3x2
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 87

Question 88.
g(x) = \(\frac{3}{4}\)x2
Answer:

Question 89.
h(x) = -5x2
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 10 Radical Functions and Equations 10.3 a 89

Lesson 10.4 Inverse of a Function

Essential Question How are a function and its inverse related?

EXPLORATION 1

Exploring Inverse Functions
Work with a partner. The functions f and g are inverses of each other. Compare the tables of values of the two functions. How are the functions related?
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 1

EXPLORATION 2

Exploring Inverse Functions
Work with a partner.
a. Plot the two sets of points represented by the tables in Exploration 1. Use the coordinate plane below.
b. Connect each set of points with a smooth curve.
c. Describe the relationship between the two graphs.
d. Write an equation for each function.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 2

Communicate Your Answer

Question 3.
How are a function and its inverse related?
Answer:

Question 4.
A table of values for a function f is given. Create a table of values for a function g, the inverse of f.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 3
Answer:

Question 5.
Sketch the graphs of f(x) = x + 4 and its inverse in the same coordinate plane. Then write an equation of the inverse of f. Explain your reasoning.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 4
Answer:

Monitoring Progress

Find the inverse of the relation.
Question 1.
(-3, -4), (-2, 0), (-1, 4), (0, 8), (1, 12), (2, 16), (3, 20)
Answer:

Question 2.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 5
Answer:

Solve y = f(x) for x. Then find the input when the output is 4.
Question 3.
f(x) = x – 6
Answer:

Question 4.
f(x) = \(\frac{1}{2}\)x + 3
Answer:

Question 5.
f(x) = 4x2
Answer:

Find the inverse of the function. Then graph the function and its inverse.
Question 6.
f(x) = 6x
Answer:

Question 7.
f(x) = -x + 5
Answer:

Question 8.
f(x) = \(\frac{1}{4}\)x – 1
Answer:

Find the inverse of the function. Then graph the function and its inverse.
Question 9.
f(x) = -x2, x ≤ 0
Answer:

Question 10.
f(x) = 4x2 + 3, x ≥ 0
Answer:

Question 11.
Is the inverse of f(x) = \(\sqrt{2x-1}\) a function? Find the inverse.
Answer:

Inverse of a Function 10.4 Exercises

Vocabulary and Core ConceptCheck

Question 1.
COMPLETE THE SENTENCE
A relation contains the point (-3, 10). The ____________ contains the point (10, -3).
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Consider the function f represented by the graph. Which is different? Find “both” answers.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 6
Answer:

In Exercises 3–8, find the inverse of the relation.
Question 3.
(1, 0), (3, -8), (4, -3), (7, -5), (9, -1)
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 3

Question 4.
(2, 1), (4, -3), (6, 7), (8, 1), (10, -4)
Answer:

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 7
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 5

Question 6.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 8
Answer:

Question 7.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 9
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 7

Question 8.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 10
Answer:

In Exercises 9–14, solve y = f(x) for x. Then find the input when the output is 2.
Question 9.
f(x) = x + 5
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 9

Question 10.
f(x) = 2x – 3
Answer:

Question 11.
f(x) = \(\frac{1}{4}\)x – 1
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 11

Question 12.
f(x) = \(\frac{2}{3}\)x + 4
Answer:

Question 13.
f(x) = 9x2
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 13

Question 14.
f(x) = \(\frac{1}{2}\)x2 – 7
Answer:

In Exercises 15 and 16, graph the inverse of the function by reflecting the graph in the line y = x. Describe the domain and range of the inverse.
Question 15.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 11
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 15

Question 16.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 12
Answer:

In Exercises 17–22, find the inverse of the function. Then graph the function and its inverse.
Question 17.
f(x) = 4x – 1
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 17

Question 18.
f(x) = -2x + 5
Answer:

Question 19.
f(x) = -3x – 2
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 19

Question 20.
f(x) = 2x + 3
Answer:

Question 21.
f(x) =\(\frac{1}{3}\)x + 8
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 21

Question 22.
f(x) = – \(\frac{3}{2}\)x + \(\frac{7}{2}\)
Answer:

In Exercises 23–28, find the inverse of the function. Then graph the function and its inverse.
Question 23.
f(x) = 4x2, x ≥ 0
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 23.1
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 23

Question 24.
f(x) = \(\frac{2}{25}\)x2, x ≤ 0
Answer:

Question 25.
f(x) = -x2 + 10, x ≤ 0
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 25.1
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 25.2

Question 26.
f(x) = 2x2 + 6, x ≥ 0
Answer:

Question 27.
f(x) = \(\frac{1}{9}\)x2 + 2, x ≥ 0
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 27.1
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 27.2

Question 28.
f(x) = -4x2 – 8, x ≤ 0
Answer:

In Exercises 29–32, use the Horizontal Line Test to determine whether the inverse of f is a function.
Question 29.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 13
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 29

Question 30.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 14
Answer:

Question 31.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 15
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 31

Question 32.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 16
Answer:

In Exercises 33–42, determine whether the inverse of f is a function. Then find the inverse.
Question 33.
f(x) = \(\sqrt{x+3}\)
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 33.1
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 33.2

Question 34.
f(x) = \(\sqrt{x-5}\)
Answer:

Question 35.
f(x) = \(\sqrt{2x-6}\)
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 35.1
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 35.2

Question 36.
f(x) = \(\sqrt{4x+1}\)
Answer:

Question 37.
f(x) = 3\(\sqrt{x-8}\)
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 37.1
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 37.2

Question 38.
f(x) = –\(\frac{1}{4} \sqrt{5 x+2}\)
Answer:

Question 39.
f(x) = –\(\sqrt{3x+5}\) – 2
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 39

Question 40.
f(x) = 2\(\sqrt{x-7}\) + 6
Answer:

Question 41.
f(x) = 2x2
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 41.1
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 41.2

Question 42.
f(x) = |x|
Answer:

Question 43.
ERROR ANALYSIS
Describe and correct the error in finding the inverse of the function f(x) = – 3x + 5.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 17
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 43

Question 44.
ERROR ANALYSIS
Describe and correct the error in finding and graphing the inverse of the function f(x) = \(\sqrt{x-3}\).
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 18.1
Answer:

Question 45.
MODELING WITH MATHEMATICS
The euro is the unit of currency for the European Union. On a certain day, the number E of euros that could be obtained for D U.S. dollars was represented by the formula shown.
E = 0.74683D
Solve the formula for D. Then find the number of U.S. dollars that could be obtained for 250 euros on that day.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 45

Question 46.
MODELING WITH MATHEMATICS
A crow is flying at a height of 50 feet when it drops a walnut to break it open. The height h (in feet) of the walnut above ground can be modeled by h = -16t2 + 50, where t is the time (in seconds) since the crow dropped the walnut. Solve the equation for t. After how many seconds will the walnut be 15 feet above the ground?
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 18
Answer:

MATHEMATICAL CONNECTIONS In Exercises 47 and 48, s is the side length of an equilateral triangle. Solve the formula for s. Then evaluate the new formula for the given value.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 19
Answer:
47.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 47

In Exercises 49–54, find the inverse of the function. Then graph the function and its inverse.
Question 49.
f(x) = 2x3
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 49.1 Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 49.2

Question 50.
f(x) = x3 – 4
Answer:

Question 51.
f(x) = (x – 5)3
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 51.1
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 51.2

Question 52.
f(x) = 8(x + 2)3
Answer:

Question 53.
f(x) = 4\(\sqrt [3]{ x }\)
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 53.1
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 53.2

Question 54.
f(x) = –\(\sqrt [3]{ x-1 }\)
Answer:

Question 55.
MAKING AN ARGUMENT
Your friend says that the inverse of the function f(x) = 3 is a function because all linear functions pass the Horizontal Line Test. Is your friend correct? Explain.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 55

Question 56.
HOW DO YOU SEE IT?
Pair the graph of each function with the graph of its inverse.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 20
Answer:

Question 57.
WRITING
Describe changes you could make to the function f(x) = x2 – 5 so that its inverse is a function. Describe the domain and range of the new function and its inverse.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 57

Question 58.
CRITICAL THINKING
Can an even function with at least two values in its domain have an inverse that is a function? Explain.
Answer:

Question 59.
OPEN-ENDED
Write a function such that the graph of its inverse is a line with a slope of 4.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 59

Question 60.
CRITICAL THINKING
Consider the function g(x) = -x.
a. Graph g(x) = -x and explain why it is its own inverse.
b. Graph other linear functions that are their own inverses. Write equations of the lines you graph.
c. Use your results from part (b) to write a general equation that describes the family of linear functions that are their own inverses.
Answer:

Question 61.
REASONING
Show that the inverse of any linear function f(x) = mx + b, where m ≠ 0, is also a linear function. Write the slope and y-intercept of the graph of the inverse in terms of m and b.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 61

Question 62.
THOUGHT PROVOKING
The graphs of f(x) = x3 – 3x and its inverse are shown. Find the greatest interval -a ≤ x ≤ a for which the inverse of f is a function. Write an equation of the inverse function.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 21
Answer:

Question 63.
REASONING
Is the inverse of f(x) = 2|x + 1|a function? Are there any values of a, h, and k for which the inverse of f(x) = a |x – h| + k is a function? Explain your reasoning.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 63

Maintaining Mathematical Proficiency

Find the sum or difference.(Section 7.1)
Question 64.
(2x – 9) – (6x + 5)
Answer:

Question 65.
(8y + 1) + (-y – 12)
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 65

Question 66.
(t2 – 4t – 4) + (7t2 + 12t + 3)
Answer:

Question 67.
(-3d2 + 10d – 8) – (7d2 – d – 6)
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 67

Graph the function. Compare the graph to the graph of f(x) = x2. (Section 8.2)
Question 68.
g(x) = x2 + 6
Answer:

Question 69.
h(x) = -x2 – 2
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 69

Question 70.
p(x) = -4x2 + 5
Answer:

Question 71.
q(x) = \(\frac{1}{3}\)x2 – 1
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 a 71

Radical Functions and Equations Performance Task: Medication and the Mosteller Formula

10.3–10.4What Did YouLearn?

Core Vocabulary
radical equation, p. 560
inverse relation, p. 568
inverse function, p. 569

Core Concepts
Lesson 10.3
Squaring Each Side of an Equation, p. 560
Identifying Extraneous Solutions, p. 562

Lesson 10.4
Inverse Relation, p. 568
Finding Inverses of Functions Algebraically, p. 570
Finding Inverses of Nonlinear Functions, p. 570
Horizontal Line Test, p. 571

Mathematical Practices
Question 1.
Could you also solve Exercises 37–44 on page 565 by graphing? Explain.
Answer:

Question 2.
What external resources could you use to check the reasonableness of your answer in Exercise 45 on page 573?
Answer:

Performance Task: Medication and the Mosteller Formula

When taking medication, it is critical to take the correct dosage. For children in particular, body surface area (BSA) is a key component in calculating that dosage. The Mosteller Formula is commonly used to approximate body surface area. How will you use this formula to calculate BSA for the optimum dosage?
To explore the answers to this question and more, go to
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations 10.4 22

Radical Functions and Equations Chapter Review

10.1 Graphing Square Root Functions (pp. 543–550)

Graph the function. Describe the domain and range. Compare the graph to the graph of f(x) = \(\sqrt{x}\).
Question 1.
g(x) = \(\sqrt{x}\) + 7
Answer:

Question 2.
h(x) = \(\sqrt{x-6}\)
Answer:

Question 3.
r(x) = –\(\sqrt{x+3}\) – 1
Answer:

Question 4.
Let g(x) = \(\frac{1}{4} \sqrt{x-6}\) + 2. Describe the transformations from the graph of f(x) = \(\sqrt{x}\) to the graph of g. Then graph g.
Answer:

10.2 Graphing Cube Root Functions (pp. 551–556)

Graph the function. Compare the graph to the graph of f(x) = \(\sqrt [3]{ x }\).
Question 5.
g(x) = \(\sqrt [3]{ x }\) + 4
Answer:

Question 6.
h(x) = -8\(\sqrt [3]{ x }\)
Answer:

Question 7.
s(x) = \(\sqrt[3]{-2(x-3)}\)
Answer:

Question 8.
Let g(x) = -3\(\sqrt [3]{ x+2 }\) – 1. Describe the transformations from the graph of f(x) = \(\sqrt [3]{ x }\) to the graph of g. Then graph g.
Answer:

Question 9.
The graph of cube root function r is shown. Compare the average rate of change of r to the average rate of change of p(x) = \(\sqrt[3]{\frac{1}{2} x}\) over the interval x = 0 to x = 8.
Answer:

10.3 Solving Radical Equations (pp. 559-566)

Solve the equation. Check your solution(s).
Question 10.
8 + \(\sqrt{x}\) = 18
Answer:

Question 11.
\(\sqrt [3]{ x-1 }\) = 3
Answer:

Question 12.
\(\sqrt{5x-9}\) = \(\sqrt{4x}\)
Answer:

Question 13.
x = \(\sqrt{3x+4}\)
Answer:

Question 14.
8\(\sqrt{x-5}\) + 34 = 58
Answer:

Question 15.
\(\sqrt{5x}\) + 6 = 5
Answer:

Question 16.
The radius r of a cylinder is represented by the function r = \(\sqrt{\frac{V}{\pi h}}\), where V is the volume and h is the height of the cylinder. What is the volume of the cylindrical can?
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations cr 16
Answer:

10.4 Inverse of a Function (pp. 567–574)

Find the inverse of the relation.
Question 17.
(1, -10), (3, -4), (5, 4), (7, 14), (9, 26)
Answer:

Question 18.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations cr 18
Answer:

Find the inverse of the function. Then graph the function and its inverse.
Question 19.
f(x) = -5x + 10
Answer:

Question 20.
f(x) = 3x2 – 1, x ≥ 0
Answer:

Question 21.
f(x) = \(\frac{1}{2} \sqrt{2 x+6}\)
Answer:

Question 22.
Consider the function f(x) = x2 + 4. Use the Horizontal Line Test to determine whether the inverse of f is a function.
Answer:

Question 23.
In bowling, a handicap is an adjustment to a bowler’s score to even out differences in ability levels. In a particular league, you can find a bowler’s handicap h by using the formula h = 0.8(210 – a), where a is the bowler’s average. Solve the formula for a. Then find a bowler’s average when the bowler’s handicap is 28.
Big Ideas Math Answer Key Algebra 1 Chapter 10 Radical Functions and Equations cr 23
Answer:

Radical Functions and Equations Chapter Test

Find the inverse of the function.
Question 1.
f(x) = 5x – 8 2.
Answer:

Question 2.
f(x) = 2\(\sqrt{x+3}\) – 1
Answer:

Question 3.
f(x) = –\(\frac{1}{3}\)x2 + 4, x ≥ 0
Answer:

Graph the function f. Describe the domain and range. Compare the graph of f to the graph of g.
Question 4.
f(x) = –\(\sqrt{x+6}\); g(x) = \(\sqrt{x}\)
Answer:

Question 5.
f(x) = \(\sqrt{x-3}\) + 2; g(x) = \(\sqrt{x}\)
Answer:

Question 6.
f(x) = \(\sqrt [3]{ x }\) – 5; g(x) = \(\sqrt [3]{ x }\)
Answer:

Question 7.
f(x) = -2\(\sqrt [3]{ x+1 }\); g(x) = \(\sqrt [3]{ x }\)
Answer:

Solve the equation. Check your solution(s).
Question 8.
9 – \(\sqrt{x}\) = 3
Answer:

Question 9.
\(\sqrt{2x-7}\) – 3 = 6
Answer:

Question 10.
\(\sqrt{8x-21}\) = \(\sqrt{18-5x}\)
Answer:

Question 11.
x + 5 = \(\sqrt{7x+53}\)
Answer:

Question 12.
When solving the equation x – 5 = \(\sqrt{ax+b}\), you obtain x = 2 and x = 8. Explain why at least one of these solutions must be extraneous.
Answer:

Describe the transformations from the graph of f(x) = \(\sqrt [3]{ x }\) to the graph of the given function. Then graph the given function.
Question 13.
h(x) = 4\(\sqrt [3]{ x-1 }\) + 5
Answer:

Question 14.
w(x) = –\(\sqrt [3]{ x+7 }\) – 2
Answer:

Question 15.
The velocity v (in meters per second) of a roller coaster at the bottom of a hill is given by v = \(\sqrt{19.6h}\), where h is the height (in meters) of the hill. (a) Use a graphing calculator to graph the function. Describe the domain and range. (b) How tall must the hill be for the velocity of the roller coaster at the bottom of the hill to be at least 28 meters per second? (c) What happens to the average rate of change of the velocity as the height of the hill increases?
Answer:

Question 16.
The speed s (in meters per second) of sound through air is given by s = 2\(\sqrt{T+273}\), where T is the temperature (in degrees Celsius).
Big Ideas Math Answers Algebra 1 Chapter 10 Radical Functions and Equations c 16
a. What is the temperature when the speed of sound through air is 340 meters per second?
b. How long does it take you to hear the wolf howl when the temperature is -17°C?
Answer:

Question 17.
How can you restrict the domain of the function f(x) = (x – 3)2 so that the inverse of f is a function?
Answer:

Question 18.
Write a radical function that has a domain of all real numbers less than or equal to 0 and a range of all real numbers greater than or equal to 9.
Answer:

Radical Functions and Equations Cumulative Assessment

Question 1.
Fill in the function so that it is represented by the graph.
Big Ideas Math Answers Algebra 1 Chapter 10 Radical Functions and Equations ca 1
Answer:

Question 2.
Consider the equation y = mx + b. Fill in values for m and b so that each statement is true.
a. When m = ______ and b = ______, the graph of the equation passes through the point (-1, 4).
b. When m = ______ and b = ______, the graph of the equation has a positive slope and passes through the point (-2, -5).
c. When m = ______ and b = ______, the graph of the equation is perpendicular to the graph of y = 4x – 3 and passes through the point (1, 6).
Answer:

Question 3.
Which graph represents the inverse of the function f(x) = 2x + 4?
Big Ideas Math Answers Algebra 1 Chapter 10 Radical Functions and Equations ca 3
Answer:

Question 4.
Consider the equation x = \(\sqrt{ax+b}\). Student A claims this equation has one real solution. Student B claims this equation has two real solutions. Use the numbers to answer parts (a)–(c).
Big Ideas Math Answers Algebra 1 Chapter 10 Radical Functions and Equations ca 4
a. Choose values for a and b to create an equation that supports Student A’s claim.
b. Choose values for a and b to create an equation that supports Student B’s claim.
c. Choose values for a and b to create an equation that does not support either student’s claim.
Answer:

Question 5.
Which equation represents the nth term of the sequence 3, 12, 48, 192, . . .?
A. an = 3(4)n-1
B. an = 3(9)n-1
C. an = 9n – 6
D. an = 9n + 3
Answer:

Question 6.
Consider the function f(x) = \(\frac{1}{2} \sqrt[3]{x+3}\). The graph represents function g. Select all the statements that are true.
Big Ideas Math Answers Algebra 1 Chapter 10 Radical Functions and Equations ca 6
Answer:

Question 7.
Place each function into one of the three categories.
Big Ideas Math Answers Algebra 1 Chapter 10 Radical Functions and Equations ca 7
Answer:

Question 8.
You are making a tabletop with a tiled center and a uniform mosaic border.
a. Write the polynomial in standard form that represents the perimeter of the tabletop.
b. Write the polynomial in standard form that represents the area of the tabletop.
c. The perimeter of the tabletop is less than 80 inches, and the area of tabletop is at least 252 square inches. Select all the possible values of x.
Big Ideas Math Answers Algebra 1 Chapter 10 Radical Functions and Equations ca 8
Answer:

Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies

Big Ideas Math Answers Grade 2 Chapter 3

Big Ideas Math Answers 2nd Grade 3rd Chapter Addition to 100 Strategies: It is very important for the students to learn the fundamental concepts in maths. To help out the students, we are presenting the Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies Answer Key in pdf format. Download BIM Grade 2 Chapter 3 Addition to 100 Strategies PDF. This pdf will help you out to finish the assignments in time.

Big Ideas Math Book 2nd Grade Answer Key Chapter 3 Addition to 100 Strategies

The different topics covered in Big Ideas Math 3rd Chapter Addition to 100 Strategies Answer Key are Add Tens Using a Number Line, Add Tens and Ones Using a Number Line, Use Place Value to Add, Decompose to Add Tens and Ones, Use Compensation to Add, Practice Addition Strategies, Addition to 100 Strategies, and others. Know more details about the addition or sum concept using 100 strategies.

Practice all the problems included in Big Ideas Math Grade 2 Answer Key 3rd Chapter Addition to 100 Strategies to score good marks in the test. Before you start the preparation, have a look at the BIM Book Grade 2 Chapter 3 Solutions. We have covered the answers for all the problems topic-wise. Hence click on the below-attached links and complete your homework effortlessly.

Vocabulary

Lesson: 1 Add Tens Using a Number Line

Lesson: 2 Add Tens and Ones Using a Number Line

Lesson: 3 Use Place Value to Add

Lesson: 4 Decompose to Add Tens and Ones

Lesson: 5 Use Compensation to Add

Lesson: 6 Practice Addition Strategies

Lesson: 7 Problem Solving: Addition

Chapter 3: Addition to 100 Strategies

Addition to 100 Strategies Vocabulary

Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 1

Organize It
Use the review words to complete the graphic organizer.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 2
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-100-Strategies-Vocabulary-Organize-It

Define It

Use your vocabulary cards to identify the words.

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 3
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-100-Strategies-Vocabulary-Question-1

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 4
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-100-Strategies-Vocabulary-Question-2

Chapter 3 Vocabulary cards

Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 5
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 6
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Chapter-3-Vocabulary-cards

Lesson 3.1 Add Tens Using a Number Line

Explore and Grow

Color and show how you can use the hundred chart to solve.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 7
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.1-Add-Tens-Using-a-Number-Line-Explore-and-Grow
Explanation:
27 + 10 = 37
We notice from the above figure that 27 is added to 10 means from 27 mustard  yellow coloring starts for 10 numbers and stops at 36 . where 37 is the sum and it is highlighted with yellow color.
62 + 30 = 92
We notice from the above figure that 62 is added to 30 means from 62 mustard  yellow coloring starts for 30 numbers and stops at 91 . where 92 is the sum and it is highlighted with yellow color.

Show and Grow

Question 1.
70 + 30 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 8
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.1-Add-Tens-Using-a-Number-Line-Show-and-Grow-Question-1
Explanation :
The open number line start at 70 and 30 is same as 3 tens So, count on 3 tens .Then the linking jumps move to right from 70 to 100. It ends at 100 showing the sum .
So ,  70 + 30 = 100 .

Question 2.
38 + 40 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 9
Answer :
38 + 40 = 78
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.1-Add-Tens-Using-a-Number-Line-Show-and-Grow-Question-2
Explanation :
The open number line starts at 38 and 40 is the same as 4 tens So, count on 4 tens. Then the linking jumps move to the right from 38 to 78. It ends at 78 showing the sum.
So ,  38 + 40 = 78

Question 3.
22 + 60 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 10
Answer :
22 + 60 = 82
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.1-Add-Tens-Using-a-Number-Line-Show-and-Grow-Question-3
Explanation :
The open number line start at 22 and 60 is same as 6 tens So, count on 6 tens .Then the linking jumps move to right from 22 to 82. It ends at 82 showing the sum .
So ,  38 + 60 = 82

Apply and Grow: Practice

Question 4.
60 + 30 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 11
Answer :
60 + 30 = 90
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.1-Add-Tens-Using-a-Number-Line-Apply-Grow-Practice-Question-4
Explanation :
The open number line start at 60 and 30 is same as 3 tens So, count on 3 tens .Then the linking jumps move to right from 60 to 90. It ends at 90 showing the sum .
So ,  60 + 30 = 90

Question 5.
51 + 20 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 12
Answer :
51 + 20 = 71
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.1-Add-Tens-Using-a-Number-Line-Apply-Grow-Practice-Question-5
Explanation :
The open number line start at 51 and 20 is same as 2 tens So, count on 2 tens .Then the linking jumps move to right from 51 to 20. It ends at 20 showing the sum .
So , 51 + 20 = 71

Question 6.
49 + 50 = ___
Answer :
49 + 50 = 99
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.1-Add-Tens-Using-a-Number-Line-Apply-Grow-Practice-Question-6
Explanation :
The open number line start at 49 and 50 is same as 5 tens So, count on 5 tens .Then the linking jumps move to right from 49 to 50. It ends at 99 showing the sum .
So , 49 + 50 = 99

Question 7.
YOU BE THE TEACHER
Your friend used an open number line to solve 40 + 20. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 13
Answer :
Yes, he is correct.
40 + 20 = 60
Explanation :
The open number line start at 40 and 20 is same as 2 tens So, count on 2 tens .Then the linking jumps move to right from 40 to 60. It ends at 60 showing the sum .
So , 40 + 20 = 60

Think and Grow: Modeling Real Life

You have 66 berries and pick 30 more. Your friend picks 90 berries. Who has more berries?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 14
Addition equation:
Model:
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 15
Answer :
Number of berries with me = 66
Number of berries picked by me = 30
Total Number of berries with me = 66 + 30 = 96
Number of berries with my friend = 90
I have more berries than my friend .
96 > 90 .
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.1-Add-Tens-Using-a-Number-Line-Think-Grow-Modeling-Real-Life

Show and Grow

Question 8.
You have 34 feathers and collect 50 more. Your friend has 85 feathers. Who has more feathers?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 16
Answer :
Number of Feathers with me = 34
Number of features collected by me = 50
Total Number of features with me = 34 + 50 = 84
Number of features with my friend  = 85
84 < 85
My friend is having more features than me

Question 9.
DIG DEEPER!
You have 49 stickers. Your friend gives you some more. Now you have 79. How many stickers did your friend give you?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 17
Answer :
Number of stickers with me = 49
Total number of stickers with me now = 79
Number of stickers given by friend to me = 79 – 49 = 30
Therefore, Number of stickers given by friend to me = 30

Add Tens Using a Number Line Homework & Practice 3.1

Question 1.
40 + 40 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 18
Answer :
40 + 40 = 80
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Add-Tens-Using-Number-Line-Homework-Practice-3.1-Question-1
Explanation :
The open number line start at 40 and 40 is same as 4 tens So, count on 4 tens .Then the linking jumps move to right from 40 to 80. It ends at 80 showing the sum .
So , 40 + 40 = 80

Question 2.
65 + 20 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 19
Answer:
65 + 20 = 85
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Add-Tens-Using-Number-Line-Homework-Practice-3.1-Question-2
Explanation :
The open number line start at 65 and 20 is same as 2 tens So, count on 2 tens .Then the linking jumps move to right from 65 to 20. It ends at 20 showing the sum .
So , 65 + 20 = 85

Question 3.
14 + 60 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 20
Answer :
14 + 60 =74
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Add-Tens-Using-Number-Line-Homework-Practice-3.1-Question-3
Explanation :
The open number line start at 14 and 60 is same as 6 tens So, count on 6 tens .Then the linking jumps move to right from 14 to 74. It ends at 74 showing the sum .
So , 14 + 60 = 74

Question 4.
Number Sense
Write an equation that matches the number line.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 21
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Add-Tens-Using-Number-Line-Homework-Practice-3.1-Question-4
Explanation :
The open number line start at 43 and 30 is same as 3 tens So, count on 3 tens .Then the linking jumps move to right from 43 to 30. It ends at 73 showing the sum .
So , 43 + 30 = 73

Question 5.
Modeling Real Life
Newton has 31 stamps and buys 20 more. Descartes has 50 stamps. Who has more stamps?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 22
Answer :
Number of stamps with Newton = 31
Number of stamps bought by newton = 20
Total Number of stamps with newton = 31 + 20 = 51
Number of stamps with Descartes = 50
51 > 50
Newton have more stamps than Descartes .

Question 6.
DIG DEEPER!
You have 64 rocks. Your friend gives you some more. Now you have 94 rocks. How many rocks did your friend give you?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 23
Answer :
Number of rocks with me = 64
Number of rocks given by friend = x
Total Number of rocks with me now = 94
Number of rocks given by friend = 94 – 64 = 30
Therefore, Number of rocks given by friend = 30

Review & Refresh

Question 7.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 24
Answer :
15 is odd number
Explanation :
A number which is divisible by 2 and generates a remainder of 0 is called an even number.
Odd numbers are whole numbers that cannot be divided exactly into pairs. Odd numbers, when divided by 2, leave a remainder of 1. 1, 3, 5, 7, 9, 11, 13, 15 … are sequential odd numbers. Odd numbers have the digits 1, 3, 5, 7 or 9 in their ones place.

Question 8.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 25
Answer :
8 is a even number as it is exactly divisible by 2
Explanation :
A number which is divisible by 2 and generates a remainder of 0 is called an even number.
Odd numbers are whole numbers that cannot be divided exactly into pairs. Odd numbers, when divided by 2, leave a remainder of 1. 1, 3, 5, 7, 9, 11, 13, 15 … are sequential odd numbers. Odd numbers have the digits 1, 3, 5, 7 or 9 in their ones place.

Question 9.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 26
Answer :
12 is even number as it is exactly divisible by 2
Explanation :
A number which is divisible by 2 and generates a remainder of 0 is called an even number.
Odd numbers are whole numbers that cannot be divided exactly into pairs. Odd numbers, when divided by 2, leave a remainder of 1. 1, 3, 5, 7, 9, 11, 13, 15 … are sequential odd numbers. Odd numbers have the digits 1, 3, 5, 7 or 9 in their ones place.

Lesson 3.2 Add Tens and Ones Using a Number Line

Color and show how you can use the hundred chart to solve.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 27
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.2-Add-Tens-and-Ones-Using-a-Number-Line

Show and Grow

Question 1.
69 + 12 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 28
Answer :
69 + 12 = 81
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.2-Add-Tens-and-Ones-Using-a-Number-Line-Show-Grow-Question-1
Explanation :
The open number line start at 69 and 12 is same as 1 tens and 2 ones So, count on 1 tens and 2 ones.Then the linking jumps move to right from 69 to 81. It ends at 81 showing the sum .
So , 69 + 12 = 81

Question 2.
55 + 34 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 29
Answer :
55 + 34 = 89
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.2-Add-Tens-and-Ones-Using-a-Number-Line-Show-Grow-Question-2
Explanation :
The open number line start at 55 and 34 is same as 3 tens and 4 ones So, count on 3 tens and 4 ones.Then the linking jumps move to right from 55 to 89. It ends at 89 showing the sum .
So , 55 + 34 = 89

Apply and Grow: Practice

Question 3.
15 + 32 = __

Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 30
Answer:
15 + 32 = 47
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.2-Add-Tens-and-Ones-Using-a-Number-Line-Apply-and-Grow-Practice-Question-3
Explanation :
The open number line start at 15 and 32 is same as 3 tens and 2 ones So, count on 3 tens and 2 ones.Then the linking jumps move to right from 15 to 32. It ends at 47 showing the sum .
So , 15 + 32 = 47

Question 4.
52 + 26 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 31
Answer :
52 + 26 = 78
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.2-Add-Tens-and-Ones-Using-a-Number-Line-Apply-and-Grow-Practice-Question-4
Explanation :
The open number line start at 5 and 26 is same as 2 tens and 6 ones So, count on 2 tens and 6 ones. Then the linking jumps move to right from 52 to 26. It ends at 78 showing the sum .
So , 52 + 26 = 78

Question 5.
38 + 41 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 32
Answer :
38 + 41 = 79
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.2-Add-Tens-and-Ones-Using-a-Number-Line-Apply-and-Grow-Practice-Question-5
Explanation :
The open number line start at 38 and 41 is same as 4 tens and 1 ones So, count on 4 tens and 1 ones.Then the linking jumps move to right from 38 to 41. It ends at 79 showing the sum .
So , 38 + 41 =79

Question 6.
65 + 33 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 33
Answer :
65 + 33 = 98
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.2-Add-Tens-and-Ones-Using-a-Number-Line-Apply-and-Grow-Practice-Question-6
Explanation :
The open number line start at 65 and 33 is same as 3 tens and 3 ones So, count on 3 tens and 3 ones.Then the linking jumps move to right from 65 to 33. It ends at 98 showing the sum .
So , 65 + 33 = 98

Question 7.
Reasoning
Complete the number line and the equation.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 34
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.2-Add-Tens-and-Ones-Using-a-Number-Line-Question-7
Explanation:
The open number line start at 46 . we have 2 tens and ___ ones. we have find the number of ones in the above figure. after adding two tens to the 46 we get 66 . if we add 4 to 66 only then the sum becomes 70. so the number ones are 4 .
Then the linking jumps move to right from 46 to 70. It ends at 70 showing the sum .
So , 46 + 24 = 70

Question 8.
There are 31 tadpoles. Then 19 more join them. How many tadpoles are there now?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 35
Answer :
Number of tadpoles = 31
Number of tadpoles joined = 19
Total Number of tadpoles now = 31 + 19 = 50
Therefore, Total Number of tadpoles now = 50 .

Think and Grow: Modeling Real Life

Your class plants 21 trees. Your friend’s class plants 39. How many more trees does your friend’s class plant than your class?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 36
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 36.1
Answer :
Number of plants planted by my class = 21
Number of plants planted my friend’s class = 39
Number of more trees does my friend’s class plant than my class = 39 – 21 = 18

Show and Grow

Question 9.
Your friend pops 76 bubbles on a piece of bubble wrap. You pop 88. How many more bubbles do you pop than your friend?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 37
Answer :
Number of bubbles popped by my friend = 76
Number of bubbles popped by me = 88
Number of more bubbles popped by me than my friend = 88 – 76 = 12
Therefore, Number of more bubbles popped by me than my friend = 12

Question 10.
DIG DEEPER!
You have 45 bouncy balls. You find some more. Now you have 91. How many bouncy balls did you find?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 38
Answer :
Number of bouncy balls with me = 45
Total Number of bouncy balls with me now = 91
Number of bouncy balls found by me = 91 – 45 = 46
Therefore Number of bouncy balls found by me =  46

Add Tens and Ones Using a Number Line Homework & Practice 3.2

Question 1.
64 + 31 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 39
Answer :
64 + 31 = 95
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Add-Tens-Ones-Using-Number-Line-Homework-Practice-3.2-Question-1
Explanation :
The open number line start at 64 and 31 is same as 3 tens and 1 ones So, count on 3 tens and 1 ones .Then the linking jumps move to right from 64 to 95. It ends at 95 showing the sum .
So , 64 + 31 = 95

Question 2.
29 + 56 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 40
Answer :
29 + 56 = 85
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Add-Tens-Ones-Using-Number-Line-Homework-Practice-3.2-Question-2
Explanation :
The open number line start at 29 and 56 is same as 5 tens and 6 ones So, count on 5 tens and 6 Ones.Then the linking jumps move to right from 29 to 85. It ends at 85 showing the sum .
So , 29 + 56 = 85

Question 3.
18 + 62 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 41
Answer :
18 + 62 = 80
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Add-Tens-Ones-Using-Number-Line-Homework-Practice-3.2-Question-3
Explanation :
The open number line start at 18 and 62 is same as 6 tens and 8 ones So, count on 6 tens and 8 ones.Then the linking jumps move to right from 18 to 80. It ends at 80 showing the sum .
So , 18 + 62 = 80

Question 4.
72 + 11 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 42
Answer :
72 + 11 = 83
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Add-Tens-Ones-Using-Number-Line-Homework-Practice-3.2-Question-4
Explanation :
The open number line start at 72 and 11 is same as 1 tens and 1ones So, count on 1 tens and 1 ones.Then the linking jumps move to right from 72 to 83. It ends at 83 showing the sum .
So , 72 + 11 = 83

Question 5.
53 + 27 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 43
Answer :
53 + 27 = 80
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Add-Tens-Ones-Using-Number-Line-Homework-Practice-3.2-Question-5
Explanation :
The open number line start at 53 and 27 is same as 2 tens and 7 Ones So, count on 2 tens and 7 Ones. Then the linking jumps move to right from 53 to 80. It ends at 80 showing the sum .
So , 53 + 27 = 80

Question 6.
46 + 32 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 44
Answer :
46 + 32 = 78
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Add-Tens-Ones-Using-Number-Line-Homework-Practice-3.2-Question-6
Explanation :
The open number line start at 46 and 32 is same as 3 tens and 2 Ones So, count on 3 tens and 2 Ones. Then the linking jumps move to right from 46 to 78. It ends at 78 showing the sum .
So , 46 + 32 = 78

Question 7.
Structure
Show 27 + 43 two ways.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 45

Answer :
27 + 43 = 70
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Add-Tens-Ones-Using-Number-Line-Homework-Practice-3.2-Question-7-2
43 + 27 = 70
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Add-Tens-Ones-Using-Number-Line-Homework-Practice-3.2-Question-7
Explanation :
The open number line start at 27 and 43 is same as 4 tens and 3 Ones So, count on 4 tens and 3 Ones .Then the linking jumps move to right from 27 to 70. It ends at 70 showing the sum .
So , 27 + 43 = 70
The open number line start at 43 and 27 is same as 2 tens and 7 Ones So, count on 2 tens and 7 Ones .Then the linking jumps move to right from 43 to 70. It ends at 70 showing the sum .
So , 43+ 27 = 70

Question 8.
Modeling Real Life
You bake 36 muffins on Friday. You bake 24 on Saturday. How many more muffins do you bake on Friday?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 46
Answer :
Number of muffins baked on Friday = 36
Number of muffins baked on Saturday = 24
Number of muffins baked more on Friday than Saturday = 36 – 24 = 12
Therefore, 12 more muffins are baked on Friday

Question 9.
DIG DEEPER!
Newton has 61 bugs. Then he finds some more. Now he has 78. How many bugs did Newton find?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 46.1

Answer :
Number of bugs with Newton = 61
Total Number of bugs with newton now = 78
Number of bugs found by Newton = 78 -61 = 17

Review & Refresh

Question 10.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 47
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Add-Tens-Ones-Using-Number-Line-Homework-Practice-3.2-Question-10
Explanation:
From the above picture we notice, 3 stack of ten cubes and 5 cubes so the Numbers have two digits. … The left digit is the tens’ place. It tells you that there are 3 tens. The last or right digit is the ones’ place which is 5. So,
Number is 35

Question 11.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 48
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Add-Tens-Ones-Using-Number-Line-Homework-Practice-3.2-Question-11
Explanation:
From the above picture we notice, 6 stack of ten cubes and 1 cube .the Numbers have two digits. … The left digit is the tens’ place. It tells you that there are 6 tens. The last or right digit is the ones’ place which is 1. So,
Number is 61

Lesson 3.3 Use Place Value to Add

Explore and Grow

How can you use a model to solve?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 49Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 50
Answer :
37 + 15 = 52
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.3-Use-Place-Value-to-Add-Explore-Grow
Explanation :
We use place value and break the addends into tens and ones. Then add the 10 tens values and ones values . Write the number formed . So, the number is  52

Show and Grow

Use place value to break apart the addends. Then find the sum.

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 51
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.3-Use-Place-Value-to-Add-Show-Grow-Question-1
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  52

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 52
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.3-Use-Place-Value-to-Add-Show-Grow-Question-2
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  87

Question 3.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 53
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.3-Use-Place-Value-to-Add-Show-Grow-Question-3
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  64

Question 4.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 54
Answer :

Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.3-Use-Place-Value-to-Add-Show-Grow-Question-4

Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  92

Apply and Grow: Practice

Use place value to break apart the addends. Then find the sum.

Question 5.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 55
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.3-Use-Place-Value-to-Add-Question-5
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  88

Question 6.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 56

Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.3-Use-Place-Value-to-Add-Question-6
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  99

Question 7.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 57
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.3-Use-Place-Value-to-Add-Question-7
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  89

Question 8.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 58
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.3-Use-Place-Value-to-Add-Question-8
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  46

Question 9.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 59

Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.3-Use-Place-Value-to-Add-Question-9
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  80

Question 10.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 60
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.3-Use-Place-Value-to-Add-Question-10
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  63

Question 11.
Number Sense
Solve. Think: Does the same number make both equations true?
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 61
Answer :
No, it requires different numbers
16 + __________ = 27
16 + 11 = 27

____ + 27 = 48
21 + 27 = 48

Think and Grow: Modeling Real Life

You use 72 blocks to make a building. Your friend uses 14 more than you. How many blocks does your friend use?
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 62
Addition equation:
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 63
Answer :
Number of blocks used by me = 72
Number of blocks used by my friend = 72 + 14  = 86
Addition Equation :
72 + 14 = 86
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.3-Use-Place-Value-to-Add-Think-Grow-Modeling-Real-Life

Show and Grow

Question 12.
Your friend has 57 beads. You have 21 more than your friend. How many beads do you have?
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 64
Answer :
Number of beads with my friend  = 57
Number of beads with me = 57 + 21
Addition equation :
57 + 21 = 78
Explanation :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.3-Use-Place-Value-to-Add-Show-Grow-Question-12

Question 13.
You have 68 grapes. 38 are red. The rest are green. How many grapes are green?
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 65
Answer :
Number of grapes with me = 68
Number of red grapes = 38
Number of green grapes = 68 – 38 = 40
Therefore, Number of green grapes = 40 .

Use Place Value to Add Homework & practice 3.3

Use place value to break apart the addends. Then find the sum.

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 66
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Use-Place-Value-Add-Homework-practice-3.3-Question-1
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  92

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 67
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Use-Place-Value-Add-Homework-practice-3.3-Question-2
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  88

Question 3.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 68
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Use-Place-Value-Add-Homework-practice-3.3-Question-3
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  100

Question 4.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 69
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Use-Place-Value-Add-Homework-practice-3.3-Question-4
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  91

Question 5.
DIG DEEPER!
You buy two packs of pens. You buy 49 pens in all. Which color pens do you buy?
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 69.1
Answer :
Total pens = 49
Number of blue pens = 25
Number of black pens = 12
Total blue and black pens = 25 + 12 = 37
Total blue black are not equal to 49 so these are not bought .
Number of red pens = 24
Total number of blue and black pens = 25 + 24 = 49
Therefore the color pens which are bought are black and blue as there total numbers pens are equal to 49 .

Question 6.
Modeling Real Life
Newton picks 41 flowers. Descartes picks 28 more than Newton. How many flowers does Descartes pick?
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 70
Answer :
Number of flowers picked by Newton = 41
Number of flowers picked by Descartes = 28 more than Newton = 28 + 41 = 69

Question 7.
Modeling Real Life
You have 57 marbles. 36 are yellow. The rest are blue. How many blue marbles do you have?
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 71
Answer :
Number of marbles with me = 57
Number of yellow marbles = 36
Number of blue marbles = total marbles – yellow marbles = 57 – 36 = 21
Therefore, Number of blue marbles = 21 .

Review & Refresh

Question 8.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 72
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Use-Place-Value-Add-Homework-practice-3.3-Question-8
Explanation:
The number before rows tell us how many groups we have.
The number after rows tell us how many are in each group.
The number at the end, after the equal signs is the total number of items .
we add number of items in each row as many times of number of rows .

Question 9.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 73
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Use-Place-Value-Add-Homework-practice-3.3-Question-9
Explanation:
The number before rows tell us how many groups we have.
The number after rows tell us how many are in each group.
The number at the end, after the equal signs is the total number of items .
we add number of items in each row as many times of number of rows .

Lesson 3.4 Decompose to Add Tens and Ones

Explore and Grow

Use a model to solve.

Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 74

Which addend did you break apart to add ? Explain.
________________________________
________________________________
________________________________
Answer :
50 + 18 = 68

Explanation :
The addend which we break apart to add is 18 . Break 18 into tens and ones that is 10 + 8. Add the tens to the first addend that is 50 + 10 = 60. Then add the ones that is 60 + 8 = 68.
So, the sum is 68 .

Show and Grow

Use place value to break apart an addend. Then find the sum.

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 75
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.4-Decompose-Add-Tens-Ones-Show-Grow-Question-1
Explanation :
The addend which we break apart to add is 12 . Break 12 into tens and ones that is 10 + 2. Add the tens to the first addend that is 61 + 10 = 71. Then add the ones that is 71 + 2 = 73.
So, the sum is 73 .

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 76
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.4-Decompose-Add-Tens-Ones-Show-Grow-Question-2
Explanation :
The addend which we break apart to add is 24 . Break 24 into tens and ones that is 20 + 4. Add the tens to the first addend that is 74 + 20 = 94. Then add the ones that is 94 + 4 = 98.
So, the sum is 98.

Question 3.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 77
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.4-Decompose-Add-Tens-Ones-Show-Grow-Question-3
Explanation :
The addend which we break apart to add is 44 . Break 44 into tens and ones that is 40 + 4. Add the tens to the first addend that is 23 + 40 = 63. Then add the ones that is 63 + 4 = 67.
So, the sum is 67 .

Question 4.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 78
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.4-Decompose-Add-Tens-Ones-Show-Grow-Question-4
Explanation :
The addend which we break apart to add is 53 . Break 53 into tens and ones that is 50 + 3. Add the tens to the first addend that is 31 + 50 = 81. Then add the ones that is 81 + 3 = 84.
So, the sum is 84 .

Apply and Grow: Practice

Use place value to break apart an addend. Then find the sum.

Question 5.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 79
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.4-Decompose-Add-Tens-Ones-Apply-Grow-Practice-Question-5
Explanation :
The addend which we break apart to add is 11 . Break 11 into tens and ones that is 10 + 1. Add the tens to the first addend that is 71 + 10 = 81. Then add the ones that is 81 + 1 = 82.
So, the sum is 82 .

Question 6.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 80
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.4-Decompose-Add-Tens-Ones-Apply-Grow-Practice-Question-6
Explanation :
The addend which we break apart to add is 23 . Break 23 into tens and ones that is 20 + 3.  Add the tens to the first addend that is 65 + 20 = 85. Then add the ones that is 85 + 3 = 88.
So, the sum is 88

.

Question 7.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 81
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.4-Decompose-Add-Tens-Ones-Apply-Grow-Practice-Question-7
Explanation :
The addend which we break apart to add is 47 . Break 47 into tens and ones that is 40 + 7. Add the tens to the first addend that is 32 + 40 = 72. Then add the ones that is 72 + 7 = 79.
So, the sum is 79 .

Question 8.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 82
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.4-Decompose-Add-Tens-Ones-Apply-Grow-Practice-Question-8
Explanation :
The addend which we break apart to add is 43 . Break 43 into tens and ones that is 40 + 3. Add the tens to the first addend that is 25 + 40 = 65. Then add the ones that is 65 + 3 = 68.
So, the sum is 68 .

Question 9.
54 + 36 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.4-Decompose-Add-Tens-Ones-Apply-Grow-Practice-Question-9
Explanation :
The addend which we break apart to add is 36 . Break 36 into tens and ones that is 30 + 6. Add the tens to the first addend that is 54 + 30 = 84. Then add the ones that is 84 + 6 = 90.
So, the sum is 90 .

Question 10.
21 + 61 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.4-Decompose-Add-Tens-Ones-Apply-Grow-Practice-Question-10
Explanation :
The addend which we break apart to add is 61 . Break 61 into tens and ones that is 60 + 1. Add the tens to the first addend that is 21 + 60 = 81. Then add the ones that is 81 + 1 = 82.
So, the sum is 82 .

Question 11.
Precision
Which expressions have a sum of 72.
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 83
Answer :
Two Equations 41 + 31 and 11 + 61 have sum of 72 .
Explanation :
52 + 22 = 74
51 + 11 = 62
41 + 31 = 72
47 + 30 = 77
11 + 61 = 72

Think and Grow: Modeling Real Life

You have 32 coins in a piggy bank. You put in 45 more. How many coins are in the bank now?
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 84
Addition equation:
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 85
Answer :
Number of Coins = 32
Number of coins added in piggy bank = You put in 45 more = 32 + 45 = 77
Therefore, Number of coins in bank = 77

Show and Grow

Question 12.
There are 81 fans at a game. 18 more arrive. How many fans are at the game now?
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 86
Answer :
Number of fans  = 81
Number of fans arrived = 18 more
Number of fans at the game = 81 + 18 = 99 fans .

Question 13.
You have 54 songs on a music player. Your friend has 31 more than you. How many songs does your friend have ?
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 86.1
Answer :
Number of songs in my music player = 54
Number of songs my friend have = 31 more than me = 31 + 54 = 85

Decompose to Add Tens and Ones Homework & Practice 3.4

Use place value to break appart an addend. Then find the sum.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 87
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Decompose-Add-Tens-Ones-Homework-Practice-3.4-Question-1
Explanation :
The addend which we break apart to add is 34 . Break 34 into tens and ones that is 30 + 4. Add the tens to the first addend that is 53 + 30 = 83. Then add the ones that is 83 + 4 = 87.
So, the sum is 87 .

Question 2.
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 88
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Decompose-Add-Tens-Ones-Homework-Practice-3.4-Question-2
Explanation :
The addend which we break apart to add is 13 . Break 13 into tens and ones that is 10 + 3. Add the tens to the first addend that is 46 + 10 = 56. Then add the ones that is 56 + 3 = 59.
So, the sum is 59 .

Question 3.
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 89
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Decompose-Add-Tens-Ones-Homework-Practice-3.4-Question-3
Explanation :
The addend which we break apart to add is 73 . Break 73 into tens and ones that is 70 + 3. Add the tens to the first addend that is 14 + 70 = 84. Then add the ones that is 84 + 3 = 87.
So, the sum is 87 .

Question 4.
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 90

Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Decompose-Add-Tens-Ones-Homework-Practice-3.4-Question-4
Explanation :
The addend which we break apart to add is 67 . Break 67 into tens and ones that is 60 + 7. Add the tens to the first addend that is 31 + 60 = 91. Then add the ones that is 91 + 7 = 98.
So, the sum is 98 .

Question 5.
28 + 61 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Decompose-Add-Tens-Ones-Homework-Practice-3.4-Question-5
Explanation :
The addend which we break apart to add is 28 . Break 28 into tens and ones that is 60 + 1. Add the tens to the first addend that is 28 + 60 = 88. Then add the ones that is 88 + 1 = 89.
So, the sum is 89 .

Question 6.
51 + 41 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Decompose-Add-Tens-Ones-Homework-Practice-3.4-Question-6
Explanation :
The addend which we break apart to add is 41 . Break 41 into tens and ones that is 40 + 1. Add the tens to the first addend that is 51 + 40 = 91. Then add the ones that is 91 + 1 = 92.
So, the sum is 92 .

Question 7.
DIG DEEPER!
Use the numbers to complete the equation.
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 91
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Decompose-Add-Tens-Ones-Homework-Practice-3.4-Question-7

Question 8.
Modeling Real Life
You see 38 ladybugs. Then you see 21 more. How many ladybugs do you see?
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 92
Answer :
Number of lady bugs seen = 38
Number of Lady bugs seen later = 21 more = 38 + 21 = 59
Therefore, Total Number of lady bugs seen = 59

Question 9.
Modeling Real Life
You read for 18 minutes and play outside for 35 minutes. How many minutes do you read and play in all?
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 93
Answer :
Minutes used for reading = 18
Minutes used to played = 35
Number of Minutes used to read and play = 18 + 35 = 53

Review & Refresh

Question 10.
Draw a line through the shorter object.
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 94
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Decompose-Add-Tens-Ones-Homework-Practice-3.4-Question-10

Lesson 3.5 Use Compensation to Add

Explore and Grow

Match the numbers that make a ten.
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 95

How can making a ten help you add greater numbers?
_________________________________
_________________________________
_________________________________
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.5-Use-Compensation-Add-Explore-Grow
Explanation :
Making a ten help you us in adding numbers easily . It helps to divide the addend into tens part and ones part. tens parts can be easily added to any numbers then later the ones part is added .

Show and Grow

Use compensation to add.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 96
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.5-Use-Compensation-Add-Show-Grow-Question-1
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 1 is taken from addend 5 and added to 59 addend to make 59 a decade number that is 59 + 1 = 60
Then add the changed addends we get, 4 + 60 = 64 .

Question 2.
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 97
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.5-Use-Compensation-Add-Show-Grow-Question-2
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 4 is taken from addend 37 and added to 16 addend to make 16 a decade number that is 16 + 4 = 20
Then add the changed addends we get, 33 + 20 = 53 .

Apply and Grow: Practice

Use compensation to add.

Question 3.
52 + 19 = ?
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 98
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.5-Use-Compensation-Add-Apply-Grow-Practice-Question-3
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 1 is taken from addend 52 and added to 19 addend to make 19 a decade number that is 19 + 1 = 20
Then add the changed addends we get, 51 + 20 = 71 .

Question 4.
43 + 8 = ?
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 99
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.5-Use-Compensation-Add-Apply-Grow-Practice-Question-4
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 2 is taken from addend 43 and added to 8 addend to make 8 a decade number that is 8 + 2 = 10
Then add the changed addends we get, 41+ 10 = 51 .

Question 5.
26 + 35 = ?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 100
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.5-Use-Compensation-Add-Apply-Grow-Practice-Question-5
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 5 is taken from addend 26 and added to 35 addend to make 35 a decade number that is 35 + 5 = 40
Then add the changed addends we get, 21 + 40 = 61 .

Question 6.
68 + 15 = ?
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 101
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.5-Use-Compensation-Add-Apply-Grow-Practice-Question-6
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 5 is taken from addend 68 and added to 15 addend to make 15 a decade number that is 15 + 5 = 20
Then add the changed addends we get, 63+ 20 = 83 .

Question 7.
Structure
Show two different ways to use compensation to find 25 + 17.
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 102
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.5-Use-Compensation-Add-Apply-Grow-Practice-Question-7
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 3 is taken from addend 25 and added to 17 addend to make 17 a decade number that is 17 + 3= 20
Then add the changed addends we get, 22+ 20 = 42 .
Takes ones from one addend to make other addend number a decade number . Here 5 is taken from addend 17 and added to 25 addend to make 25 a decade number that is 25 + 5= 30
Then add the changed addends we get, 30+ 12= 42 .

Think and Grow: Modeling Real Life

You need 100 tickets for a prize. You have 79 and win 17 more. Do you have enough tickets for a prize?
Addition equation:
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 103
Answer :
Number of Tickets for a prize = 100
Number of tickets with me = 79
Number of tickets won = 17
Total Number of tickets with me = 79 + 17 = 96
100 > 96 .
No, I don’t have enough tickets for a prize .

Show and Grow

Question 8.
You need to set up 80 chairs. You set up 32. Your friend sets up 49. Do you and your friend set up enough chairs?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 104
Answer :
Numbers chairs required to setup = 80
Number of chairs set up by me = 32
Number of chairs set up by my friend = 49
Total chairs set up by me and my friend = 32 + 49 = 81
80 < 81
Yes, me and my friend set up enough chairs .

Question 9.
DIG DEEPER!
You need 50 containers of slime for a party. You have 28. You buy 17 more. How many more containers do you need?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 105
Answer :
Number of containers required for slime a party = 50
Number of containers with me = 28
Number of containers bought = 17
Total Number of containers with me = 28 + 17 =45
Number of containers required more = 50 – 45 = 5

Use Compensation to Add Homework & Practice 3.5

Use compensation to add.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 3 Addition to 100 Strategies 106
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Use Compensation-Add-Homework-Practice-3.5-Question-1
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 1 is taken from addend 31 and added to 46 addend to make 31 a decade number that is 31 – 1 = 30
Then add the changed addends we get, 47+ 30 = 77 .

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 107
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Use Compensation-Add-Homework-Practice-3.5-Question-2
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 5 is taken from addend 27 and added to 15 addend to make 15 a decade number that is 15 + 5 = 20
Then add the changed addends we get, 22+ 20 = 42 .

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 108
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Use Compensation-Add-Homework-Practice-3.5-Question-3
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 4 is taken from addend 34 and added to 66 addend to make 66 a decade number that is 34 – 4 = 30
Then add the changed addends we get, 70+ 30 = 100 .

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 109
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Use Compensation-Add-Homework-Practice-3.5-Question-4
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 1 is taken from addend 56 and added to 29 addend to make 29 a decade number that is 29 + 1 = 30
Then add the changed addends we get, 55+ 30 = 85 .

Use compensation to add.

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 110
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Use Compensation-Add-Homework-Practice-3.5-Question-5
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 4 is taken from addend 75 and added to 6 addend to make 6 a decade number that is 6 + 4 = 10
Then add the changed addends we get, 71+ 10 = 81 .

Question 6.
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 111
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Use Compensation-Add-Homework-Practice-3.5-Question-6
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 2 is taken from addend 44 and added to 38 addend to make 38 a decade number that is 38 + 2 = 40
Then add the changed addends we get, 42+ 40 = 84 .

Question 7.
Writing
Explain why you make one addend a decade number when using compensation to add.
________________________________
________________________________
________________________________
Answer :
Addends are numbers used in addition problems, they refers to the numbers that are added together. … When solving addition problems, it is advisable to make one of the addends a tens number because, it makes the addition operation easier to manipulate.
Compensation is a mental math strategy for multi-digit addition that involves adjusting one of the addends to make the equation easier to solve. Some students may prefer this strategy as an alternative to left-to-right addition or the breaking up the second number strategy.

Question 8.
Modeling Real Life
You want to read 50 pages. You read 18 on Monday and 27 on Tuesday. Do you reach your goal?
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 112
Answer :
Number of pages i want to read = 50
Number of pages read on Monday = 18
Number of pages read on Tuesday = 27
Number of pages read by me on Monday and Tuesday = 18 + 27 = 45
50 > 45 .
No, I didn’t reach the goal. More 5 pages required to read .

Review & Refresh

Draw one line to show the parts

Question 9.
2 squares
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 113
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Use Compensation-Add-Homework-Practice-3.5-Question-9
Explanation :
The above rectangle is divided into 2 squares by drawing a line in the middle of the rectangle. it is drawn on half of the length of the rectangle .

Question 10.
2 triangles
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 114
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Use Compensation-Add-Homework-Practice-3.5-Question-10
Explanation :
Draw any line from any of the vertexes of the triangle . It divided the given triangle into two equal triangles as shown in the above diagram .

Lesson 3.6 Practice Addition Strategies

Explore and Grow

Use any strategy to find the sum.
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 115

Compare your strategy to your partner’s strategy. How are your strategies alike? How are they different?
________________________________
________________________________
________________________________
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.6-Practice-Addition-Strategies-Explore-Grow
Explanation :
In the first method We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 76
In the Second Method The addend which we break apart to add is 41 . Break 41 into tens and ones that is 40 + 1. Add the tens to the first addend that is 35 + 40 = 75. Then add the ones that is 75 + 1 = 76.
So, the sum is 76 .

Show and Grow

Question 1.
50 + 35 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.6-Practice-Addition-Strategies-Show-Grow-Question-1
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 85

Question 2.
64 + 32 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.6-Practice-Addition-Strategies-Show-Grow-Question-2
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 96

Question 3.
19 + 26 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.6-Practice-Addition-Strategies-Show-Grow-Question-3
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 45

Question 4.
78 + 13 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.6-Practice-Addition-Strategies-Show-Grow-Question-4
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 91

Apply and Grow: Practice

Question 5.
32 + 41 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.6-Practice-Addition-Strategies-Apply-Grow-Practice-Question-5
Explanation :
The addend which we break apart to add is 41 . Break 41 into tens and ones that is 40 + 1. Add the tens to the first addend that is 32 + 40 = 72. Then add the ones that is 72 + 1 = 73.
So, the sum is 73 .

Question 6.
56 + 18 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.6-Practice-Addition-Strategies-Apply-Grow-Practice-Question-6
Explanation :
The addend which we break apart to add is 18 . Break 18 into tens and ones that is 10 + 8. Add the tens to the first addend that is 56 + 10 = 66. Then add the ones that is 66 + 8 = 74.
So, the sum is 74 .

Question 7.
47 + 36 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.6-Practice-Addition-Strategies-Apply-Grow-Practice-Question-7
Explanation :
The addend which we break apart to add is 36 . Break 36 into tens and ones that is 30 + 6. Add the tens to the first addend that is 47 + 30 =77 . Then add the ones that is 77 + 6 = 83.
So, the sum is 83 .

Question 8.
91 + 9 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.6-Practice-Addition-Strategies-Apply-Grow-Practice-Question-8
Explanation :
The addend which we break apart to add is 9 . Break 9 into tens and ones that is 10 – 1. Add the tens to the first addend that is 91 + 10 = 101. Then add the ones that is 101 – 1 = 100.
So, the sum is 100 .

Question 9.
___ = 67 + 22
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.6-Practice-Addition-Strategies-Apply-Grow-Practice-Question-9
Explanation :
The addend which we break apart to add is 22 . Break 22 into tens and ones that is 20 + 2. Add the tens to the first addend that is 67 + 20 =87 . Then add the ones that is 87 + 2 = 89.
So, the sum is 89 .

Question 10.
__ = 51 + 39
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Lesson-3.6-Practice-Addition-Strategies-Apply-Grow-Practice-Question-10
Explanation :
The addend which we break apart to add is 39 . Break 39 into tens and ones that is 30 + 9. Add the tens to the first addend that is 51 + 30 = 81  . Then add the ones that is 51 + 39 = 90.
So, the sum is 90 .

Question 11.
YOU BE THE TEACHER
Your friend breaks apart an addend to solve 24 + 56. Is your friend correct? Explain.
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 115.1

Answer :
Yes , he is correct.
Explanation :
The addend which we break apart to add is 56 . Break 56 into tens and ones that is 50 + 6. Add the tens to the first addend that is 24 + 50 = 74  . Then add the ones that is 74 + 6 = 80.
So, the sum is 80 .

Think and Grow: Modeling Real Life

You collect 23 cans of a food drive. Your friend collects 35 more than you. How many cans does your friend collect?
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 116
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 117

Answer :
Number of cans collected by me = 23
Number of cans collected by my friend = 35 more than me
Number of cans collected by my friend = 35+ 23 = 58

Show and Grow

Question 12.
You jump rope 67 times today. Tomorrow you want to jump rope 15 more times than you did today. How many times should you jump rope tomorrow?
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 118
Answer :
Number of rope jumps today by me = 67
Number of rope jumps by me for tomorrow = 15 more times than you did today
Number of rope jumps by me for tomorrow = 15 + 67 = 82

Question 13.
Your friend uses 19 fewer nails than you to build a birdhouse. Your friend uses 13 nails. How many nails do you use?
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 119
Answer :
Number of nails used by my friend = 13
Number of nails used by me = x
Your friend uses 19 fewer nails than me
x – 19 = 13
x = 13 + 19
x = 32
Therefore , Number of nails used by me = 32

Question 14.
The book club has 27 fewer students than the art club. The book club has 24 students. How many students are in the art club?
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 120
Answer :
Number of students in book club = Numbers of students in art club – 27
Number of students in book club = 24
Numbers of students in art club = 24 + 27 (From given condition ) = 51

Practice Addition Strategies Homework & Practice 3.6

Question 1.
20 + 40 = __
Answer :
20 + 40 = 60
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Practice-Addition-Strategies-Homework-Practice-3.6-Question-1

Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 60

Question 2.
60 + 26 = ___
Answer :
60 + 26 = 86
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Practice-Addition-Strategies-Homework-Practice-3.6-Question-2
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 86

Question 3.

19 + 61 = ___
Answer :
19 + 61 = 80
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Practice-Addition-Strategies-Homework-Practice-3.6-Question-3
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 80

Question 4.

43 + 28 = ___
Answer :
43 + 28 = 71
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Practice-Addition-Strategies-Homework-Practice-3.6-Question-4
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 71

Question 5.
__ = 45 + 46
Answer :
45 + 45 = 91
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Practice-Addition-Strategies-Homework-Practice-3.6-Question-5
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 91

Question 6.
__ = 33 + 59
Answer :
33 + 59 = 92
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Practice-Addition-Strategies-Homework-Practice-3.6-Question-6
Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 92

Question 7.
YOU BE THE TEACHER
Your friend uses compensation to solve 28 + 33. Is your friend correct? Explain.
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 121
Answer :
No , he is wrong .
Explanation :
After adding 2 to 28 then we should subtract 2 from 33 .then we get 28 + 2 = 30 and 33 – 2 = 31 . now add the changed addends we get 30 + 31 = 61 .
So, the sum is 61 .

Question 8.
Modeling Real Life
You have 15 blue crayons and 26 red crayons. How many crayons do you have in all?
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 122
Answer :
Number of blue crayons = 15
Number of red crayons = 26
Total number of crayons = 15 + 26 = 41

Question 9.
Modeling Real Life
The home team scores 46 points. The visiting team scores 13 more than the home team. How many points does the visiting team score?
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 123
Answer :
Score of home team = 46
Score of visiting team = 13 more than the home team = 13 + 46 = 59
Therefore, visiting team score = 59 points .

Review & Refresh

Question 10.
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 124
Answer :
Number of Drums = 5
Number of Guitars = 5 + 4 = 9
5< 9
It means the number of Guitars are greater than the number of drums.

Lesson 3.7 Problem Solving: Addition

Model the story.

Newton has 15 books. Descartes has 22 more. How many books does Descartes have?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 125
Answer :
Number of books with Newton =15
Number of books with Descartes = 22 more = 22 + 15  = 37

How many books do Newton and Descartes have in all?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 125.1
Answer :
Total Number of books with Newton and Descartes = 15 + 37 = 52

Show and Grow

Question 1.
You have 7 cherries. You pick 6 more. Your friend has 45 cherries. How many cherries do you and your friend have in all?
Answer :
Number of cherries with me = 7
Number pf cherries picked by me = 6
Total Number of cherries with me = 7 + 6 = 13
Number of cherries with my friend = 45
Total Number of cherries with me and my friend = 13 + 45 = 58

Apply and Grow: Practice

Question 2.
You have 15 blueberries. You pick 12 more. Your friend picks 32 blueberries. How many blue berries do you and your friend pick in all?
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 127
Answer :
Number of blueberries with me = 15
Number of blueberries picked by me = 12 more
Total Number of blueberries with me = 15 + 12 = 27
Number of blueberries with my friend = 32
Total Number of blueberries with me and my friend = 27 + 32 = 59

Question 3.
You have 26 green crayons. You find 10 more. You have 16 red crayons. How many crayons do you have in all?
Big Ideas Math Solutions Grade 2 Chapter 3 Addition to 100 Strategies 128
Answer :
Number of Green crayons with me = 26
Number of green crayons found = 10
Total Number of green crayons = 26 + 10 = 36
Number of red crayons = 16
Total Number of crayons = 36 + 16 = 52

Question 4.
You take 24 pictures on Friday. You take 20 more on Saturday. You and your friend take 50 pictures in all. How many pictures does your friend take?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 129
Answer :
Number of pictures taken on Friday by me= 24
Number of pictures taken on Saturday by me = 20 + 24 = 44
Number of pictures taken by me and my friend = 50
Number of pictures taken by friend = 50 – 44 = 6

Think and Grow: Modeling Real Life

There are 31 children at a craft fair. There are 20 more adults than children at the craft fair. How many people are at the craft fair in all?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 130
Circle what you know.
Underline what you need to find.
Solve:
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 131
Answer :
Number of children at craft fair= 31
Number of adults at craft fair = 20 more than children’s = 20 + 31 = 51
Total Number of people at craft fair = 31 + 51 = 82.

Show and Grow

Question 5.
Your friend has 40 pens. You have 16 more than your friend. How many pens do you and your friend have in all?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 132
Answer :
Number of pens with my friend = 40
Number of pens with me = 16 more than my friend. = 16 + 40 = 56 .
Total Number of pens with me and my friend = 40 + 56 = 96

Question 6.
You plant 28 seeds. Your friend plants 13 more than you. How many seeds do you and your friend plant in all?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 133
Answer :
Number of seeds planted by me = 28
Number of seeds planted by my friend = 13 more than me = 13 + 28 = 41
Total Number of seeds planted by me and my friend = 28 + 41 = 69

Problem Solving: Addition Homework & Practice 3.7

Question 1.
You see 14 zebras and 4 elephants at the zoo. Then you see 20 snakes. How many animals do you see in all?
Answer :
Number of zebras = 14
Number of elephants = 4
Number of snakes = 20
Total Number of animals = Number of zebras + Number of elephants + Number of snakes = 14 + 4 + 20 = 38

Question 2.
You win 14 tokens at one game and 15 at another. You and your friend have 40 tokens in all. How many tokens does your friend have?
Big Ideas Math Answers 2nd Grade Chapter 3 Addition to 100 Strategies 134
Answer :
Number of tokens won in one game = 14
Number of tokens won in another game = 15
Total Number of tokens with me = 14 + 15 = 29
Number of tokens with me and my friend = 40
Number of tokens with my friend = total tokens – number of tokens with me = 40 – 29 = 11

Question 3.
Your class sells 40 raffle tickets. Your friend’s class sells 15 more than your class. How many raffle tickets do your class and your friend’s class sell in all?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 135
Answer :
Number of raffle tickets sold by my class = 40
Number of tickets sold my friends class = 15 more than my class = 15 + 40 = 55
Total Number of tickets sold by my class and friends class = 40 + 55 = 95

Question 4.
Modeling Real Life
Your friend picks 33 apples. You pick 11 more than your friend. How many apples do you and your friend pick in all? Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 136
Answer :
Number of apples picked my friend = 33
Number of apples picked by me = 11 more than my friend = 11 + 33 = 44
Total Number of apples with me and my friend = 33 + 44 = 77 apples.

Review & Refresh

Get to 10 to subtract

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 137
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Problem-Solving-Addition-Homework-Practice-3.7-Question-5

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 138
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Problem-Solving-Addition-Homework-Practice-3.7-Question-6

Addition to 100 Strategies Performance Task 3

Question 1.
You go to an amusement park. The wait times and ride times for the rides are shown below.
a. Find the total number of minutes it takes to wait and ride each ride.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 139
Answer :
Total Time = Wait time + Ride Time
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Performance-Task 3-Question-1
b. How many more minutes do you spend at the log ride than at the bumper cars?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 140
Answer :
Number of minutes at log ride = 68 minutes.
Number of minutes at Bumper cars = 44 minutes .
Number of more minutes spend at the log ride than at the bumper cars = 68 – 44 = 24 minutes.

c. You want to ride 2 different rides in less than 90 minutes. Which rides can you ride?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 141
Answer :
The two rides which i can ride less than 90 minutes are Ferris wheel and Bumper cars.
Explanation :
Number of minutes at Ferris wheel = 42
Number of minutes at Bumper cars = 44
Total Time = 42 + 44 = 86
86 > 90 minutes.

Question 2.
At 12:30, you spend 25 minutes eating lunch and 30 minutes playing games. How many more minutes are there until the show starts?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 142
Answer :
Time spent in eating lunch = 25 minutes
Time spent in playing games = 30 minutes
Total Time = 25 + 30 = 55 minutes
From 12 : 30 total time spent is 55 minutes so the time is 1:25
The show starts at 1 : 30.
The time left to start the show is 5 minutes.

Addition to 100 Strategies Activity

Three in a Row: Addition

To Play: Players take turns. On your turn, spin both spinners. Add the two numbers and cover the sum. Continue playing until a player gets three in a row.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 143

Addition to 100 Strategies Chapter Practice 3

3.1 Add Tens Using a Number Line

Question 1.
30 + 40 = __
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 144
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.1-Add-Tens-Using-Number-Line-Question-1
Explanation :
The open number line start at 30 and 40 is same as 4 tens So, count on 4 tens .Then the linking jumps move to right from 30 to 70. It ends at 70 showing the sum .
So , 30 + 40 = 70

Question 2.
26 + 50 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 145
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.1-Add-Tens-Using-Number-Line-Question-2
Explanation :
The open number line start at 26 and 50 is same as 5 tens So, count on 5 tens .Then the linking jumps move to right from 26 to 70. It ends at 76 showing the sum .
So , 26 + 50 = 76

3.2 Add Tens and Ones Using a Number Line

Question 3.
46 + 33 = ___
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 146
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.2-Add-Tens-Ones-Using-Number-Line-Question-3
Explanation :
The open number line start at 46 and 33 is same as 3 tens and 3 ones So, count on 3 tens and 3 ones.Then the linking jumps move to right from 46 to 79. It ends at 79 showing the sum .
So , 46 + 33 = 79 .

Question 4.
Modeling Real Life
You find 57 bugs. Your friend finds 12 more than you. How many bugs does your friend find?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 147
Answer :
Number of bugs found by me = 57
Number of bugs found by my friend = 12 more than you = 57 + 12 = 69
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.2-Add-Tens-Ones-Using-Number-Line-Question-4
Explanation :
The open number line start at 57 and 12 is same as 1 tens and 2 ones So, count on 1 tens and 2 ones.Then the linking jumps move to right from 57 to 69. It ends at 69 showing the sum .
So , 57 + 12 = 69

3.3 Use Place Value o Add

Use place value to break apart the addends. Then find the sum.

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 148
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.3-Use-Place-Value-Add-Question-5

Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is  87

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 149
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.3-Use-Place-Value-Add-Question-6

Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 86

Question 7.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 150
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.3-Use-Place-Value-Add-Question-7

Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 96

Question 8.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 151
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.3-Use-Place-Value-Add-Question-8

Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 81

3.4 Decompose to Add Tens and Ones

Use place value to break apart an addend. Then find the sum.

Question 9.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 152
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.4-Decompose-Add-Tens-Ones-Question-9

Explanation :
The addend which we break apart to add is 42 . Break 42 into tens and ones that is 40 + 2. Add the tens to the first addend that is 37 + 40 = 77. Then add the ones that is 77 + 2 = 79.
So, the sum is 79 .

Question 10.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 153
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.4-Decompose-Add-Tens-Ones-Question-10

Explanation :
The addend which we break apart to add is 26 . Break 26 into tens and ones that is 20 + 6. Add the tens to the first addend that is 51 + 20 = 71. Then add the ones that is 71 + 6 = 77.
So, the sum is 77 .

Question 11.
Precision
Which expressions have a sum of 95?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 154
Answer :
Expressions : 43 + 52 , 31 + 64  and 73 + 22 have sum of 95
Explanation :
43 + 52 = 95
25 + 30 = 55
31 + 64 = 95
73 + 22 = 95
71 + 12 = 83

3.5 Use Compensation to Add

Use compensation to add.

Question 12.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 155
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.5-Use-Compensation-Add-Question-12
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 5 is taken from addend 66 and added to 25 addend to make 25 a decade number that is 25 + 5 = 30
Then add the changed addends we get, 61 + 30 = 91 .

Question 13.
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 156
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.5-Use-Compensation-Add-Question-13
Explanation :
Takes ones from one addend to make other addend number a decade number . Here 1 is taken from addend 53 and added to 39 addend to make 39 a decade number that is 39+ 1 = 40
Then add the changed addends we get, 52 + 40 = 92 .

Question 14.
Modeling Real Life
You and your friend want to find 90 seashells. You find 45. Your friend finds 51. Do you and your friend reach your goal?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 157
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 158
Answer :
Total Numbers of seashells to be found = 90
Number of shells found by me = 45
Number of shells found by my friend = 51
Total Number of shells found by me and my friend = 45 + 51 = 96
90 < 96 .
Yes, we reached the goal .

3.6 Practice Addition Strategies

Question 15.
47 + 36 = __
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.6-Practice-Addition-Strategies-Question-15

Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 83

Question 16.
50 + 19 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.6-Practice-Addition-Strategies-Question-16

Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 69

Question 17.
32 + 55 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.6-Practice-Addition-Strategies-Question-17

Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 87

Question 18.

54 + 26 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-3-Addition-to-100-Strategies-Addition-to-100-Strategies-Chapter-Practice-3-3.6-Practice-Addition-Strategies-Question-18

Explanation :
We use place value and break the addends into tens and ones. Then add the tens values and ones values .Write the number formed . So, the number is 80

3.7 Problem Solving: Addition

Question 19.
Newton has 23 rocks. He finds 7 more. Descartes has 51 rocks. How many rocks do Newton and Descartes have in all?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 159
Answers :
Number of rocks with Newton = 23
Number of rocks found by newton = 7
Total number of rocks with newton = 23 + 7 = 30
Number of rocks with Descartes = 51
Total Number of rocks with Newton and Descartes = 30 + 51 = 81

Question 20.
Your friend has 22 bracelets. You have 16 more. How many bracelets do you and your friend have in all?
Big Ideas Math Answer Key Grade 2 Chapter 3 Addition to 100 Strategies 160
Answer :
Number of bracelets with my friend = 22
Number of bracelets with me = 16
Total Number of bracelets with me and my friend = 22 + 16 = 38 Bracelets
Therefore, Total Number of bracelets with me and my friend = 38 Bracelets.

Conclusion:

I hope that the detailed solutions provided at Big Ideas Math Book Grade 2 Chapter 3 Addition to 100 Strategies Answer Key are useful for the students. By referring to this solution key, you can solve exercise problems on your own. In this way, you can assess your strengths and weaknesses and concentrate on the areas you are lagging in. Students can stay in touch with us to find the solutions for all Big Ideas Math Answer Key for Grade 2 Chapters.

Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers

Big Ideas Math Answers Grade 3 Chapter 8

Are you searching for the solutions of Big Ideas Math Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers questions? If yes, then you are at the correct place. We are giving the solution and explanation for each and every question mentioned at BIM 3rd Grade 8th Chapter Add and Subtract Multi-Digit Numbers Book. Students who want to complete their homework in time can download the Big Ideas Math  Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers Solutions PDF.

Big Ideas Math Book 3rd Grade Answer Key Chapter 8 Add and Subtract Multi-Digit Numbers

BIM 3rd Grade 8th Chapter Add and Subtract Multi-Digit Numbers Answers are not only useful for the children but also for teachers and parents. If parents or teachers find any question difficult to solve, then they can refer to Big Ideas Math Book Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers Solution Key.

The topics included in the answer key are Identify Addition Properties, Use Number Lines to Addition, Use Mental Math to Add, Use Partial Sums to Add, Add Three-Digit Numbers, Add Three or More Numbers, Use Number Lines to Subtract, Use Mental Math to Subtract, Subtract Three-Digit Numbers, Relate Addition and Subtraction. The problem solving or performance task sections mentioned after these lessons are helpful to check skills.

Lesson 1 – Identify Addition Properties

Lesson 2 – Use Number Lines to Addition

Lesson 3 –  Use Mental Math to Add

Lesson 4 – Use Partial Sums to Add

Lesson 5 – Add Three-Digit Numbers

Lesson 6 – Add Three or More Numbers

Lesson 7 – Use Number Lines to Subtract

Lesson 8 – Use Mental Math to Subtract

Lesson 9 – Subtract Three-Digit Numbers

Lesson 10 – Relate Addition and Subtraction

Lesson 11 – Problem Solving: Addition and Subtraction

Performance Task

Lesson 8.1 Identify Addition Properties

Explore and Grow

Use the addition table to write all of the addition equations that have a sum of 13.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 1
What do you notice now?
Answer:
By adding a vertical number and the horizontal number we can find the sum of 13.
0 + 0 = 0
0 + 1 = 0
Add blue line and yellow line.
Add 10 in the blue column and 3 in the yellow column.
10 + 3 = 13
Thus the addition equation has the sum of 13.

Structure
Use the addition table to write all of the equations that have a sum of 12. What do you notice?

Answer:
Add blue line and yellow line.
Add 10 in the blue column and 2 in the yellow column.
10 + 2 = 12
Thus the addition equation has the sum of 12.

Think and Grow: Addition Properties
Changing the order of addends does not change the sum.
3 + 5 = 5 + 3
Associative Property of Addition
Changing the grouping of addends does not change the sum.
(7 + 6) + 4 = 7 + (6 + 4)
Addition Property of Zero
The sum of any number and 0 is that number.
9 + 0 = 9
Example
Identify the property.
56 + 0 = ___
12 + 29 = 29 + 12 _____
(24 + 17) + 23 = 24 + (17 + 23) ____

Answer:
56 + 0 = 56
It shows Addition Property of Zero. The Addition Property of Zero defines the sum of any number and 0 is that number.
12 + 29 = 29 + 12
Changing the order of addends does not change the sum.
(24 + 17) + 23 = 24 + (17 + 23)
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Show and Grow

Identify the property.

Question 1.
16 + (14 + 19) = (16 + 14) + 19

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 2.
11 + 54 = 54 + 11

Answer:
Associative Property of Addition defines the changing the order of addends does not change the sum.

Question 3.
0 + 43 = 43

Answer: Addition Property of Zero
The Addition Property of Zero defines the sum of any number and 0 is that number.

Question 4.
(27 + 18) + 22 = 27 + (18 + 22)

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Apply and Grow: Practice

Identify the property.

Question 5.
(28 + 16) + 14 = 28 + (16 + 14)

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 6.
12 + 35 = 35 + 12

Answer:
Associative Property of Addition defines the changing the order of addends does not change the sum.

Question 7.
36 + 0 = 36

Answer:
The Addition Property of Zero defines the sum of any number and 0 is that number.

Question 8.
11 + (9 + 57) = (11 + 9) + 57

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Find the missing number.

Question 9.
23 + 45 = 45 + ___

Answer: 23

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
23 + 45 = 45 + x
23 + 45 = 45 + 23
Thus the missing number is 23.

Question 10.
(13 + 12) + __ = 13 + (12 + 45)

Answer: 45

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
(13 + 12) + x = 13 + (12 + 45)
Thus the missing number is 45.

Question 11.
4 + (76 + 10) = (___ + 76) + 10

Answer: 4

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
4 + (76 + 10) = (x + 76) + 10
4 + (76 + 10) = (4 + 76) + 10
Thus the missing number is 4.

Question 12.
98 + ___ = 98

Answer: 0

Explanation:
Let the missing number be x.
The Addition Property of Zero defines the sum of any number and 0 is that number.
98 + x = 98
x = 98 – 98
x = 0
Thus the missing number is 0.

Question 13.
(___ + 0) + 32 = 6 + 32

Answer: 6

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
(x + 0) + 32 = 6 + 32
x + 32 = 38
x = 38 – 32
x = 6
Thus the missing number is 6.

Question 14.
64 + (5 + 23) = (23 + ___) + 64

Answer: 5

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
64 + (5 + 23) = (23 + x) + 64
64 + 28 = 23 + x + 64
82 = x + 87
x = 87 – 82
x = 5
Thus the missing number is 5.

Question 15.
DIG DEEPER!
Use the numbers 24, 54, and 11 to write an equation that shows the Associative Property of Addition.

Answer:
We can write the equation by using the Associative Property of Addition.
24 + (54 + 11) = (24 + 54) + 11

Writing
Use a property to find the sum. Which property did you use? Why?

Question 16.
54 + 0 = __

Answer: 54

Explanation:
We can find the sum of the 54 + 0 by using the addition property of zero.
The Addition Property of Zero defines the sum of any number and 0 is that number.
54 + 0 = 54

Question 17.
(46 + 17) + 33 = ___

Answer: 96

Explanation:
We can find the sum of the (46 + 17) + 33 by using the Associative Property of Addition.
(46 + 17) + 33 = 96

Question 8.
20 + 63 = ___

Answer: 83

Explanation:
We can find the sum of the Associative Property of Addition.
20 + 63 = 83

Think and Grow: Modeling Real Life

Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 2
A tourist visits 13 museums, 19 memorials, and 11 monuments. Explain how to use a property to find the total number of sites the tourist visits.
(13 + 19) + 11 = ?
Explain:
The tourist visits ___ sites.

Answer:
Given,
A tourist visits 13 museums, 19 memorials, and 11 monuments.
We can find the total number of sites the tourist visits.
(13 + 19) + 11 = 13 + (19 + 11) = 43
Thus the tourist visits 43 sites.

Show and Grow

Question 19.
A farmer sells 34 cucumbers, 48 ears of corn, and 26 bell peppers at a farmer’s market. Explain how to use properties to find the total number of vegetables the farmer sells.
(34 + 48) + 26 = ?

Answer:
Given,
A farmer sells 34 cucumbers, 48 ears of corn, and 26 bell peppers at a farmer’s market.
We can find the sum of the Associative Property of Addition.
(34 + 48) + 26 = 34 + (48 + 26)
34 + 48 + 26 = 108
Thus the total number of vegetables the farmer sells is 108.

Question 20.
How many people go on the field trip?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 3

Answer:
The number of adults = 20
The number of grade 2 students = 47
The number of grade 3 students = 53
20 + 47 + 53 = 120
Thus 120 people go on the field trip.

DIG DEEPER!
The Grade 2 and Grade 3 students are divided into 10 equal groups. How many students are in each group? Explain.

Answer:
Given that,
The Grade 2 and Grade 3 students are divided into 10 equal groups.
The number of grade 2 students = 47
The number of grade 3 students = 53
47 + 53 = 100
100/10 = 10
Thus there are 10 students in each group.

Identify Addition Properties Homework & Practice 8.1

Identify the property.

Question 1.
(79 + 12) + 13 = 79 + (12 + 13)

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 2.
24 + 63 = 63 + 24

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 3.
0 + 64 = 64

Answer: Associative property of zero
The Addition Property of Zero defines the sum of any number and 0 is that number.

Question 4.
37 + (43 + 19) = (37 + 43) + 19

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 5.
17 + 38 = 38 + 17

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 6.
18 + 48 = 48 + 18

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Find the missing number.

Question 7.
36 + __ = 36

Answer: 0

Explanation:
Let the missing number be x.
36 + x = 36
x = 36 – 36
x = 0
Thus the missing number is 0.

Question 8.
25 + __ + 11 = 25 + 11

Answer: 0

Explanation:
Let the missing number be y.
25 + y + 11 = 25 + 11
y + 36 = 36
y = 36 – 36
y = 0
Thus the missing number is 0.

Question 9.
0 + 43 = __ + 0

Answer: 43

Explanation:
Let the missing number be t.
0 + 43 = t + 0
43 = t
Thus the missing number is 43.

Question 10.
(22 + 19) + 28 = 19 + (___ + 28)

Answer: 22

Explanation:
Let the missing number be p.
22 + 19 + 28 = 19 + p + 28
69 = 47 + p
p = 69 – 47
p = 22
Thus the missing number is 22.

Question 11.
Number Sense
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 4
Newton uses two properties. Identify the properties he uses.

Answer:
Given the expression (18 + 27) + 12 = 27 + (18 + 12)
By seeing the above expression we can say that Newton used Associative Property of Addition and Commutative Property of Addition.

Question 12.
Open-Ended
Write an equation that shows the Commutative Property of Addition.

Answer: 4 + 2 = 2 + 4
Commutative property of addition: Changing the order of addends does not change the sum.

Question 13.
Structure
Explain how the Associative Property of Addition and the Associative Property of Multiplication are alike and how they are different.

Answer:
Associative property explains that addition and multiplication of numbers are possible regardless of how they are grouped.
Example:
2 × (3 × 5) = (2 × 3) × 5
2 + (3 + 5) = (2 + 3) + 5
The method is same but the solution for both the equations are different.

Question 14.
Modeling Real Life
A florist uses 11 roses, 12 lilies, and 19 daisies to make bouquets. How many flowers does he use?

Answer:
Given that,
A florist uses 11 roses, 12 lilies, and 19 daisies to make bouquets.
Add all the flowers to find the total number of flowers he used.
11 + 12 + 19 = 42
Thus he used 42 flowers.

DIG DEEPER!
The florist uses 6 flowers for each bouquet. How many bouquets does he make? Explain.

Answer:
Given,
The florist uses 6 flowers for each bouquet.
Total flowers = 42
Divide the total number of flowers by the number of flowers in each bouquet
42/6 = 7
Thus he made 7 bouquets.

Review & Refresh

Find the product.

Question 15.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 400

Answer:
Multiply the two numbers 0 and 3.
0 × 3 = 0

Question 16.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 5

Answer:
Multiply the two numbers 7 and 7.
7 × 7 = 49

Question 17.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 6

Answer:
Multiply the two numbers 10 and 4.
10 × 4 = 40

Question 18
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 7

Answer:
Multiply the two numbers 7 and 6.
7 × 6 = 42

Lesson 8.2 Use Number Lines to Addition

Explore and Grow

Color to find 33 + 25. Then model your jumps on the number line.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 8

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-8
First, you have to color the two numbers 33 and 25.
Now add 33 and 25
33 + 25 = 58
Now try to solve the problem by using the number line.
Start at 33. Count by twenties, then by 5s.
33 + 20 + 5 = 58

Reasoning

How can finding 33 + 25 help you find 533 + 25?

Answer:
We can find the sum of 533 and 25 with the help of the above problem.
33 + 25 = 58
533+ 25 = 538
Just add 5 on the left side to get the sum.

Think and Grow: Adding on a Number Line

Example
Find 245 + 38
One Way: Use the count on strategy. Start at 245. Count on by tens, then by ones.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 9
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-9

Another Way: Use the make a ten strategy. Start at 245. Count on to the nearest ten. Then count on by tens and by ones.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 10
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-10

Show and Grow

Question 1.
Use the count on strategy to find 368 + 24.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 11

Answer: 392
Use the count on strategy. Start at 368. Count on by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-11
368 + 10 + 10 + 2 + 2 = 392

Question 2.
Use the count on strategy to find 57 + 179.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 12

Answer: 236
Use the count on strategy. Start at 57. Count on by hundreds, tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-12

Apply and Grow: Practice

Question 3.
Use the count on strategy to find 47 + 216.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 13

Answer: 263
Use the count on strategy. Start at 47. Count on by hundreds, tens, then by ones.
47 + 100 + 100 + 10 + 6 = 263
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-13

Question 4.
Use the make a ten strategy to find 478 + 64.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 14

Answer: 542
Use the count on strategy. Start at 478. Count on by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-14

Find the sum

Question 5.
395 + 85 = ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 15

Answer: 480
Use the count on strategy. Start at 395. Count on by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-15

Question 6.
653 + 109 = ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 16

Answer: 762
Use the count on strategy. Start at 353. Count on by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-15

Question 7.
Humans have 24 rib bones and 182 other types of bones. How many bones do humans have in all?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 17

Answer:
Given,
Humans have 24 rib bones and 182 other types of bones.
182 + 24 = 206
Thus Humans have 206 bones.

Question 8.
Structure
Write the equation shown by the number line.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 18

Answer:
436 + 4 + 20 + 4
440 + 20 + 4 = 464

Question 9.
Structure
Show two different ways to find 225 + 39 using a number line.

Answer:
Use the count on strategy. Start at 225. Count on by tens, then by ones.
225 + 30 + 9 = 264
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-15

Think and Grow: Modeling Real Life

Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 19
The Leaning Tower of Pisa has 294 steps. A visitor climbs 156 steps, takes a break, and then climbs 78 more steps. Does the visitor reach the top of the tower?
Addition equation:
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 20
The visitor ___ reach the top of the tower.

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-16
156 + 78 = 234
294 – 234 = 60
Thus the visitor does not reach the top of the tower.

Show and Grow

Question 10.
A book has 216 pages. You have already read 167 pages. You read 49 more pages. Do you finish reading the book?

Answer:
Given that,
A book has 216 pages. You have already read 167 pages. You read 49 more pages.
167 + 49 = 216 pages
216 – 216 = 0
Thus you finish reading the book.

Question 11.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 21
DIG DEEPER!
A puzzle has 350 pieces. You put 95 pieces together. Your friend puts 185 pieces together. Do you and your friend complete the puzzle? If not, how many pieces are left?

Answer:
Given that,
A puzzle has 350 pieces. You put 95 pieces together. Your friend puts 185 pieces together.
95 + 185 = 280 pages
350 – 280 = 70 pages
No you and your friend do not complete the puzzle.
70 pieces are left to complete the puzzle.

Question 12.
A music library has 483 songs. You listen to162 different songs one week and 171 more songs the next week. How many songs are left?

Answer:
Given that,
A music library has 483 songs.
You listen to162 different songs one week and 171 more songs the next week.
162 + 171 = 333 songs
483 – 333 = 150 songs
Thus 150 songs are left.

Use Number Lines to Addition Homework & Practice 8.2

Question 1.
Use the count on strategy to find 402 + 39.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 22

Answer:
Use the count on strategy. Start at 402. Count on by tens, then by ones.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-22

Question 2.
Use the make a ten strategy to find 81 + 647.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 23

Answer:
Use the count on strategy. Start at 81. Count on by hundreds, tens, then by ones.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-23

Question 3.
Find the sum.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 24

Answer:

Question 4.
718 + 226 = __
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 25

Answer:
Use the count on strategy. Start at 718. Count on by hundreds, tens, then by ones.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-25
718 + 226 = 944

Question 5.
YOU BE THE TEACHER
Your friend uses a number line to find 435 + 27. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 26

Answer:
Your friend uses a number line to find 435 + 27.
435 + 27 = 462
Your friend is not correct.

Question 6.
YOU BE THE TEACHER
Your friend says she can find 64 + 691 by starting at 691 on a number line because of the Associative Property of Addition. Is your friend correct? Explain.

Answer:
Your friend says she can find 64 + 691 by starting at 691 on a number line because of the Associative Property of Addition.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-25
Yes, your friend is correct.

Question 7.
Modeling Real Life
Your cousin needs to write a 400-word essay. She writes 318 words during class. She finishes her essay by writing 94 words at home. Does your cousin’s essay have enough words?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 401

Answer:
Given,
Your cousin needs to write a 400-word essay.
She writes 318 words during class. She finishes her essay by writing 94 words at home.
318 + 94 = 412
412 – 400 = 12
Yes your cousin’s essay have enough words.

Question 8.
DIG DEEPER!
There are 275 apartments in an apartment building. There are 203 two-bedroom apartments rented, and 56 one-bedroom not apartments rented. How many apartments are rented?

Answer:
Given that,
There are 275 apartments in an apartment building.
There are 203 two-bedroom apartments rented, and 56 one-bedroom not apartments rented.
203 + 56 = 259
Thus 259 apartments are rented.

Find the quotient

Question 9.
18 ÷ 6 = ___

Answer: 3

Explanation:
Divide the two numbers 18 and 6.
18/6 = 3
Thus the quotient is 3.

Question 10.
32 ÷ 8 = ___

Answer: 4

Explanation:
Divide the two numbers 32 and 8.
32/4 = 4
Thus the quotient is 4.

Question 11.
72 ÷ 9 = __

Answer: 8

Explanation:
Divide the two numbers 72 and 9.
72/9 = 8
Thus the quotient is 8.

Lesson 8.3 Use Mental Math to Add

Explore and Grow

What addition problem is shown? How can you find the sum?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 27

Answer: 366 + 504 = 870

Explanation:
By seeing the above figure we can find the sum.
First count the number of blocks
There are 100 blocks in each figure
There are three 100 blocks, six 10 blocks, and 6 blocks.
300 + 60 + 6 = 366
There are five 100 blocks and 4 blocks.
500 + 4 = 504
Now add both
366 + 504 = 870

Reasoning
Show how to find 402 + 248.

Answer:
You can find the sum of 402 and 248 by using the above arrays.
Take 10 × 10 block
402 – Four 10 × 10 blocks, 2 blocks
248 – Two 10 × 10 blocks, Four 10 blocks, and eight blocks
402 + 248 = 650

Think and Grow: Mental Math Strategies for Addition

Example
Find 247 + 328.
Use compensation to add.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 28

Answer:
You can find the sum of 247 and 328 by using mental math strategies.
247 + 3 = 250
328 – 3 = 325
250
+325
575
Thus the sum of 247 and 328 is 575.

Example
Find 119 + 163.
Make a ten and count on to add.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 29

Answer:
119 + 163 = 119 + (1 + 100 + 60 + 2)
(119 + 1) + 100 + 60 + 2
120 + 100 + 60 + 2
382

Show and Grow

Question 1.
Use compensation to find 322 + 158.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 30

Answer:
322 – 2 = 320
158 + 2 = 160
320
+160
480
So, 322 + 158 = 480

Question 2.
Make a ten and count on to find 695 + 187.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 31

Answer:
695 + 187 = 695 + (5 + 100 + 80 + 2)
(695 + 5) + (100 + 80 + 2)
700 + 100 + 80 + 2
882
So, 695 + 187 = 882

Apply and Grow: Practice

Question 3.
Use compensation to find the sum.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 32

Answer:
604 – 4 = 600
275 + 4  = 279
600
+279
879

Question 4.
Make a ten and count on to find the sum.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 33

Answer:
359 + 1 = 360
360 + 300 = 660
660 + 17 = 677
So, 359 + 318 = 677

Use mental math to find the sum.

Question 5.
436 + 248 = __

Answer:
436 + 248 = 436 + (8 + 40 + 200)
436 + (4 + 4 + 40 + 200)
(436 + 4) + 4 + 40 + 200
440 + 244
684
So, 436 + 248 = 684

Question 6.
795 + 102 = ___

Answer:
102 – 2 = 100
795 + 2 = 797
797
+100
897
So, 795 + 102 = 897

Question 7.
503 + 71 = ___

Answer:
503 – 3 = 500
71 + 3 = 74
500
+74
574
So, 503 + 71 = 574

Question 8.
589 + 407 = ___

Answer:
589 + 1 = 590
407 – 1 = 406
590
+406
996

Question 9.
734 + 97 = ___

Answer:
734 – 3 = 731
97 + 3 = 100
731
+100
831

Question 10.
352 + 164 = ___

Answer:
352 – 2 = 350
164 + 2 = 166
350
+166
516

Question 11.
297 + 211 = ___

Answer:
297 + 3 = 300
211 – 3 = 208
300
+208
508

Question 12.
426 + 364 = ___

Answer:
426 + 4 = 430
364 – 4 = 360
430
+360
790

Question 13.
159 + 104 = ___

Answer:
159 + 1 = 160
104 – 1 = 103
160
+103
263

Question 14.
A community shelter collects 101 shirts and 109 pairs of pants from a clothing drive. How many clothing items does the community shelter collect?

Answer:
Given,
A community shelter collects 101 shirts and 109 pairs of pants from a clothing drive.
101 – 1 = 100
109 + 1 = 110
100
+110
210
Thus the community shelter collect 210 clothing items.

Question 15.
Number Sense
Descartes wants to use compensation to find 238 + 127. Which numbers could he use?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 34

Answer:
Descartes wants to use compensation to find 238 + 127.
The number near 238 is 240, 127 is 130.
Thus he could use 240 and 130

Think and Grow: Modeling Real Life

A company manager has $900. Does he have enough money to buy the laptop and the cell phone?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 35
Addition equation:
Compare:
The manager ___ have enough money.

Answer:
Given,
A company manager has $900.
The cost of the laptop is $595
The cost of the mobile is $249.
595 + 5 = 600
249 – 5 = 244
600
+244
844
900 – 844 = 56
Thus the manager does not have enough money.

Show and Grow

Question 16.
A USB drive holds 600 photos. You have 279 photos on a digital camera and 337 photos on a cell phone. Can the USB drive hold all of your photos?

Answer:
Given,
A USB drive holds 600 photos. You have 279 photos on a digital camera and 337 photos on a cell phone.
279 + 1 = 280
337 – 1 = 336
280
+336
616
Yes the USB drive can hold all of your photos

Question 17.
A school board president has $1,000. Which two items can she buy?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 36

DIG DEEPER!
The president buys the two items. How much money does she have left?

Answer:
A school board president has $1,000.
The cost of a swing set is $648
The cost of the seesaw is $372
The cost of a Dome Climber is $498
498
+372
870
1000 – 870 = 130
Thus she has left $130.

Use Mental Math to Add Homework & Practice 8.3

Use compensation to find the sum.

Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 37

Answer:
248 + 2 = 250
137 – 2 = 135
250
+135
385
So, 248 + 137 = 385

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 38

Answer:
401 – 1 = 400
165 + 1 = 166
400
+166
566

Make a ten and count on to find the sum

Question 3.
506 + 369 = ?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 39

Answer:
506 + 369 = 506 + (4 + 300 + 60 + 5)
(506 + 4) + 300 + 60 + 5
510 + 365
875
So, 506 + 369 = 875

Question 4.
617 + 348 = ?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 40

Answer:
617 + 348 = 617 + (3 + 300 + 40 + 5)
(617 + 3) + 300 + 40 + 5
620 + 345
965
617 + 348 = 965

Use mental math to find the sum.

Question 5.
478 + 219 = ___

Answer:
478 + 2 = 480
219 – 2 = 217
480
+217
697

Question 6.
503 + 64 = ___

Answer:
503 – 3 = 500
64 + 3 = 67
500
+67
567

Question 7.
288 + 242 = ___

Answer:
288 + 2 = 290
242 – 2 = 240
290
+240
530

Question 8.
Structure
Explain how to make a ten to find the sum.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 41

Answer:
738 + 126 = 738 + (6 + 20 + 100)
(738 + 2) + 4 + 20 + 100
740 + 124
864

Question 9.
Writing
How is compensation make a ten similar to the strategy? How is it different?

Answer:
Compensation is a mental math strategy for multi-digit addition that involves adjusting one of the addends to make the equation easier to solve. The methods are different but the solutions are the same.

Question 10.
Modeling Real Life
A binder holds 250 sheets of paper. You have 107 science papers and 142 math papers. Can the binder hold all of your papers?

Answer:
Given,
A binder holds 250 sheets of paper. You have 107 science papers and 142 math papers.
107 – 2 = 105
142 + 2 = 144
105 + 144 = 249
Yes, the binder can hold all of your papers.

Question 11.
Modeling Real Life
A school nurse has $450. Which two items can she buy?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 42

DIG DEEPER!
The school nurse buys the two items. How much money does she have left?

Answer:
A school nurse has $450.
The cost of a Stethoscope = $119
The cost of thermometer = $ 308
119 + 1 = 120
308 – 1 = 307
120
+307
427
450 – 427 = 23
She has left $23

Review & Refresh

Question 12.
It costs $1 to get into each football game. Newton buys a chicken wrap for $2 and a drink for $1 each game. How much money does Newton spend in 4 games?

Answer:
Given,
It costs $1 to get into each football game. Newton buys a chicken wrap for $2 and a drink for $1 each game.
1 + 2 + 1 = 4
4 × $4 = $8
Thus Newton spent $8 for 4 games.

Lesson 8.4 Use Partial Sums to Add

Explore and Grow

Model each number. Draw to show each model.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 43
How can you use the models to find 341 + 227?

Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-43
You can find the sum of 341 and 227 is

300 + 40 + 1
200 + 20 + 7
500 + 60 + 8 = 568

Reasoning
How can breaking apart addends help you add three-digit numbers?

Answer: Breaking apart addends helps you add three-digit numbers easily. Mind math is possible with this breaking apart addends.

Think and Grow: Use Partials Sums to Add

Example
Find 356 + 408.
Step 1: Write each number in expanded form.
Step 2: Find the partial sums.
Step 3: Add the partial sums.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 44

Answer: 764

Show and Grow

Use partial sums to add.

Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 45

Answer:
319 = 300 + 10 + 9
234 = 200 + 30 + 4
553 = 500 + 40 + 13

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 46

Answer:
568 = 500 + 60 + 8
173 = 100 + 70 + 3
741 = 600 + 130 + 11

Question 3.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 47

Answer:
424 = 400 + 20 + 4
450 = 400 + 50 + 0
874 = 800 + 70 + 4

Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 48

Answer:
281 = 200 + 80 + 1
365 = 300 + 60 + 5
646 = 500 + 140 + 6

Question 5.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 49

Answer:
127 = 100 + 20 + 7
609 = 600 + 0 + 9
736 = 700 + 20 + 16

Question 6.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 50

Answer:
276 = 200 + 70 + 6
39 =  000 + 30 + 9
315 = 200 + 100 + 15

Apply and Grow: Practice

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 51

Answer:
759 = 700 + 50 + 9
202 = 200 + 00 + 2
961 = 900 + 50 + 11

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 52

Answer:
864 = 800 + 60 + 4
131 = 100 + 30 + 1
995 = 900 + 90 + 5

Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 53

Answer:
387 = 300 + 80 + 7
94 = 000 + 90 + 4
481 = 300 + 170 + 11

Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 54

Answer:
560 = 500 + 60 + 0
273 = 200 + 70 + 3
833 = 700 + 130 + 3

Question 11.
498 + 375 = ___

Answer:
498 = 400 + 90 + 8
375 = 300 + 70 + 5
873 = 700 + 160 + 13

Question 12.
209 + 158 = ___

Answer:
209 = 200 + 00 + 9
158 = 100 + 50 + 8
367 = 300 + 50 + 17

Think and Grow: Modeling Real Life

A giant panda weighs 696 pounds less than a polar bear. How much does the polar bear weigh?
Addition equation:
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 55

Answer:
Given,
A giant panda weighs 696 pounds less than a polar bear.
696 – 263 = 433
Thus the polar bear weighs 433 pounds.

Show and Grow

Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 56
Question 15.
A herd of wildebeests has 258 fewer members than a herd of zebras has. There are 335 wildebeests in the herd. How many zebras are in their herd?

Answer:
Given,
A herd of wildebeests has 258 fewer members than a herd of zebras has. There are 335 wildebeests in the herd.
335 – 258 = 77
Therefore there are 77 zebras in their herd.

Question 16.
There are three candidates in an election. Candidate A receives 184 fewer votes than Candidate B. Who wins the election?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 57

Answer:
Given,
There are three candidates in an election. Candidate A receives 184 fewer votes than Candidate B.
Number of votes candidate A receives = 347
347 + 184 = 631
Thus Candidate B wins the election.

Question 17.
DIG DEEPER!
You, your friend, and your cousin play a video game. Your friend scores 161 fewer points than you. Your cousin scores 213 more points than your friend. What is each player’s score? Who wins?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 58

Answer:
Your friend score is 579
Your friend scores 161 fewer points than you.
579 + 161 = 740
Your score is 740
Your cousin scores 213 more points than your friend.
213 + 579 = 592
Your cousin’s score is the highest among all three.
So, your cousin wins the game.

Use Partial Sums to Add Homework & Practice 8.4

Use partials sums to add

Question 1.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 59

Answer:
We can find the sum by using the partial sum model.
479 = 400 + 70 + 9
356 = 300 + 50 + 6
835 = 700 + 120 + 15
So, 479 + 356 = 835

Question 2.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 60

Answer:
We can find the sum by using the partial sum model.
674 = 600 + 70 + 4
321 = 300 + 20 + 1
995 = 900 + 90 + 5

Question 3.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 61

Answer:
We can find the sum by using the partial sum model.
396 = 300 + 90 + 6
278 = 200 + 70 + 8
674 = 500 + 160 + 14

Question 4.
564 + 218 = ___

Answer:
We can find the sum by using the partial sum model.
564 = 500 + 60 + 4
218 = 200 + 10 + 8
782 = 700 + 70 + 12

Question 5.
190 + 123 = ___

Answer:
We can find the sum by using the partial sum model.
190 = 100 + 90 + 0
123 = 100 + 20 + 3 
313 = 200 + 110 + 3

Question 6.
YOU BE THE TEACHER
Your friend uses partial sums to find 205 + 124. Is your friend correct? Explain.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 62

Answer: Your friend is incorrect.
205 = 200 + 00 + 5
124 = 100 + 20 + 4
329 = 300 + 20 + 9
The sum of 205 + 124 = 329

Question 7.
Patterns
Write and solve the next problem in the pattern.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 63

Answer:
The addend is increased by 100.
So, the next pattern is
516
+178
694

Question 8.
Modeling Real Life
There are worker bees and drone bees in a beehive. A hive has 268 fewer drones than workers. There are 351 drone bees. How many worker bees are there?
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 64

Answer:
Given,
There are worker bees and drone bees in a beehive. A hive has 268 fewer drones than workers. There are 351 drone bees.
351
-268
83
Therefore there are 83 worker bees.

Question 9.
Modeling Real Life
Three athletes compete in Olympic weight lifting. Weight lifter A lifts 104 fewer pounds than Weight lifter B. Who lifts the most weight?
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 65

Answer:
Three athletes compete in Olympic weight lifting. Weight lifter A lifts 104 fewer pounds than Weight lifter B.
Weight lifter A lifts 368 pounds
368 + 104 = 472
Therefore, Weight Lifter B lifts the most weight.

Review & Refresh

Circle the value of the underlined digit.

Question 10.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 66

Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-66
In the given value 2 is in the ones place so the answer is 2.

Question 11.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 67

Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-67
In the given value 4 is in the hundreds place so the answer is 400.

Question 12.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 68

Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-68
In the given value 0 is in tens place so the answer is 0.

Lesson 8.5 Add Three-Digit Numbers

Explore and Grow

Model the equation. Draw your model. Then find the sum.
195 + 308 = ___

Answer:
195 = 100 + 90 + 5
308 = 300 + 00 + 8
503 = 400 + 90 + 13

Reasoning
How can you use an estimate to check whether your answer reasonable?

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
195 = 200
308 = 300
200
+300
500
Ths sum is about 500.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
195
+308
503
503 is close to 500. So, the answer is reasonable.

Think and Grow: Add Three-Digit Numbers

Example
Find 236 + 378. Check whether your answer is reasonable.
Step 1: Estimate. Round each addend to the nearest hundred.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 69

Show and Grow

Find the sum. Check whether your answer is reasonable.

Question 11.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 70

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
457 – 500
133 – 100
500
+100
600
The sum is about 600.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
457
+133 
590
590 is close to 600. So, the answer is reasonable.

Question 2.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 71

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
269 – 300
354 – 300
300
+300
300
Step 2: Find the sum. Add the ones, tens, then the hundreds.
269
+354
623
623 is close to 600. So, the answer is reasonable.

Question 3.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 72

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
560 – 600
274 – 300
600
+300
900
Step 2: Find the sum. Add the ones, tens, then the hundreds.
560
+274 
834
834 is close to 900. So, the answer is reasonable.

Question 4.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 73

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
345 – 300
286 – 300
300
+300
600
Step 2: Find the sum. Add the ones, tens, then the hundreds.
345
+286
631
631 is close to 600. So, the answer is reasonable.

Question 5.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 74

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
129 – 100
668 – 700
100
+700
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
129
+668
797
797 is close to 800. So, the answer is reasonable.

Question 6.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 75

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
383 – 400
539 – 500
400
+500
900
Step 2: Find the sum. Add the ones, tens, then the hundreds.
383
+539
922
922 is close to 900. So, the answer is reasonable.

Apply and Grow: practice

Find the sum. Check whether your answer is reasonable.

Question 7.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 76

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
803 – 800
179 – 200
800
+200
1000
Step 2: Find the sum. Add the ones, tens, then the hundreds.
803
+179
982
982 is close to 1000. So, the answer is reasonable.

Question 8.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 77

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
608 – 600
239 – 200
600
+200
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
608
+239
847
847 is close to 800. So, the answer is reasonable.

Question 9.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 78

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
427 – 400
385 – 400
400
+400
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
427
+385
812
812 is close to 800. So, the answer is reasonable.

Question 10.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 79

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
401 – 400
109 – 100
400
+100
500
Step 2: Find the sum. Add the ones, tens, then the hundreds.
401
+109
510
510 is close to 500. So, the answer is reasonable.

Question 11.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 80

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
265 – 300
157 – 100
300
+100
400
Step 2: Find the sum. Add the ones, tens, then the hundreds.
265
+157
422
422 is close to 400. So, the answer is reasonable.

Question 12.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 81

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
375 – 400
64 – 100
400
+100
500
Step 2: Find the sum. Add the ones, tens, then the hundreds.
375
+64
439
439 is close to 500. So, the answer is reasonable.

Question 13.
Estimate: ___
469 + 284 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
469 – 500
284 – 300
500
+300
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
469
+284 
753
753 is close to 800. So, the answer is reasonable.

Question 14.
Estimate: ___
580 + 246 = __

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
580 – 600
246 – 200
600
+200
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
580
+246
826
826 is close to 800. So, the answer is reasonable.

Question 15.
Estimate: ___
796 + 135 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
796 – 800
135 – 100
800
+100
900
Step 2: Find the sum. Add the ones, tens, then the hundreds.
796
+135
931
931 is close to 900. So, the answer is reasonable.

Question 16.
A truck driver travels 428 miles on Monday. He travels 473 miles on Tuesday. How many miles does he travel in all on Monday and Tuesday?

Answer:
Given that,
A truck driver travels 428 miles on Monday. He travels 473 miles on Tuesday.
428
+473
901
Thus he travels 901 miles on Monday and Tuesday.

Question 17.
Reasoning
Your friend finds a sum. Is her answer reasonable? If not, describe her mistake.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 85

Answer:
119
+ 187
306
Your answer is not reasonable.

Think and Grow: Modeling Real Life

A construction team builds an 825-meter-long boardwalk on a beach. The team builds 408 meters one week and 377 meters the next week. Is the boardwalk complete?
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 86
Addition equation:
The boardwalk ___ complete.

Answer:
Given,
A construction team builds an 825-meter-long boardwalk on a beach. The team builds 408 meters one week and 377 meters the next week.
408
+377
785
The boardwalk did not complete.

Show and Grow

Question 18.
A road crew repaves the road on a 547-meter-long bridge. The crew repaves 318 meters the first day and 229 meters the second day. Is the road on the bridge completely repaved?

Answer:
Given,
A road crew repaves the road on a 547-meter-long bridge.
The crew repaves 318 meters on the first day and 229 meters on the second day.
318
+229
547
Yes, the road on the bridge completely repaved.

Question 19.
A family drives from St. Louis to Orlando for a vacation. The family drives 363 miles the first day and 386 miles the second day. How many miles does the family have left to drive?

Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 87
Answer:
Given,
A family drives from St. Louis to Orlando for a vacation. The family drives 363 miles the first day and 386 miles the second day.
363 + 386 = 749 miles
922 miles – 749 miles = 173 miles
173 miles the family have left to drive.

Question 20.
Which booth had more visitors in all?
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 88

Answer:
Day 1 – 468 + 527 = 995
Day 2 – 416 + 374 = 790
995 – 790 = 205
Thus day 1 has more visitors in all.

Add Three-Digit Numbers Homework & Practice 8.5

Find the sum. Check whether your answer is answerable.

Question 1.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 89

Answer: 630

Explanation:
The estimated number of 493 is 490
The estimated number of 142 is 140.
490
+ 140
630

Question 2.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 90

Answer: 820

Explanation:
The estimated number of 763 is 760.
The estimated number of 58 is 60.
760
+ 60
820

Question 3.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 91

Answer: 580

Explanation:
The estimated number of 308 is 310
The estimated number of 273 is 270
310
+270
580

Question 4.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 92

Answer: 420

Explanation:
The estimated number of 276 is 280
The estimated number of 138 is 140
280
+ 140
420

Question 5.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 93

Answer: 700

Explanation:
The estimated number of 532 is 530
The estimated number of 167 is 170.
530
+170
700

Question 6.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 94

Answer: 980

Explanation:
The estimated number of 680
The estimated number of 296 is 300.
680
+300
980

Question 7.
Estimate: ___
595 + 280 = ___

Answer: 880

Explanation:
The estimated number of 595 is 600
The estimated number of 280
600
+ 280
880

Question 8.
Estimate: ___
419 + 295

Answer: 720

Explanation:
The estimated number of 419  is 420
The estimated number of 295 is 300
420
+ 300
720

Question 9.
Estimate: ___
498 + 305 = ___

Answer: 800

Explanation:
The estimated number of 498 is 500
The estimated number of 305 is 300
500
+300
800

Question 10.
Open-Ended
Complete the addends so you need to regroup to add. Then find the sums.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 95

Answer:
Let the addend be 3
479
+283
762
If you take 3 as addend then you need to regroup to find the sum.
Let the addend be 9
697
+135
832
If you take 9 as addend then you need to regroup to find the sum.

Question 11.
DIG DEEPER!
Find the missing digits.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 96

Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-96
You will get 466 if you add 107 and 359.
You will get 982 if you add 748 and 234
You will get 962 if you add 670 and 292.
You will get 982 if you add 809 and 173.

Question 12.
Modeling Real Life
Newton wants to complete a 770-mile hike in 2 months. He hikes 423 miles the first month and 347 miles the second month. Does he complete the hike?

Answer:
Given,
Newton wants to complete a 770-mile hike in 2 months.
He hikes 423 miles the first month and 347 miles the second month
423
+347
770
Thus he Newton completed the hike.

Question 13.
Modeling Real Life
You ship a package 750 miles from San Diego to Salt Lake City. The package is now in Las Vegas. How many miles are left until your package is delivered?
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 97

Answer:
Given,
You ship a package 750 miles from San Diego to Salt Lake City. The package is now in Las Vegas.
121 + 270 = 391 miles until Las Vegas.
750 – 391 = 359 miles
Therefore 359 miles are left until your package is delivered.

Review & Refresh

Find the quotient.

Question 14.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 98

Answer: 10

Explanation:
Divide the two numbers 100 and 10.
100/10 = 10
Thus the quotient is 10.

Question 15.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 99

Answer: 9

Explanation:
Divide the two numbers 45 and 5.
45/5 = 9
Thus the quotient is 9.

Question 16.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 100

Answer: 7

Explanation:
Divide the two numbers 14 and 7.
14/2 = 7
Thus the quotient is 7.

Question 17.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 101

Answer: 8

Explanation:
Divide the two numbers 10 and 80.
80/10 = 8
Thus the quotient is 8.

Question 18.
Divide 25 by 5.

Answer: 5

Explanation:
Divide the two numbers 5 and 25.
25/5 = 5
Thus the quotient is 5.

Question 19.
Divide 30 by 10.

Answer: 3

Explanation:
Divide the two numbers 10 and 30.
30/10 = 3
Thus the quotient is 3.

Question 20.
Divide 8 by 2.

Answer: 4

Explanation:
Divide the two numbers 8 and 2.
8/2 = 4
Thus the quotient is 4.

Lesson 8.6 Add Three or More Numbers

Find the sum of the numbers. Which two numbers should you add first?
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 102

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
348 = 350
478 = 480
152 = 150
350 + 480 + 150 = 980
The sum is about 980.
517 = 520
117 = 120
283 = 280
520 + 120 + 280 = 920
The sum is about 920
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
348
+478
+152
978
978 is close to 980. So, the answer is reasonable.
517
+117
+283
917
917 is close to 920, So, the answer is reasonable.

Reasoning
Why did you choose those numbers? Compare your strategy to your partner’s strategy.

Answer:
348 = 350
478 = 480
152 = 150
I choose these numbers because they are nearest to ten. This strategy will help to improve the mental math. After solving the problem you can verify the answer with the actual sum.

Think and Grow: Add Three or More Numbers

Example
Find 138 + 221 + 176 + 92. Check whether your answer is reasonable.
Step 1: Estimate. Round each addend to the nearest ten.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 103

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
140 + 220 + 180 + 90 = 630
The sum is about 630.
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
138
+221
+176
+92
627
627 is close to 630. So, the answer is reasonable.

Show and Grow

Find the sum. Check whether your answer is reasonable.

Question 1.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 104

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
342 = 340
73 = 70
267 = 270
340 + 70 + 270 = 680
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
342
+73
+267
682
682 is close to 680. So, the answer is reasonable.

Question 2.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 105

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
191 = 190
452 = 450
206 = 210
190 + 450 + 210 = 850
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 105
849
849 is close to 850. So, the answer is reasonable.

Question 3.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 106

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
65 = 60
98 = 100
637 = 640
640
+100
+60
800
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
637
+98
+65
800
800 is close to 800. So, the answer is reasonable.

Question 4.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 107

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
241 = 240
394 = 390
85 = 80
193 = 190
390
240
190
+80
900
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 107
913
913 is close to 900. So, the answer is reasonable.

Question 5.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 108

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
136 = 140
51 = 50
64 = 60
410 = 410
140
410
+60
+50
660
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 108
661
661 is close to 660. So, the answer is reasonable.

Question 6.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 109

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
105 = 100
113 = 110
222 = 220
307 = 310
100 + 110 + 220 + 310 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 109
747
747 is close to 740. So, the answer is reasonable.

Apply and Grow: Practice

Find the sum. Check whether your answer is answerable.

Question 7.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 110

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
557 = 560
79 = 80
283 = 280
560 + 80 + 280 = 920
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 110
919
919 is close to 920. So, the answer is reasonable.

Question 8.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 111

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
382 = 380
357 = 360
160 + 380 + 360 = 900
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
160
382
357
899
899 is close to 900. So, the answer is reasonable.

Question 9.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 112

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
35 = 30
68 = 70
827 = 830
30 + 70 + 830 = 930
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
35 + 68 + 827 = 930
930 is close to 930. So, the answer is reasonable.

Question 10.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 113

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
153 = 150
235 = 230
458 = 460
67 = 70
150 + 230 + 460 + 70 = 910
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
153 + 235 + 458 + 67 = 913
913 is close to 910. So, the answer is reasonable.

Question 11.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 114

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
108 = 110
172 = 170
200 = 200
263 = 260
110 + 170 + 200 + 260 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
108 + 172 + 200 + 263 = 743
743 is close to 740. So, the answer is reasonable.

Question 12.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 115

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
181 = 180
629 = 630
140 = 140
23 = 20
180 + 630 + 140 + 20 = 970
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
181 + 629 + 140 + 23 = 973
973 is close to 970. So, the answer is reasonable.

Question 13.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 116

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
213 = 210
208 = 210
462 = 460
111 = 110
210 + 210 + 460 + 110 = 990
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 116
994
994 is close to 990. So, the answer is reasonable.

Question 14.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 117

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
108 = 110
172 = 170
200 = 200
263 = 260
110 + 170 + 200 + 260 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 117
743
743 is close to 740. So, the answer is reasonable.

Question 15.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 118

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
259 = 260
233 = 230
223 = 220
147 = 150
260 + 230 + 220 + 150 = 860
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 118
862
862 is close to 860. So, the answer is reasonable.

Question 16.
Number Sense
Use the Associative Property of Addition to find (345 + 234) + 206.

Answer:
We can find the (345 + 234) + 206 by using the Associative Property of Addition.
(a + b) + c = a + (b + c)
(345 + 234) + 206 = 345 + (234 + 206)
345 + 440 = 785

Question 17.
YOU BE THE TEACHER
Your friend finds 364 + 109 + 27. Is your friend correct? Explain.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 119

Answer:
Your friend is incorrect. The order of the sum is wrong. 27 is placed in the wrong pattern. 2 and 7 must be places on tens place and ones place.
364
109
+27
503

Think and Grow: Modeling Real Life

An elevator has a weight limit of 1,000 pounds. A 186-pound man has three 265-pound boxes to deliver. Can he bring all 3 boxes on the elevator at once?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 120

Understand the problem:
Make a plan:
Solve:
He ___ bring all 3 boxes on the elevator at once.

Answer:
Given that,
An elevator has a weight limit of 1,000 pounds.
A 186-pound man has three 265-pound boxes to deliver.
265 + 265 + 265 + 186 = 981
1000 – 981 = 19 pounds
Thus he can bring all 3 boxes on the elevator at once.

Show and Grow

Question 18.
An auditorium has 650 seats. 175 students from each of 3 schools compete in a math competition. 68 teachers assist. Are there enough seats for all of the students and teachers?

Answer:
Given,
An auditorium has 650 seats. 175 students from each of 3 schools compete in a math competition. 68 teachers assist.
175 + 175 + 175 + 68 = 593
Thus the seats are enough for all of the students and teachers.

Question 19.
DIG DEEPER!
Four students at a school organize a petition for more lunch food options. They need 500 signatures. How many more signatures do they need?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 121

Answer:
Add the number of signatures of all the students
A  + B + C + D = 77 + 108 + 112 + 96 = 393
500 – 393 = 107
Thus they need 107 signatures more.

Add Three or More Numbers Homework & Practice 8.6

Find the sum. Check whether your answer is reasonable

Question 1.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 122

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
862 is close to 860. So, the answer is reasonable.

Question 2.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 123

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
65 = 60
41 = 40
786 = 790
60 + 40 + 790 = 890
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 123
892
892 is close to 890. So, the answer is reasonable.

Question 3.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 124

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
409 = 410
87 = 90
463 = 460
410 + 90 + 460 = 960
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 124
959
959 is close to 960. So, the answer is reasonable.

Question 4.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 125

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
302 = 300
253 = 250
169 = 170
18 = 20
300 + 250 + 170 + 20 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 125
742
742 is close to 740. So, the answer is reasonable.

Question 5.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 126

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
353 = 350
121 = 120
154 = 150
116 = 120
350 + 120 + 150 + 120 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 126
744
744 is close to 740. So, the answer is reasonable.

Question 6.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 127

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
213 = 210
251 = 250
139 = 140
210 + 270 + 250 + 140 = 870
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 127
873
873 is close to 870. So, the answer is reasonable.

Question 7.
Structure
Which problem can you solve without regrouping?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 128

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-128

Question 8.
Reasoning
You add 602 + 125 + 231. Your friend adds 231 + 602 + 125. Do you both get the same answer? Use an additional property to explain.

Answer:
There are four mathematical properties that involve addition. The properties are the commutative, associative, additive identity and distributive properties. Additive Identity Property: The sum of any number and zero is the original number.
602 + 125 + 231 = 958
231 + 602 + 125 = 958
The sum will be the same irrespective of the change of order.

Question 9.
Modeling Real Life
A firefighter’s ladder has a weight limit of 750 pounds. One firefighter weighs 196 pounds. Another firefighter weighs 243 pounds. They each have67 pounds of gear. If both firefighters wear their gear, can they climb the ladder at the same time?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 129

Answer:
A firefighter’s ladder has a weight limit of 750 pounds.
One firefighter weighs 196 pounds. Another firefighter weighs 243 pounds. They each have 67 pounds of gear.
196 + 243 + 67 + 67 = 573
750 – 573 = 177 pounds
If both firefighters wear their gear, they can climb the ladder at the same time.

Question 10.
DIG DEEPER!
Your principal agrees to make a lip-sync video if the school’s social media page reaches 1,000 likes in 5 days. How many more likes does the school’s page need?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 130

Answer:
Given,
Your principal agrees to make a lip-sync video if the school’s social media page reaches 1,000 likes in 5 days.
Add all the number of likes
573 + 168 + 201 + 47 = 989
1000 – 989 = 19 likes
Thus the school’s page needs 19 likes more.

Review & Refresh

Question 11.
Find the area of the shape
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 131

Answer:
From the figure, we can observe that each block = 1 sq. cm
We have to find the area of the shaded part.
The shaded region is in the form of a rectangle.
So, we have to find the area of the rectangle.
A = l × b
A = 7 × 3
A = 21 sq.cm
Thus the area of the shape is 21 sq. cm

Lesson 8.7 Use Number Lines to Subtract

Explore and Grow

Color to find 79 – 47. Then model your jumps on the number line.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 132

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-132 (1)

Reasoning
How can finding 79 – 47 help you find 379 – 47?

Answer:
You can subtract 47 from 79 by using the number line.
79 – 47 = 32
Just add 300 to 79 or add 3 to the left and subtract 47 from 379.
379 – 47 = 332

Think and Grow: Subtracting on a Number Line

Example
Find 358 – 82.
One Way:
Use the count back strategy. Start at 358. Count back by tens, then by ones.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 133

Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-133

Another Key:
Use the count on strategy. Start at 82. Count on until you reach 358.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 134
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-134

Show and Grow

Question 1.
Use the count back strategy to find 273 – 36.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 135

Answer:
Use the count back strategy. Start at 273. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-135

Question 2.
Use the strategy to find 124 – 45.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 136

Answer:
Use the count on strategy. Start at 45. Count on until you reach 124.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-136

Apply and Grow: Practice

Question 3.
Use the count back strategy to find 961 – 38.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 137

Answer:
Use the count back strategy. Start at 961. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-137

Question 4.
Use the count back strategy to find 853 – 77.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 138

Answer:
Use the count back strategy. Start at 853. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-138

Find the difference.

Question 5.
316 – 24 = ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 139

Answer:
Use the count back strategy. Start at 316. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-139

Question 6.
548 – 113 = ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 140

Answer:
Use the count back strategy. Start at 548. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-140

Question 7.
Your friend knows 154 words in Italian. You want to know just as many words as your friend. So far, you have learned 73 words. How many words do you have left to learn?

Answer:
Given,
Your friend knows 154 words in Italian. You want to know just as many words as your friend. So far, you have learned 73 words.
154 – 73 = 81
Thus 81 words are left to learn.

Question 8.
Structure
Write the equation shown by the number line.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 141

Answer:
By seeing the above number line we can find the subtraction equation.
Use the count on strategy. Start at 36. Count on until you reach 407.
407 – 36 = 371

Think and Grow: Modeling Real Life

Each member of a marching band and a football team is awarded a ribbon. The marching band has 123 members. The football team has 66 members. How many more ribbons are needed for the marching band than for the football team?
Subtraction equation:
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 142
___ more ribbons are needed for the marching band.

Answer:
Given,
Each member of a marching band and a football team is awarded a ribbon. The marching band has 123 members. The football team has 66 members.
123 – 66 = 57
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-142
Thus 57 ribbons are needed for the marching band than for the football team.

Show and Grow

Question 9.
A marine biologist feeds 435 pounds of fish to an orca and 50 pounds of fish to a sea lion. How many more pounds did the orca eat than the sea lion?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 143

Answer:
Given,
A marine biologist feeds 435 pounds of fish to an orca and 50 pounds of fish to a sea lion.
435 – 50 = 385
Thus 385 more pounds the orca eat than the sea lion.

Question 10.
There are 620 paper lanterns for a festival. Some are let go. There are 42 left. How many paper lanterns were let go?

Answer:
Given,
There are 620 paper lanterns for a festival. Some are let go. There are 42 left.
620 – 42 = 578
Therefore 578 paper lanterns were let go.

Question 11.
DIG DEEPER!
There are some guests at an amusement park. 387 of them leave when it rains. 474 of them stay. How many guests were there before it rained?

Answer:
Given that,
There are some guests at an amusement park. 387 of them leave when it rains. 474 of them stay.
387 + 474 = 861
Thus 861 guests were there before it rained.

Use Number Lines to Subtract Homework & Practice 8.7

Question 1.
Use the count back strategy to find 232 – 53.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 144

Answer:
Use the count back strategy. Start at 232. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-144

Question 2.
Use the count back strategy to find 796 – 81.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 145

Answer:
Use the count back strategy. Start at 796. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-145

Find the difference

Question 3.
474 – 19 = ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 146

Answer:
Use the count back strategy. Start at 474. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-146

Question 4.
615 – 204 = ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 147

Answer:
Use the count back strategy. Start at 615. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-147

Question 5.
Writing
Write and solve a subtraction word problem using 995 and 238.

Answer:
There are 995 guests at an amusement park. 238 of them leave when it rains. How many guests were there before it rained?
995 – 238
We can solve the problem by using the number line.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-146

Question 6.
DIG DEEPER!
Which number lines can you use to find 734 – 308?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 148

Answer:
Among all the number lines option ii is used to find 734 – 308.

Question 7.
Modeling Real Life
You take 107 pictures on a field trip to a zoo. Your friend takes 73 pictures. How many more pictures do you take than your friend?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 149

Answer:
Given that,
You take 107 pictures on a field trip to a zoo. Your friend takes 73 pictures.
107 – 73 = 34
You take 34 pictures than your friend.

Question 8.
Modeling Real Life
An author has 350 copies, not of her book. Some are signed. 115 copies are signed. How many copies are signed?

Answer:
Given,
An author has 350 copies, not of her book. Some are signed. 115 copies are signed.
350 – 115 = 235 copies
Therefore 235 copies are to be signed.

Review & Refresh

Find the quotient.

Question 9.
Divide 25 by 5.

Answer: 5

Explanation:
Divide the two numbers 25 and 5.
25/5 = 5
Thus the quotient is 5.

Question 10.
Divide 40 by 4.

Answer: 10

Explanation:
Divide the two numbers 40 and 4.
40/4 = 10
Thus the quotient is 10.

Question 11.
Divide 72 by 8.

Answer: 9

Explanation:
Divide the two numbers 72 and 8.
72/8 = 9
Thus the quotient is 9.

Lesson 8.8 Use Mental Math to Subtract

Explore and Grow

750 – 300 = ___
650 – 200 = ____
750 – 300 = ___
700 – 250 = ____
750 – 300 = ___
740 – 290 = ____

What patterns do you notice? Explain.

Answer:
750 – 300 = 450
650 – 200 = 450
750 – 300 = 450
700 – 250 = 450
750 – 300 = 450
740 – 290 = 450
Here you notice that all the answers are 450 for different patterns.

Think and Grow: Mental Math Strategies for Subtraction

Example
Find 433 – 198.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 150
One Way: Use compensation to change both numbers.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 151
433 + 2 = 435
198 + 2 = 200
435
-200
235
So, 433 – 198 = 235

Another Way: Use compensation to change one number.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 152
198 is close to 200 and it is easier to subtract 200.
433
-198 + 2 = 200
233
233 + 2 = 235
So, 433 – 198 = 235

Show and Grow

Use compensation to find the difference

Question 1.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 153

Answer:
Use compensation to change both numbers.
Add 4 to both numbers and use mental math strategy.
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-153

Question 2.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 154

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-154

Question 3.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 155

Answer:
Use compensation to change one number.
219 is close to 220 and it is easier to subtract 220.
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-155

Apply and Grow: Practice

Use compensation to find the difference

Question 4.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 156

Answer:
Use compensation to change both numbers.
Add 1 to both numbers and use mental math strategy.
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-156

Question 5.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 157

Answer:
117 is close to 120. So it is easier to subtract 120.
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-157

Use mental math to find the difference

Question 6.
643 – 115 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
643 + 5 = 648
115 + 5 = 120
648
-120
528
So, 643 – 115 = 528

Question 7.
863 – 257 = ___

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
863 + 3 = 866
257 + 3 = 260
866
-260
606

Question 8.
768 – 543 = ___

Answer:
Use compensation to change both numbers.
Add 2 to both numbers and use mental math strategy.
768 + 2 = 770
543 + 2 = 545
770
-545
225

Question 9.
688 – 414 = ___

Answer:
Use compensation to change both numbers.
Add 1 to both numbers and use mental math strategy.
688 + 1 = 689
414 + 1 = 415
689
-415
274

Question 10.
499 – 106 = ___

Answer:
Use compensation to change both numbers.
Add 4 to both numbers and use mental math strategy.
499 + 4 = 503
106 + 4 = 110
503
-110
393

Question 11.
495 – 162 = ___

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
495 + 3 = 498
162 + 3 = 165
498
-165
333

Question 12.
874 – 515 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
874 + 5 = 879
515 + 5 = 520
879
-520
359

Question 13.
637 – 228 = ___

Answer:
Use compensation to change both numbers.
Add 2 to both numbers and use mental math strategy.
637 + 2 = 639
228 + 2 = 230
639
-230
409

Question 14.
986 – 432 = ___

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
986 + 3 = 989
432 + 3 = 435
989
-435
554

Question 15.
A movie theater has 225 seats. 108 seats are not taken. How many seats are not taken?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 158

Answer:
Given that,
A movie theater has 225 seats. 108 seats are not taken.
225 – 108 = 117
Therefore 117 seats are not taken.

Question 16.
Reasoning
Your friend starts to find 741 – 295. What is the next step? Explain.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 159

Answer:
The next step is to subtract the original numbers.
741
-295
446

Think and Grow: Modeling Real Life

A softball coach has $325 for new equipment. She buys the catching gear. Does she have enough money left to buy the bat?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 402

Subtraction equation:
Compare:
The coach ___ has enough money to buy the bat.

Answer:
Given A softball coach has $325 for new equipment. She buys the catching gear.
325 + 1 = 326
219 + 1 = 220
326
-220
106
The cost of the bat is $109.
Thus she does not have enough money.

Show and Grow

Question 17.
A store owner has 550 T-shirts. He sells 333 of them. Then he receives an order for 168 T-shirts. Does he have enough T-shirts to complete the order?

Answer:
Given that,
A store owner has 550 T-shirts. He sells 333 of them.
550 – 333 = 217
Then he receives an order for 168 T-shirts.
217 – 168 = 49
Yes, he has enough T-shirts to complete the order.

Question 18.
The manager of a gaming center has $700 for new electronics. She buys the game system. Does she have enough money left for either of the other two items? If so, which one?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 160

Answer:
Given that,
The manager of a gaming center has $700 for new electronics. She buys the game system.
$399 + $169 = $568
700 – 568 = $132
Thus she has enough money left for either of the other two items.
She can buy the game system and a bundle of games.

DIG DEEPER!
How much more money does the manager need to buy both the television and the bundle of games?

Answer:
The cost of television is $379
The cost of the bundle of games = $169
379 + 169 = 548
Thus the manager need $548 to buy both the television and the bundle of games

Use Mental Math to Subtract Homework & Practice 8.8

Use compensation to find the difference.

Question 1.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 161

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
596 + 3 = 599
317 + 3 = 320
599
-320
279
So, 596 – 317 = 279

Question 2.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 162

Answer:
Add 6 to 214 to make the subtraction easier.
685
-214 + 6 = 220
685
-220
465
So, 685 – 214 = 465

Use mental math to find the difference.

Question 3.
782 – 489 = ___

Answer:
Use compensation to change both numbers.
Add 1 to both numbers and use mental math strategy.
782 + 1 = 783
489 + 1 = 490
783
-490
293
So, 782 – 489 = 293

Question 4.
672 – 266 = ___

Answer:
Use compensation to change both numbers.
Add 4 to both numbers and use mental math strategy.
672 + 4 = 676
266 + 4 = 270
676
-270
406
So, 672 – 266 = 406

Question 5.
983 – 155 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
983 + 5 = 988
155 + 5 = 160
988
-160
820
So, 983 – 155 = 820

Question 6.
744 – 125 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
744 + 5 = 749
125 + 5 = 130
749
130
619
So, 744 – 125 = 619

Question 7.
967 – 619 = ___

Answer:
Use compensation to change both numbers.
Add 1 to both numbers and use mental math strategy.
967 + 1 = 968
619 + 1 = 620
968
-620
348
So, 967 – 619 = 348

Question 8.
854 – 517 = ___

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
854 + 3 = 857
517 + 3 = 520
857
-520
337
So, 854 – 517 = 337

Question 9.
472 – 215 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
472 + 5 = 479
215 + 5 = 220
479
220
259
So, 472 – 215 = 259

Question 10.
883 – 335 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
883 + 5 = 888
335 + 5 = 340
888
-340
548
So, 883 – 335 = 548

Question 11.
575 – 198 = ___

Answer:
Use compensation to change both numbers.
Add 2 to both numbers and use mental math strategy.
575 + 2 = 577
198 + 2 = 200
577
-200
377
So, 575 – 198 = 377

Question 12.
Reasoning
To find 765 – 246, Newton adds 5 to each number and then subtracts. To find the difference, Descartes adds 4 to each number, and then subtracts. Will they both get the correct answer? Explain.

Answer:
Given,
To find 765 – 246, Newton adds 5 to each number and then subtracts.
765 + 5 = 770
246 + 5 = 251
770 – 251 = 519
To find the difference, Descartes adds 4 to each number, and then subtracts.
765 + 4 = 769
246 + 4 = 250
769 – 250 = 519
Yes they both get the correct answer.

Question 13.
Modeling Real Life
A custodian has 350 desks to clean. She cleans 124 desks on the first floor and 147 desks on the second floor. Does she clean all of the desks?

Answer:
Given that,
A custodian has 350 desks to clean.
She cleans 124 desks on the first floor and 147 desks on the second floor.
124 + 147 = 271
350
-271
79
No, she did not clean all of the desks.

Question 14.
Modeling Real Life
A fashion designer has $ 725 to spend on new supplies. She buys the sewing machine. Does she have enough money left for either of the other two items? If so, which one?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 163

Answer:
Given,
A fashion designer has $ 725 to spend on new supplies. She buys the sewing machine.
725 – 495 = 230
Yes she has enough money left for either of the other two items
She can buy a Mannequin.

DIG DEEPER!
How much more money does the fashion designer need to buy both the mannequin and the fashion design software?

Answer:
Given,
The cost of the Mannequin is $129
The cost of the fashion design software is $329
129 + 329 = 458
458 – 230 = 228
Thus she needs 228 to buy both the mannequin and the fashion design software.

Review & Refresh

Draw equal groups. Then complete the equations.

Question 15.
3 groups of 6
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 164

Answer:
3 groups of 6 means 3 times 6.
6 + 6 + 6 = 18
3 × 6 = 18

Question 16.
4 groups of 9
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 165

Answer:
4 groups of 9 means 4 times 9.
9 + 9 + 9 + 9 = 36
4 × 9 = 36

Lesson 8.9 Subtract Three-Digit Numbers

Explore and Grow

Model the equation. Draw to show your model. Then find the difference.
694 – 418 = ___

Reasoning
How can you use an estimate to check whether your answer is reasonable?

Answer:
Find 694 – 418. Check whether your answer is reasonable.
694 = 700
418 = 400
700
-400
300
The difference is about 300.
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup
Big-ideas-math-answers-grade-3-chapter-8-img-1
Step 3: Check 276 is close to 300, so the answer is reasonable.

Think and Grow: Subtract Three-Digit Numbers

Example
Find 604 – 215. Check whether your answer is reasonable.
Step 1: Estimate. Round each number to the nearest hundred.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 166
600
-200
400
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 167
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-167
Step 3: Check 389 is close to 400, so the answer is reasonable.

Show and Grow

Find the difference. Check whether your answer is reasonable.

Question 1.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 168

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
300
200
100
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-Grade-3-Chapter-8-Add-and-subtract-digit-numbers-img-2
Step 3: Check 136 is close to 100, so the answer is reasonable.

Question 2.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 169

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
538 = 500
371 = 400
500 – 400 = 100
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers img-3
Step 3: Check 167 is close to 100, so the answer is reasonable.

Question 3.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 170

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
500 – 300 = 200
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Bigideas-math-answer-key-3rd-grade-chapter-8-add-and-subtract-multi-digit-numbers-img-4
Step 3: Check 238 is close to 200, so the answer is reasonable.

Question 4.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 171

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
963 = 1000
429 = 400
1000 – 400 = 600
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-answers-3rd-grade-chapter-8-add-and-subtract-multi-digit-numbers-img-5
Step 3: Check 534 is close to 600, so the answer is reasonable.

Question 5.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 172

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
641 = 600
287 = 300
600 – 300 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-Book-Grade-3-chapter-8-add-and-subtract-multi-digit-numbers-img-6
Step 3: Check 354 is close to 300, so the answer is reasonable.

Question 6.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 173

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
832 = 800
359 = 300
800 – 300 = 500
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-answers-3rd-grade-chapter-8-img-7
Step 3: Check 473 is close to 500, so the answer is reasonable.

Apply and Grow: Practice

Find the difference. Check whether your answer is reasonable.

Question 7.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 174

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
518 = 500
232 = 200
500 – 200 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-14
Step 3: Check 286 is close to 300, so the answer is reasonable.

Question 8.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 175

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
971 = 1000
320 = 300
1000 – 300 = 700
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-15
Step 3: Check 651 is close to 700, so the answer is reasonable.

Question 9.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 176

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
565 = 600
289 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-16
Step 3: Check 276 is close to 300, so the answer is reasonable.

Question 10.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 177

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
546 = 500
341 = 300
500 – 300 = 200
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-17
Step 3: Check 205 is close to 200, so the answer is reasonable.

Question 11.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 178

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
707 = 700
453 = 400
700 – 400 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-18
Step 3: Check 254 is close to 300, so the answer is reasonable.

Question 12.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 179

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
406 = 400
77 = 100
400 – 100 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-answers-chapter-8-add-and-subtract-multi-digit-numbers-img-19
Step 3: Check 329 is close to 300, so the answer is reasonable.

Question 13.
Estimate: ____
552 – 381 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
552 = 600
381 = 400
600 – 400 = 200
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-answers-chapter-8-add-and-subtract-multi-digit-numbers-img-20
Step 3: Check 171 is close to 200, so the answer is reasonable.

Question 14.
Estimate: ___
725 – 146 = ____

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
725 = 700
146 = 100
700 – 100 = 600
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-answers-chapter-8-add-and-subtract-multi-digit-numbers-img-21
Step 3: Check 579 is close to 600, so the answer is reasonable.

Question 15.
Estimate: ___
800 – 486 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
486 = 500
800 – 500 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-answers-chapter-8-add-and-subtract-multi-digit-numbers-img-22
Step 3: Check 314 is close to 300, so the answer is reasonable.

Question 16.
The number of rings on a tree is equal to its age. A redwood tree has 473 rings. A bristlecone pine tree has 806 rings. How much older is the bristlecone pine tree than the redwood tree?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 180

Answer:
Given that,
The number of rings on a tree is equal to its age. A redwood tree has 473 rings. A bristlecone pine tree has 806 rings
806
-473
333
Thus 333 older is the bristlecone pine tree than the redwood tree.

Question 17.
Writing
Explain how to regroup 408 to subtract 259.

Answer:
Working each column from right to left
9 is greater than 8 so you must regroup:
Take 1 from 4, so 4 becomes 3.
Add 10 to 0, so 0 becomes 10.
Take 1 from 10, so 10 becomes 9.
Add 10 to 8, so 8 becomes 18.
18 minus 9 is 9.
9 minus 5 is 4.
3 minus 2 is 1.
408
-259
149

Think and Grow: Modeling Real Life

How many more pennies does the class need to reach the goal?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 500
Subtraction equation:
The class needs to collect ___ more pennies.

Answer:
800
-444
356
The class needs to collect 356 more pennies.

Show and Grow

Question 18.
How many more campers can attend the summer camp?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 501

Answer:
Total summer camp openings = 405
Filled = 316
400 – 316 = 84
84 more campers can attend the summer camp.

Question 19.
A musician wants to buy a set of speakers that costs $672. She saves $224 each month for 2 months. How much money does she still need to save?

Answer:
Given that,
A musician wants to buy a set of speakers that costs $672.
She saves $224 each month for 2 months.
224 + 224 = 448
672 – 448 = 224
Thus she still need to save $224.

Question 20.
DIG DEEPER!
Newton has 442 packages to deliver. Descartes has 464. Newton delivers 174 packages, and Descartes delivers 188. Who is closer to finishing his deliveries?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 502

Answer:
Given,
Newton has 442 packages to deliver. Descartes has 464. Newton delivers 174 packages, and Descartes delivers 188.
442 – 174 = 268
464 – 188 = 276
Thus Newton is closer to finishing his deliveries.

Subtract Three-Digit Numbers Homework & Practice 8.9

Find the difference. Check whether your answer is reasonable.

Question 1.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 503

Answer:
The estimated number for 571 is 600.
The estimated number for 220 is 200.
600
-200
400
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-Grade-3-Answer-Key-Chapter-8-Add-&-Subtract-Multi-Digit-Numbers-img-8
352 is close to 400. So the answer is reasonable.

Question 2.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 504

Answer:
The estimated number for 421 is 400.
The estimated number for 277 is 300.
400
-300
100
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-3rd-Grade-Solutions-Chapter-8-Add-and-subtract-multi-digit-numbers-img-9
144 is close to 100. So the answer is reasonable.

Question 3.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 505

Answer:
The estimated number for 534 is 500.
The estimated number for 186 is 200
500
-200
300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Bigideas-math-answers-3rd-grade-chapter-8-img-10
348 is close to 300. So the answer is reasonable.

Question 4.
Estimate: ___
690 – 298 = ___

Answer:
The estimated number for 690 is 700
The estimated number for 298 is 300.
700
-300
400
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-grade-3-solution-key-chapter-8-add-and-subtract-multi-digit-numbers-img-11
392 is close to 400. So the answer is reasonable.

Question 5.
Estimate: ___
613 – 472 = ___

Answer:
The estimated number for 613 is 600.
The estimated number for 472 is 500
600
-500
100
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-answer-key-grade-3-8th-chapter-add-&-subtract-multi-digit-numbers-img-12
141 is close to 100. So the answer is reasonable.

Question 6.
Estimate: ___
835 – 189 = ___

Answer:
The estimated number for 835 is 800.
The estimated number for 189 is 200.
800
-200
600
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-Solution-Key-Grade-3-Chapter-8-img-13
646 is close to 600. So the answer is reasonable.

Question 7.
YOU BE THE TEACHER
Your friend says you have to regroup every time you subtract from a number that has a zero. Is your friend correct? Explain.

Answer:
Your friend says you have to regroup every time you subtract from a number that has a zero.
Yes, your friend is correct. Because whenever you subtract a number from 0 you have to regroup.

Question 8.
Modeling Real Life
How many more soup can labels does the school need to reach the goal?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 506

Answer:
To find how many more soup can label does the school need to reach the goal you have to subtract label collected from school soup.
1000 – 638 = 362

Question 9.
Modeling Real Life
Newton wants to buy a couch that costs $594. He saves $198 each month for 2 months. How much money does he still need to save?

Answer:
Given,
Newton wants to buy a couch that costs $594. He saves $198 each month for 2 months.
198 × 2 = $396
594 – 396 = -198
Therefore Newton need to save $198 to buy the couch.

Question 10.
DIG DEEPER!
Find the missing digits.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 507

Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-507

Review & Refresh

Round the number to the nearest ten and to the nearest hundred.

Question 11.
64
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 508

Answer:
The number 64 nearest ten is 60.
The number nearest hundred to 64 is 60.

Question 12.
411
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 509

Answer:
The number 411 nearest ten is 410
The number nearest hundred to 411 is 400.

Lesson 8.10 Relate Addition and Subtraction

Explore and Grow

How can you find the missing number? How do you know you are correct?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 510

Reasoning
How are addition and subtraction related?

Answer:
497 – 358 = 136
Check the answer by using the addition model.
136
+358
497
So, the answer is reasonable.

Think and Grow: Relate Addition and Subtraction

Inverse operations are operations that “undo” each other. Addition and subtraction are inverse operations.
Example
Find 846 – 283. Use the inverse operation to check.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 511

Answer:
First, subtract 283 from 846 and then use the inverse operation to check the solution.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-511

Example
Find 355 + 437. Use the inverse operation to check.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 512

Answer:
First, add 355 from 437 and then use the inverse operation to check the solution.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-512

Show and Grow

Find the sum or difference. Use the inverse operation to check.

Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 183

Answer:
First, subtract 682 from 419 and then use the inverse operation to check the solution.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-183
So, the answer is reasonable.

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 184

Answer:
First, add 169 from 745 and then use the inverse operation to check the solution.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-184
So, the answer is reasonable.

Question 3.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 513

Answer:
First, add 376 from 238 and then use the inverse operation to check the solution.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-513
So, the answer is reasonable.

Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 514

Answer:
First, subtract 547 from 285 and then use the inverse operation to check the solution.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-514
So, the answer is reasonable.

Question 5.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 515

Answer:
First, add 463 from 349 and then use the inverse operation to check the solution.
463
+349
812
Use the inverse operation to check the solution
812
-349
463
So, the answer is reasonable.

Question 6.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 516

Answer:
First, subtract 790 from 317 and then use the inverse operation to check the solution.
790
-317
473
Use the inverse operation to check the solution
473
+317
790
So, the answer is reasonable.

Apply and Grow: Practice

Find the sum or difference. Use the inverse operation to check.

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 517

Answer:
First, subtract 857 from 567 and then use the inverse operation to check the solution.
857
-567
290
Use the inverse operation to check the solution
290
+567
857
So, the answer is reasonable.

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 518

Answer:
First, add 762 from 143 and then use the inverse operation to check the solution.
762
+143
905
Use the inverse operation to check the solution
905
-143
762
So, the answer is reasonable.

Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 519

Answer:
First, add 653 from 217 and then use the inverse operation to check the solution.
653
+217
870
Use the inverse operation to check the solution
870
-217
653
So, the answer is reasonable.

Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 520

Answer:
First, subtract 294 from 156 and then use the inverse operation to check the solution.
294
-156
138
Use the inverse operation to check the solution
138
+156
294
So, the answer is reasonable.

Question 11.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 521

Answer:
First, add 475 from 438 and then use the inverse operation to check the solution.
475
+438
913
Use the inverse operation to check the solution
913
-438
475
So, the answer is reasonable.

Question 12.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 522

Answer:
First, subtract 514 from 386 and then use the inverse operation to check the solution.
514
-386
128
Use the inverse operation to check the solution
128
+386
514
So, the answer is reasonable.

Question 13.
Which one Doesn’t Belong? Which equation does not belong with the other three?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 523

Answer:
The fourth equation does not belong with the other three. Because it is not using the inverse operation of addition and subtraction.

Question 14.
Open-Ended
Write a subtraction equation that has a difference of 381.

Answer: 901 – 520 = 381
Take the number on your own and write the subtraction equation with the difference of 381.

Think and Grow: Modeling Real Life

A kayak costs $321. A customer pays $196 for the kayak after using a gift card. How much money is the gift card worth?
The gift card is worth $___.
Check:
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 524

Answer:
Given that,
A kayak costs $321. A customer pays $196 for the kayak after using a gift card.
321 – 196 = 125
Thus the gift card worth $125.

Show and Grow

Question 15.
You print 600 flyers for an event. You hand out some of them. There are 237 left. How many flyers did you hand out?

Answer:
Given that,
You print 600 flyers for an event. You hand out some of them. There are 237 left.
600 – 237 = 363
Thus you hand out 363 flyers.
363 + 237 = 600

Question 16.
A building has 163 floors. You start on the 28th floor. You go up in the elevator 126 floors. Then you go down 145 floors. On which floor do you end?

Answer:
Given that,
A building has 163 floors. You start on the 28th floor. You go up in the elevator 126 floors. Then you go down 145 floors.
126 – 28 = 98 floors
145 – 98 = 47
Thus you end at 47th floor.

Question 17.
DIG DEEPER!
A bus travels from Boston to Washington, D.C. On the way back, the bus stops in New York City. How many miles has the bus traveled in all? How many miles does the bus have left to travel?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 525

Answer:
Given that,
A bus travels from Boston to Washington, D.C. On the way back, the bus stops in New York City.
We have to find How many miles has the bus traveled in all
441 + 225 = 666 miles
Thus a bus travels 666 miles.
666 – 225 = 441 miles

Relate Addition and Subtraction Homework & Practice 8.10

Find the sum or difference. Use the inverse operation to check.

Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 526

Answer:
931
-544
387
Now you have to do the inverse operation.
387
+544
931

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 527

Answer:
623
+285
908
Now you have to do the inverse operation.
908
-285
623

Question 3.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 528

Answer:
523
+237
760
Now you have to do the inverse operation.
760
-237
523

Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 529

Answer:
403
-252
151
Now you have to do the inverse operation.
151
+252
403

Question 5.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 530

Answer:
612
+387
999
Now you have to do the inverse operation.
999
-387
612

Question 6.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 531

Answer:
511
-371
140
Now you have to do the inverse operation.
140
+371
511

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 532

Answer:
437
+156
593
Now you have to do the inverse operation.
593
-156
437

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 533

Answer:
726
-362
364
Now you have to do the inverse operation.
364
+362
726

Question 9.
YOU BE THE TEACHER
Your friend uses an inverse operation to check her answer. Is your friend correct? Explain.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 510

Answer:
No, your friend is incorrect.
380
-159
221
Now you have to do the inverse operation.
221
+159
380

Question 10.
Which One Doesn’t Belong? Which does not belong with the other three?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 535

Answer:
208 + 475 = 683
Now you have to do the inverse operation.
683 – 475 = 208
The second figure does not belong to the other three expressions.

Question 11.
Modeling Real Life
A telescope costs $169. A customer pays $119 for the telescope after using a gift card. How much money is the gift card worth?

Answer:
Given that,
A telescope costs $169. A customer pays $119 for the telescope after using a gift card.
169
-119
50
The cost of the gift card is $50.

Question 12.
DIG DEEPER!
A train travels from Dallas to San Antonio. On the way back, the train stops in Austin. How many miles has the train travel? How many miles does the train have left to travel?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 536

Answer:
Given,
A train travels from Dallas to San Antonio. On the way back, the train stops in Austin.
274 – 79 = 195 miles
The train has traveled 195 miles.
79 miles left to travel from Austin to San Antonio.

Review & Refresh

Question 13.
Use the Distributive Propertytofind the area of the rectangle.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 537

Answer:
4 × 8 = 4 × (4 + 4)
4 × 8 = (4 × 4) + (4 × 4)
4 × 8 = 16 +16
4 × 8 = 32
Thus the area of the rectangle = 32 square foot.

Lesson 8.11 Problem Solving: Addition and Subtraction

Explore and Grow

You read 150 pages in three weeks.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 538

what does p represent?
p = ___

Answer:
Let 9 be the number of pages read
56 + 47 + p = 150
103 + p = 150
p = 150 – 103
p = 47

Construct Arguments
Explain to your partner how to find what n represents.
250 + n = 580

Answer:
Given the expression 250 + n = 580
n = 580 – 250
n = 330

Think and Grow: Using the Problem-Solving Plan
Example
Newton has 368 baseball cards. He gives away 139 of them. He buys 26 more. How many cards does he have now?

Understand the Problem

  • Newton has ___ cards.
  • He gives away of them.
  • He buys ___ more.
  • You need to find how many ___ he has now.

Answer:

  • Newton has 368 cards.
  • He gives away of them.
  • He buys 26 more.
  • You need to find how many cards he has now.

Make a Plan

How will you solve?

  • Subtract ___ from ___ to find how many ___ he has left after he gives some away.
  • Then add ___ to the difference to find how many he has now.

Answer:

  • Subtract 139 from 368 to find how many cards he has left after he gives some away.
  • Then add 26 to the difference to find how many he has now.

Solve
Draw a part-part-whole model and write an equation.
Use a letter to represent the unknown number.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 539

Newton has __ cards now.

Answer:
Step 1:
c is the unknown difference.
368 – 139 = c
c = 229
Step 2:
c = 229
229 + 26 = n
n = 225
Thus the unknown sum is 225.

Show and Grow

Question 1.
Explain how you can check whether your answer above is reasonable.

Answer:
You can check the answer by using addition and subtraction.
368
-139
229
Now check whether the answer is correct or not.
229
+139
368
So, the answer is reasonable.

Apply and Grow: Practice

Write equations to solve. Use letters to represent the unknown numbers. Check whether your answer is reasonable.

Question 2.
A baker makes 476 muffins. He sells 218 of them. Then he makes 390 more. How many muffins does the baker have now?

Answer:
Given,
A baker makes 476 muffins. He sells 218 of them.
476
-218
258
Then he makes 390 more.
390
+258
648
Thus the baker has 348 muffins now.

Question 3.
Newton knocks down 146 pins in his first bowling game. He knocks down 19 more pins in his second game than in his first game. How many pins does he knock down in all?

Answer:
Given that,
Newton knocks down 146 pins in his first bowling game. He knocks down 19 more pins in his second game than in his first game.
146
+19
165
He knocks down 165 pins in his second game.
To find the total number of pins we have to the points in the first game and second game.
146
+165
311
Thus he knockdowns 311 pins in all.

Question 4.
You are traveling to a campground that is 243 miles away. You travel 155 miles in the morning and 59 miles in the afternoon. How many more miles do you need to travel before you get to the campground?

Answer:
Given that,
You are traveling to a campground that is 243 miles away.
You travel 155 miles in the morning and 59 miles in the afternoon.
155
+59
214
243
-214
029
Thus you need to travel 29 miles to get to the campground.

Question 5.
There are 205 lawn tickets and 585 bleacher tickets sold for a concert. There are 680 fewer VIP tickets sold than lawn and bleacher tickets combined. How many VIP tickets are sold?

Answer:
Given,
There are 205 lawn tickets and 585 bleacher tickets sold for a concert.
205 + 585 = 790
There are 680 fewer VIP tickets sold than lawn and bleacher tickets combined.
790
-680
110
Thus 110 VIP tickets are sold.

Think and Grow: Modeling Real Life

How many more people went to see the movie on Friday than on Thursday and Saturday combined?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 542
Understand the problem:
Make a plan:
Solve:
__ more people went to see the movie on Friday than on Thursday and Saturday combined.

Answer:
Thursday – 346 people
Saturday – 512 people
346 + 512 = 858 people
897
-858
39
39 more people went to see the movie on Friday than on Thursday and Saturday combined.

Show and Grow

Question 6.
How many more people used the ferry on Friday than on Saturday and Sunday combined?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 543
Explain how you can check whether your answer is reasonable.

Answer:
Friday – 903 passengers
Saturday – 624 passengers
Sunday – 255 passengers
624
+255
879
Now subtract 879 from 903.
903
-879
024

Problem Solving: Addition and Subtraction Homework & Practice 8.11

Write equations to solve. Use letters to represent the unknown numbers. Check whether your answer is reasonable.

Question 1.
Newton has 387 tokens, and Descartes has 295. They use a total of 461 tokens. How many tokens do they have now?

Answer: 222 tokens

Explanation:
Given that,
Newton has 387 tokens, and Descartes has 295.
387 + 295 = 682
They use a total of 461 tokens.
682 – 461 = 222 tokens
Thus they have 222 tokens now.

Question 2.
There are 125-second graders and 118 third graders at a museum. There are 249 more adults than students at the museum. How many adults are at the museum?

Answer:
Given that,
There are 125-second graders and 118 third graders at a museum.
125 + 118 = 243 graders
There are 249 more adults than students at the museum.
243 + 249 = 492
Therefore 492 adults are at the museum.

Question 3.
You received 171 votes in a coloring contest. Your friend received 24 fewer votes than you. How many people voted for you and your friend in all?

Answer:
Given,
You received 171 votes in a coloring contest. Your friend received 24 fewer votes than you.
171 + 24 = 195 votes
The number of votes for your friend is 195.
171+ 195 = 366 votes
Thus 366 people voted for you and your friend.

Question 4.
Writing
Write and solve a two-step problem that can be solved using addition or subtraction.

Answer:
You bought 10 packs of sketches and your friend bought 4 packs less than you. Each pack contains 10 sketches. Find how many sketches you and your friend bought in all.
Sol: You bought 10 packs of sketches and your friend bought 4 packs less than you.
10 – 4 = 6 packs
Your friend bought 6 packs.
Each pack contains 10 sketches.
10 + 6 = 16
16 × 10 = 160
Thus you and your friend bought 160 sketches.

Question 5.
Modeling Real Life
How many more fish were caught on Sunday than on Friday and Saturday combined?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 544
Explain how you can check whether your answer is reasonable.

Answer:
Number of fish caught on Friday and Saturday = 127 + 244 = 371
Number of fish caught on Sunday = 564
564 – 371 = 193
193 more fish were caught on Sunday than on Friday and Saturday combined.

Review & Refresh

Question 6.
Use the multiplication table.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 545
Describe the pattern in the shaded row and column.
What property explains this pattern?

Answer: The pattern shows that it is the multiple of 5.
5 × 1 = 5
5 × 2 = 10
5 × 3 = 15
5 × 4 = 20
5 × 5 = 25

Add and Subtract Multi-Digit Numbers Performance Task

Your school holds a talent show.

Question 1.
You and your friend hand out programs to guests before the show. You each start with 250 programs. There are 114 programs left. How many programs did you and your friend hand out?

Answer:
250 – 114 = 136 programs
Thus you and your friend hand out 136 programs.
136 + 114 = 250 programs

Question 2.
75 students wait backstage to perform in the show. There are 336 children, 125 adults, and 14 teachers in the audience.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 216
a. How many people are at the talent show in all? Explain how to use addition properties to find the sum.

Answer:
Given,
75 students wait backstage to perform in the show. There are 336 children, 125 adults, and 14 teachers in the audience.
75 + 336 + 125 + 14 = 550
Thus there are 550 people are in the talent show.

b. Four students perform in each of the first 4 acts. How many students still need to perform?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 217

Answer: 16 students

Explanation:
Given,
Four students perform in each of the first 4 acts.
4 × 4 = 16
Thus 16 students need to perform.

c. Each performer is given a juice box backstage. Juice boxes come in packages of 10. How many packages did the teachers buy? How many juice boxes are left?

Answer:
Given,
Each performer is given a juice box backstage. Juice boxes come in packages of 10.
1 box – 10 packages
16 × 1 = 16 boxes
16 × 10 = 160 packages

Add and Subtract Multi-Digit Numbers Activity

Three in a Row: Addition and Subtraction

Directions:
1. Players take turns.
2. On your turn, spin both spinners. Add or subtract the two numbers. Cover the sum or difference.
3. If the sum or difference is already covered, then you lose your turn.
4. The first player to get three counters in a row, horizontally, vertically, or diagonally, wins!
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 218

Answer:
Sums:
547 + 107 = 654
547 + 338 = 885
547 + 262 = 809
I got the three counters vertically.

Add and Subtract Multi-Digit Numbers Chapter Practice

8.1 Identify Addition Properties

Identify the property.

Question 1.
59 + 0 = 59

Answer:
It shows the Addition Property of Zero. The Addition Property of Zero defines the sum of any number and 0 is that number.

Question 2.
(14 + 32) + 6 = 14 + (32 + 6)

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 3.
27 + 51 = 51 + 27

Answer:
Commutative Property of addition Changing the grouping of addends does not change the sum.

Question 4.
Structure
Which equations show the Commutative Property of Addition?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 219

Answer:
64 + 12 = 12 + 64 – Commutative Property of addition changing the grouping of addends does not change the sum.
71 + 0 = 71 – It shows the Addition Property of Zero. The Addition Property of Zero defines the sum of any number and 0 is that number.
(56 + 21) + 34 = 56 + (21 + 34) – It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.
26 + (41 + 4) = 4 + (26 + 41) – It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

8.2 Use Number Lines to Add

Question 5.
Find 648 + 37.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 220

Answer:
Use the count on strategy. Start at 648. Count on by tens, then by ones.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-220

8.3 Use Mental Math to Add

Use mental math to find the sum.

Question 6.
192 + 107 = ___

Answer:
You can find the sum of 192 and 107 by using mental math strategies.
192 – 2 = 190
107 + 2 = 109
190
+109
299

Question 7.
676 + 114 = ___

Answer:
You can find the sum of 676 and 114 by using mental math strategies.
676 + 4 = 680
114 – 4 = 110
680
+110
790

Question 8.
716 + 279 = ___

Answer:
You can find the sum of 716 and 279 by using mental math strategies.
716 – 1 = 715
279 + 1 = 280
715
+280
995

Question 9.
501 + 468 = ___

Answer:
You can find the sum of 501 and 468 by using mental math strategies.
501 – 1 = 500
468 + 1 = 469
500
+469
969

Question 10.
527 + 343 = ___

Answer:
You can find the sum of 527 and 343 by using mental math strategies.
527 + 3 = 530
343 – 3 = 340
530
+340
870

Question 11.
441 + 189 = ___

Answer:
You can find the sum of 441 and 189 by using mental math strategies.
441 – 1 = 440
189 + 1 = 190
440
+190
630

8.4 Use Partial Sums to Add

Use partial sums to add.

Question 12.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 221

Answer:
586 = 500 + 80 + 6
107 = 100 + 00 + 7
693 = 600 + 80 + 13

Question 13.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 222

Answer:
647 = 600 + 40 + 7
293 = 200 + 90 + 3
940 = 800 + 130 + 10

Question 14.
Modeling Real Life
On Earth, your cousin weighs 207 pounds less than he would on Jupiter. Your cousin weighs 135 pounds on Earth. How much would he weigh on Jupiter?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 223

Answer:
Given,
On Earth, your cousin weighs 207 pounds less than he would on Jupiter. Your cousin weighs 135 pounds on Earth.
207 + 135 = 342
Thus your cousin would weigh 342 pounds on Jupiter.

8.5 Add Three-Digit Numbers

Find the sum. Check whether your answer is reasonable.

Question 15.
Estimate: ___
326 + 490 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
326 = 300
490 = 500
300 + 500 = 800
The sum is about 800.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
326
+490
816
816 is close to 800. So, the answer is reasonable.

Question 16.
Estimate: ___
657 + 189 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
657 = 700
189 = 200
600 + 200 = 800
The sum is about 800.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
657 + 189 = 846
846 is close to 800. So, the answer is reasonable.

Question 17.
Estimate: ___
543 + 261 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
543 = 500
261 = 300
500 + 300 = 800
The sum is about 800.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
543 + 261 = 804
804 is close to 800. So, the answer is reasonable.

8.6 Add Three or More Numbers

Question 18.
Estimate: ___
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 224

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
78 = 100
433 = 400
367 = 400
100 + 400 + 400 = 900
The sum is about 900.

Question 19.
Estimate: ___
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 225

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
194 = 200
151 = 200
244 = 200
231 = 200
200 + 200 + 200 + 200 = 800
The sum is about 800.

Question 20.
Estimate: ___
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 226

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
373 = 400
329 = 300
118 = 100
61 = 100
400 + 300 + 100 + 100 = 900
The sum is about 900.

8.7 Use Number Lines to Subtract

Question 21.
Find 856 – 29.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 511

Answer:
Use the count back strategy. Start at 856. Count back by tens, then by ones.
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-511

Question 22.
Structure
Write the equation shown by the number line.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 512

Answer:
By seeing the above number we can find the subtraction equation.
764 – 50 = 714
714 – 4 = 710
710 – 3 = 707
The subtraction equation is 764 – 57 = 707

8.8 Use Mental Math to Subtract

Use mental math to find the difference.

Question 23.
957 – 619 = ___

Answer:
957 – 7 = 950
619 + 7 = 626
950
-626
324
The difference is 324.

Question 24.
831 – 415 = ___

Answer:
831 – 1 = 830
415 + 1 = 416
830
-416
414
The difference is 414.

Question 25.
876 – 366 = ___

Answer:
876 – 6 = 870
366 + 6 = 372
870 – 372 = 498

Question 26.
636 – 317 = ___

Answer:
636 + 6 = 642
317 – 6 = 311
642 – 311 = 331
The difference is 331.

Question 27.
965 – 528 = ___

Answer:
528 – 8 = 520
965 + 8 = 973
973 – 520 = 453
The difference is 453.

Question 28.
384 – 118 = ____

Answer:
684 + 4 = 688
118 – 4 = 114
688 – 114 = 574
The difference is 574.

8.9 Subtract Three-Digit Numbers

Find the difference. Check whether your answer is reasonable.

Question 29.
Estimate: ___
963 – 51 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
963 = 1000
51 = 100
1000 – 100 = 900
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
963 -51 = 912
912 is close to 900. So, the answer is reasonable.

Question 30.
Estimate: ___
878 – 594 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
878 = 900
594 = 600
900 – 600 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
878 – 594 = 284
284 is close to 300. So, the answer is reasonable.

Question 31.
Estimate: ___
766 – 297 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
766 = 800
297 = 300
800 – 300 = 500
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
766 – 297 = 469
469 is close to 500. So, the answer is reasonable.

Question 32.
YOU BE THE TEACHER
Your friend finds 760 – 482. Is your friend correct? Explain.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 513

Answer: No your friend is not correct.
BIM Answer Key for Grade 3 Chapter 8 img-13

8.10 Relate Addition and Subtraction

Find the sum or difference. Use the inverse operation to check.

Question 33.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 514

Answer:
649
+227
876
Now you have to do the inverse operation of the sum.
876
-227
649

Question 34.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 515

Answer:
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 515
288
Now you have to do the inverse operation of the difference.
288
+517
805

8.11 Problem Solving: Addition and Subtraction

Question 35.
There are 532 dogs enrolled in police academies. 246 dogs graduate in July, and 187 dogs graduate in August. How many dogs still need to graduate?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 516

Answer:
Given that,
There are 532 dogs enrolled in police academies.
246 dogs graduate in July, and 187 dogs graduate in August.
246 + 187 = 433
532
-433
99
Thus 99 dogs are still needed to graduate.

Add and Subtract Multi-Digit Numbers Cumulative Practice 1 – 8

Question 1.
Which numbers round to 300 when rounded to the nearest hundred?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 517

Answer: The numbers round to 300 when rounded to the nearest hundred is
298, 309, 347
Thus the correct answer is the option a, b, c.

Question 2.
You buy 18 cups of yogurt. The yogurt is sold in packs of 6 cups. How many packs of yogurt do you buy?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 519

Answer:
Given,
You buy 18 cups of yogurt. The yogurt is sold in packs of 6 cups.
18/6 = 3 packs
Thus the correct answer is option c.

Question 3.
A bedroom floor is 9 feet long and 8 feet wide. What is the area of the bedroom floor?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 520

Answer:
Given,
A bedroom floor is 9 feet long and 8 feet wide.
The area of the rectangle = l × w
A = 9ft × 8ft
A = 72 sq. ft
Thus the area of the bedroom floor is 72 sq. ft.

Question 4.
Your friend says 458 – 298 = 160. How can you use inverse operations to check your friend’s answer? Is your friend correct?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 544

Answer:
Your friend says 458 – 298 = 160.
Your friend is correct.
The correct answer is option a.

Question 5.
Your friend says she needs (9 × 3) + (3 × 9) = 27 × 27 = 54 tiles to make the design. Why is her thinking incorrect?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 545

Answer:
Your friend says she needs (9 × 3) + (3 × 9) = 27 × 27 = 54 tiles to make the design.
Her thinking is incorrect because she needs to add 27 and 27 but she multiplied.
(9 × 3) + (3 × 9) = 27 + 27 = 54

Question 6.
Part A
What is the least number that can be made with the digits 7, 9, and 8 using each digit only once?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 546
Your friend says the greatest number he can make with the digits 7, 9, and 8 using each digit only once is 879. Is he correct? If not, correct his answer. Explain.

Answer:
Given,
Your friend says the greatest number he can make with the digits 7, 9, and 8 using each digit only once is 879
Your friend is incorrect because the greatest number with the digits 7, 9, and 8 is 987.

Question 7.
Which equation is shown by the number line?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 547

Answer:
The count starts from 0.
The count jumps from 0 and skips for every 3s.
3 × 7 = 21
Thus the correct answer is option c.

Question 8.
Find the sum.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 548

Answer:
548
+372
920

Question 9.
Which equations show the Associative Property of Addition?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 549

Answer:
According to the associative property of addition, the sum of three or more numbers remains the same regardless of how the numbers are grouped.
Options A and D show the equation for the Associative Property of Addition.

Question 10.
A teacher takes 7 students on a field trip. Each student pays $5. How much money does the teacher collect in all?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 550

Answer:
Given,
A teacher takes 7 students on a field trip. Each student pays$5.
7 × $5 = $35
Thus the teacher collects $35 in all.
The correct answer is option c.

Question 11.
What is the area of the shape?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 551

Answer:
You can divide the figure into two parts.
Figure 1:
l = 4m
b = 3m
A = l × b
A = 4 × 3 = 12 sq. m
Figure 2:
l = 8m
b = 3m
A = l × b
A = 8 × 3 = 24 sq. m
Add the area of both the figures 12 + 24 = 36 sq.m
Thus the correct answer is option D.

Question 12.
There are 459 girls and 552 boys in a school. How many more boys are there than girls?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 552

Answer:
Given that,
There are 459 girls and 552 boys in school.
Subtract the number of girls from the number of boys.
552
-459
93
Thus the correct answer is option b.

Question 13.
Look at the pattern. What rule was used to make the pattern?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 553

Answer:
1, 3, 9, 27 are the multiples of 3.
Thus the correct answer is option c.

Question 14.
A smoothie shop sells 368 smoothies in July and 205 smoothies in August. About how many more smoothies did the shop sell in July than in August?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 554

Answer:
Given that,
A smoothie shop sells 368 smoothies in July and 205 smoothies in August.
368
-205
163
Thus the shop sell 163 smoothies in July than in August.

Question 15.
Which shape does not have an area of 16 square units?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 555

Answer:
There are 16 counters in the first figure
2 × 8 = 16
There are 16 counters in the second figure
4 × 4 = 16
There are 15 counters in the third figure
5 × 3 = 15
Thus the correct answer is option c.

Complete the table

Question 16.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 556

Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-556
You can complete the table by multiplying the rows and columns.

Question 17.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 557

Answer:
You can complete the table by multiplying the rows and columns.
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-557

Add and Subtract Multi-Digit Numbers Steam Performance Task 1-8

Question 1.
The carpeting in the third-grade classrooms of an elementary school is being replaced. One roll of carpet covers 100 square yards. The map shows the classrooms that will receive the new carpet.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 560
a. Explain two different ways to find the area of Classroom A.

Answer:
You can find the area of classroom A by using the composite figure.
The shape of classroom A is square.
a = 10 yd
Area of the classroom is a × a
A = 10 × 10 = 100 sq.yd
Thus the area of the classroom A is 100 sq. yd
Another way:
l = 10 yd
w = 3 yd
A = 10 × 3 = 30 sq. yd
l = 10 yd
A = 8 × 4 = 32 sq. yd
l = 10 yd
w = 3 yd
A = 10 × 3 = 30 sq. yd

b. Find the total area of all of the classrooms in square yards.

Answer:
Area of the classroom is a × a
A = 10 × 10 = 100 sq.yd
Thus the area of the classroom A is 100 sq. yd
Area of Classroom D = 10 × 7 = 70 sq.yd
Area of Classroom B = 11 × 7 = 77 sq.yd
Area of Classroom C = 12 × 7 = 84 sq. yd, 8 × 3 = 24 sq.yd
Area of Classroom C = 84 + 24 = 108 sq. yd
Total area of the classrooms = 100 + 70 + 77 + 108 = 355 sq. yd

c. Estimate the number of rolls of carpet needed for the classrooms. Explain.

Answer:
There are 4 classrooms. So, the estimated number of rolls of carpet is 4.

d. Find the area of the hallway in square yards. Is there enough carpet for the hallway?

Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 561
Answer:
First, divide the hallway into 3 parts.
i. It is in the form of a square.
a = 3 yd
A = 3 × 3 = 9 sq.yd
ii. It is in the form of rectangle
A = l × b
A = 5 × 4 = 20 sq.yd
iii. It is in the form of rectangle
A = l × b
A = 5 × 3 = 15 sq.yd
The area of the hallway in square yards = 9 + 20 + 15 = 44 sq.yd

Question 2.
Each school keeps a record of the total number of students in each class and grade.
a. Use the number of students in your class to estimate the total number of students in your grade. Explain.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 562

Answer:
Let the number of students in your class is 47.
By this, we can estimate the total number of students in your grade i.e., 50.
The total number of students in your grade is 50.

b. Use the table to write the number of students in each grade of your school.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 563

Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-563
You can prepare the table by estimating the number of students in your school.

c. How does your estimate compare to the actual number of students in your grade? Explain.

Answer: You can compare the number of students in the above table with the actual number of students in your grade.

d. What is the total number of students in your school?

Answer:
Add the number of students of all the grades
40+ 50 + 52 + 47 + 47 + 50 = 286
Thus there are 286 students in your school.

e. Write and answer a question using the information from the table above.

Answer:
Compare the number of students in grade 3 with the actual number of students in your school in grade 3?
The estimated number of students in grade 3 in the above table is 50.
The actual number of students in your school is 48.
50 – 48 = 2

f. What is one reason your principal may want to know the total number of students in your class, grade, or school?

Answer: Shaping a vision of academic success for all students.

Conclusion:

I wish that the information given here regarding Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers is helpful for you. This pdf will help to score good marks in the exam. Get the solutions of BIM Book Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers from here. Stay tuned to our page to know the solution and brief explanation for other chapters of grade 3.

Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties

Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties

Follow a free step-by-step guide to Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties available here. There are various topics involved in Big Ideas Math Answers like Properties of Addition and Multiplication, Writing Expressions, Algebraic Expressions, and so on. Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties will aid you to enhance your math skill and problem-solving techniques. Know all the problems and solutions and tips to solve them in an easy manner. Go through the below sections to know the properties and example problems.

Big Ideas Math Book 6th Grade Answer Key Chapter 5 Algebraic Expressions and Properties

To ease your learning and make it quick by using Big Ideas Math Book 6th Grade Solution Key Chapter 5 Algebraic Expressions and Properties pdf. Algebraic Expressions and Properties pdf concepts are explained by proficient math experts. With the help of the Big Ideas Book 6th Grade Answer Key, you can get the solutions to every problem. Just click on the links and start practicing all the problems for the exams and score well in the exam. Follow all the problems and practice them to improve time management skills and accuracy.

Algebraic Expressions and Properties STEAM Video/Performance

Lesson: 1 Algebraic Expressions

Lesson: 2 Writing Expressions

Lesson: 3 Properties of Addition and Multiplication

Lesson: 4 The Distributive Property

Lesson: 5 Factoring Expressions

Chapter: 5 – Algebraic Expressions and Properties

Algebraic Expressions and Properties STEAM Video/Performance

Shadow Drawings
Expressions can be used to represent the growth of living things over time. Can you think of any other real-life situations in which you would want to use an expression to represent a changing quantity?
Watch the STEAM Video “Shadow Drawings.” Then answer the following questions.

Question 1.
Tory traces the shadow of a plant each week on the same day of the week and at the same time of day. Why does she need to be so careful about the timing of the drawing?

Answer:
Tory traces the shadow of a plant each week on the same day of the week and at the same time of day.
Because she needs to represent the growth of the plant over time. Thus she needs to be very careful about the timing of the drawing.

Question 2.
The table shows the height of the plant each week for the first three weeks. About how tall was the plant after 1.5 weeks? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties 1

Answer: 10.5 inches

Explanation:
The above table represents the height of the plant for 3 weeks.
1 week = 7 inches
0.5 week = 7/2 = 3.5 inches
1.5 week = 7 + 3.5 = 10.5 inches

Question 3.
Predict the height of the plant when Tory makes her next three weekly drawings.

Answer: 42 inches (approx)

Explanation:
The height of the plant is increased by 7 inches every week.
The height of the plant is increased to 22 inches for the first three weeks.
7 × 6 = 42 inches
Thus we predict the height of the plant when Tory makes her next three weekly drawings is 42 inches.

Performance Task

Describing Change

After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given data sets for the following real-life situations.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 2
You will be asked to use given data to write expressions and make predictions. Do the expressions provide accurate predictions far into the future?
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 3

Getting Ready for Chapter 5

Question 1.
Work with a partner.
a. You baby sit for 3 hours. You receive $24. What is your hourly wage?

  • Write the problem. Underline the important numbers and units you need to solve the problem.
  • Read the problem carefully a second time. Circle the key phrase for the question.
    Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 4
  • Write each important number or phrase, with its units, on a piece of paper. Write +, −, ×, ÷, and = on five other pieces of paper.
    Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 5
  • Arrange the pieces of paper to answer the question, “What is your hourly wage?”
  • Evaluate the expression that represents the hourly wage.
    Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 6

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-5-Algebraic-Expressions-and-Properties-6
According to the given details, Babysit receives $24 for 3 hours.
3 hours = $24
1 hour = 24/3 = $8
Thus your hourly wage is $8

b. How can you use your hourly wage to find how much you will receive for any number of hours worked?

Answer:
You can multiply by $8 in given any number of hours worked to get the total receive.

Vocabulary

The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 7

Lesson 5.1 Algebraic Expressions

EXPLORATION 1
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 8
Work with a partner. Identify any missing information that is needed to answer each question. Then choose a reasonable quantity and write an expression for each problem. After you have written the expression, evaluate it using mental math or some other method.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 9
a. You receive $24 for washing cars. How much do you earn per hour?

Answer:
Let the number of hours 8.
24 ÷ 8 = 3
Thus you earn $3 per hour.

b. You buy 5 silicone baking molds at a craft store. How much do you spend?
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 10

Answer:
Let the cost of each silicone baking molds is $3
5 × $3 = $15
Thus you spend $15 to buy 5 silicone baking molds.
c. You are running in a mud race. How much farther do you have to go after running 2000 feet?
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 11

Answer:
You are running a 4500-foot race.
4500 – 2000 = 2500 feet
Thus you have to run 2500 feet more.

d. A rattlesnake is 25 centimeters long when it hatches. The snake grows at a rate of about 1.6 centimeters per month for several months. What is the length of the rattlesnake?
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 12

Answer:
Given,
A rattlesnake is 25 centimeters long when it hatches. The snake grows at a rate of about 1.6 centimeters per month for several months.
Let the number of months be m
25 + 1.6m
If months = 12
1.6 × 12 = 19.2 cms
25 + 19.2 = 44.2 centimeters

5.1 Lesson

Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 13

An algebraic expression is an expression that may contain numbers, operations, and one or more variables. A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 14

Try It

Identify the terms, coefficients, and constants in the expression.

Question 1.
12 + 10c

Answer:
12 – constant
10 – coefficient
c – variable or term
A term without a variable is called a constant.
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Question 2.
15 + 3w + \(\frac{1}{2}\)

Answer:
15 and \(\frac{1}{2}\) – constant
3 – coefficient
w – variable or term
A term without a variable is called a constant.
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Question 3.
z2 + 9z

Answer:
2 – exponent
9 – coefficient
z – variable or term
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

write the expression using exponents.

Question 4.
j . j . j . j . j . j

Answer:
j raised to the sixth power
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor

Question 5.
9 . k . k . k . k . k

Answer:
9 – constant
k raised to the fifth power
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor

Question 6.
Evaluate 24 + c when c = 9.

Answer: 33

Explanation:
Given equation
24 + c
and also given c is equal to 9
substitute c value in the equation we get
24 + 9 = 33

Question 7.
Evaluate d − 17 when d = 30.

Answer: 13

Explanation:
Given equation
d – 17
and also given d is equal to 30
substitute d value in the equation we get
30 – 17 = 13

Question 8.
Evaluate 18 ÷ q when q = \(\frac{1}{2}\)

Answer: 36

Explanation:
Given equation
18 ÷ q
and also given q is equal to \(\frac{1}{2}\)
substitute q value in the equation we get
18 ÷ \(\frac{1}{2}\) = 36

Evaluate the expression when p = 24 and q = 8.

Question 9.
p ÷ q

Answer:
p = 24
q = 8
p ÷ q
Substitute the value of p and q in the expression
24 ÷ 8 = 3

Question 10.
q + p

Answer:  32

Explanation:
Given,
p = 24
q = 8
q + p
Substitute the value of p and q in the expression
8 + 24 = 32
Hence we get the answer is 32

Question 11.
p – q

Answer: 16

Explanation:
Given,
p = 24
q = 8
p – q
Substitute the value of p and q in the expression
24 – 8 = 16
Hence we get the answer is 16

Question 12.
p . q

Answer: 192

Explanation:
Given,
p = 24
q = 8
p . q
Substitute the value of p and q in the expression
24 . 8  by multiplying 24 with 8 we get 192
Hence  the answer is 192

Evaluate the expression when y = 6.

Question 13.
5y + 1

Answer: 31

Explanation:
Given equation
5y + 1
and also given y = 6
Now substitute 6 in the given equation
5.6 + 1 is
5 multiply with 6 and then add with 1 we get 30 + 1 is 31
hence answer is 31

Question 14.
30 – 24 ÷ y

Answer: 1

Explanation:
Given equation
30 – 24 ÷ y
and also given y = 6
Now substitute 6 in the given equation
30 – 24 ÷ y is
30 – 24 ÷6
6 ÷ 6
six divided by six
we get 1
Hence answer is 1

Question 15.
y2 – 7

Answer: 29

Explanation:
Given equation
y² – 7
and also given y = 6
Now substitute 6 in the given equation
y2 – 7
6² – 7
6 × 6 – 7
36 – 7
29
Hence answer is 29

Question 16.
1.5 + y2

Answer: 37.5

Explanation:
Given equation
1.5 + y²
and also given y = 6
Now substitute 6 in the given equation
1.5 + 6²
1.5 +36
37.5
Hence answer is 37.5

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 17.
WHICH ONE DOESN’T BELONG?
Which expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 15

Answer: The expression 3(4) + 5 does not belong to the other three. Because it does not contain any variable in the expression.

Question 18.
ALGEBRAIC EXPRESSIONS
Identify the terms, coefficients, and constants in the expression 9h + 1.

Answer:
1 – constant
9 – coefficient
h – variable or term
A term without a variable is called a constant.
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

EVALUATING EXPRESSIONS
Evaluate the expression when m = 8.

Question 19.
m – 7

Answer: 1
Given equation
m – 7
given m value is 8
Now substitute m value 8  in the given equation we get
m – 7
8 – 7
1
Hence the answer is 1

Question 20.
5m + 4

Answer: 44
Given equation
5m + 4
given m value is 8
Now substitute m value 8  in the given equation we get
5 . 8 + 4
40 + 4
44
Hence the answer is 44

Question 21.
NUMBER SENSE
Does the value of the expression 20 − x increase, decrease, or stay the same as x increases? Explain.

Answer:
When the value of x increases, the value of 20 – x decreases. And when the value of x does not change, 20 – x remains the same. When greater values of x are subtracted from 20, you will have a lower value left. Therefore, as x increases, the value of the expression 20 – x will decrease.

Question 22.
OPEN-ENDED
Write an algebraic expression using more than one operation. When you evaluate the expression, how do you know which operation to perform first?

Answer:
Example  40 – 2(6 – 4)²
40 – 2 (2)²
40 – 2(4)
40 – 8
32

When evaluating an expression, proceed in this order:

parentheses are done first.
exponents are done next.
multiplication and division are done as they are encountered from left to right.
addition and subtraction are done as they are encountered from left to right.

Question 23.
STRUCTURE
Is the expression 8.2 ÷ m . m . m . m the same as the expression 8.2 ÷ m4? Explain your reasoning.

Answer: Yes
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor
Hence the expression 8.2 ÷ m . m . m . m the same as the expression 8.2 ÷ m⁴

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 24.
The expression 12.25m + 29.99 gives the cost (in dollars) of a gym membership for m months. You have $180 to spend on a membership. Can you buy a one-year membership?

Answer: Yes.
Given
The expression 12.25m + 29.99
gym membership for m months
You have $180 to spend on a membership
Let’s solve your equation step-by-step.
12.25m+29.99=180
Step 1: Subtract 29.99 from both sides.
12.25m+29.99−29.99=180−29.99
12.25m=150.01
Step 2: Divide both sides by 12.25.
12.25m/12.25=150.01/12.25
m=12.245714
The expression 12.25m + 29.99= 12.25(12.24)+29.99 = 179.93

Question 25.
DIG DEEPER!
The expression p −15 gives the amount (in dollars) you pay after using the coupon when the original amount of a purchase is p dollars. The expression 30 + 6n gives the amount of money (in dollars) you save after n weeks. A jacket costs $78. Can you buy the jacket after 6 weeks? Explain.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 16

Answer:
The amount you pay using coupon = p (purchase) -15
so we simply plug in our value into this equation
$78- $15 = $63
so $63 is the amount you pay after using the coupon on an original purchase of $78
n = 6 weeks
30 + 6n
30 + 6(6)
30 + 36 = 66
No, you cannot buy the jacket after 6 weeks.

Algebraic Expressions Homework & Practice 5.1

Review & Refresh

You ask 40 students which of three items from the cafeteria they like the best. You record the results on the piece of paper shown.

Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 500

Question 1.
What percent of students answered salad?

Answer: 25%

Explanation:
Total number of students = 40
The number of students who like a salad in the Cafeteria food is 10 students
40/10 = 4
The percentage of the students is
(10/40) × 100 = 25%
Thus the percentage of students who answered salad is 25%.

Question 2.
How many students answered pizza?

Answer: 12 students

Explanation:
Total number of students = 40
The number of students who like a salad in the Cafeteria food is 10 students
The number of students who answered pasta is 18
40 – 18 – 10 = 12
Thus the number of students who answered pizza is 12.

Question 3.
What percent of students answered pasta?

Answer: 45%

Explanation:
Total number of students = 40
The number of students who answered pasta is 18
(18/40) × 100 = 45
Thus the percentage of students who answered pasta is 45%

Find the missing quantity in the double number line.

Question 4.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 17

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-5-Algebraic-Expressions-and-Properties-17
The ratio of 2 and 10 is 1:5
The equivalent ratio of 1:5 is 6:30
So, the missing quantity is 6.

Question 5.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 18

Answer:
Based on the ratios we can find the missing quantity.
The cost of 1 pastry is $2.5
The cost of 9 pastries is $22.5
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-5-Algebraic-Expressions-and-Properties-18

Divide. Write the answer in simplest form.

Question 6.
1\(\frac{3}{8}\) ÷ \(\frac{3}{4}\)

Answer: \(\frac{1}{2}\)
To simplify a fraction, divide the top and bottom by the highest number that
can divide into both numbers exactly.
Simplifying (or reducing) fractions means making the fraction as simple as possible

Question 7.
2\(\frac{7}{9}\) ÷ 2

Answer: \(\frac{7}{9}\)
To simplify a fraction, divide the top and bottom by the highest number that
can divide into both numbers exactly.
Simplifying (or reducing) fractions means making the fraction as simple as possible

Question 8.
4 ÷ 4\(\frac{2}{5}\)

Answer: \(\frac{5}{2}\)
To simplify a fraction, divide the top and bottom by the highest number that
can divide into both numbers exactly.
Simplifying (or reducing) fractions means making the fraction as simple as possible

Question 9.
3\(\frac{2}{3}\) ÷ 1\(\frac{2}{7}\)

Answer: latex]\frac{14}{3}[/latex]
To simplify a fraction, divide the top and bottom by the highest number that
can divide into both numbers exactly.
Simplifying (or reducing) fractions means making the fraction as simple as possible

Concepts, Skills, & Problem Solving

EVALUATING EXPRESSIONS
Write and evaluate an expression for the problem. (See Exploration 1, p. 201.)

Question 10.
The scores on your first two history tests are 82 and 95. By how many points did you improve on your second test?

Answer: 13

Explanation:
The scores on your first two history tests are 82 and 95.
95 – 82 = 13
Thus you have to improve 13 points on your second test.

Question 11.
You buy a hat for $12 and give the cashier a $20 bill. How much change do you receive?

Answer: $8

Explanation:
Given,
You buy a hat for $12 and give the cashier a $20 bill.
20 – 12 = 8
Thus you receive $8 change.

Question 12.
You receive $8 for raking leaves for 2 hours. What is your hourly wage?

Answer: $4

Explanation:
Given,
You receive $8 for raking leaves for 2 hours.
2 hour = $8
1 hour = ?
8/2 = 4
Thus the hourly wage is $4.

Question 13.
Music lessons cost $20 per week. How much do 6 weeks of lessons cost?

Answer: $120

Explanation:
Given,
Music lessons cost $20 per week.
1 week = $20
$20 × 6 = $120
Thus the cost of 6 weeks is $120.

ALGEBRAIC EXPRESSIONS
Identify the terms, coefficients, and constants in the expression.

Question 14.
7h + 3

Answer:
3 – constant
7 – coefficient
h – variable or term
A term without a variable is called a constant.
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Question 15.
g + 12 + 9g

Answer:
12 – constant
9 – coefficient
g – variable or term
A term without a variable is called a constant.
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Question 16.
5c2 + 7d

Answer:
2 – exponent
5 and 7 – coefficient
c and d – variable or term
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Question 17.
2m2 + 15 + 2p2

Answer:
15 – constant
2 – exponent
2 – coefficient
m and p – variable or term
A term without a variable is called a constant.
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Question 18.
6 + n2 + \(\frac{1}{2}\)d

Answer:
6 – constant
\(\frac{1}{2}\)– exponent
2 – coefficient
n and d – variable or term
A term without a variable is called a constant.
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Question 19.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 19

Answer:
2 – exponent
8 and \(\frac{1}{3}\)– coefficient
x – variable or term
A term without a variable is called a constant.
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Question 20.
YOU BE THE TEACHER
Your friend finds the terms, coefficients, and constants in the algebraic expression 2x2y. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 20

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-5-Algebraic-Expressions-and-Properties-20
A term without a variable is called a constant.
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Question 21.
PERIMETER
You can use the expression 2ℓ + 2w to find the perimeter of a rectangle, where ℓ is the length and w is the width.
a. Identify the terms, coefficients, and constants in the expression.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 21

Answer:
Given Expression 2ℓ + 2w
The coefficients of the expression are 2.
terms – l, w
b. Interpret the coefficients of the terms.

Answer: The coefficients of the terms are 2, 2
2 + 2 = 4

USING EXPONENTS
Write the expression using exponents.

Question 22.
b . b . b

Answer:
b raised to the third power
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor

Question 23.
g . g . g . g . g

Answer:
g raised to the fifth power
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor

Question 24.
8 . w . w . w . w

Answer:
8 – constant
w raised to the fourth power
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor

Question 25.
5 . 2 . y . y . y

Answer:
5 and 2 – constant
y raised to the third power
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor

Question 26.
a . a . c . c

Answer:
a and c  raised to the two power
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor

Question 27.
2 . 1 . x . z . z .z . z

Answer:
2 and 1 – constant
x raised to the 1 power
z raised to the fourth power
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor

Question 28.
YOU BE THE TEACHER
Your friend writes the product using exponents. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 22

Answer: correct
5  – constant
n  raised to the fourth power
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor

Question 29.
AREA
Write an expression using exponents that represents the area of the square.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 23

Answer:
Side = 5d
Area of the square = s × s
A = 5d × 5d
A = 25d²

Question 30.
REASONING
Suppose the man in the St. Ives poem has x wives, each wife has x sacks, each sack has x cats, and each cat has x kits. Write an expression using exponents that represent the total number of kits, cats, sacks, and wives.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 24

Answer:
Suppose the man in the St. Ives poem has x wives, each wife has x sacks, each sack has x cats, and each cat has x kits.
x = 7
7 × 7 × 7 × 7

EVALUATING EXPRESSIONS
Evaluate the expression when a = 3, b = 2, and c = 12.

Question 31.
6 + a

Answer: 9

Explanation:
Given equation
6 + a
given data
a = 3
now substitute given data in the equation we get
6 + 3 = 9
hence the answer is 9

Question 32.
b.5

Answer: 10

Explanation:
Given equation
b.5
given data
b = 2
now substitute given data in the equation we get
2 . 5= 10 (2 multiply by 5)
hence the answer is 10

Question 33.
c – 1

Answer: 11

Explanation:
Given equation
c – 1
given data
c = 12
now substitute given data in the equation we get
12 – 1 = 11
hence the answer is 11

Question 34.
27 ÷ a

Answer: 9

Explanation:
Given equation
27 ÷a
given data
a = 3
now substitute given data in the equation we get
27 ÷ 3 = 9
Hence the answer is 9

Question 35.
12 – b

Answer: 10

Explanation:
Given equation
12 -b
given data
b = 2
now substitute given data in the equation we get
12 – 2 = 10
Hence the answer is 10

Question 36.
c + 5

Answer: 17

Explanation:
Given equation
c + 5
given data
c = 12
now substitute given data in the equation we get
12+ 5 = 17
Hence the answer is 17

Question 37.
2a

Answer: 6

Explanation:
Given equation
2a
given data
a = 3
now substitute given data in the equation we get
2(3) = 6
Hence the answer is 6

Question 38.
c ÷ 6

Answer: 2

Explanation:
Given equation
c ÷ 6
given data
c = 12
now substitute given data in the equation we get
12 ÷ 6 = 2
Hence the answer is 2

Question 39.
a + b

Answer: 5

Explanation:
Given equation
a + b
given data
a = 3, b = 2
now substitute given data in equation we get
3 + 2 = 5
hence the answer is 5

Question 40.
c + a

Answer: 15

Explanation:
Given equation
a + b
given data
a = 3, b = 2
now substitute given data in equation we get
3 + 2 = 5
hence the answer is 5

Question 41.
c – a

Answer: 11

Explanation:
Given equation
c – a
given data
a = 3, c = 12
now substitute given data in equation we get
12 – 3 = 11
hence the answer is 11

Question 42.
a – b

Answer: 1

Explanation:
Given equation
a – b
given data
a = 3, b = 2
now substitute given data in equation we get
3 – 2 = 1
hence the answer is 1

Question 43.
\(\frac{c}{a}\)

Answer: 4

Explanation:
Given equation
\(\frac{c}{a}\)
given data
a = 3, c = 12
now substitute given data in equation we get
\(\frac{12}{3}\) = 4
hence the answer is 4

Question 44.
\(\frac{c}{b}\)

Answer: 6

Explanation:
Given equation
\(\frac{c}{b}\)
given data
b = 2, c = 12
now substitute given data in equation we get
\(\frac{12}{2}\) = 6
hence the answer is 6

Question 45.
b.c

Answer: 24

Explanation:
Given equation
b . a
given data
b = 2, c = 12
now substitute given data in equation we get
2(12) = 24
hence the answer is 24

Question 46.
c(a)

Answer: 36

Explanation:
Given equation
c(a)
given data
a = 3, c = 12
now substitute given data in equation we get
12(3) = 36
hence the answer is 36

Question 47.
PROBLEM SOLVING
You earn 15n dollars for mowing n lawns. How much do you earn for mowing 1 lawn? 7 lawns?

Answer:
Given,
You earn 15n dollars for mowing n lawns.
n = 1
15 × 1 = 15
n = 7
15 × 7 = 105
Thus you earn $15 for mowing 1 lawn and $105 for mowing 7 lawns.

EVALUATING EXPRESSIONS
Copy and complete the table.

Question 48.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 25

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-5-Algebraic-Expressions-and-Properties-25
Explanation:
Given data from the table
x = 3 ,6, and 9
substitute x values in the given equation x . 8
3(8) = 24
6(8) = 48
9(8) = 72

Question 49.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 26

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-5-Algebraic-Expressions-and-Properties-26
Explanation:
Given data from the table
x = 2 ,4, and 8
substitute x values in the given equation 64 ÷ x = 24
64 ÷ 2 = 32
64 ÷ 4 = 16
64 ÷ 8 = 8

Question 50.
MODELING REAL LIFE
Due to gravity, an object falls 16t2 feet in t seconds. You drop a rock from a bridge that is 75 feet above the water. Will the rock hit the water in 2 seconds? Explain.

Answer:  No
Given
Due to gravity, an object falls 16t2 feet in t seconds. You drop a rock from a bridge that is 75 feet above the water
d = 16t² (d in ft, t in sec)
Set d = 75 ft and solve for t.
75 = 16t²
t = √(75/16) sec ≅ 2.17 sec
It hits the water in 2.17 sec.

EVALUATING EXPRESSIONS
Evaluate the expression when a = 10, b = 9, and c = 4.

Question 51.
2a + 3

Answer: 23
Given equation
2a + 3
given data
a = 10,
now substitute given data in equation we get
2(10)+3
= 20 + 3
= 23
hence the answer is 23

Question 52.
4c – 7.8

Answer:
Given equation
4c – 7.8
given data
a = 10,
now substitute given data in equation we get
4(4) – 7.8
= 16 – 7.8
= 8.2
Hence the answer is 8.2

Question 53.
\(\frac{a}{4}\) + \(\frac{1}{3}\)

Answer: \(\frac{17}{6}\) or 2.83
Given equation
\(\frac{a}{4}\) + \(\frac{1}{3}\)
given data
a = 10,
now substitute given data in equation we get
\(\frac{10}{4}\) + \(\frac{1}{3}\)
= \(\frac{17}{6}\)
= 2.83
hence the answer is \(\frac{17}{6}\) or 2.83

Question 54.
\(\frac{24}{b}\) + 8

Answer:
\(\frac{32}{3}\) or 10.66
Given equation
\(\frac{24}{b}\) + 8
given data
b = 9,
now substitute given data in equation we get
\(\frac{24}{9}\) + 8
= \(\frac{8}{3}\) + 8
= \(\frac{32}{3}\)
= 10.66
Hence the answer is \(\frac{32}{3}\) or 10.66

Question 55.
c2 + 6

Answer: 22
Given equation
c² + 6
given data
c = 4,
now substitute given data in equation we get
c² + 6
= 4² + 6
= (4 × 4) + 6
= 16 + 6
= 22
Hence the answer is 22

Question 56.
a2 – 18

Answer: 82
Given equation
a² – 18
given data
a = 10,
now substitute given data in equation we get
a² – 18
= 10² – 18
= (10 × 10) – 18
= 100 – 18
= 82
Hence the answer is 82

Question 57.
a + 9c

Answer: 40
Given equation
a  + 9c
given data
a = 10, c = 4
now substitute given data in equation we get
10 + 9(4)
= 10 + 36
= 46
Hence the answer is 46

Question 58.
bc + 12.3

Answer: 48.3
bc + 12.3
given data
b = 9, c = 4
now substitute given data in equation we get
9(4) + 12.3
= 36 + 12.3
= 48.3
Hence the answer is 48.3

Question 59.
3a + 2b – 6c

Answer: 24
3a + 2b – 6c
given data
a = 10, b = 9, c = 4
now substitute given data in equation we get
3(10) + 2(9) – 6(4)
= 30 + 18 – 24
= 24
Hence the answer is 24

Question 60.
YOU BE THE TEACHER
Your friend evaluates the expression when m = 8. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 27

Answer:  incorrect.

Explanation:
Given the expression 5m + 3
m = 8
5 × 8 + 3
40 + 3 = 43
By this we can say that your friend is incorrect.

Question 61.
PROBLEM SOLVING
After m months, the height of a plant is (10 + 3m) millimeters. How tall is the plant after 8 months? 3 years?

Answer:
8 months = 34 millimeters
EXPLANATION:
You have to plug in the number 8 as “m” in the equation then solve. 10 + 3(8)
3 years = 118 millimeters
3 years converted to months would be 12×3 which equals 36. Then put it into the equation 10 + 3(36) and you do the multiplication first.

Question 62.
STRUCTURE
You use a video streaming service to rent x new releases and y standard rentals. Which expression tells you how much money you will need?
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 28
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 29

Answer:
4x+3y

Explanation:
It is given that,
The number of new releases are represented by ‘x’ and number of standard rentals are represented by ‘y’.
Also, the cost for one new release = $4 and the cost of one standard rental = $3.
Thus, the cost of ‘x’ new releases = $4x and the cost of ‘y’ new releases = $3y.
So, the total cost = cost of ‘x’ new releases + cost of ‘y’ standard rentals.
i.e. Total cost = 4x + 3y in dollars.
Thus, the expression to tell the money required is 4x+3y.

Question 63.
OPEN-ENDED
You float 2000 feet along a lazy river water ride. The ride takes less than 10 minutes. Give two examples of possible times and speeds.

Answer:
if you travel at 200 feet per minute you can make in ten minutes

Question 64.
DIG DEEPER!
The expression 20a + 13c is the cost (in dollars) for a adults and c students to enter a science center.
a. How much does it cost for an adult? a student? Explain your reasoning.

Answer:
It costs $20 per adult. If this is a cost function,
which it is because the wording is “the cost (in dollars) for adults and c students”, the adult is the cost for 1 adult, 1a, is 20.
That relates the number of adults to the cost of 1 adult.
It costs $13 per student. Again, this is a cost function, so since the student is c,
the cost for 1 student, 1c, is 13. That relates the number of students to the cost of 1 student.

b. Find the total cost for 4 adults and 24 students.

Answer:
The total cost for 4 adult and 24 students looks like this:
20(4) + 13(24) which is 80 + 312 = $392

c. You find the cost for a group. Then the numbers of adults and students in the group both double. Does the cost double? Explain your answer using an example.

Answer:
If you have 3 adults and 3 students in your group, the cost is 20(3) + 13(3) which is $99.
If you double the number of each, let’s see if the cost doubles.
We will “up” the numbers to 6 each. 20(6) + 13(6) = $198. Is $198 the double of $99.
Yes it is. Let’s do it again to check. Let’s double the 6.
20(12) + 13(12) = $396, and $198 doubled does in fact equal $396

d. In part(b), the number of adults is cut in half, but the number of students doubles. Is the cost the same? Explain your answer.

Answer:
20(12) + 13(12) = $396, and $198 doubled does in fact equal $396.

Question 65.
REASONING
The volume of the cube (in cubic inches) is equal to four times the area of one of its faces (in square inches). What is the volume of the cube?
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 30

Answer:
Given,
The volume of the cube (in cubic inches) is equal to four times the area of one of its faces (in square inches).
We know that,
The volume of a cube = x³
Multiply 3 sides = x × x × x = x³

Lesson 5.2 Writing Expressions

EXPLORATION 1
Writing Expressions
Work with a partner. You use a $20 bill to buy lunch at a café. You order a sandwich from the menu board shown.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 31
a. Complete the table. In the last column, write a numerical expression for the amount of change you receive.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 32

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-5-Algebraic-Expressions-and-Properties-32
b. Write an algebraic expression that represents the amount of change you receive when you order any sandwich from the menu board.

Answer:
You have only $20 for lunch, so if you order any sandwich from the menu
the board then the amount will be change
if you ordered chicken salad then your changing amount is
price = $ 4.95
change Received = $20 – $4.95 = $15.05
change Received = $15.05

c. The expression 20 − 4.65 represents the amount of change one customer receives after ordering from the menu board. Explain what each part of the expression represents. Do you know what the customer ordered? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 32.1

Answer:
calculated changing amount for chicken salad in part (b),
so comparison between chicken salad expression and beef expression
For Chicken salad
Price = $4.95
change Recived = $20 – $4.95 = $15.05
change Recieved = $15.05
For Roast Beef
price = $6.75
Change Recived = $20 – $6.75 = $13.25
change Recived = $13.25

5.2 Lesson

Some words can imply math operations.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 33

Try It

Write the phrase as an expression.

Question 1.
the sum of 18 and 35

Answer: 53
Given
the sum of 18 and 35
so by add 18 with 35
we get 18+35
=53
So the answer is 53

Question 2.
6 times 50

Answer: 300
given
6 times 50
so multiply 6 with 50 we get
300
Hence the answer is 300

Write the phrase as an expression.

Question 3.
25 less than a number b

Answer: b – 25

Question 4.
a number x divided by 4

Answer:  \(\frac{x}{4}\)

Question 5.
the total of a number t and 11

Answer: t + 11

Question 6.
100 decreased by a number k.

Answer: 100 – k

Try It

Question 7.
Your friend has 5 more than twice as many game tokens as you. Let t be the number of game tokens you have. Write an expression for the number of game tokens your friend has.

Answer: 5+2t
given
Your friend has 5 more than twice as many game tokens as you
5+2t because 5 more then is adding 5 and twice as many is doubling or multiplying by 2

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

WRITING EXPRESSIONS
Write the phrase as an expression.

Question 8.
the sum of 7 and 11

Answer: 7 + 11 we get 18

Question 9.
5 subtracted from 9

Answer: 9 – 5 we get 4

Question 10.
DIFFERENT WORDS, SAME QUESTION
Which is different? Write “both” expressions.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 34

Answer: 12 more than x and x increased by 12 has different words but the question is same.

Question 11.
PRECISION
Your friend says that the phrases below have the same meaning. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 35

Answer: No your friend is incorrect.
“the difference of number x and 12” is x – 12
“the diffrence of 12 and number x” is 12 – x

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 12.
A company rents paddleboards by charging a rental fee plus an hourly rate. Write an expression that represents the cost (in dollars) of renting a paddleboard for h hours. How much does an eight-hour rental cost?
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 36

Answer:
The independent variable is that whose values do not take into account the values of other variables.
That is the time in hours for this item. Then, for the dependent variable, the answer would be the cost of renting.
The value of the dependent variable is based on the changes done in the values of the independent variable.

Question 13.
DIG DEEPER!
A county fair charges an entry fee of $7 and $0.75 for each ride token. You have $15. Write an expression that represents the amount (in dollars) you have left after entering the fair and purchasing n tokens. How many tokens can you purchase? How much money do you have left after purchasing 6 tokens?
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 37

Answer:
Given,
A county fair charges an entry fee of $7 and $0.75 for each ride token.
You have $15.
15.00 – 7.00 = 8.00
6 x .75 = 4.50
8.00- 4.5 = 3.50

Writing Expressions Homework & Practice 5.2

Review & Refresh

Identify the terms, coefficients, and constants in the expression.

Question 1.
4f + 8

Answer:
8 – constant
4 – coefficient
f- variable or term
A term without a variable is called a constant.
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Question 2.
\(\frac{4}{5}\) + 3s + 2

Answer:
\(\frac{4}{5}\) and 2 – constant
3 – coefficient
s- variable or term
A term without a variable is called a constant.
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Question 3.
9h2 + \(\frac{8}{9}\)p + 1

Answer:
1 – constant
2 – exponent
9 and \(\frac{8}{9}\)  – coefficient
h and p – variable or term
An exponent is a number or letter written above and to the right of a mathematical expression called the base
The exponent tells us how many times the base is used as a factor
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Copy and complete the statement.

Question 4.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 38

Answer: 7.5 gallon per hour

Explanation:
Convert from cup per minute to gallon per hour
1 Cup per minute = 3.75 gallon per hour
2 cup per minute = 2 × 3.75 = 7.5 gallon per hour

Question 5.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 39

Answer: 2362.2 feet per minute

Explanation:
Convert from meter per second to feet per minute
1 meter per second = 196.85 feet per minute
12 minute per second = 2362.2 feet per minute

Question 6.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 40

Answer: 4898.79 kg per hour

Explanation:
Convert from lb per second to kg per hour
1 lb per second = 1632.93 kg per hour
3 lb per second = 4898.79 kg per hour

Divide. Write the answer in simplest form.

Question 7.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 41

Answer:  \(\frac{8}{9}\) or 0.533

Explanation:
by dividing \(\frac{1}{2}\) with \(\frac{5}{8}\)
we get \(\frac{1}{2}\) ÷ \(\frac{5}{8}\)
\(\frac{8}{9}\)
0.533

Question 8.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 42

Answer: \(\frac{4}{9}\)

Explanation:
by dividing \(\frac{1}{3}\) with \(\frac{3}{4}\)
we get \(\frac{1}{3}\) ÷ \(\frac{3}{4}\)
\(\frac{4}{9}\)
0.44

Question 9.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 43

Answer: \(\frac{2}{15}\) or 0.133

Explanation:
by dividing \(\frac{2}{5}\) with \(\frac{3}{1}\)
we get \(\frac{2}{5}\) ÷ \(\frac{3}{1}\)
\(\frac{2}{15}\)
0.133

Question 10.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 44

Answer: \(\frac{7}{2}\) or 3.5

Explanation:
by dividing \(\frac{3}{1}\) with \(\frac{6}{7}\)
we get \(\frac{3}{1}\) ÷ \(\frac{6}{7}\)
\(\frac{7}{2}\)
3.5

Concepts, Skills, & Problem Solving

STRUCTURE
The expression represents the amount of change you receive after buying n sandwiches. Explain what each part of the expression represents. (See Exploration 1, p. 209.)

Question 11.
10 – 5.25n

Answer:
Let us assume that you buy 1 sandwich.
10 – 5.25(1)
= 10 – 5.25
= 4.75

Question 12.
20 – 4.95n

Answer:
Let us assume that you buy 2 sandwiches.
20 – 4.95(2)
20 – 9.9 = 10.1

Question 3.
100 – 6.75n

Answer:
Let us assume that you buy 6 sandwiches.
100 – 6.75(6)
= 100 – 40.50
= 59.50

WRITING EXPRESSIONS
Write the phrase as an expression.

Question 14.
5 less than 8

Answer: 5 < 8

Explanation:
The phrase “less than” indicates < symbol.
So, the expression is 5 < 8

Question 15.
the product of 3 and 12

Answer: 3 × 12

Explanation:
The phrase “product” indicates multiplication.
So, the expression is 3 × 12

Question 16.
28 divided by 7

Answer: 28 ÷ 7

Explanation:
The phrase “divided by” indicates ÷ symbol
So, the expression is 28 ÷ 7

Question 17.
the total of 6 and 10

Answer: 6 + 10 = 16

Explanation:
The phrase “total” indicates ‘+’ symbol.
So, the expression is 6 + 10 = 16

Question 18.
3 fewer than 18

Answer: 18 – 3

Explanation:
The phrase fewer than indicates ‘subtraction’
So, the expression is 18 – 3

Question 19.
17 added to 15

Answer: 17 + 15

Explanation:
Given 17 added to 15
so add 17 with 15

Question 20.
13 subtracted from a number x

Answer: 13 – x

Explanation:
The number represents the variable x
So here we have to subtract number x from 13
That gives the expression 13 – x

Question 21.
5 times a number d

Answer: 5d

Explanation:
The number represents variable d.
The word “times” represents ×
So the expression is 5d

Question 22.
the quotient of 18 and a number a

Answer: 18 ÷ a

Explanation:
a represents the number
So, the expression would be 18 ÷ a

Question 23.
the difference of a number s and 6

Answer: s – 6

Question 24.
7 increased by a number w

Answer: 7 + w

Question 25.
a number t cubed

Answer: t³

YOU BE THE TEACHER
Your friend writes the phrase as an expression. Is your friend correct? Explain your reasoning.

Question 26.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 45

Answer: incorrect

Explanation:
Given,
The quotient of 8 and a number is y is 8 ÷ y
By this, we can say that your friend is incorrect.

Question 27.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 46

Answer: correct

Explanation:
Given,
16 decreased by a number x
It means we have to subtract x from 16.
Thus the expression is 16 – x.

Question 28.
NUMBER SENSE
Five friends share the cost of a dinner equally.
a. Write an expression that represents the cost (in dollars) per person.

Answer: 600 ÷ p

b. Make up a reasonable total cost and test your expression.

Answer:
The total cost of dinner is $600.
Now divide the cost per person.
There are 5 friends.
The expression is 600 ÷ p
p = 5
600 ÷ 5 = 12
The cost per person is $120

Question 29.
MODELING REAL LIFE
A biologist analyzes 15 bacteria samples each day.
a. Copy and complete the table.
b. Write an expression that represents the total number of samples analyzed after n days.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 47

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-5-Algebraic-Expressions-and-Properties-47
The expression that represents the total number of samples analyzed after n days is 15n

Question 30.
PROBLEM SOLVING
To rent a moving truck for the day, it costs $33 plus $1 for each mile driven.
a. Write an expression that represents the cost (in dollars) to rent the truck.

Answer: 33 + d
b. You drive the truck 300 miles. How much do you pay?

Answer:
Use the expression 33 + d to find how much do you pay.
33 + 300 = 333
Thus you pay $333 for 300 miles.

WRITING PHRASES
Give two ways to write the expression as a phrase.

Question 31.
n + 6

Answer:
According to the given details, we can write the expression in two ways
a number n more than 6
the sum of a number n and 6

Question 32.
4w

Answer:
According to the given details, we can write the expression in two ways
a number w is four times
The product of 4 and a number w

Question 33.
15 – b

Answer:
According to the given details we can write the expression in two ways.
15 decreased by a number d
The difference of 15 and a number b

Question 34.
14 – 3z

Answer:
According to the given details, we can write the expression in two ways.
The product of 3 and z subtracted from 14.
The difference of 14 and product of 3 and a number z.

EVALUATING EXPRESSIONS
Write the phrase as an expression. Then evaluate the expression when x = 5 and y = 20.

Question 35.
3 less than the quotient of a number y and 4

Answer:
given 3 less than the quotient of a number y and 4
The quotient of a number y and 4
Now we are given that 3 less than the quotient of a number y and 4
So, \(\frac{y}{4}\) – 3
Hence the expression becomes \(\frac{y}{4}\) – 3

Question 36.
the sum of a number x and 4, all divided by 3

Answer:
given
a sum, x and 4, so that becomes x + 4
All divided by denotes that we need to do the addition before division, so we need to put it in parentheses:
(x + 4)
And it is divided by 3.
So you can write the answer down in a few different ways:
(x + 4) ÷ 3
(x + 4) / 3
\(\frac{x + 4}{3}\)

Question 37.
6 more than the product of 8 and a number x

Answer:
A Product is an answer to a multiplication.
The product of 8 and a number is 8×n=8n
6 more than that means to add on 6
So the expression “6 more than the product of 8 and a number” is
8n+6

Question 38.
the quotient of 40 and the difference of a number y and 16

Answer:
as we know that
The expression ” the quotient of and the difference of a number y and ” is equivalent to the algebraic equation
\(\frac{40}{y-16}\)
For  y = 20  given problem
substitute the value of y in the equation
\(\frac{40}{20-16}\)  = \(\frac{40}{4}\) = 10

Question 39.
MODELING REAL LIFE
It costs $3 to bowl a game and $2 for shoe rental.
a. Write an expression that represents the total cost (in dollars) of g games.
b. Use your expression to find the total cost of 8 games.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 48

Answer:
a. If shoes are for all games in row, P=2+3g
b. If not, P=5g, with g the number of games
Explanation:
The price depends if shoes are included for the whole games, you just have to pay them once and then you have to pay g games, so P(g)=2+3g
Else, you have to pay for g shoes and g bowl party, so P(g)=2g+3g⇔P(g)=5g

Question 40.
MODELING REAL LIFE
Florida has 8 less than 5 times the number of counties in Arizona. Georgia has 25 more than twice the number of counties in Florida.
a. Write an expression that represents the number of counties in Florida.
b. Write an expression that represents the number of counties in Georgia.
c. Arizona has 15 counties. How many do Florida and Georgia have?

Answer:
Number of countries in Florida=5×a-8
Number of countries in Georgia=2×f+25
Explanation:
Given that the number of countries in Florida is denoted by f,
The number of countries in Arizona is given by a and the number of countries in Georgia is given by g.
as Florida has 8 less than five times the number of countries in Arizona
So f=5×a-8
Georgia has 25 more than twice the number of countries in Florida
So g=2×f+25
Now it is given that Arizona has 15 countries i.e. a=15
So f=5×15-8
f=67
g=2×67+25
g=159
Hence, the number of countries in Florida is:67
number of countries in Georgia is:159

Question 41.
PATTERNS
There are 140 people in a singing competition. The graph shows the results for the first five rounds.
a. Write an expression that represents the number of people after each round.
b. Assuming this pattern continues, how many people compete in the ninth round? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 49

Answer:
a) The expression is y = 140 – 15 x
b ) There are 5 participants in the 9th round

Step-by-step explanation:
Step 1 :
From the graph we can see that,
Number of people in the singing competition = 140
Number of people in the first round = 125
Number of people in the second round = 110
Number of people in the third round = 95
Number of people in the fourth round = 80
Number of people in the fifth round = 65
Step 2 :
From the given data, We can get that the difference between the number of people participating in each round is 15 less than the previous round .
The first round has 15 people less than the total number of 140
Let x represent the number of the round and y represent the number of people participating in each round .
Then the expression to represent this would be
y = 140 – 15 x
Step 3 :
To find the number of participants in the 9th round given the same pattern continues.
For the 9th round x = 9, as x represents the number of the round
Substituting this in the equation obtained in step 2, we get
y = 140 -15 (9) = 140 – 135 = 5
There are 5 participants in the 9th round

Question 42.
NUMBER SENSE
The difference between two numbers is 8. The lesser number is a. Write an expression that represents the greater number.

Answer:
Given The difference between the two numbers is 8. The lesser number is a.
b-a=8
b is the greater number; b>a
So, b=a+8

Question 43.
NUMBER SENSE
One number is four times another. The greater number is x. Write an expression that represents the lesser number

Answer: y = x/4

Let us assume the smaller number y.
We know that four times the smaller number is equal to x. So, four times y is equal to x.
Turning this into an expression, you get 4y = x.
This means, to get y on its own, we need to divide by 4 on both sides, giving us the answer of y = x/4

Lesson 5.3 Properties of Addition and Multiplication

EXPLORATION 1

Identifying Equivalent Expressions
Work with a partner.
a. Choose four values for a variable x. Then evaluate each expression for each value of x. Are any of the expressions equivalent? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 49.1

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-5-Algebraic-Expressions-and-Properties-49.1
The equation x + 8, 4 + 4 + x, x + 4 + 4 are equal.
The equation 16x, 4.(x.4), (4.x).x are equal.
b. You have used the following properties in a previous course. Use the examples to explain the meaning of each property.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 50
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 51
Are these properties true for algebraic expressions? Explain your reasoning.

Answer:
Commutative property of addition: The commutative property of addition says that changing the order of addends does not change the sum. Here’s an example: 5 + 2 = 7 or 2 + 5 = 7
Commutative property of multiplication: The commutative property of multiplication tells us that we can multiply a string of numbers in any order. Basically: 2 x 3 x 5 will create the same answer as 3 x 5 x 2, or 2 x 5 x 3
Associative property of addition: The associative property of addition says that changing the grouping of the addends does not change the sum.
Example: 2 + (7 + 5) = (2 + 7) + 5
Associative Property of Multiplication: The associative property is a math rule that says that the way in which factors are grouped in a multiplication problem does not change the product.
Example: 2 × (7 × 5) = (2 × 7) × 5

5.3 Lesson

Expressions that result in the same number for any value of each variable are equivalent expressions. You can use the Commutative and Associative Properties to write equivalent expressions.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 52

Try It

Simplify the expression. Explain each step.

Question 1.
10 + (a + 9)

Answer: a+19
Step 1: Eliminate redundant parentheses
10+(a+9)=10+a+9
Step 2:Add the numbers
Step 3: Rearrange terms
a+19

Question 2.
(c + \(\frac{2}{3}\)) + \(\frac{1}{2}\)

Answer: c + \(\frac{7}{6}\)
(c + \(\frac{2}{3}\)) + \(\frac{1}{2}\)
c+ \(\frac{7}{6}\)

Question 3.
5(4n)

Answer: 20n
Multiply 5 with 4n
we get 20n

Simplify the expression. Explain each step.

Question 4.
12.b.0

Answer: 0
step 1
12 multiply with a number b
and then multiply with 0
we get 0

Question 5.
1.m.24

Answer: 24m
step 1
1 multiply with a number m
and then multiply with 24
we get 24m

Question 6.
(t + 15) + 0

Answer: t + 15
step 1
Add t + 15 with 0
and then
we get t + 15

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

USING PROPERTIES
Simplify the expression. Explain each step.

Question 7.
(7 + c) + 4

Answer:
Step 1: Eliminate redundant parentheses
4+(7+c)= 4+7+c
Step 2:Add the numbers
Step 3: Rearrange terms
11 + c

Question 8.
4(b.6)

Answer: 24b
Step 1: Eliminate redundant parentheses
4 × b× 6 = 4 × 6× b
multiply 4 with 6 we get 24 and then with b
so we get 24b

Question 9.
0.b.9

Answer: 0
Step 1: Eliminate redundant parentheses
0 × b× 9 = 0 × 9× b
multiply 0 with 9 we get 0 and then with b
so we get 0

Question 10.
WRITING
Explain what it means for expressions to be equivalent. Then give an example of equivalent expressions.

Answer:
equivalent expressions are algebraic expressions that, although they look different, turn out to really be the same.

Example:
Let’s consider this algebraic expression: 2(x^2 + x). If we substitute 1 for the variable, the expression equals 4. But what about the expression
2x^2 + 2x? If, again, we substitute 1 for the variable x, we still get 4. How does this happen?
What we really did was simplify the original expression by distributing the 2 into the part in parentheses. So we really haven’t changed the expression at all – all we’ve done is rewrite it in a different form.
Because these two expressions are really the same, no matter what number we substitute for x, the results will always be identical. If we use 0, both expressions come out to 0. If we use 10, both expressions come out to 220. If we use 100, both expressions come out to 20,200. We get the same result no matter how large or small the number we use for x.

Question 11.
OPEN-ENDED
Write an algebraic expression that can be simplified using the Associative Property of Multiplication and the Multiplication Property of One.

Answer:
Let us consider an expression 4 × 30.
We can simplify the expression by using the Associative Property of Multiplication
4 × 30 = 4 × (5 × 6)
(4 × 5) × 6 = 4 × (5 × 6)
20 × 6 = 4 × 30
120 = 120

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 12.
You and five friends form a team for an outdoor adventure race. Your team needs to raise money to pay for $130 of travel fees, x dollars for each team member’s entry fee, and $85.50 for food. Use an algebraic expression to find the total amount your team needs to raise when the entry fee is $25.50 per person.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 53

Answer: (130+85)+25.5p=total amount

Step-by-step explanation:
130+85 is how much you’d need just for the fees and food, for the entry fee its per person
so it would be $25.5 for every person that is part of your team.

Question 13.
You have $50 and a $15 gift card to spend online. You purchase a pair of headphones for $34.99 and 8 songs for x dollars each. Use an algebraic expression to find the amount you have left when each song costs $1.10.

Answer: $21.21
Given
You have $50 and a $15 gift card to spend online
purchase a pair of headphones for $34.99 and 8 songs for x dollars each.
Use an algebraic expression to find the amount you have left when each song costs $1.10.
so
50 + 15 -34.99 – 8x
65 – 34.99 – 8(1.10)
30.01 – 8.8
$21.21

Properties of Addition and Multiplication Homework & Practice 5.3

Review & Refresh

Write the phrase as an expression.

Question 1.
10 added to a number p

Answer:  p + 10
First consider the expression for
10 added to a number p
we get p + 10
An algebraic expression is a mathematical phrase that contains a combination of numbers, variables and operational symbols.
A variable is a letter that can represent one or more numbers.

Question 2.
the product of 6 and a number m

Answer: 6m
First consider the expression for
the product of 6 and a number m
so multiply 6 with m
we get 6m
An algebraic expression is a mathematical phrase that contains a combination of numbers, variables and operational symbols.
A variable is a letter that can represent one or more numbers.

Question 3.
the quotient of a number b and 15

Answer: \(\frac{b}{15}\)
First consider the expression for
the quotient of a number b and 15
so divide b by 15
we get \(\frac{b}{15}\)
An algebraic expression is a mathematical phrase that contains a combination of numbers, variables and operational symbols.
A variable is a letter that can represent one or more numbers.

Question 4.
7 fewer than a number s

Answer:  s – 7
First consider the expression for
7 fewer than a number s
so subtract a number s with 7
we get s – 7
An algebraic expression is a mathematical phrase that contains a combination of numbers, variables and operational symbols.
A variable is a letter that can represent one or more numbers.

Write the prime factorization of the number.

Question 5.
36

Answer:
The number 36 can be written as a product of primes as
36 = 2² x 3².
The expression 2² x 3² is said to be the prime factorization of 36

Question 6.
144

Answer:
The prime factor of the 144 is 24 x 32.
144 = 2 x 2 x 2 x 2 x 3 x 3

Question 7.
147

Answer:
Factors of 147: 1, 3, 7, 21, 49, 147.
Prime factorization: 147 = 3 x 7 x 7,
which can also be written
147 = 3 x (7²)

Question 8.
205

Answer:
the prime factors of the number 205.
If we put all of it together we have the factors 5 x 41 = 205.
it can also be written in exponential form as 5¹ x 41¹.

Evaluate the expression.

Question 9.
8.092 + 3.5

Answer: 11.592
Given expression 8.092 + 3.5
now add 8.092 with 3.5 we get
11.592

Question 10.
16.78 – 12.237

Answer: 4.543
Given expression 16.78 – 12.237
now subtract  16.75 with 12.237  we get
4.543

Question 11.
9.17 + 1.83 + 2.641

Answer:  13.641
Given expression 9.17 + 1.83 + 2.641
now add 9.17  with 1.83  we get  11 then add with 2.641
we get 13.641

Question 12.
8.43 – 6.218 + 4.2

Answer:
Given expression 8.43 – 6.218 + 4.2
now subtract  8.43 with 6.218  we get 2.212  then add with 4.2
we get  6.412

Represent the ratio relationship using a graph.

Question 13.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 54

Answer:
Big Ideas Math Grade 6 Chapter 5 Algebraic Expressions and Properties Answers img_1

Question 14.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 55

Answer:
Big Ideas Math Grade 6 Chapter 5 Algebraic Expressions and Properties Answers img_2

Concepts, Skills, & Problem Solving
MATCHING
Match the expression with an equivalent expression. (See Exploration 1, p. 215.)

Question 15.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 56

Answer: B

Explanation:
3 + 3 + y is equivalent to y + 3 + 3
So, the correct answer is option B

Question 16.
Big Ideas Math Answers 6th Grade Chapter 5 Algebraic Expressions and Properties 57

Answer: C

Explanation:
(y.y).3 = y(3 . y)
So, the correct answer is option C.

Question 17.
Big Ideas Math Answers 6th Grade Chapter 5 Algebraic Expressions and Properties 58

Answer: A

Explanation:
3 . 1 . y = y . 3
So, the correct answer is option A.

Question 18.
Big Ideas Math Answers 6th Grade Chapter 5 Algebraic Expressions and Properties 59

Answer: D

Explanation:
(3 + 0) + (y + y)
(3 + y) + y
So, the correct answer is option D.

IDENTIFYING PROPERTIES
Tell which property the statement illustrates.

Question 19.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 60

Answer: commutative property of multiplication

Explanation:
The commutative property of multiplication tells us that we can multiply a string of numbers in any order.
5 . p = p . 5

Question 20.
2 + (12 + r) = (2 + 12) + r

Answer: associative property of addition

Explanation:
Associative property of addition: Changing the grouping of addends does not change the sum.
2 + (12 + r) = (2 + 12) + r

Question 21.
4 . (x . 10) = (4 . x) . 10

Answer: associative property of multiplication

Explanation:
The associative property is a math rule that says that the way in which factors are grouped in a multiplication problem does not change the product.

Question 22.
x + 7.5 = 7.5 + x

Answer: commutative property of addition

Explanation:
The commutative property of addition says that changing the order of addends does not change the sum.

Question 23.
(c + 2) + 0 = c + 2

Answer: Additive Identity

Explanation:
Additive identity is a number, which when added to any number, gives the sum as the number itself. It means that additive identity is “0” as adding 0 to any number, gives the sum as the number itself.

Question 24.
a . 1 = a

Answer: Multiplicative Identity

Explanation:
According to the multiplicative identity property of 1, any number multiplied by 1, gives the same result as the number itself. It is also called the Identity property of multiplication because the identity of the number remains the same.

Question 25.
YOU BE THE TEACHER
Your friend states the property that the statement illustrates. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 61

Answer: correct

Explanation:
Associative property of addition: Changing the grouping of addends does not change the sum.
(7 + x) + 3 = (x + 7) + 3
Thus we can say that your friend is correct.

USING PROPERTIES
Simplify the expression. Explain each step.

Question 26.
6 + (5 + x)

Answer: 11 + x
Step 1: Eliminate redundant parentheses
6 + (5 + x) = 6+5+x
Step 2:Add the numbers
Step 3: Rearrange terms
11 + x

Question 27.
(14 + y) + 3

Answer: 17 + y
Step 1: Eliminate redundant parentheses
(14 + y) + 3 = 14 + 3 + y
Step 2:Add the numbers
Step 3: Rearrange terms
17 + y

Question 28.
6(2b)

Answer: 12b
Step 1: Eliminate redundant parentheses
6(2b) = 6 × 2 ×b
Step 2:multiply  the numbers
Step 3: Rearrange terms
we get 12b

Question 29.
7(9w)

Answer: 63w
Step 1: Eliminate redundant parentheses
7(9w) = 7 ×  ×w
Step 2:multiply  the numbers
Step 3: Rearrange terms
we get 63w

Question 30.
3.2 + (x + 5.1)

Answer: 8.3 + x
Step 1: Eliminate redundant parentheses
3.2 + (x + 5.1) = 3.2 + 5.1 + x
Step 2:Add the numbers
Step 3: Rearrange terms
8.3 + x

Question 31.
(0 + a) + 8

Answer: 8 + a
Step 1: Eliminate redundant parentheses
(0 + a) + 8 = 0 + 8 +a
Step 2:Add the numbers
Step 3: Rearrange terms
8 + a

Question 32.
9 . c . 4

Answer: 36c
Step 1: Eliminate redundant parentheses
9 . c . 4 = 9 . 4 . c
Step 2: multiply the numbers
Step 3: Rearrange terms
36c

Question 33.
(18.6 . d) . 1

Answer: 18.6 d
Step 1: Eliminate redundant parentheses
(18.6 . d) . 1 = (18.6 . 1) d
Step 2: multiply the numbers
Step 3: Rearrange terms
18.6 d

Question 34.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 62

Answer:
Step 1: Eliminate redundant parentheses
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 62 = 3k + (4\(\frac{1}{5}\) + 8\(\frac{3}{5}\))
Step 2: add the numbers
Step 3: Rearrange terms
3k + 12\(\frac{4}{5}\)

Question 35.
(2.4 + 4n) + 9

Answer:
Step 1: Eliminate redundant parentheses
(2.4 + 4n) + 9 = 2.4 + 9 + 4n
Step 2:Add the numbers
Step 3: Rearrange terms
11.4 + 4n

Question 36.
(3s) . 8

Answer: 24s
Step 1: Eliminate redundant parentheses
(3s) . 8 = (3 . 8) s
Step 2: multiply the numbers
Step 3: Rearrange terms
24s

Question 37.
z . 0 . 12

Answer: 0
Step 1: Eliminate redundant parentheses
z . 0 . 12 = z × 0 × 12
Step 2: multiply the numbers
Step 3: Rearrange terms
0

Question 38.
GEOMETRY
The expression 12 + x + 4 represents the perimeter of a triangle. Simplify the expression.

Answer: x+16

Explanation:
your simplifying so you only combine like terms.
12 and 4 dont have any variables following them so you add the two.
12+4=16
there is only 1 x in the expression so just add it on.
16+x

Question 39.
PRECISION
A case of scout cookies has 10 cartons. A carton has 12 boxes. The amount you earn on a whole case is 10(12x) dollars.
a. What does x represent?
b. Simplify the expression.

Answer:
Given
A case of scout cookies has 10 cartons. A carton has 12 boxes
A. x represents the money made per box
B. 10(12x) = 120x

Question 40.
MODELING REAL LIFE
A government estimates the cost to design new radar technology over a period of m months. The government estimates $840,000 for equipment, $15,000 for software,and $40,000 per month for wages. Use an algebraic expression to find the total cost the government estimates when the project takes 16 months to complete.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 63

Answer: $1,495,000.

Explanation:
The cost that is affected by the number of months mm is the wages while the first two costs are fixed. Hence, you multiply $40,000 by the number of months so that
840,000+15,000+40,000m
840,000+15,000+40,000m
For m=16m=16 (16 months),
=840,000+15,000+40,000(16)
=840,000+15,000+40,000(16)
=840,000+15,000+640,000
=840,000+15,000+640,000
=1,495,000
=1,495,000
So, the total cost is $1,495,000.

WRITING EXPRESSIONS
Write the phrase as an expression. Then simplify the expression.

Question 41.
7 plus the sum of a number x and 5

Answer:
7+(x+5)  sum means addition and it says x and 5. then it says 7 plus that so you put 7.
In reality, you wouldn’t need parenthesis,
but if you want it to match the phrase, you can put it in parenthesis.

Question 42.
the product of 8 and a number y, multiplied by 9

Answer: 8y x 9 ⇒ 72y
Given
the product of 8 and a number y ⇒ 8 × y = 8y
and also given multiplied by 9 ⇒ 8y × 9
so we get 72 y

USING PROPERTIES
Copy and complete the statement using the specified property.

Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 64

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-5-Algebraic-Expressions-and-Properties-64

Question 48.
GEOMETRY
Five identical triangles form the trapezoid shown.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 65
a. What is the perimeter of the trapezoid?

Answer:
The perimeter of the trapezoid is a + b + c + d
P = sum of four sides
P = 3x + 7 + 7 + 2x
P = 5x + 14
Thus the perimeter of the trapezoid is 5x + 14
b. How can you use some or all of the triangles to form a new trapezoid with a perimeter of 3x +14? Explain your reasoning.

Answer:
You can reduce the size of the triangle to form a new trapezoid with a perimeter of 3x +14
The perimeter of the trapezoid is a + b + c + d
P = sum of four sides
P = 2x + 7 + 7 + 1x
P = 3x + 14

Question 49.
DIG DEEPER!
You and a friend sell hats at a fair booth. You sell 16 hats on the first shift and 21 hats on the third shift. Your friend sells x hats on the second shift.
a. The expression 37(14) + 10x represents the amount (in dollars) that you both earn. How can you tell that your friend is selling the hats for a lower price?
b. You earn more money than your friend. What can you say about the value of x?

Answer:
37+x

Explanation:
1st shift: 16 hats were sold
2nd shift: x hats were sold
3rd shift: 21 hats were sold
The total is
16+x+21
Simplifying
37+x
another step
16 + 21 + x = y
y = the number of hats sold
16 = hats sold on first shift
21 = hats on the third shift
x = hats on the second shift

Lesson 5.4 The Distributive Property

EXPLORATION 1
Using Models to Simplify Expressions
Work with a partner.
a. Use the models to simplify the expressions. Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 66

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-5-Algebraic-Expressions-and-Properties-66
b. In part(a), check that the original expressions are equivalent to the simplified expressions.
c. You used the Distributive Property in a previous course. Use the example to explain the meaning of the property.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 67
Distributive Property: 6(20 + 3) = 6(20) + 6(3)
Is this property true for algebraic expressions? Explain your reasoning.

Answer:
The distributive property of multiplication over addition can be used when you multiply a number by a sum.
6(20 + 3) = 6(20) + 6(3)
6(23) = 120 + 18
138 = 138
Yes this property true for algebraic expressions

Try It

Use the Distributive Property to simplify the expression.

Question 1.
7(a + 2)

Answer: 7(a) + 7(2)

Explanation:
Given,
The distributive property of multiplication over addition can be used when you multiply a number by a sum.
7(a + 2) = (7 × a) + (7 × 2)
7a + 14
Thus 7(a + 2) = 7a + 14

Question 2.
3(d – 11)

Answer: 3(d) – 3(11)

Explanation:
Given,
The distributive property of multiplication over addition can be used when you multiply a number by a sum.
3(d – 11) = (3 × d) – (3 × 11)
3d – 33
Thus 3(d – 11) = 3d – 33

Question 3.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 68

Answer: 12a + 12\(\frac{2}{3}\)b

Explanation:
Given,
The distributive property of multiplication over addition can be used when you multiply a number by a sum.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 68= (12 × a) + (12 × \(\frac{2}{3}\)b)
= 12a + 12\(\frac{2}{3}\)b
Thus Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 68= 12a + 12\(\frac{2}{3}\)b

Question 4.
7(2 + 6 – 4d)

Answer: 7(2) + 7(6) – 7(4d)

Explanation:
Given,
The distributive property of multiplication over addition can be used when you multiply a number by a sum.
7(2 + 6 – 4d) = 7(8 – 4d)
56 – 28d
Thus 7(2 + 6 – 4d) = 56 – 28d

Simplify the expression.

Question 5.
8 + 3z – z

Answer: 8 + 2z

Explanation:
Combine the like terms and simplify the expression.
8 + 3z – z
8 + z(3 – 1)
8 + 2z
So, 8 + 3z – z = 8 + 2z

Question 6.
3(b + 5) + b + 2

Answer: 4b + 17

Explanation:
Combine the like terms and simplify the expression.
3(b + 5) + b + 2
= 3b + 15 + b + 2
= 3b + b + 15 + 2
= 4b + 17
Thus 3(b + 5) + b + 2 = 4b + 17

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
WRITING
One meaning of the word distribute is to give something to each member of a group. How can this help you remember the Distributive Property?

Answer:
You must distribute or give the number outside the parenthesis to all the numbers inside the parenthesis

SIMPLIFYING EXPRESSIONS
Use the Distributive Property to simplify the expression.

Question 8.
3(x + 10)

Answer: 3(x) + 3(10)

Explanation:
We can simplify the expression by using the Distributive Property
The distributive property of multiplication over addition can be used when you multiply a number by a sum.
3(x + 10) = (3 × x) + (3 × 10)
= 3x + 30
So, 3(x + 10) = 3x + 30

Question 9.
15(4n – 2)

Answer: 15(4n) – 15(2)

Explanation:
We can simplify the expression by using the Distributive Property
The distributive property of multiplication over addition can be used when you multiply a number by a sum.
15(4n – 2) = (15 × 4n) – (15 × 2)
= 60n – 30
So, 15(4n – 2) = 60n – 30

Question 10.
2w + 4 + 13w + 1

Answer:15w + 5 = 5(3w + 1)

Explanation:
We can simplify the expression by using the Distributive Property
The distributive property of multiplication over addition can be used when you multiply a number by a sum.
2w + 4 + 13w + 1
Combine the like terms
=2w + 13w + 4 + 1
=15w + 5
=5(3w + 1)

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 11.
You purchase a remote-controlled drone for d dollars. Your friend purchases a drone that costs $35 more than your drone. Your brother purchases a drone that costs three times as much as your friend’s drone. Write and simplify an expression that represents the cost (in dollars) of your brother’s drone.

Answer:
Brother = 105+3d

Explanation:
Given
Drone = d
Friend = $35 more expensive than yours.
Brother = 3 times as much as your friend
Required
Write an expression for the cost of your brother’s drone
I’ll solve by analyzing the question one sentence after the other
You: Drone = d

Your friend:
Friend = $35 more expensive than yours.
This means
Brother = 3 × Friend
Substitute 35 + d for Friend
Brother = 3 × (35+d)
Open bracket
Brother = 3×35+3×d
Brother=105+3d

Question 12.
Write and simplify an expression that represents the total cost (in dollars) of buying the items shown for each member of a baseball team.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 68.1

Answer: 9(10 + x)

Explanation:
Number of members in the baseball team = 9
Cost of the pant = $10
Cost of a belt = $x
The expression that represents the total cost (in dollars) of buying the items shown for each member of a baseball team is 9(10 + x)

Question 13.
DIG DEEPER!
One molecule of caffeine contains x oxygen atoms, twice as many nitrogen atoms as oxygen atoms, 4 more carbon atoms than nitrogen atoms, and 1.25 times as many hydrogen atoms as carbon atoms. Write and simplify an expression that represents the number of hydrogen atoms in one molecule of caffeine.

Answer:
Given,
One molecule of caffeine contains x oxygen atoms, twice as many nitrogen atoms as oxygen atoms, 4 more carbon atoms than nitrogen atoms, and 1.25 times as many hydrogen atoms as carbon atoms.
One molecule of caffeine contains x oxygen atoms = x
twice as many nitrogen atoms as oxygen atoms = 2 . x
4 more carbon atoms than nitrogen atoms = 4 . x
1.25 times as many hydrogen atoms as carbon atoms = 1.25 × x
x + 2x + 6x + 1.25(6x)

The Distributive Property Homework & Practice 5.4

Review & Refresh

Simplify the expression. Explain each step.

Question 1.
(s + 4) + 8

Answer: s + 12

Explanation:
According to the distributive property, multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together.
(s + 4) + 8
= s + 4 + 8
= s + 12

Question 2.
(12 + x) + 2

Answer: 14 + x

Explanation:
According to the distributive property, multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together.
12 + x + 2
x + 12 + 2
x + 14

Question 3.
3(4n)

Answer: 12n

Explanation:
According to the distributive property, multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together.
3 × 4n = 12n

You are given the difference of the numbers of boys and girls in a class and the ratio of boys to girls. How many boys and how many girls are in the class?

Question 4.
3 more boys; 5 for every 4

Answer:
Let the total number of boys in the class be x.
The ratio of boys and girls is 5 : 4
Number of boys = 5x
Number of girls = 4x
5x – 4x = 3
x = 3
Number of boys = 5(3) = 15
15 + 3 = 18
Number of girls = 4(3) = 12
To find the total number of boys and girls in the class we have to add them
18 + 12 = 30
Thus there are 30 boys and girls in the class.

Question 5.
8 more girls; 3 for every 2

Answer:
Let the total number of boys in the class be x.
The ratio of boys and girls is 3 : 2
Number of boys = 3x
Number of girls = 2x
3x – 2x = 8
x = 8
Number of boys = 3x = 3(8) = 24
Number of girls = 2x = 2(8) = 16
More number of girls = 16 + 8 = 24
To find the total number of boys and girls in the class we have to add them
24 + 24 = 48
Thus there are 48 boys and girls in the class.

Question 6.
4 more girls; 9 for every 13

Answer:
Let the total number of boys in the class be x.
The ratio of girls and boys is 13 : 9
Number of boys = 9x
Number of girls = 13x
13x – 9x = 4
4x = 4
x = 1
Number of boys = 9x = 9(1) = 9
Number of girls = 13x = 13(1) = 13
13 + 4 = 17
To find the total number of boys and girls in the class we have to add them
9 + 17 = 26
Thus there are 26 boys and girls in the class.

Question 7.
6 more boys; 7 for every 4

Answer:
Let the total number of boys in the class be x.
The ratio of girls and boys is 7 : 4
Number of boys = 7x
Number of girls = 4x
7x – 4x = 6
3x = 6
x = 2
Number of boys = 7(2) = 14
Number of girls = 4(2) = 8
15 + 6 = 21
To find the total number of boys and girls in the class we have to add them
21 + 8 = 29
Thus there are 29 boys and girls in the class.

Divide.

Question 8.
301 ÷ 7

Answer: 43
So divide 301 with 7
then we get 43

Question 9.
1722 ÷ 14

Answer: 123
So divide 1722 with 14
then we get 123

Question 10.
629 ÷ 12

Answer: 52.41
So divide 629 with 12
then we get 52.14

Question 11.
8068 ÷ 31

Answer: 260.25
So divide 8068 with 31
then we get 260.25

Concepts, Skills, & Problem Solving
USING MODELS
Use the model to simplify the expression. Explain your reasoning. (See Exploration 1, p. 221.)

Question 12.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 69

Answer: 5z + 30

Explanation:
The shape of the above figure is a rectangle.
Area of the rectangle = l × b
A = (z + 6) × 5
A = 5z + 30

Question 13.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 70

Answer: 6s

Explanation:
Combine the like terms
There are 4s in the first block
There are 2s in the second block
4s + 2s = 6s

SIMPLIFYING EXPRESSIONS
Use the Distributive Property to simplify the expression.

Question 14.
3(x + 4)

Answer: 3x + 12

Explanation:
We can use the distributive property to simplify the expression.
3(x + 4) = (3 × x) + (3 × 4)
= 3x + 12

Question 15.
10(b – 6)

Answer: 10b – 60

Explanation:
We can use the distributive property to simplify the expression.
10(b – 6) = (10 × b) – (10 × 6)
= 10b – 60

Question 16.
6(s – 9)

Answer: 6s – 54

Explanation:
We can use the distributive property to simplify the expression.
6(s – 9) = (6 × s) – (6 × 9)
= 6s – 54

Question 17.
7(8 + y )

Answer: 56 + 7y

Explanation:
We can use the distributive property to simplify the expression.
7(8 + y) = (7 × 8) + (7 × y)
= 56 + 7y

Question 18.
8(12 + a)

Answer: 96 + 8a

Explanation:
We can use the distributive property to simplify the expression.
8(12 + a) = (8 × 12) + (8 × a)
= 96 + 8a

Question 19.
9(2n + 1)

Answer: 18n + 9

Explanation:
We can use the distributive property to simplify the expression.
9(2n + 1) = (9 × 2n) + (9 × 1)
= 18n + 9

Question 20.
12(6 – k)

Answer: 72 – 12k

Explanation:
We can use the distributive property to simplify the expression.
12(6 – k) = (12 × 6) – (12 × k)
= 72 – 12k

Question 21.
18(5 – 3w)

Answer: 90 – 54w

Explanation:
We can use the distributive property to simplify the expression.
18(5 – 3w) = (18 × 5) – (18 × 3w)
= 90 – 54w

Question 22.
9(3 + c + 4)

Answer: 63 + 9c

Explanation:
We can use the distributive property to simplify the expression.
9(3 + c + 4) = 9(7 + c)
(9 × 7) + ( 9 × c)
63 + 9c

Question 23.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 71

Answer: 3 + x/4

Explanation:
We can use the distributive property to simplify the expression.
1/4(8 + x + 4) = 1/4(12 + x)
= 3 + x/4

Question 24.
8(5g + 5 – 2)

Answer: 40g + 24

Explanation:
We can use the distributive property to simplify the expression.
8(5g + 5 – 2) = 8(5g) + 8(5) – 8(2)
40g + 40 – 16
40g + 24

Question 25.
6(10 + z + 3)

Answer: 78 + 6z

Explanation:
We can use the distributive property to simplify the expression.
6(10 + z + 3) = 6(10) + 6(z) + 6(3)
= 60 + 6z + 18
= 78 + 6z

Question 26.
4(x + y)

Answer: 4x + 4y

Explanation:
We can use the distributive property to simplify the expression.
4(x + y) = (4 × x) + (4 × y)
= 4x + 4y

Question 27.
25(x – y)

Answer: 25x – 25y

Explanation:
We can use the distributive property to simplify the expression.
25(x – y) = (25 × x) – (25 × y)
= 25x – 25y

Question 28.
7(p + q + 9)

Answer: 7p + 7q + 63

Explanation:
We can use the distributive property to simplify the expression.
7(p + q + 9) = (7 × p) + (7 × q) + (7 × 9)
= 7p + 7q + 63

Question 29.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 72

Answer: n + 3m + 2

Explanation:
We can use the distributive property to simplify the expression.
1/2 (2n + 4 + 6m) = 1/2 (2n) + 1/2 (4) + 1/2 (6m)
= n + 2 + 3m

MATCHING
Match the expression with an equivalent expression.

Question 30.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 73

Answer: C

Explanation:
Given,
6(n + 4) = 6n + 24
Now take 3 as common factor
3(2n + 8)
Thus the equivalent expression is option C.

Question 31.
Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties 74

Answer: A

Explanation:
Given,
2(3n + 9) = 6n + 18
Now take 3 as common factor
6n + 18 = 3(2n + 6)
Thus the equivalent expression is option A.

Question 32.
Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties 75

Answer: D

Explanation:
Given,
6(n + 2) = 6n + 12
Thus the equivalent expression is option D.

Question 33.
Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties 76

Answer: B

Explanation:
Given,
3(2n + 3) = 6n + 9
Thus the equivalent expression is option B.

Question 34.
STRUCTURE
Each day, you run on a treadmill for r minutes and lift weights for 15 minutes. Which expressions can you use to find how many minutes of exercise you do in 5 days? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties 77

Answer: 5(r + 15)

Explanation:
Each day, you run on a treadmill for r minutes and lift weights for 15 minutes.
They are both expressions with r as the variable.
They are not equivalent expressions.
They are equivalent expressions.
Both expressions contain the terms 5, r, and 15.
5(15 + r) = 5(15) + 5(r)

Question 35.
MODELING REAL LIFE
A cheetah can run 103 feet per second. A zebra can run x feet per second. Write and simplify an expression that represents how many feet farther the cheetah can run in 10 seconds.
Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties 78

Answer:
Given,
A cheetah can run 103 feet per second. A zebra can run x feet per second.
We are given that the rate of cheetah Rc is:
Rc = 103 ft / sec
And the rate of zebra Rz is:
Rz = x ft / sec
We are to find the distance between the two after 10 seconds, that is:
distance = (103 – x) 10
distance = 1030 – 10 x

COMBINING LIKE TERMS
Simplify the expression.

Question 36.
6(x + 4) + 1

Answer: 6x + 25

Explanation:
Given the expression
6(x + 4) + 1
= (6 × x) + (6 × 4) + 1
= 6x + 24 + 1
= 6x + 25
Thus, 6(x + 4) + 1 = 6x + 25

Question 37.
5 + 8(3 + x)

Answer: 29 + 8x

Explanation:
Given the expression
5 + 8(3 + x)
= 5 + (8 × 3) + (8 × x)
= 5 + 24 + 8x
= 29 + 8x
Thus 5 + 8(3 + x) = 29 + 8x

Question 38.
x + 3 + 5x

Answer: 6x + 3

Explanation:
Given the expression
x + 3 + 5x
Combine the like terms
x + 5x + 3
6x + 3
So, x + 3 + 5x = 6x + 3

Question 39.
7y + 6 – 1 + 12y

Answer: 19y + 5

Explanation:
Given the expression
7y + 6 – 1 + 12y
Combine the like terms
7y + 12y + 6 – 1
= 19y + 5
So, 7y + 6 – 1 + 12y = 19y + 5

Question 40.
4d + 9 – d – 8

Answer: 3d + 1

Explanation:
Given the expression
4d + 9 – d – 8
Combine the like terms
4d – d + 9 – 8
3d + 1
So, 4d + 9 – d – 8 = 3d + 1

Question 41.
n + 3(n – 1)

Answer: 4n – 3

Explanation:
Given the expression
n + 3(n – 1)
n + (3 × n) – (3 × 1)
n + 3n – 3
4n – 3
So, n + 3(n – 1) = 4n – 3

Question 42.
2v + 8v – 5v

Answer: 5v

Explanation:
Given the expression
2v + 8v – 5v
Combine the like terms
v(2 + 8 – 5)
v(5) = 5v
Thus 2v + 8v – 5v = 5v

Question 43.
5(z + 4) + 5(2 – z)

Answer: 30

Explanation:
Given the expression
5(z + 4) + 5(2 – z)
= (5 × z) + (5 × 4) + (5 × 2) – (5 × z)
= 5z + 20 + 10 – 5z
Now combine the like terms
5z + 20 + 10 – 5z = 30

Question 44.
2.7(w – 5.2)

Answer: 2.7w – 14.04

Explanation:
Given the expression
2.7(w – 5.2)
= (2.7 × w) – (2.7 × 5.2)
= 2.7w – 14.04
Thus 2.7(w – 5.2) = 2.7w – 14.04

Question 45.
Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties 79

Answer:1 \(\frac{1}{2}\)y

Explanation:
Given the expression
\(\frac{2}{3}\)y + \(\frac{1}{6}\)y + y
y(\(\frac{2}{3}\) + \(\frac{1}{6}\) + 1)
= 1 \(\frac{1}{2}\)y
So, Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties 79 = 1 \(\frac{1}{2}\)y

Question 46.
Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties 80

Answer: 2\(\frac{3}{4}\)z + \(\frac{3}{10}\)

Explanation:
Given the expression
Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties 80
\(\frac{3}{4}\)z + \(\frac{3}{10}\) + 2z
2\(\frac{3}{4}\)z + \(\frac{3}{10}\)

Question 47.
7(x + y) – 7x

Answer: 7y

Explanation:
Given the expression
7(x + y) – 7x
= (7 × x) + (7 × y) – 7x
= 7x + 7y – 7x
= 7y
So, 7(x + y) – 7x = 7y

Question 48.
4x + 9y + 3(x + y)

Answer: 7x + 12y

Explanation:
Given the expression
4x + 9y + 3(x + y)
=4x + 9y + 3x + 3y
=4x + 3x + 9y + 3y
=7x + 12y
Thus 4x + 9y + 3(x + y) = 7x + 12y

Question 49.
YOU BE THE TEACHER
Your friend simplifies the expression. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties 81

Answer: your friend is incorrect

Explanation:
8x – 2x + 5x
= 6x + 5x
= 11x
By this we can say that your friend is incorrect.

Question 50.
REASONING
Evaluate each expression by(1) using the Distributive Property and (2) evaluating inside the parentheses first. Which method do you prefer? Is your preference the same for both expressions? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties 82

Answer:
a. 2(3.22 – 0.12)
We can solve this by using the Distributive Property
= 2(3.22) – 2(0.12)
= 6.44 – 0.24
= 6.20
2(3.22 – 0.12) = 6.20
b. 12(\(\frac{1}{2}\) + \(\frac{2}{3}\))
We can solve this by using the Distributive Property
(12 × \(\frac{1}{2}\)) + (12 × \(\frac{2}{3}\))
= 6 + 8
= 14
12(\(\frac{1}{2}\) + \(\frac{2}{3}\)) = 14

Question 51.
DIG DEEPER!
An art club sells 42 large candles and 56 small candles.
a. Write and simplify an expression that represents the profit.
b. A large candle costs $5, and a small candle costs $3. What is the club’s profit?
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 83

Answer:
For large candles, the selling price is $10 and its making cost is $x.
So, by selling a large candle the profit is $(10 – x)
Again for small candles, the selling price is $5 and its making cost is $y.
So, by selling a small candle the profit is $(5 – y)
Therefore, in a sale of 42 large candles and 56 small candles, the total profit will be, P = 42 (10 – x) + 56 (5 – y)
P = 420 – 42x + 280 – 56y
P = $(700 – 42x – 56y

Question 52.
REASONING
Find the difference between the perimeters of the rectangle and the hexagon. Interpret your answer.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 84

Answer:
Rectangle:
Perimeter of the Rectangle = 2l + 2w
P = 2(2x + 7) + 2(2x)
P = 4x + 14 + 4x
P = 8x + 14
Hexagon:
Perimeter of the Hexagon = 6a
P = x + x + 2x + 2x + x + 6 + x + 8
P = 8x + 14
The difference between the perimeters of the rectangle and the hexagon is (8x + 14) – (8x + 14) = 0

Question 53.
PUZZLE
Add one set of parentheses to the expression 7 . x + 3 + 8 . x + 3 . x + 8 − 9 so that it is equivalent to 2(9x + 10).

Answer:
7×(X+3)+8×X+3×X+8-9
=7×(X+3)+11×X-1
=7×X+21+11×X-1
=18×X+20
=2(9X+10)

Lesson 5.5 Factoring Expressions

EXPLORATION 1
Finding Dimensions
Work with a partner.
a. The models show the area (in square units) of each part of a rectangle. Use the models to find missing values that complete the expressions. Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 85
b. In part(a), check that the original expressions are equivalent to the expressions you wrote. Explain your reasoning.
c. Explain how you can use the Distributive Property to rewrite a sum of two whole numbers with a common factor.

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-5-Algebraic-Expressions-and-Properties-85

Try It

Factor the expression using the GCF.

Question 1.
9 + 15

Answer: 3

Explanation:
The factors of 9 are: 1, 3, 9
The factors of 15 are: 1, 3, 5, 15
The number does not contain any common variable factors.
Then the greatest common factor is 3.

Question 2.
60 + 45

Answer: 15

Explanation:
The factors of 45 are: 1, 3, 5, 9, 15, 45
The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
The number does not contain any common variable factors.
Then the greatest common factor is 15.

Question 3.
30 – 20

Answer: 10

Explanation:
The factors of 20 are: 1, 2, 4, 5, 10, 20
The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30
Then the greatest common factor is 10.

Factor the expression using the GCF.

Question 4.
7x + 49

Answer: 7

Explanation:
Since 7x, 49 contain both numbers and variables, there are two steps to find the GCF.
The factors of 7 are: 1, 7
The factors of 49 are: 1, 7, 49
Then the greatest common factor is 7.

Question 5.
8y – 44

Answer: 4

Explanation:
Since 8y, 44 contain both numbers and variables, there are two steps to find the GCF.
The factors of 8 are: 1, 2, 4, 8
The factors of 44 are: 1, 2, 4, 11, 22, 44
Then the greatest common factor is 4.

Question 6.
25a + 10b

Answer: 5

Explanation:
Since 25a, 10b contain both numbers and variables, there are two steps to find the GCF.
The factors of 10 are: 1, 2, 5, 10
The factors of 25 are: 1, 5, 25
Then the greatest common factor is 5.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

FACTORING EXPRESSIONS
Factor the expression using the GCF.

Question 7.
16 + 24

Answer: 8

Explanation:
The factors of 16 are: 1, 2, 4, 8, 16
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
Then the greatest common factor is 8.

Question 8.
49 – 28

Answer: 7

Explanation:
The factors of 28 are: 1, 2, 4, 7, 14, 28
The factors of 49 are: 1, 7, 49
Then the greatest common factor is 7.

Question 9.
8y + 14

Answer: 2

Explanation:
Since 8y, 14 contain both numbers and variables, there are two steps to find the GCF.
The factors of 8 are: 1, 2, 4, 8
The factors of 14 are: 1, 2, 7, 14
Then the greatest common factor is 2.

Question 10.
WHICH ONE DOESN’T BELONG?
Which expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 86

Answer: 6(4n + 3)

Explanation:
i. 3(8n + 12)
= 24n + 36
ii. 4(6n + 9)
24n + 36
iii. 6(4n + 3)
24n + 18
iv. 12(2n + 3)
24n + 36
Thus the third expression does not belong to the other three.

Question 11.
REASONING
Use what you know about factoring to explain how you can factor the expression 18x + 30y + 9z. Then factor the expression.

Answer: 3 (6x + 10y + 3z)

Explanation:
Factoring is a mathematical representation of the expression with the help of common factors.
In this expression, we have given terms and their coefficients. Since we cannot factorize the terms, they will be intact in the process.
We can factorize this expression with the help of prime numbers:
2 × 3 × 3 × x + 2 × 3 × 5 × y + 3 × 3 × z
And finally, we will be able to obtain common factor from this expression:
3 ( 6x + 10y + 3z)

Question 12.
CRITICAL THINKING
Identify the GCF of the terms (x . x) and (4 . x). Explain your reasoning. Then use the GCF to factor the expression x2 + 4x.

Answer:
x2 + 4x
= x(x + 4)
The factors of 1 are: 1
The factors of 4 are: 1, 2, 4
Then the greatest common factor is 1.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 13.
A youth club receives a discount on each pizza purchased for a party. The original price of each pizza is x dollars. The club leader purchases 8 pizzas for a total of (8x − 32) dollars. Factor the expression. What can you conclude about the discount?

Answer:
Given that,
A youth club receives a discount on each pizza purchased for a party.
The original price of each pizza is x dollars.
The club leader purchases 8 pizzas for a total of (8x − 32) dollars.

Question 14.
Three crates of food are packed on a shuttle departing for the Moon. Each crate weighs x pounds. On the Moon, the combined weight of the crates is (3x − 81) pounds. What can you conclude about the weight of each crate on the Moon?
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 87

Answer:
Three crates of food are packed on a shuttle departing for the Moon.
Each crate weighs x pounds. On the Moon, the combined weight of the crates is (3x − 81) pounds.
The factors of 3 are: 1, 3
The factors of 81 are: 1, 3, 9, 27, 81
Then the greatest common factor is 3.

Factoring Expressions Homework & Practice 5.5

Review & Refresh

Use the Distributive Property to simplify the expression.

Question 1.
2(n + 8)

Answer: 2n + 16

Explanation:
Given the expression 2(n + 8)
The distributive property explains that multiplying two numbers (factors) together will result in the same thing as breaking up one factor into two addends, multiplying both addends by the other factor, and adding together both products.
(2 × n) + (2 × 8)
= 2n + 16

Question 2.
3(4 + m)

Answer: 12 + 3m

Explanation:
The distributive property explains that multiplying two numbers (factors) together will result in the same thing as breaking up one factor into two addends, multiplying both addends by the other factor, and adding together both products.
3(4 + m)
= (3 × 4) + (3 × m)
= 12 + 3m

Question 3.
7(b – 3)

Answer: 7b – 21

Explanation:
The distributive property explains that multiplying two numbers (factors) together will result in the same thing as breaking up one factor into two addends, multiplying both addends by the other factor, and adding together both products.
7(b – 3)
= (7 × b) – (7 × 3)
= 7b – 21

Question 4.
10(4 – w)

Answer: 40 – 10w

Explanation:
The distributive property explains that multiplying two numbers (factors) together will result in the same thing as breaking up one factor into two addends, multiplying both addends by the other factor, and adding together both products.
10 (4 – w)
= (10 × 4) – (10 × w)
= 40 – 10w

Write the phrase as an expression.

Question 5.
5 plus a number p

Answer: 5 + p

Explanation:
The phrase plus indicates addition.
The Express would be 5 + p

Question 6.
18 less than a number r

Answer: 18 < r

Explanation:
The phrase less than indicates < symbol.
So the expression would be 18 < r

Question 7.
11 times a number d

Answer: 11d

Explanation:
The phrase times indicates × symbol.
So, the expression would be 11d

Question 8.
a number c divided by 25

Answer: c ÷ 25

Explanation:
The phrase divided by indicates ÷
So, the expression would be c ÷ 25

Decide whether the rates are equivalent.

Question 9.
84 feet in 12 seconds
217 feet in 31 seconds

Answer: The rates are equivalent

Explanation:
Given,
84 feet in 12 seconds
217 feet in 31 seconds
Here we have to check whether the rates are equivalent or not.
84 feet in 12 seconds = 7 : 1
217 feet in 31 seconds = 7 : 1
Thus the rates are equivalent.

Question 10.
12 cups of soda for every 54 cups of juice
8 cups of soda for every 36 cups of juice

Answer: the rates are equivalent

Explanation:
Given,
12 cups of soda for every 54 cups of juice
8 cups of soda for every 36 cups of juice
Here we have to check whether the rates are equivalent or not.
12 cups of soda for every 54 cups of juice = 2 : 9
8 cups of soda for every 36 cups of juice = 2 : 9
Thus the rates are equivalent.

Match the decimal with its equivalent percent.

Question 11.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 88

Answer: C

Explanation:
The fraction form of the decimal 0.36 is \(\frac{36}{100}\)
\(\frac{36}{100}\) × 100 = 36%
Thus the equivalent percent is option C.

Question 12.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 89

Answer: B

Explanation:
The fraction form of the decimal 3.6 is \(\frac{36}{10}\)
\(\frac{36}{10}\) × 100 = 36 × 10 = 360%
Thus the equivalent percent is option B.

Question 13.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 90

Answer: A

Explanation:
The fraction form of the decimal 0.0036 is \(\frac{36}{10000}\)
\(\frac{36}{10000}\) × 100 = \(\frac{36}{100}\) = 0.36%
Thus the equivalent percent is option A.

Question 14.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 91

Answer: B

Explanation:
Explanation:
The fraction form of the decimal 0.0036 is \(\frac{36}{1000}\)
\(\frac{36}{1000}\) × 100 = \(\frac{36}{10}\) =3.6%
Thus the equivalent percent is option D.

Concepts, Skills, & Problem Solving
FINDING DIMENSIONS
The model shows the area (in square units) of each part of a rectangle. Use the model to find missing values that complete the expression. Explain your reasoning. (See Exploration 1, p. 227.)

Question 15.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 92

Answer:4(3 + 4)

Explanation:
We can find the area of the rectangle by using the distributive property.
Let us take 4 as the common factor
(12 + 16) = 4(3 + 4)

Question 16.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 93

Answer: 8(6 + 4)

Explanation:
We can find the area of the rectangle by using the distributive property.
Let us take 8 as a common factor
48 + 32 = 8 (6 + 4)

FACTORING NUMERICAL EXPRESSIONS
Factor the expression using the GCF.

Question 17.
7 + 14

Answer: 7

Explanation:
The factors of 7 are: 1, 7
The factors of 14 are: 1, 2, 7, 14
Then the greatest common factor is 7.

Question 18.
12 + 42

Answer: 6

Explanation:
The factors of 12 are: 1, 2, 3, 4, 6, 12
The factors of 42 are: 1, 2, 3, 6, 7, 14, 21, 42
Then the greatest common factor is 6.

Question 19.
22 + 11

Answer: 11

Explanation:
The factors of 11 are: 1, 11
The factors of 22 are: 1, 2, 11, 22
Then the greatest common factor is 11.

Question 20.
70 + 95

Answer: 5

Explanation:
The factors of 70 are: 1, 2, 5, 7, 10, 14, 35, 70
The factors of 95 are: 1, 5, 19, 95
Then the greatest common factor is 5.

Question 21.
60 – 36

Answer: 12

Explanation:
The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36
The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Then the greatest common factor is 12.

Question 22.
100 – 80

Answer: 20

Explanation:
The factors of 80 are: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80
The factors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, 100
Then the greatest common factor is 20.

Question 23.
84 + 28

Answer: 28

Explanation:
The factors of 28 are: 1, 2, 4, 7, 14, 28
The factors of 84 are: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Then the greatest common factor is 28.

Question 24.
48 + 80

Answer: 16

Explanation:
The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
The factors of 80 are: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80
Then the greatest common factor is 16.

Question 25.
19 + 95

Answer: 19

Explanation:
The factors of 19 are: 1, 19
The factors of 95 are: 1, 5, 19, 95
Then the greatest common factor is 19.

Question 26.
44 – 11

Answer: 11

Explanation:
The factors of 11 are: 1, 11
The factors of 44 are: 1, 2, 4, 11, 22, 44
Then the greatest common factor is 11.

Question 27.
18 – 12

Answer: 6

Explanation:
The factors of 12 are: 1, 2, 3, 4, 6, 12
The factors of 18 are: 1, 2, 3, 6, 9, 18
Then the greatest common factor is 6.

Question 28.
48 + 16

Answer: 16

Explanation:
The factors of 16 are: 1, 2, 4, 8, 16
The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Then the greatest common factor is 16.

Question 29.
98 – 70

Answer: 14

Explanation:
The factors of 70 are: 1, 2, 5, 7, 10, 14, 35, 70
The factors of 98 are: 1, 2, 7, 14, 49, 98
Then the greatest common factor is 14.

Question 30.
58 + 28

Answer: 2

Explanation:
The factors of 28 are: 1, 2, 4, 7, 14, 28
The factors of 58 are: 1, 2, 29, 58
Then the greatest common factor is 2.

Question 31.
72 – 39

Answer: 3

Explanation:
The factors of 39 are: 1, 3, 13, 39
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Then the greatest common factor is 3.

Question 32.
69 + 84

Answer: 3

Explanation:
The factors of 69 are: 1, 3, 23, 69
The factors of 84 are: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Then the greatest common factor is 3.

Question 33.
REASONING
The whole numbers a and b are divisible by c, where b is greater than a. Is a + b divisible by c ? Is b − a divisible by c ? Explain your reasoning.

Answer:
Let the whole numbers a and b be 12 and 54
Let c be 6.
In this case b is greater than a.
Now let us check whether and b are divisible by c.
a + b ÷ c:
12 + 54 ÷ 6
66 ÷ 6 = 11
b – a ÷ c:
54 – 12 ÷ 6
42 ÷ 6 = 7

Question 34.
MULTIPLE CHOICE
Which expression is not equivalent to 81x + 54?
A. 27(3x + 2)
B. 3(37x + 18)
C. 9(9x + 6)
D. 6(13x + 9)

Answer: B. 3(37x + 18), D. 6(13x + 9)

Explanation:
Given the expression 81x + 54
A. 27(3x + 2)
= (27 × 3x) + (27 × 2)
= 81x + 54
B. 3(37x + 18)
= 111x + 54
C. 9(9x + 6)
= 81x + 54
D. 6(13x + 9)
78x + 54
Thus the expression is not equivalent to 81x + 54 are option B and D.

FACTORING ALGEBRAIC EXPRESSIONS
Factor the expression using the GCF.

Question 35.
2x + 10

Answer: 2

Explanation:
Since 2x, 10 contain both numbers and variables, there are two steps to find the GCF.
The factors of 2 are: 1, 2
The factors of 10 are: 1, 2, 5, 10
Then the greatest common factor is 2.

Question 36.
15x + 6

Answer: 3

Explanation:
Since 15x, 6 contain both numbers and variables, there are two steps to find the GCF.
The factors of 6 are: 1, 2, 3, 6
The factors of 15 are: 1, 3, 5, 15
Then the greatest common factor is 3.

Question 37.
26x – 13

Answer: 13

Explanation:
Since 26x, 13 contain both numbers and variables, there are two steps to find the GCF.
The factors of 13 are: 1, 13
The factors of 26 are: 1, 2, 13, 26
Then the greatest common factor is 13

Question 38.
50x – 60

Answer: 10

Explanation:
Since 50x, 60 contain both numbers and variables, there are two steps to find the GCF.
The factors of 50 are: 1, 2, 5, 10, 25, 50
The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Then the greatest common factor is 10.

Question 39.
36x + 9

Answer: 9

Explanation:
Since 36x, 9 contain both numbers and variables, there are two steps to find the GCF.
The factors of 9 are: 1, 3, 9
The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36
Then the greatest common factor is 9

Question 40.
14x – 98

Answer: 14

Explanation:
Since 14x, 98 contain both numbers and variables, there are two steps to find the GCF.
The factors of 14 are: 1, 2, 7, 14
The factors of 98 are: 1, 2, 7, 14, 49, 98
Then the greatest common factor is 14.

Question 41.
18p + 26

Answer: 2

Explanation:
Since 18p, 26 contain both numbers and variables, there are two steps to find the GCF.
The factors of 18 are: 1, 2, 3, 6, 9, 18
The factors of 26 are: 1, 2, 13, 26
Then the greatest common factor is 2.

Question 42.
16m + 40

Answer: 8

Explanation:
Since 16m, 40 contain both numbers and variables, there are two steps to find the GCF.
The factors of 16 are: 1, 2, 4, 8, 16
The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40
Then the greatest common factor is 8.

Question 43.
24 + 72n

Answer: 24

Explanation:
Since 24, 72n contain both numbers and variables, there are two steps to find the GCF.
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Then the greatest common factor is 24.

Question 44.
50 + 65h

Answer: 5

Explanation:
Since 50, 65h contain both numbers and variables, there are two steps to find the GCF.
The factors of 50 are: 1, 2, 5, 10, 25, 50
The factors of 65 are: 1, 5, 13, 65
Then the greatest common factor is 5.

Question 45.
76d – 24

Answer: 4

Explanation:
Since 76d, 24 contain both numbers and variables, there are two steps to find the GCF.
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
The factors of 76 are: 1, 2, 4, 19, 38, 76
Then the greatest common factor is 4.

Question 46.
27 – 45c

Answer: 9

Explanation:
Since 27, 45c contains both numbers and variables, there are two steps to find the GCF.
The factors of 27 are: 1, 3, 9, 27
The factors of 45 are: 1, 3, 5, 9, 15, 45
Then the greatest common factor is 9.

Question 47.
18t + 38x

Answer: 2

Explanation:
Since 18t, 38x contains both numbers and variables, there are two steps to find the GCF.
The factors of 18 are: 1, 2, 3, 6, 9, 18
The factors of 38 are: 1, 2, 19, 38
Then the greatest common factor is 2.

Question 48.
90y + 65z

Answer: 5

Explanation:
Since 90y, 65z contains both numbers and variables, there are two steps to find the GCF.
The factors of 65 are: 1, 5, 13, 65
The factors of 90 are: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
Then the greatest common factor is 5.

Question 49.
10x – 25y

Answer: 5

Explanation:
Since 10x, 25y contains both numbers and variables, there are two steps to find the GCF.
The factors of 10 are: 1, 2, 5, 10
The factors of 25 are: 1, 5, 25
Then the greatest common factor is 5.

Question 50.
24y + 88x

Answer: 8

Explanation:
Since 24y, 88x contains both numbers and variables, there are two steps to find the GCF.
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
The factors of 88 are: 1, 2, 4, 8, 11, 22, 44, 88
Then the greatest common factor is 8.

Question 51.
OPEN-ENDED
Use the Distributive Property to write two expressions that are equivalent to 8x + 16.

Answer: 8(x + 2), 4(2x + 4)

Explanation:
By using the Distributive Property we can write 8x + 16 in two ways.
8x + 16
Take 8 as a common factor
8(x + 2)
Next take 4 as a common factor
4(2x + 4)
The expressions 8(x + 2) and 4(2x + 4) are equivalent to 8x + 16

MATCHING
Match the expression with an equivalent expression.

Question 52.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 94

Answer: C

Explanation:
Given the expression 8x + 16y
Let us take 4 as a common factor.
4(2x + 4y)
Thus the equivalent expression is option C.

Question 53.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 95

Answer: B

Explanation:
Given the expression 4x + 8y
Let us take 2 as a common factor.
4x + 8y = 2(4y + 2x)
Thus the equivalent expression is option B.

Question 54.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 96

Answer: D

Explanation:
Given the expression 16x + 8y
Let us take 8 as a common factor.
16x + 8y = 8(2x + y)
This can also be written as 8(y + 2x)
Thus the equivalent expression is option D.

Question 55.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 97

Answer: A

Explanation:
Given the expression 8x + 4y
Let us take 4 as a common factor.
8x + 4y = 4(2x + y)
Thus the equivalent expression is option A.

Question 56.
YOU BE THE TEACHER
Your friend factors the expression 24x + 56. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 98

Answer: your friend is incorrect

Explanation:
Given the expression 24x + 56
Take 8 as the common factor from the expression.
24x + 56 = 8(3x + 7)
By this, we can say that your friend is incorrect.

Question 57.
MODELING REAL LIFE
You sell soup mixes for a fundraiser. For each soup mix you sell, the company that makes the soup receives x dollars, and you receive the remaining amount. You sell 16 soup mixes for a total of (16x + 96) dollars. How much money do you receive for each soup mix that you sell?

Answer: $6

Explanation:
Here, the cost price of each soup = x dollars
The cost price of 16 soup = 16 x
The selling price of 16 soup = 16 x + 96
Since, the total money received for 16 soup = The selling price of 16 soup – The cost price of 16 soup
= 16 x + 96 – 16 x
= 96
Thus, the total money received for 16 soup = 96 dollars
⇒ The total money received for 1 soup = 96/16 dollars
⇒ The total money received for 1 soup = 6 dollars
Hence, for each soup 6 dollars is received.

Question 58.
PROBLEM SOLVING
A clothing store is having a sale on holiday socks. Each pair of socks costs x dollars. You leave the store with 6 pairs of socks and spend a total of (6x − 14) dollars. You pay with $40. How much change do you receive? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 99

Answer: $14

Explanation:
Given,
A clothing store is having a sale on holiday socks. Each pair of socks costs x dollars.
You leave the store with 6 pairs of socks and spend a total of (6x − 14) dollars.
You pay with $40.
Since, the cost of 1 pair of socks = x
The cost of 6 pair of socks = 6 x
Total amount spend = $ 40
According to the question,
6 x – 14 = 40
6 x = 40 + 14
6 x = 54
x = 9
Thus, the original price of one pair of socks = $ 9
The original price of 6 pairs of socks = $ 54
Hence, the total change = original price of socks – the price of socks in the sale = 54 – 40 = $ 14

Question 59.
STRUCTURE
You buy 37 concert tickets for $8 each, and then sell all 37 tickets for $11 each. The work below shows two ways you can determine your profit. Describe each solution method. Which do you prefer? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 100

Answer:
Given,
You buy 37 concert tickets for $8 each, and then sell all 37 tickets for $11 each.
I prefer the second method which is simplified using the distributive property.
Number of concert tickets = 37
Actual price = $8
selling price = $11
Profit = Number of tickets (selling price – cost price)
P = 37 (11 – 8)
P = 37(3)
P = $111

Question 60.
NUMBER SENSE
The prime factorizations of two numbers are shown, where a and b represent prime numbers. Write the sum of the two numbers as an expression of the form Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 101. Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 101.1

Answer:
Let us consider a and b are 2 and 5
Number 1:
2 × 11 × 5 × a
Number 2:
7 × b × 3 × 3
14 (2 + 5) = 140

Algebraic Expressions and Properties Connecting Concepts

Connecting Concepts
Using the Problem-Solving Plan

Question 1.
A store sells 18 pairs of the wireless earbuds shown. Customers saved a total of $882 on the earbuds. Find the original price of the earbuds.
Understand the problem.
You know the percent discount on a pair of wireless earbuds, the number of pairs of earbuds sold, and the total amount of money that customers saved. You are asked to find the original price of the earbuds.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 102
Make a plan
First, write an expression that represents the total amount of money that customers pay for the earbuds. Then factor the expression to find the discount (in dollars) on each pair of earbuds. Finally, solve a percent problem to find the original price.
Solve and check.
Use the plan to solve the problem. Then check your solution.

Answer:
Number of pairs = 18
Amount saved = $882
Let d be the cost of the wireless earbuds
So, the expression would be 18d + 882

Question 2.
All of the weight plates in a gym are labeled in kilograms. You want to convert the weights to pounds. Write an expression to find the number of pounds in z kilograms. Then find the weight in pounds of a plate that weighs 20.4 kilograms.
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 103

Answer:
All of the weight plates in a gym are labeled in kilograms. You want to convert the weights to pounds
Convert from kilograms to pounds
1 kg = 2.20 pounds
The expression would be 2.20z
Now we have to find the weight of 20.4 kilograms to pounds.
20.4kg = 2.20 × 20.4
= 44.97 pounds

Question 3.
You buy apple chips and banana chips in the ratio of 2 : 7.
a. How many ounces of banana chips do you buy when you buy n ounces of apple chips? Explain.

Answer:
The ratio of apple chips and banana chips = 2 : 7
Apple chips = 2
Banana chips = 7
2 + 7 = 9
n/2 = 9
n = 18
18 ounces of apple chips
n/7 = 9
n = 9 × 7
n = 63
63 ounces of banana chips for 18 ounces of apple chips.
b. You buy 12 ounces of apple chips. How many ounces of banana chips do you buy?

Answer:
You buy 12 ounces of apple chips.
The ratio of apple chips and banana chips = 2 : 7
Apple chips = 2x
Banana chips = 7x
2 + 7 = 9
n = 12
For 12 ounces of apple chips you buy 42 ounces of banana chips

Performance Task

Describing Change

At the beginning of this chapter, you watched a STEAM video called “Shadow Drawings.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 104

Algebraic Expressions and Properties Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 105

Graphic Organizers

You can use an Example and Non-Example Chartto list examples and non-examples of a concept. Here is an Example and Non-Example Chart for the CommutativeProperty of Addition.
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 106

Choose and complete a graphic organizer to help you study the concept.

  1. algebraic expressions
  2. variable
  3. Commutative Property of Multiplication
  4. Associative Property of Addition
  5. Associative Property of Multiplication
  6. Addition Property of Zero
  7. Multiplication Property of Zero
  8. Multiplication Property of One
  9. Distributive Property

Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 107

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 108

5.1 Algebraic Expressions (pp. 201 – 208)

Identify the terms, coefficients, and constants in the expression.

Question 1.
9x + 2 + 8y

Answer:
2 – constant
9, 8 – coefficient
x, y – variable or term
A term without a variable is called a constant.
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Question 2.
3x2 + x + 7

Answer:
7 – constant
3, 1 – coefficient
x – variable or term
2 is the exponent
A term without a variable is called a constant.
The numerical factor of a term that contains a variable is called a coefficient.
An exponent refers to the number of times a number is multiplied by itself.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Question 3.
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 109

Answer:
1 – constant
7, 1/4 – coefficient
q – variable or term
A term without a variable is called a constant.
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Evaluate the expression when x = 20, y = 4, and z = 7.

Question 4.
x ÷ 5

Answer: 4

Explanation:
Given expression is x ÷ 5
Where x = 20
20 ÷ 5 = 4
Thus x ÷ 5 = 4

Question 5.
12 – z

Answer: 5

Explanation:
Given expression is 12 – z
Where z = 7
12 – z
= 12 – 7
= 5
Thus 12 – z = 5

Question 6.
4y

Answer: 16

Explanation:
The given expression is 4y
Where
y = 4
4y = 4(4) = 16

Question 7.
y + x

Answer: 24

Explanation:
Given expression is y + x
Where
y = 4
x = 20
y + x
= 4 + 20 = 24
Thus y + x = 24

Question 8.
x . z

Answer: 80

Explanation:
Given expression is x . z
Where
x = 20
z = 7
x . z = 20 × 7 = 140
Thus x . z = 140

Question 9.
x – y

Answer: 16

Explanation:
Given expression is x – y
Where
y = 4
x = 20
20 – 4 = 16
Thus x – y = 16

Question 10.
3z + 8

Answer: 29

Explanation:
Given expression is 3z + 8
Where
z = 7
3z + 8 = 3(7) + 8
= 21 + 8
= 29
Thus 3z + 8 = 29

Question 11.
8y – x

Answer: 12

Explanation:
Given expression is 8y – x
Where
y = 4
x = 20
8y – x = 8(4) – 20
= 32 – 20
= 12
Thus 8y – x = 12

Question 12.
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 110

Answer:

Explanation:
Given expression is Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 110
Where
y = 4
x = 20
(20)²/4
= (20 × 20)/4
= 5 × 20
= 100

Question 13.
The amount earned (in dollars) for recycling pounds of copper is 2p. How much do you earn for recycling 28 pounds of copper?

Answer: 56

Explanation:
Given,
The amount earned (in dollars) for recycling pounds of copper is 2p.
Substitute the value of p in the expression.
p = 28
2p = 2(28) = 56

Question 14.
While playing a video game, you score p game points and b triple bonus points. An expression for your score is p + 3b. What is your score when you earn 245 game points and 20 triple bonus points?
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 111

Answer:
Given,
While playing a video game, you score p game points and b triple bonus points. An expression for your score is p + 3b.
p = 245
b = 20
Substitute the value of p and b in the given expression
p + 3b = 245 + 3(22)
245 + 66 = 311

Question 15.
Tickets for a baseball game cost a dollars for adults and c dollars for children. The expression 2a + 3c represents the cost (in dollars) for a family to go to the game. What is the cost for the family when an adult ticket is $17 and a child ticket is $12?

Answer:
Given,
Tickets for a baseball game cost a dollars for adults and c dollars for children.
The expression 2a + 3c represents the cost (in dollars) for a family to go to the game.
a = 17
c = 12
2a + 3c
Substitute the value of a and c in the given expression
= 2(17) + 3(12)
= 34 + 36
= 70

Question 16.
Add one set of parentheses to the expression 2x2 + 4 − 5 so that the value of the expression is 75 when x = 6.

Answer:
2x2 + 4 − 5
x = 6
2(6)2 + 4 − 5
= 2(36) – 1
= 72 – 1
= 71
Add 4 to the expression to get the value 75
2x2 + 4 − 5 + 4
x = 6
2(6)2 + 4 − 5 + 4
= 2(36) – 1 + 4
= 72 – 1 + 4
= 71 + 4
= 75

5.2 Writing Expressions (pp. 209–214)

Write the phrase as an expression.

Question 17.
9 fewer than 23

Answer: 9 – 23

Explanation:
The phrase “fewer than” represents – symbol
Thus the expression would be 9 – 23

Question 18.
6 more than the quotient of 15 and 3

Answer: 6 + (15 ÷ 3)

Explanation:
The phrase more than represents ‘+’ and quotient represents ‘÷’
Thus the expression would be 6 + (15 ÷ 3)

Question 19.
the product of a number d and 32

Answer: 32d

Explanation:
The phrase product represents ‘×’ symbol.
Thus the expression would be 32d

Question 20.
a number t decreased by 17

Answer: d – 17

Explanation:
The phrase “decreased” represents ‘-‘ symbol.
Thus the expression would be d – 17

Question 21.
Your basketball team scored 4 fewer than twice as many points as the other team.
a. Write an expression that represents the number of points your team scored.

Answer:
Your basketball team scored 4 fewer than twice as many points, x, as the other team.
n = 2x – 4
b. The other team scored 24 points. How many points did your team score?

Answer: 14

Explanation:
The other team scored 24 points i.e. n=24.
24 = 2x – 4
2x = 24 + 4
2x = 28
x = 28/2
x = 14
Thus the team score 14 points.

Question 22.
The boiling temperature (in degrees Celsius) of platinum is 199 more than four times the boiling temperature (in degrees Celsius) of zinc.
a. Write an expression that represents the boiling temperature (in degrees Celsius) of platinum.

Answer: P = 199 + 4z
b. The boiling temperature of zinc is 907 degrees Celsius. What is the boiling temperature of platinum?

Answer:
The boiling temperature of zinc is 907 degrees Celsius.
P = 199 + 4z
P = 199 + 4(907)
P = 3827

Question 23.
Write an algebraic expression with two variables, x and y, that has a value of 50 when x = 3 and y = 5. (pp. 215–220)

Answer: 5x + 7y = 50

Explanation:
Let us assume that the algebraic expression is 5x + 7y = 50
Where x = 3 and y = 5
5(3) + 7(5) = 50
15 + 35 = 50
50 = 50
Our assumption is correct.

5.3 Properties of Addition and Multiplication

Simplify the expression. Explain each step.

Question 24.
10 + (2 + y)

Answer: 12 + y

Explanation:
Given the expression 10 + (2 + y)
Combine the like terms
10 + 2 + y = 12 + y

Question 25.
(21 + b) + 1

Answer: 22 + b

Explanation:
Given the expression (21 + b) + 1
Combine the like terms
21+ 1 + b
22 + b

Question 26.
3(7x)

Answer: 21x

Explanation:
Given the expression 3(7x)
3 × 7x = 21x

Question 27.
1(3.2w)

Answer: 3.2w

Explanation:
Given the expression 1(3.2w)
1 × 3.2w = 3.2w

Question 28.
5.3 + (w + 1.2)

Answer: 6.5 + w

Explanation:
Given the expression 5.3 + (w + 1.2)
Combine the like terms
5.3 + 1.2 + w
= 6.5 + w

Question 29.
(0 + t) + 9

Answer: t + 9

Explanation:
Given the expression (0 + t) + 9
Combine the like terms
(0 + t) + 9 = t + 9

Question 30.
The expression 7 + 3x+ 4 represents the perimeter of the triangle. Simplify the expression.
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 112

Answer:
Given,
The expression 7 + 3x+ 4 represents the perimeter of the triangle.
Combine the like terms
7 + 3x+ 4 = 11 + 3x

Question 31.
Write an algebraic expression that can be simplified using the Associative Property of Addition.

Answer:
Associative property of addition: Changing the grouping of addends does not change the sum. For example, ( 2 + 3 ) + 4 = 2 + ( 3 + 4 ) (2 + 3) + 4 = 2 + (3 + 4) (2+3)+4=2+(3+4) left parenthesis, 2, plus, 3, right parenthesis, plus, 4, equals, 2, plus, left parenthesis, 3, plus, 4, right parenthesis.

5.4 The Distributive Property

Use the Distributive Property to simplify the expression.

Question 32.
2(x + 12)

Answer:
We can simplify the expression by using the Distributive Property.
The distributive property explains that multiplying two numbers together will result in the same thing as breaking up one factor into two addends, multiplying both addends by the other factor, and adding together both products.
2(x + 12) = 2 × x + 2 × 12
= 2x + 24

Question 33.
11(4b – 3)

Answer:
Given the expression 11(4b – 3)
We can simplify the expression by using the Distributive Property.
The distributive property explains that multiplying two numbers together will result in the same thing as breaking up one factor into two addends, multiplying both addends by the other factor, and adding together both products.
11(4b – 3) = 11 × 4b – 11 × 3
= 44b – 33

Question 34.
8(s – 1)

Answer:
Given the expression 8(s – 1)
We can simplify the expression by using the Distributive Property.
The distributive property explains that multiplying two numbers together will result in the same thing as breaking up one factor into two addends, multiplying both addends by the other factor, and adding together both products.
8(s – 1) = 8 × s – 8 × 1
= 8s – 8

Question 35.
6(6 + y)

Answer:
Given the expression 6(6 + y)
We can simplify the expression by using the Distributive Property.
The distributive property explains that multiplying two numbers together will result in the same thing as breaking up one factor into two addends, multiplying both addends by the other factor, and adding together both products.
6(6 + y) = 6 × 6 + 6 × y
36 + 6y

Simplify the expression.

Question 36.
5(n + 3) + 4n

Answer:
Given the expression 5(n + 3) + 4n
Combine the like terms
5(n + 3) + 4n
= 5 × n + 5 × 3 + 4n
= 5n + 15 + 4n
= 9n + 15

Question 37.
t + 2 + 6t

Answer:
Given the expression t + 2 + 6t
Combine the like terms
t + 2 + 6t
7t + 2

Question 38.
3z + 14 + 5z – 9

Answer:
Given the expression 3z + 14 + 5z – 9
Combine the like terms
3z + 14 + 5z – 9
8z – 5

Question 39.
A family of three goes to a salon. Each person gets a haircut and highlights. The cost of each haircut is $15, and the cost per person for highlights is x dollars. Write and simplify an expression that represents the total cost (in dollars) for the family at the salon.

Answer: 45 + 3(x)

Explanation:
Given,
A family of three goes to a salon. Each person gets a haircut and highlights.
The cost of each haircut is $15, and the cost per person for highlights is x dollars.
The expression would be 45 + 3x
Each person gets a haircut = 3 × 15 = 45

Question 40.
Each day, you take vocal lessons for v minutes and trumpet lessons for 30 minutes. Write and simplify an expression to find how many minutes of lessons you take in 4 days.
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 113

Answer:
Given,
Minutes of vocal lessons I take each day =
Minutes of trumpet lessons I take each day = 30
To write and simplify how many minutes of lessons I take in 4 days.
Solution:
Total number of minutes of lesson I take each day can be calculated by adding the minutes of lessons taken each day and can be given as:
= v + 30
The expression would be (v + 30)
So in 4 days I will take = 4(v + 30)
= 4v + 120
Thus, total minutes of lessons I take in 4 days = 4v + 120

5.5 Factoring Expressions (pp. 227 – 232)

Factor the expression using the GCF.

Question 41.
42 – 12

Answer: 6

Explanation:
The factors of 12 are: 1, 2, 3, 4, 6, 12
The factors of 42 are: 1, 2, 3, 6, 7, 14, 21, 42
Then the greatest common factor is 6.

Question 42.
15 + 35

Answer: 5

Explanation:
The factors of 15 are: 1, 3, 5, 15
The factors of 35 are: 1, 5, 7, 35
Then the greatest common factor is 5.

Question 43.
36x – 28

Answer: 4

Explanation:
The factors of 28 are: 1, 2, 4, 7, 14, 28
The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36
Then the greatest common factor is 4.

Question 44.
24 + 64x

Answer: 8

Explanation:
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
The factors of 64 are: 1, 2, 4, 8, 16, 32, 64
Then the greatest common factor is 8.

Question 45.
60 – 150x

Answer: 30

Explanation:
The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
The factors of 150 are: 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150
Then the greatest common factor is 30.

Question 46.
16x + 56y

Answer: 8

Explanation:
The factors of 16 are: 1, 2, 4, 8, 16
The factors of 56 are: 1, 2, 4, 7, 8, 14, 28, 56
Then the greatest common factor is 8.

Question 47.
A soccer team receives a discount on each jersey purchased. The original price of each jersey is x dollars. The team buys 18 jerseys for a total of (18x − 36) dollars. What can you conclude about the discount?

Answer:
Given,
A soccer team receives a discount on each jersey purchased.
The original price of each jersey is x dollars.
The team buys 18 jerseys for a total of (18x − 36) dollars.
Multiply 18 with the original price after that you decrease 36 dollars.
By this, we can conclude that $36 is the discount

Question 48.
You sell apple cider for a fundraiser. For each gallon of cider you sell, the company that makes the cider receives x dollars, and you receive the remaining amount. You sell 15 gallons of cider for (15x + 45) dollars. How much money do you receive for each gallon of cider that you sell?
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 113.1

Answer:
Given,
You sell apple cider for a fundraiser. For each gallon of cider you sell, the company that makes the cider receives x dollars, and you receive the remaining amount.
You sell 15 gallons of cider for (15x + 45) dollars
Let x = 1
15(1) + 45 = 60
x = 2
15(2) + 45 = 30 + 45 = 75

Algebraic Expressions and Properties Practice Test

Question 1.
Identify the terms, coefficients, and constants of \(\frac{q}{3}\) + 6 + 9q.

Answer:
6 – constant
9,  \(\frac{1}{3}\) – coefficient
q – variable or term
A term without a variable is called a constant.
The exponent tells us how many times the base is used as a factor
The numerical factor of a term that contains a variable is called a coefficient.
A variable is a symbol that represents one or more numbers. Each number or variable by itself, or product of numbers and variables in an algebraic expression, is called a term.

Question 2.
Evaluate 4b – a when a = 12 and b = 7.

Answer: 16

Explanation:
Given the expression 4b – a
where a = 12 and b = 7
4(7) – 12
28 – 12 = 16
Thus 4b – a = 16

Write the phrase as an expression.

Question 3.
25 more than 50

Answer: The phrase for the expression 25 more than 50 is 25 + 50

Question 4.
6 less than the quotient of 32 and a number y.

Answer: The phrase for the expression 6 less than the quotient of 32 and a number y is 6 – 32 ÷ y

Simplify the expression. Explain each step.

Question 5.
3.1 + (8.6 + m)

Answer:
Given the expression 3.1 + (8.6 + m)
Combine the like terms
3.1 + (8.6 + m)
3.1 + 8.6 + 3
11.7 + m

Question 6.
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 114

Answer:
Given the expression (\(\frac{2}{3}\) × t) × 1 \(\frac{1}{2}\)
First, convert the mixed fraction to the improper fraction
1 \(\frac{1}{2}\) = \(\frac{3}{2}\)
\(\frac{3}{2}\) × \(\frac{2}{3}\) × t
= t
Thus (\(\frac{2}{3}\) × t) × 1 \(\frac{1}{2}\) = t

Question 7.
4(x + 8)

Answer:
Given the expression 4(x + 8)
= 4 × x + 4 × 8
= 4x + 32

Question 8.
4t + 7 + 2t – 2

Answer: 6t + 5

Explanation:
Combine the like terms
4t + 7 + 2t – 2
6t + 5
Thus 4t + 7 + 2t – 2 = 6t + 5

Factor the expression using the GCF.

Question 9.
18 + 24

Answer:
Find the GCF for 18, 24
The factors of 18 are 1, 2, 3, 6, 9, 18
The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24
The common factors are 1, 2, 3, 6.
Thus the GCF of 18 + 24 is 6.

Question 10.
15x + 20

Answer:
Since 15x, 20 contain both numbers and variables, there are two steps to find the GCF.
Find the GCF fo the numerical part 15, 20
The factors of 20 are 1, 2, 4, 5, 10, 20
The factors of 15 are 1, 3, 5, 15
The number do not contain any common variable factors.
The greatest common factor is 5

Question 11.
32x – 40y

Answer: 8

Explanation:
Since 32x, 40y contain both numbers and variables, there are two steps to find the GCF.
Find the GCF for the numerical part 32, 40
The factors of 32 are 1, 2, 4, 8, 16, 32
The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40
The number do not contain any common variable factors.
Then the greatest common factor is 8.

Question 12.
Playing time is added at the end of a soccer game to make up for stoppages. An expression for the length (in minutes) of a 90-minute soccer game with minutes of stoppage time is 90 + x. How long is a game with 4 minutes of stoppage time?
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 114.1

Answer: 94

Explanation:
Given,
Playing time is added at the end of a soccer game to make up for stoppages.
An expression for the length (in minutes) of a 90-minute soccer game with minutes of stoppage time is 90 + x.
x = 4
Substitute the value of x in the expression
90 + 4 = 94

Question 13.
The expression 15 . x . 6 represents the volume of a rectangular prism with a length of 15, a width of x, and a height of 6. Simplify the expression.

Answer:
Given,
The expression 15 . x . 6 represents the volume of a rectangular prism with a length of 15, a width of x, and a height of 6.
The formula for the volume of a rectangular prism is lwh
V = 15 × x × 6
V = 90x

Question 14.
The Coiling Dragon Cliff Skywalk in China is 128 feet longer than the length x (in feet) of the Tianmen Skywalk in China. The world’s longest glass-bottom bridge, located in China’s Zhangjiajie National Park, is about 4.3 times longer than the Coiling Dragon Cliff Skywalk. Write and simplify an expression that represents the length (in feet) of the world’s longest glass-bottom bridge.
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 115

Answer:
The Coiling Dragon Cliff Skywalk in China is 128 feet longer than the length x (in feet) of the Tianmen Skywalk in China.
128 + x
The world’s longest glass-bottom bridge, located in China’s Zhangjiajie National Park, is about 4.3 times longer than the Coiling Dragon Cliff Skywalk.
(128 + x) × 4.3
128 × 4.3 + 4.3 × x
550.4 + 4.3x
Thus the expression is 4.3x + 550.4ft

Question 15.
A youth group is making and selling sandwiches to raise money. The cost to make each sandwich is dollars. The group sells 150 sandwiches for a total of (150h + 450) dollars. How much profit does the group earn for each sandwich sold?

Answer:
We are given that
Cost to make each sandwich= h dollar
The total selling price of 150 sandwiches=(150h+450 )dollars
We have to find the profit earn by the group on each sandwich sold.
Total cost for making 150 sandwiches= 150 × h
Total cost for making 150 sandwiches=150 h
Now,
Profit made by the group on 150 sandwiches
=150h+450-150h
=450 dollars
Profit earn by group for each sandwich sold= 450/150
Profit earn by group for each sandwich sold=$ 3

Question 16.
You make party favors for an event. You tie 9 inches of ribbon around each party favor. Write an expression for the number of inches of ribbon needed for n party favors. The ribbon costs $3 for each yard. Write an expression for the total cost (in dollars) of the ribbon.

Answer:
Need 9n inches of ribbon Cost is $0.75n
Since you need 9 inches per party favor and you have n party favors, the amount of ribbon you need is 9 times n inches.
So R = 9n inches
Since the ribbon costs $3 per yard and our current equation is inches, we need to do a bit of conversion.
There are 36 inches per yard and 9 inches per ribbon, so each yard of ribbon can make 36/9 = 4 ribbons for the party favors.
So the cost per party favor will be one-fourth of the cost per yard of ribbon per party favor.
So $3.00 / 4 = $0.75, making the cost equation: C = $0.75n

Algebraic Expressions and Properties Cumulative Practice

Question 1.
The student council is organizing a school fair. Council members are making signs to show the prices for admission and for each game a person can play.
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 116

Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 117

Let x represent the number of games. Which expression can you use to determine the total amount (in dollars) a person pays for admission and playing x games?
A. 2.25
B. 2.25x
C. 2 + 0.25x
D. 2x + 0.25

Answer: 2 + 0.25x

Explanation:
Given the data,
The student council is organizing a school fair. Council members are making signs to show the prices for admission and for each game a person can play.
Admission fee is $2
Price per game is $0.25
Let x represent the number of games.
The expression would be the sum of admission and product of price and number of games
That is 2 + 0.25x
Thus the correct answer is option C.

Question 2.
Which ratio relationship is represented in the graph?
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 118
F. 2 cups of flour for every cup of sugar \(\frac{1}{2}\) up of sugar
G. 6 cups of flour for every 3 cups of sugar
H. 1 cup of flour for every 4 cups of sugar
I. \(\frac{1}{2}\) cup of flour for every 1 cup of sugar

Answer: 2 cups of flour for every cup of sugar \(\frac{1}{2}\) up of sugar

Explanation:
The graph represents that 2 cups of flour for every cup of sugar.
Thus the correct answer is option F.

Question 3.
At a used bookstore, you can purchase two types of books.
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 119
You can use the expression 3h + 2p to find the total cost (in dollars) for h hardcover books and p paperback books. What is the total cost (in dollars) for 6 hardcover books and 4 paperback books?
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 120

Answer:
Given expression 3h + 2p
h represents hardcover books
p represents paperback books
The cost of hardcover books is $3
The cost of paperback books is $2
3h + 2p = 3(3) + 2(2)
= 9 + 4
= 13
We have to find the total cost (in dollars) for 6 hardcover books and 4 paperback books.
The equation will be 6h + 4p
= 6(3) + 4(5)
= 18 + 20
= 38
The total cost (in dollars) for 6 hardcover books and 4 paperback books is $38.

Question 4.
Your friend divided two decimal numbers. Her work is shown in the box below. What should your friend change in order to divide the two decimal numbers correctly?
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 121

Answer: C

Explanation:
0.07 ÷ 14.56 can be written as
7 ÷ 0.1456
Thus the correct answer is option C.

Question 5.
What is the value of 4.391 + 5.954?
F. 9.12145
G. 9.245
H. 9.345
I. 10.345

Answer: 10.345

Explanation:
Add two decimal numbers 4.391 and 5.954
4.391
+5.954
10.345
Thus the correct answer is option I.

Question 6.
The circle graph shows the eye color of students in a sixth-grade class. Nine students in the class have brown eyes. How many students are in the class?
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 122
A. 4 students
B. 18 students
C. 20 students
D. 405 students

Answer: 20 students

Explanation:
By seeing the above figure we can find the number of students in the class.
The circle graph shows the eye color of students in a sixth-grade class. Nine students in the class have brown eyes.
45/9 = 5
Each student is equal to 5%
35/5 = 7 students
20/5 = 4 students
9 + 7 + 4 = 20 students
Thus the correct answer is option C.

Question 7.
Properties of Addition and Multiplication are used to simplify an expression.
Big Ideas Math Solutions Grade 6 Chapter 5 Algebraic Expressions and Properties 123
What number belongs in place of the x?

Answer:
36 × 23 + 33 × 64 = 36 × 23 + 64 × 33
= 36 × 23 + 64 × (23 + 10)
= 36 × 23 + 64 × 23 + 64 × 10
= 23(36 + 64) + 64 × 10
= 100 × 23 + 64 × 10
x = 100
Thus the number that belong to x is 100.

Question 8.
What is the prime factorization of 1350?
F. 10 . 135
G. 2 . 3 . 5
H. 6 . 225
I. 2 . 33 . 52

Answer:
The prime factorisation of 1350 is
1350 = 2 × 3 × 3 × 3 × 5 × 5
1350 = 2 . 33 . 52
Thus the correct answer is option I.

Question 9.
A horse gallops at a speed of 44 feet per second. What is the speed of the horse in miles per hour?
A. \(\frac{1}{2}\) mile per hour
B. 30 miles per hour
C. 64\(\frac{8}{15}\) miles per hour
D. 158,400 miles per hour

Answer: 30 miles per hour

Explanation:
Given,
A horse gallops at a speed of 44 feet per second.
Convert from feet per second to miles per hour.
1 foot per second = 0.68 miles per hour
44 feet per second = 30 miles per hour
Thus the correct answer is option B.

Question 10.
Which equation correctly demonstrates the Distributive Property?
F. a(b + c) = ab + ac
G. a(b + c) = ab + ac
H. a + (b + c) = (a + b) + (a + c)
I. a + (b + c) = (a + b) . (a + c)

Answer: F

Explanation:
The distributive property explains that multiplying two numbers (factors) together will result in the same thing as breaking up one factor into two addends, multiplying both addends by the other factor, and adding together both products.
a(b + c) = ab + ac
Thus the correct answer is option F.

Question 11.
Which number is equivalent to 2\(\frac{4}{5}\) . 1\(\frac{2}{7}\) ?
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 124

Answer: C

Explanation:
Convert the mixed fraction to the improper fraction.
2\(\frac{4}{5}\) = \(\frac{14}{5}\)
1\(\frac{2}{7}\) = \(\frac{9}{7}\)
Now multiply both the fractions
\(\frac{14}{5}\) × \(\frac{9}{7}\) = \(\frac{126}{35}\)
Now convert from improper fraction to the mixed fractions.
\(\frac{126}{35}\) = 3 \(\frac{3}{5}\)
Thus the correct answer is option C.

Question 12.
Which pair of numbers does not have a least common multiple of 24?
F. 2, 12
G. 3, 8
H. 6, 8
I. 12, 24

Answer: 6, 8
The least common multiple of 24 is 24.
2, 12 is 24
3, 8 is 24
6, 8 is 48
Thus the correct answer is option H.

Question 13.
Use the Properties of Multiplication to simplify the expression in an efficient way. Show your work and explain how you used the Properties of Multiplication.
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 125
Big Ideas Math Answer Key Grade 6 Chapter 5 Algebraic Expressions and Properties 126

Answer:
We can use the associative property of multiplication to simplify the expression.
The associative property is a math rule that says that the way in which factors are grouped in a multiplication problem does not change the product.
(25 × 18) × 4 = 25 × (18 × 4)
450 × 4 = 1800

Question 14.
Which number is not a perfect square?
A. 64
B. 81
C. 96
D. 100

Answer: 96

Explanation:
A. 64 = 8 × 8
It is a perfect square.
B. 81 = 9 × 9
It is a perfect square.
C. 96 = 12 × 8
96 is not a perfect square.
D. 100 = 10 × 10
It is a perfect square.
Thus the correct answer is option C.

Conclusion:

Big Ideas Math Answers Grade 6 Chapter 5 Algebraic Expressions and Properties is here along with the solutions. Check all the problems and solutions in the above sections along with the examples. For all the queries, you can question us in the below comment box. Share the above article and pdf links with your friends and peers to overcome all the issues in maths. To clear all your doubts and questions, check all the links given in the above sections.

Big Ideas Math Answers Grade 6 Chapter 6 Equations

Big Ideas Math Answers Grade 6 Chapter 6 Equations

Download Big Ideas Math Answers Grade 6 Chapter 6 Equations Pdf for free of cost. If you are browsing for various questions of equations, then here is the one-stop solution. Know the benefit of referring to Big Ideas Math Book 6th Grade Answer Key Chapter 6 Equations. You can learn the tips and simple methods to solve all the problems. With the below-given material, it acts as a guide to get better marks in the exam. Get a free step-by-step solution from this ultimate guide ie., BIM 6th Grade Solutions Ch 6 Equations. Check the below sections to get various details and information.

Big Ideas Math Book 6th Grade Answer Key Chapter 6 Equations

It is necessary for the candidates to understand the concept in maths. Concept is important than scoring marks in the exam. Therefore relate the questions with real-time problems and understand the concept in depth. With all the factors into consideration, BIM 6th Grade Answer Key or Equations Pdf is prepared. Click on the links in the next sections and start practicing the problems. Follow the various topics, Steam videos and Solved problems available in BIM Equations book and pdf.

Performance Task

Lesson 1: Writing Equations in One Variable

Lesson: 2 Solving Equations Using Addition or Subtraction

Lesson: 3 Solving Equations Using Multiplication or Division

Lesson: 4 Writing Equations in Two Variables

Chapter 6 – Equations

Equations STEAM Video/ Performance Task

STEAM Video

Rock Climbing
Equations can be used to solve many different kinds of problems in real life, such as estimating the amount of time it will take to climb a rock wall. Can you think of any other real-life situations where equations are useful?
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 1
In rock climbing, a pitch is a section of a climbing route between two anchor points. Watch the STEAM Video “Rock Climbing.”en answer the following questions.
1. How can you use pitches to estimate the amount of time it will take to climb a rock wall?
2. Are there any other methods you could use to estimate the amount of time it will take to climb a rock wall? Explain.
3. You know two of the three pieces of information below. Explain how you can find the missing piece of information.
Average climbing speed
Height of rock wall
Time to complete climb

Performance Task

Planning the Climb
After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given information about two rock-climbing routes.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 2
Route 1: 500 feet, 125 feet per pitch
Route 2: 1200 feet, 8 pitch
You will find the average speed of the climbers on Route 1 and the amount of time it takes to complete Route 2. Will the average speed of the climbers on Route 1 provide accurate predictions for the amount of time it takes to climb other routes? Explain why or why not.

Equations Getting Ready for Chapter 6

Chapter Exploration

Work with a partner. Every equation that has an unknown variable can be written as a question. Write a question that represents the equation. Then answer the question.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 3

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-6-Equations-3

Work with a partner. Write an equation that represents the question. Then answer the question.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 4

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-6-Equations-4

Vocabulary
The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
equation
independent variable
inverse operations
dependent variable
equation in two variables.

Lesson 6.1 Writing Equations in One Variable

EXPLORATION 1

Writing Equations
Work with a partner. Customers order sandwiches at a cafe from the menu board shown.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 1
a. The equation 6.75x =20.25 represents the purchase of one customer from the menu board. What does the equation tell you about the purchase? What cannot be determined from the equation?
b. The four customers in the table buy multiple sandwiches of the same type. For each customer, write an equation that represents the situation. Then determine how many sandwiches each customer buys. Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 2
Answer:

Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 3

An equation
is a mathematical sentence that uses an equal sign, =, to show that two expressions are equal.
Expressions
4 + 8
x + 8
Equations
4 + 8 = 12
x + 8 = 12
To write a word sentence as an equation, look for key words or phrases such as is, the same as, or equals to determine where to place the equal sign.

Try It

Write the word sentence as an equation.
Question 1.
9 less than a number be equals 2.
Answer: 9-x=2

Explanation:
We have to write the equation for the word sentence
The phrase “less than” indicates -.
let the number be x.
9 – x = 2

Question 2.
The product of a number g and 5 is 30.
Answer: 5 × g=30

Explanation:
We have to write the equation for the word sentence
The phrase “product” indicates ‘×’
g × 5 = 30

Question 3.
A number k increased by 10 is the same as 24.
Answer: k + 10 = 24

Explanation:
We have to write the equation for the word sentence
The phrase “increased” indicates ‘+’
The equation is k + 10 = 24

Question 4.
The quotient of a number q and 4 is 12.
Answer: q ÷ 4 = 12

Explanation:
We have to write the equation for the word sentence
The phrase quotient indicates ‘÷’
The equation is q ÷ 4 = 12

Question 5.
2\(\frac{1}{2}\) is the same as the sum of a number w and \(\frac{1}{2}\).
Answer: 2 \(\frac{1}{2}\) = w + \(\frac{1}{2}\)

Explanation:
We have to write the equation for the word sentence
The phrase sum indicates ‘+’
The equation is 2 \(\frac{1}{2}\) = w + \(\frac{1}{2}\)

Question 6.
WHAT IF?
Each server decorates one table. Which equation can you use to find c?
Answer: We can use the multiplication equation to find c.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
VOCABULARY
How are expressions and equations different?
Answer: An expression is a number, a variable, or a combination of numbers and variables and operation symbols. An equation is made up of two expressions connected by an equal sign.

Question 8.
DIFFERENT WORDS, SAME QUESTION
Which is different? Write “both” equations.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 4
Answer: n-4=8
4<8

Question 9.
OPEN-ENDED
Write a word sentence for the equation 28 −n= 5.
Answer: 28 less than a number n is equals to 5.

Question 10.
WRITING
You purchase x items for $4 each. Explain how the variable in the expression 4x and the variable in the equation 4x= 20 are similar. Explain how they are different.
Answer:
You purchase x items for $4 each
4x = 20
x = 20/4
x = 5

Question 11.
After four rounds, 74 teams are eliminated from a robotics competition. There are 18 teams remaining. Write and solve an equation to find the number of teams that started the competition.
Answer:
Given,
After four rounds, 74 teams are eliminated from a robotics competition. There are 18 teams remaining.
Let x be 74 teams
let y be 18 teams
The equation would be
x + y = 92
74 + 18 = 92
Thus the total number of teams are 92.

Question 12.
The mass of the blue copper sulfate crystal is two-thirds the mass of the red fluorite crystal. Write an equation you can use to find the mass (in grams) of the blue copper sulfate crystal.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 5
Answer: blue copper sulfate crystal = 2/3 (red fluorite crystal)

Question 13.
DIG DEEPER!
You print photographs from a vacation. Find the number of photographs you can print for $3.60.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 6
Answer: We can print 15 photographs for $3.60

Explanation:
Cost of each print = $0.24
The total cost for photographs is $3.60
3.60/0.24 = 15
Thus We can print 15 photographs for $3.60

Writing Equations in One Variable Homework & Practice 6.1

Review & Refresh

Factor the expression using the GCF.
Question 1.
6 + 27
Answer: 3 (2 + 9)

Explanation:
Given the expression 6 + 27
Take 3 as the common factor
3(2 + 9)

Question 2.
9w + 72
Answer: 9(w + 8)

Explanation:
Given the expression 9w + 72
Take 9 as the common factor
9w + 72 = 9(w + 8)

Question 3.
42 + 24n
Answer: 6(7 + 4n)

Explanation:
Given the expression 42 + 24n
Take 6 as the common factor
42 + 24n = 6(7 + 4n)

Question 4.
18h + 30k
Answer: 6(3h + 5k)

Explanation:
Given the expression 18h + 30k
Take 6 as the common factor
18h + 30k = 6(3h + 5k)

Question 5.
Which number is not equal to 225%?
A. 2\(\frac{1}{4}\)
B. \(\frac{9}{4}\)
C. \(\frac{50}{40}\)
D. \(\frac{45}{20}\)
Answer: C
225% is not equal to \(\frac{50}{40}\)

Evaluate the expression when a = 7.
Question 6.
6 + a
Answer: 13

Explanation:
Given the expression 6 + a
where a = 7
Substitute the value of a in the expression
6 + 7 = 13

Question 7.
a – 4
Answer: 3

Explanation:
Given the expression a – 4
where a = 7
Substitute the value of a in the expression
a – 4
7 – 4 = 3

Question 8.
4a
Answer: 28

Explanation:
Given the expression 4a
where a = 7
Substitute the value of a in the expression
4 × 7 = 28

Question 9.
\(\frac{35}{a}\)
Answer: 5

Explanation:
Given the expression \(\frac{35}{a}\)
where a = 7
Substitute the value of a in the expression
\(\frac{35}{7}\) = 5

Find the perimeter of the rectangle.
Question 10.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 7
Answer:
l = 8 ft
Area = 40 sq ft
We know that,
Area of rectangle = l × w
40 sq. ft = 8 ft × w
w = 40/8 = 5 ft
Thus the width of the above rectangle is 5 ft.

Question 11.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 8
Answer:
l = 13 cm
w = ?
A = 52 sq. cm
We know that,
Area of rectangle = l × w
52 sq. cm = 13 cm × w
w = 52/13
w = 4 cm
Thus the width of the above rectangle is 4 cm.

Question 12.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 9
Answer:
A = 224 sq. miles
l = 14 miles
We know that,
Area of rectangle = l × w
224 sq. miles = 14 × w
w = 224/14
w = 16 miles
Thus the width of the above figure is 16 miles.

Concepts, Skills, & Problem Solving

WRITING EQUATIONS A roast beef sandwich costs $6.75. A customer buys multiple roast beef sandwiches. Write an equation that represents the situation. Then determine how many sandwiches the customer buys. (See Exploration 1, p. 245.)
Question 13.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 10
Answer:
Given,
A roast beef sandwich costs $6.75.
Amount used for payment = $50.
Change Received = $16.25
The total number of sandwich the customer buys = 5

Question 14.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 11
Answer:
Given,
A roast beef sandwich costs $6.75.
Amount used for payment = $80.
Change Received = $19.25
Amount used for payment – Change Received
= $ 80 – $19.25
= $60.75
1 sandwich = $6.75
The total number of sandwich the customer buys = 9

WRITING EQUATIONS Write the word sentence as an equation.
Question 15.
A number y decreased by 9 is 8.
Answer: y – 9 = 8

Explanation:
We have to write the word sentence in the equation form.
y – 9 = 8

Question 16.
The sum of a number x and 4 equals 12.
Answer: x + 4 = 12

Explanation:
We have to write the word sentence in the equation form.
x + 4 = 12

Question 17.
9 times a number b is 36.
Answer: 9b = 36

Explanation:
We have to write the word sentence in the equation form.
The phrase times indicates ‘×’
The equation would be 9b = 36

Question 18.
A number w divided by 5 equals 6.
Answer: w ÷ 5 = 6

Explanation:
We have to write the word sentence in the equation form.
The phrase divided by indicates ‘÷’
The equation would be w ÷ 5 = 6

Question 19.
54 equals 9 more than a number t.
Answer: 54 = 9 + t

Explanation:
We have to write the word sentence in the equation form.
The phrase more than indicates ‘+’
The equation would be 54 = 9 + t

Question 20.
5 is one-fourth of a number c.
Answer: 5 = 1/4 c

Explanation:
We have to write the word sentence in the equation form.
The phrase of indicates ‘×’
The equation would be 5 = 1/4 c

Question 21.
9.5 less than a number n equals 27.
Answer: 9.5 – n = 27

Explanation:
We have to write the word sentence in the equation form.
The phrase less than indicates ‘-‘
The equation would be 9.5 – n = 27

Question 22.
11\(\frac{3}{4}\) is the quotient of a number y and 6\(\frac{1}{4}\).
Answer: 11\(\frac{3}{4}\) = y ÷ 6\(\frac{1}{4}\)

Explanation:
We have to write the word sentence in the equation form.
The phrase quotient indicates ‘÷’
The equation would be 11\(\frac{3}{4}\) = y ÷ 6\(\frac{1}{4}\)

Question 23.
YOU BE THE TEACHER
Your friend writes the word sentence as an equation. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 12
Answer:
Given the word sentence, 5 less than a number n is 12.

Question 24.
MODELING REAL LIFE
Students and faculty raise $6042 for band uniforms. The faculty raised $1780. Write an equation you can use to find the amount a (in dollars) the students raised.
Answer:
Given,
Students and faculty raise $6042 for band uniforms (x).
The faculty raised $1780 (y)
The students raised be z
z = x – y
z = 6042 – 1780
z = 4262

Question 25.
MODELING REAL LIFE
You hit a golf ball 90 yards. It travels three-fourths of the distance to the hole. Write an equation you can use to find the distance d (in yards) from the tee to the hole.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 13
Answer:
Given,
You hit a golf ball 90 yards. It travels three-fourths of the distance to the hole.
3/4 × D = 90
D = 360/3
D = 120

GEOMETRY Write an equation you can use to find the value of x.
Question 26.
Perimeter of triangle: 16 in.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 14
Answer:
side of the triangle = x
Perimeter of triangle 16 in
P = a + b + c
16 in = x + x + x
3x = 16
x = 16/3
x = 5.3
Thus the side of the triangle is 5.3 inches.

Question 27.
Perimeter of square: 30 mm
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 15
Answer:
4x = 30
x = 30/4
x = 7.5 mm

Question 28.
MODELING REAL LIFE
You sell instruments at a Caribbean music festival. You earn $326 by selling 12 sets of maracas,6 sets of claves, and x djembe drums. Find the number of djembe drums you sold.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 16
Answer:
Let the price of maracas be m
Let the price of claves be c
Let the price of djembe drums be x
Number of maracas = 12 sets
Number of claves = 6 sets
Number of djembe drums = xx
Total earned amount = $326
The equation would be
12m + 6c + dxx = 326
The cost for 1 maracas is $14
For 12 sets = 12 × 14 = $168
The cost for 1 clave = $5
For 6 sets = 6 × 5 = $30
The cost for 1 djembe drums is $16
For x sets = 16x
12m + 6c + dxx = 326
168 + 30 + 16x = 326
16x = 128
x = 128 ÷ 16
x = 8

Question 29.
PROBLEM SOLVING
Neil Armstrong set foot on the Moon 109.4 hours after Apollo 11departed from the Kennedy Space Center. Apollo 11landed on the Moon about 6.6 hours before Armstrong’s first step. How many hours did it take for Apollo 11 to reach the Moon?
Answer:
Given,
Neil Armstrong set foot on the Moon 109.4 hours after Apollo 11 departed from the Kennedy Space Center.
Apollo 11landed on the Moon about 6.6 hours before Armstrong’s first step.
To find how many hours did it take for Apollo 11 to reach the Moon we have to subtract 6.6 hours from 109.4 hours
109.4 – 6.6 = 102.8 hours
Thus it took 102.8 hours for Apollo 11 to reach the Moon.

Question 30.
LOGIC
You buy a basket of 24 strawberries. You eat them as you walk to the beach. It takes the same amount of time to walk each block. When you are halfway there, half of the berries are gone. After walking 3 more blocks, you still have 5 blocks to go. You reach the beach 28 minutes after you began. One-sixth of your strawberries are left.
a. Is there enough information to find the time it takes to walk each block? Explain.

Answer:
Yes, you are given enough information to find the time to walk each block
To find the total number of block you, add
3 + 5 + 8 = 16
Also the time it takes to walk the 16 blocks is given 28 minutes.

b. Is there enough information to find how many strawberries you ate while walking the last block? Explain.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 17
Answer:
No, there is not enough information to find how many strawberries ate while walking the last block.
You are only given the amount of strawberries you started with 24 and what you have left (1/6) with 5 blocks to go. Therefore you can only be given how many strawberries were eaten walking the last block.

Question 31.
DIG DEEPER!
Find a sales receipt from a store that shows the total price of the items and the total amount paid including sales tax.
a. Write an equation you can use to find the sales tax rate r.
b. Can you use r to find the percent for the sales tax? Explain.
Answer:
Total amount paid = total price + (total price × sales tax rate)
sample equation
14.20 = 13.27 + (13.27 × 0.07)
Yes, you can use r to find the percent for the sales tax.
Multiplying r by 100 gives the percent for the sales tax.

Question 32.
GEOMETRY
A square is cut from a rectangle. The side length of the square is half of the unknown width w. The area of the shaded region is 84 square inches. Write an equation you can use to find the width (in inches).
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations 6.1 18
Answer:
Given,
A square is cut from a rectangle. The side length of the square is half of the unknown width w.
The area of the shaded region is 84 square inches.
84 square inches divided by 14 inches equals 6
84 divided by 14 = s
84 ÷ 14 = s

Lesson 6.2 Solving Equations Using Addition or Subtraction

EXPLORATION 1

Solving an Equation Using a Tape Diagram
Work with a partner. A student solves an equation using the tape diagrams below.
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 1
a. What equation did the student solve? What is the solution?

Answer: x + 4 = 12

Explanation:
By seeing step 1 we can say that the equation for the above tape diagram x + 4 = 12

b. Explain how the tape diagrams in Steps 2 and 3 relate to the equation and its solution.
Answer:
By seeing the steps 2 and 3 we can say
8 + 4 = 12
x + 4 = 12
x = 12 – 4
x = 8

EXPLORATION 2
Solving an Equation Using a Model
Work with a partner.
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 2
a. How are the two sides of an equation similar to a balanced scale?
b. When you add weight to one side of a balanced scale, what can you do to balance the scale? What if you subtract weight from one side of a balanced scale? How does this relate to solving an equation?
c. Use a model to solve x + 2 = 7. Describe how you can solve the equation algebraically.
Answer:
x + 2 = 7
x = 7 – 2
x = 5

Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 3

Try It

Tell whether the given value is a solution of the equation.
Question 1.
a + 6 = 17; a = 9
Answer: not a solution

Explanation:
Given the equation a + 6 = 17
when a = 9
9 + 6 = 17
15 ≠ 17
Thus the equation is not a solution.

Question 2.
9 – g = 5; g = 3
Answer: not a solution

Explanation:
Given the equation 9 – g = 5
where g = 3
9 – 3 = 5
6 ≠ 5
Thus the equation is not a solution.

Question 3.
35 – 7n; n = 5
Answer: solution

Explanation:
Given the equation 35 – 7n
where n = 5
35 – 7(5)
35 – 35 = 0
Thus the equation is a solution.

Question 4.
\(\frac{q}{2}\) = 28; q = 14
Answer: not a solution

Explanation:
Given the equation \(\frac{q}{2}\) = 28
where q = 14
\(\frac{14}{2}\) = 28
7 ≠ 28
Thus the equation is not a solution.

You can use inverse operations to solve equations. Inverse operations “undo” each other. Addition and subtraction are inverse operations.

Solve the equation. Check your solution.
Question 5.
k – 3 = 1
Answer: k = 4

Explanation:
Given the equation k – 3 = 1
k = 1 + 3
k = 4

Question 6.
n – 10 = 4
Answer: n = 14

Explanation:
Given the equation n – 10 = 4
n = 4 + 10
n = 14

Question 7.
15 = r – 6
Answer: r = 21

Explanation:
Given the equation 15 = r – 6
15 + 6 = r
r = 21

Question 8.
s + 8 = 17
Answer: s = 9

Explanation:
Given the equation s + 8 = 17
s = 17 – 8
s = 9

Question 9.
9 = y + 6
Answer: y = 3

Explanation:
Given the equation 9 = y + 6
9 – 6 = y
y = 3

Question 10.
13 + m = 20
Answer: m = 7

Explanation:
Given the equation 13 + m = 20
m = 20 – 13
m = 7

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

CHECKING SOLUTIONS Tell whether the given value is a solution of the equation.
Question 11.
n + 8 = 42; n = 36
Answer: not a solution

Explanation:
Given the equation n + 8 = 42
where n = 36
36 + 8 = 44
44 ≠ 42
Thus the value is not a solution.

Question 12.
g – 9 = 24; g = 35
Answer: not a solution

Explanation:
Given the equation g – 9 = 24
where g = 35
35 – 9 = 24
26 ≠ 24
Thus the value is not a solution.

SOLVING EQUATIONS Solve the equation. Check your solution.
Question 13.
x – 8 = 12
Answer: 20

Explanation:
Given the equation x – 8 = 12
x = 12 + 8
x = 20

Question 14.
b + 14 = 33
Answer: 19

Explanation:
Given the equation b + 14 = 33
b = 33 – 14
b = 19

Question 15.
WRITING
When solving x + 5 =16, why do you subtract 5 from the left side of the equation? Why do you subtract 5 from the right side of the equation?
Answer:
To solve the equation we have to subtract 5.
x + 5 = 16
x = 16 – 5
x = 11

Question 16.
REASONING
Do the equations have the same solution? Explain your reasoning.
x – 8 = 6
x – 6 = 8
Answer:
i. x – 8 = 6
x = 6 + 8
x = 14
ii. x – 6 = 8
x = 8 + 6
x = 14
Yes both the equations has same solutions.

Question 17.
STRUCTURE
Just by looking at the equation x + 6 + 2x = 2x + 6 + 4, find the value of x. Explain your reasoning.
Answer:
x + 6 + 2x = 2x + 6 + 4
3x + 6 = 2x + 10
3x – 2x = 10 – 6
x = 4

Question 18.
An emperor penguin is 45 inches tall. It is 24 inches taller than a rockhopper penguin. Write and solve an equation to find the height (in inches) of a rockhopper penguin. Is your answer reasonable? Explain.
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 4
Answer:
Given,
An emperor penguin is 45 inches tall. It is 24 inches taller than a rockhopper penguin.
45 inches – 24 inches = 21 inches
Thus the height of the rockhopper penguin is 21 inches.

Question 19.
DIG DEEPER!
You get in an elevator and go up 2 floors and down8 floors before exiting. Then you get back in the elevator and go up 4 floors before exiting on the 12th floor. On what floors did you enter the elevator?
Answer: The answer to your question is 14 floor

Explanation:
To solve this problem start from the end changing the sense if it says up, then consider the action as down, etc.
Last floor = 12
Go down 4 floors = 12 – 4 = 8
Go up 8 floors = 8 + 8 = 16
Go down 2 floors = 16 – 2 = 14

Solving Equations Using Addition or Subtraction Homework & Practice 6.2

Review & Refresh

Write the word sentence as an equation.
Question 1.
Th sum of a number x and 9 is 15.
Answer: x + 9 = 15

Explanation:
We have to write the equation for the word sentence
The phrase sum indicates ‘+’
The equation would be x + 9 = 15

Question 2.
12 less than a number m equals 20.
Answer: 12 – m = 20

Explanation:
We have to write the equation for the word sentence
The phrase less than indicates ‘-‘
The equation would be 12 – m = 20

Question 3.
The product of a number d and 7 is 63.
Answer: d7 = 63

Explanation:
We have to write the equation for the word sentence
The phrase product indicates ‘×’
The equation would be d × 7 = 63

Question 4.
18 divided by a number s equals 3.
Answer: 18 ÷ s = 3

Explanation:
We have to write the equation for the word sentence
The phrase divided by indicates ‘÷’
The equation would be 18 ÷ s = 3

Divide. Write the answer in simplest form.
Question 5.
\(\frac{1}{2}\) ÷ \(\frac{1}{4}\)
Answer: 2

Explanation:
Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second fraction.
Take the reciprocal of the second fraction by flipping the numerator and denominator and changing the operation to multiplication. Then the equation becomes
\(\frac{1}{2}\) × \(\frac{4}{1}\) = 2

Question 6.
12 ÷ \(\frac{3}{8}\)
Answer: 32

Explanation:
Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second fraction.
Take the reciprocal of the second fraction by flipping the numerator and denominator and changing the operation to multiplication. Then the equation becomes
12 × \(\frac{8}{3}\)
= 4 × 8
= 32

Question 7.
8 ÷ \(\frac{4}{5}\)
Answer: 10

Explanation:
Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second fraction.
Take the reciprocal of the second fraction by flipping the numerator and denominator and changing the operation to multiplication. Then the equation becomes
8 × \(\frac{5}{4}\)
= 2 × 5
= 10

Question 8.
\(\frac{7}{9}\) ÷ \(\frac{3}{2}\)
Answer: \(\frac{14}{27}\)

Explanation:
Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second fraction.
Take the reciprocal of the second fraction by flipping the numerator and denominator and changing the operation to multiplication. Then the equation becomes
\(\frac{7}{9}\) × \(\frac{2}{3}\) = \(\frac{14}{27}\)

Question 9.
Which ratio is not equivalent to 72 : 18?
A. 36 : 9
B. 18 : 6
C. 4 : 1
D. 288 : 72
Answer: B. 18 : 6

Explanation:
72 : 18 = 36:9, 4 : 1, 288 : 72
18 : 6 is not equivalent to 72 : 18
Thus the correct answer is option B.

Evaluate the expression.
Question 10.
(2 + 52) ÷ 3
Answer: 9

Explanation:
Given the expression (2 + 52) ÷ 3
(2 + 25) ÷ 3
27 ÷ 3 = 9

Question 11.
6 + 23 . 3 – 5
Answer: 25

Explanation:
Given the expression 6 + 23 . 3 – 5
6 + 8 . 3 – 5
6 + 24 – 5
6 + 19
25

Question 12.
4 . [3 + 3(20 – 42 – 2)]
Answer: 36

Explanation:
Given the expression 4 . [3 + 3(20 – 42 – 2)]
4(3 + 3(20 – 16 – 2))
4(3 + 3(2))
4 (3 + 6)
4(9)
36

Question 13.
Find the missing values in the ratio table. Then write the equivalent ratios.
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 5
Answer:
Big-Ideas-Math-Answers-6th-Grade-Chapter-6-Equations-6.2-5

Concepts, Skills, & Problem Solving

CHOOSE TOOLS Use a model to solve the equation. (See Explorations 1 and 2, p. 251.)
Question 14.
n + 7 = 9
Answer: n = 2

Explanation:
n + 7 = 9
n = 9 – 7
n = 2

Question 15.
t + 4 = 5
Answer: t = 1

Explanation:
t + 4 = 5
t = 5 – 4
t = 1

Question 16.
c + 2 = 8
Answer: c = 6

Explanation:
c + 2 = 8
c = 8 – 2
c = 6

CHECKING SOLUTIONS Tell whether the given value is a solution of the equation.
Question 17.
x + 42 = 85; x = 43
Answer: solution

Explanation:
x + 42 = 85
Substitute the value of x in the equation
x = 43
43 + 42 = 85

Question 18.
8b = 48; b = 6
Answer: solution

Explanation:
8b = 48
Substitute the value of b in the equation
b = 6
8(6) = 48
48 = 48

Question 19.
19 – g = 7; g = 15
Answer: not a solution

Explanation:
19 – g = 7
Substitute the value of g in the equation
g = 15
19 – 15 = 7
2 ≠ 7
This is not a solution

Question 20.
\(\frac{m}{4}\) = 16; m = 4
Answer: not a solution

Explanation:
\(\frac{m}{4}\) = 16
Substitute the value of m in the equation
\(\frac{4}{4}\) = 16
1 ≠ 16
This is not a solution

Question 21.
w + 23 = 41; w = 28
Answer: not a solution

Explanation:
w + 23 = 41
Substitute the value of w in the equation
28 + 23 = 41
51 ≠ 41
This is not a solution

Question 22.
s – 68 = 11; s = 79
Answer: solution

Explanation:
Given,
s – 68 = 11
Substitute the value of s in the equation
s = 79
79 – 68 = 11
11 = 11
This is a solution

SOLVING EQUATIONS Solve the equation. Check your solution.
Question 23.
y – 7 = 3
Answer:
Given the equation
y – 7 = 3
y = 3 + 7
y = 10

Question 24.
z – 3 = 13
Answer:
Given the equation
z – 3 = 13
z = 13 +3
z = 16

Question 25.
8 = r – 14
Answer:
Given the equation
8 = r – 14
r = 8 + 14
r = 22

Question 26.
p + 5 = 8
Answer:
Given the equation
p + 5 = 8
p = 8 – 5
p = 3

Question 27.
k + 6 = 18
Answer:
Given the equation
k + 6 = 18
k = 18 – 6
k = 12

Question 28.
64 = h + 30
Answer:
Given the equation
64 = h + 30
h = 64 – 30
h = 34

Question 29.
f – 27 = 19
Answer:
Given the equation
f – 27 = 19
f = 19 +27
f = 46

Question 30.
25 = q + 14
Answer:
Given the equation
25 = q + 14
q = 25 – 14
q = 11

Question 31.
\(\frac{3}{4}\) = j – \(\frac{1}{2}\)
Answer:
Given the equation
\(\frac{3}{4}\) = j – \(\frac{1}{2}\)
\(\frac{3}{4}\) + \(\frac{1}{2}\) = j
j = 1 \(\frac{1}{4}\)

Question 32.
x + \(\frac{2}{3}\) = \(\frac{9}{10}\)
Answer:
Given the equation
x + \(\frac{2}{3}\) = \(\frac{9}{10}\)
x = \(\frac{9}{10}\) – \(\frac{2}{3}\)
x = \(\frac{7}{30}\)

Question 33.
1.2 = m – 2.5
Answer:
Given the equation
1.2 = m – 2.5
m = 1.2 + 2.5
m = 3.7

Question 34.
a + 5.5 = 17.3
Answer:
Given the equation
a + 5.5 = 17.3
a = 17.3 – 5.5
a = 11.8

YOU BE THE TEACHER Your friend solves the equation. Is your friend correct? Explain your reasoning.
Question 35.
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 6
Answer:
x + 7 = 13
x = 13 – 7
x = 4
Your friend is incorrect

Question 36.
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 7
Answer:
34 = y – 12
y – 12 = 34
y = 34 + 12
y = 46

Question 37.
MODELING REAL LIFE
The main span of the Sunshine SkywayBridge is 366 meters long. The bridge’s main span is 30 meters shorter than the main span of the Dames Point Bridge. Write and solve an equation to find the length (in meters) of the main span of the Dames Point Bridge.
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 8
Answer:
Given,
The main span of the Sunshine SkywayBridge is 366 meters long.
The bridge’s main span is 30 meters shorter than the main span of the Dames Point Bridge.
336 – 30 = 306
Let the main span of the Sunshine SkywayBridge be x
Let the main span of the Dames Point Bridge be y
x – y = 306

Question 38.
PROBLEM SOLVING
A park has 22 elm trees. Elm leaf beetles have been attacking the trees. After removing several of the diseased trees, there are 13 healthy elm trees left. Write and solve an equation to find the number of elm trees that were removed.
Answer:
Given,
A park has 22 elm trees. Elm leaf beetles have been attacking the trees.
After removing several of the diseased trees, there are 13 healthy elm trees left.
x – y = 9
22 – 13 = 9
Thus the number of trees removed 9.

Question 39.
PROBLEM SOLVING
The area of Jamaica is 6460 square miles less than the area of Haiti. Find the area (in square miles) of Haiti.
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 9
Answer:
Given,
The area of Jamaica is 6460 square miles less than the area of Haiti.
Y = X – 6460
Y = Haiti
X = area of Jamaica

Question 40.
REASONING
The solution of the equation x+ 3 = 12 is shown. Explain each step. Use a property, if possible.
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 10
Answer:
The sum of a number x and 3 is 12
x + 3 = 12
x = 12 – 3
x = 9

WRITING EQUATIONS Write the word sentence as an equation. Then solve the equation.
Question 41.
13 subtracted from a number w is 15.
Answer: w – 13 = 15

Explanation:
We have to write the equation for the word sentence
The phrase subtracted indicates ‘-‘
The equation would be w – 13 = 15

Question 42.
A number k increased by 7 is 34.
Answer: K + 7 = 34

Explanation:
We have to write the equation for the word sentence
The phrase increased indicates ‘+’
The equation would be K + 7 = 34

Question 43.
9 is the difference of a number n and 7.
Answer: n – 7 = 9

Explanation:
We have to write the equation for the word sentence
The phrase difference indicates ‘-‘
The equation would be n – 7 = 9

Question 44.
93 is the sum of a number g and 58.
Answer: g + 58 = 93

Explanation:
We have to write the equation for the word sentence
The phrase sum indicates ‘+’
The equation would be g + 58 = 93

Question 45.
11 more than a number k equals 29.
Answer: 11 + k = 29

Explanation:
We have to write the equation for the word sentence
The phrase more than indicates ‘+’
The equation would be 11 + k = 29

Question 46.
A number p decreased by 19 is 6.
Answer: p – 19 = 6

Explanation:
We have to write the equation for the word sentence
The phrase decreased indicates ‘-‘
The equation would be p – 19 = 6

Question 47.
46 is the total of 18 and a number d.
Answer: 18 + d = 46

Explanation:
We have to write the equation for the word sentence
The phrase total indicates ‘+’
The equation would be 18 + d = 46

Question 48.
84 is 99 fewer than a number c.
Answer: 84 = 99 – c

Explanation:
We have to write the equation for the word sentence
The phrase fewer than indicates ‘-‘
The equation would be 84 = 99 – c

SOLVING EQUATIONS Solve the equation. Check your solution.
Question 49.
b + 7 + 12 = 30
Answer:
Given the equation
b + 7 + 12 = 30
b = 30 – 19
b = 11

Question 50.
y + 4 − 1 = 18
Answer:
Given the equation
y + 4 − 1 = 18
y + 3 = 18
y = 18 – 3
y = 15

Question 51.
m + 18 + 23 = 71
Answer:
Given the equation
m + 18 + 23 = 71
m + 41 = 71
m = 71 – 41
m = 30

Question 52.
v − 7 = 9 + 12
Answer:
Given the equation
v − 7 = 9 + 12
v – 7 = 21
v = 21 + 7
v = 28

Question 53.
5 + 44 = 2 + r
Answer:
Given the equation
5 + 44 = 2 + r
49 = 2 + r
r = 49 – 2
r = 47

Question 54.
22 + 15 = d− 17
Answer:
Given the equation
22 + 15 = d− 17
37 = d – 17
d = 37 + 17
d = 54

GEOMETRY Solve for x.
Question 55.
Perimeter = 48 ft
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 11
Answer:
P = a + b + c
48 ft = x + 20 + 12
x = 48 – 32
x = 16 ft

Question 56.
Perimeter = 132 in.
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 12
Answer:
P = a + b + c + d
132 = 34 + 16 + 34 + x
132 – 84 = x
x = 50 in

Question 57.
Perimeter = 93 ft
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 13
Answer:
P = 8(a + b + c + d + e)
93 ft = 8(18 + 18 + 15 + d + 15)
d = 93/528
d = 0.17

Question 58.
SIMPLIFYING AND SOLVING Compare and contrast the two problems.
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 14
Answer:
2(x + 3) – 4
= 2x + 6 – 4
= 2x + 2

Question 59.
PUZZLE
In a magic square, the sum of the numbers in each row, column, and diagonal is the same. Find the values of a, b, and c. Justify your answers.
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 15
Answer:
The sum of rows and columns is 53.
a = 22
b = 0
c = 0

Question 60.
REASONING
On Saturday, you spend $33, give $15 to a friend, and receive $20 for mowing your neighbor’s lawn. You have $21 left. Use two methods to find how much money you started with that day.
Answer:
Given,
On Saturday, you spend $33, give $15 to a friend, and receive $20 for mowing your neighbor’s lawn.
You have $21 left.
x = a + b + c – d
x = 33 + 15 + 20 – 21
x = 68 – 21
x = 47

Question 61.
DIG DEEPER!
You have $15.
Big Ideas Math Answers 6th Grade Chapter 6 Equations 6.2 16
a. How much money do you have left if you ride each ride once?
b. Do you have enough money to ride each ride twice? Explain.
Answer:
a. bumper cost : $1.75
super pendulum : $1.25 + $1.50= $2.75
giant slide :  $1.75-$0.50= $1.25
ferris wheels : $1.50+$0.50=$2
total money spent=$7.75
money left=$7.75-$15=$7.25

b. No,
money required to ride once =$7.75
total money required to ride twice=$7.75+$7.75=$15.5

Question 62.
CRITICAL THINKING
Consider the equation 15 − y = 8. Explain how you can solve the equation using the Addition and Subtraction Properties of Equality.
Answer:
15 − y = 8
15 = 8 + y
8 + y = 15
y = 15 – 8
y = 7

Lesson 6.3 Solving Equations Using Multiplication or Division

EXPLORATION 1

Solving an Equation Using a Tape Diagram
Work with a partner. A student solves an equation using the tape diagrams below.
Big Ideas Math Answers Grade 6 Chapter 6 Equations 6.3 1
a. What equation did the student solve? What is the solution?

Answer:
ax = b
4x = 20
x = 20/4
x = 5

b. Explain how the tape diagrams in Steps 2 and 3 relate to the equation and its solution.
Answer:
Step 2 and step 3 shows that x = 5

EXPLORATION 2

Solving an Equation Using a Model
Work with a partner. Three robots go out to lunch. They decide to split the $12 bill evenly. The scale represents the number of robots and the price of the meal.
Big Ideas Math Answers Grade 6 Chapter 6 Equations 6.3 2
a. How much does each robot pay?

Answer:
Three robots go out to lunch.
They decide to split the $12 bill evenly.
12/3 = 4
Thus each robot pay $4.

b. When you triple the weight on one side of a balanced scale, what can you do to balance the scale? What if you divide the weight on one side of a balanced scale in half? How does this relate to solving an equation?
c. Use a model to solve 5x = 15. Describe how you can solve the equation algebraically.
Answer:
5x = 15
x = 15/5
x = 3

Try It

Solve the equation. Check your solution.
Question 1.
\(\frac{a}{8}\) = 6
Answer: a = 48

Explanation:
Given the equation
\(\frac{a}{8}\) = 6
a = 6 × 8
a = 48

Question 2.
14 = \(\frac{2y}{5}\)
Answer: y = 35

Explanation:
Given the equation
14 = \(\frac{2y}{5}\)
14 × 5 = 2y
2y = 70
y = 70/2
y = 35

Question 3.
3z ÷ 2 = 9
Answer: z = 6

Explanation:
Given the equation
3z ÷ 2 = 9
3z = 9 × 2
3z = 18
z = 18/3
z = 6

Question 4.
p . 3 = 18
Answer: p = 6

Explanation:
Given the equation
p . 3 = 18
p = 18/3
p = 6

Question 5.
12q = 60
Answer: q = 5

Explanation:
Given the equation
12q = 60
q = 60/12
q = 5

Question 6.
81 = 9r
Answer: r = 9

Explanation:
Given the equation
81 = 9r
r = 81/9
r = 9

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

SOLVING EQUATIONS Solve the equation. Check your solution.
Question 7.
6 = \(\frac{2y}{3}\)
Answer: y = 9

Explanation:
Given the equation
6 = \(\frac{2y}{3}\)
6 × 3 = 2y
2y = 18
y = 18/2
y = 9

Question 8.
8s = 56
Answer: s = 7

Explanation:
Given the equation
8s = 56
s = 56/8
s = 7

Question 9.
WHICH ONE DOESN’T BELONG?
Which equation does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 6 Equations 6.3 3
Answer: \(\frac{1}{4}\)x = 27 does not belong with the other three.
Because
3x= 36
x = 36/3
x = 12
3/4 x = 9
3x = 36
x = 36/3
x = 12
4x = 48
x = 48/4
x = 12

STRUCTURE Just by looking at the equation, find the value of x. Explain your reasoning.
Question 10.
5x + 3x = 5x + 18
Answer:
Given the equation
x(5+3)=5x+18
8x=5x+18
8x-5x=18
3x=18
x=18/3
x=6

Question 11.
8x + \(\frac{x}{2}\) = 8x + 6
Answer:
Given the equation
8x + \(\frac{x}{2}\) = 8x + 6
x (8 + \(\frac{1}{2}\) ) = 8x + 6
8.5x = 8x + 6
8.5x – 8x = 6
0.5x = 6
x = 6/0.5
x = 12

Question 12.
The area of the screen of the smart watch is shown. What are possible dimensions for the length and the width of the screen? Justify your answer.
Big Ideas Math Answers Grade 6 Chapter 6 Equations 6.3 4
Answer:
Given,
Area = 1625 sq.mm
L = 65 mm
W = 25 mm
We know that,
Area of the rectangle = l × w
1625 = 65 × 25
Thus the length and width of the smart watch is 65 mm and 25 mm.

Question 13.
A rock climber climbs at a rate of 720 feet per hour. Write and solve an equation to find the number of minutes it takes for the rock climber to climb 288 feet.
Answer:
Given,
A rock climber climbs at a rate of 720 feet per hour.
The equation is y = 12x
It takes 24 minutes for the rock climber to get 288 feet
288 = 12x
288/12 = 12x/12
24 = x
Now we have time in minutes that it takes to get 288 feet.

Question 14.
DIG DEEPER!
A gift card stores data using a black, magnetic stripe on the back of the card. Find the width w of the stripe.
Big Ideas Math Answers Grade 6 Chapter 6 Equations 6.3 5
Answer:
Given,
Area = 46 \(\frac{3}{4}\) sq. cm
L = 8 \(\frac{1}{2}\) cm
w = 4 cm + x+ \(\frac{2}{3}\) cm
We know that,
Area of the rectangle = l × w
46 \(\frac{3}{4}\) = 4 cm + x+ \(\frac{2}{3}\) × 8 \(\frac{1}{2}\)
= (4 + 0.6 + x) × 8.5
46.75 = 34 + 5.4 + 8.5x
8.5x = 46.75 – 34 – 5.1
8.5x = 7.6
x = 7.6/8.5
x = 0.89
Thus the width is 0.89 cm

Solving Equations Using Multiplication or Division Homework & Practice 6.3

Review & Refresh
Solve the equation. Check your solution.
Question 1.
y – 5 = 6
Answer: y = 11

Explanation:
Given the equation
y – 5 = 6
y = 6 + 5
y = 11

Question 2.
m + 7 = 8
Answer: 1

Explanation:
Given the equation
m + 7 = 8
m = 8 – 7
m = 1

Question 3.
\(\frac{7}{8}\) = \(\frac{1}{4}\) + 9
Answer:
\(\frac{1}{4}\) + 9 = \(\frac{9}{4}\)
\(\frac{7}{8}\) ≠ \(\frac{9}{4}\)
not a solution

Question 4.
What is the value of a3 when a= 4?
A. 12
B. 43
C. 64
D. 81
Answer: 64

Explanation:
a3 when a= 4
4 × 4 × 4 = 64
Thus the correct answer is option C.

Multiply. Write the answer in simplest form.
Question 5.
\(\frac{1}{5}\) . \(\frac{2}{9}\)
Answer:
For fraction multiplication, multiply the numerators and then multiply the denominators to get
\(\frac{1}{5}\) . \(\frac{2}{9}\) = \(\frac{2}{45}\)

Question 6.
\(\frac{5}{12}\) × \(\frac{4}{7}\)
Answer:
For fraction multiplication, multiply the numerators and then multiply the denominators to get
\(\frac{5}{12}\) × \(\frac{4}{7}\) = \(\frac{5}{21}\)

Question 7.
2\(\frac{1}{3}\) . \(\frac{3}{10}\)
Answer:
For fraction multiplication, multiply the numerators and then multiply the denominators to get
2\(\frac{1}{3}\) = \(\frac{7}{3}\)
\(\frac{7}{3}\) × \(\frac{3}{10}\) = \(\frac{21}{30}\)

Question 8.
1\(\frac{3}{4}\) × 2\(\frac{2}{3}\)
Answer:
1\(\frac{3}{4}\) = \(\frac{7}{4}\)
2\(\frac{2}{3}\) = \(\frac{8}{4}\)
\(\frac{7}{4}\) × \(\frac{8}{4}\) = \(\frac{56}{16}\)

Multiply.
Question 9.
Big Ideas Math Answers Grade 6 Chapter 6 Equations 6.3 6
Answer: 0.36

Explanation:
Multiply the two decimals
0.4 × 0.9 = 0.36

Question 10.
Big Ideas Math Answers Grade 6 Chapter 6 Equations 6.3 7
Answer: 0.39

Explanation:
Multiply the two decimals
0.78 × 0.5 = 0.39

Question 11.
2.63 × 4.31
Answer: 11.3353

Explanation:
Multiply the two decimals
2.63 × 4.31 = 11.3353

Question 12.
1.115 × 3.28
Answer: 69.2

Explanation:
Multiply the two decimals
1.115 × 3.28 = 69.2

Concepts, Skills, &Problem Solving

CHOOSE TOOLS Use a model to solve the equation. (See Explorations 1 and 2, p. 259.)
Question 13.
8x = 8
Answer: 1

Explanation:
Given the equation
8x = 8
x = 8/8
x = 1

Question 14.
9 = 3y
Answer: 3

Explanation:
Given the equation
9 = 3y
y = 9/3
y = 3

Question 15.
2z = 14
Answer: 7

Explanation:
Given the equation
2z = 14
z = 14/2
z = 7

SOLVING EQUATIONS Solve the equation. Check your solution.
Question 16.
\(\frac{s}{10}\) = 7
Answer: 70

Explanation:
Given the equation
\(\frac{s}{10}\) = 7
s = 7 × 10
s = 70

Question 17.
6 = \(\frac{t}{s}\)
Answer: 6s = t

Explanation:
Given the equation
6 = \(\frac{t}{s}\)
t = 6s

Question 18.
5x ÷ 6 = 20
Answer:
Given the equation
5x ÷ 6 = 20
5x = 20 × 6
5x = 120
x = 120/5
x = 24

Question 19.
24 = \(\frac{3}{4}\)r
Answer:
Given the equation
24 = \(\frac{3}{4}\)r
24 × 4 = 3r
96 = 3r
r = 32

Question 20.
3a = 12
Answer: 4

Explanation:
Given the equation
3a = 12
a = 12/3
a = 4

Question 21.
5 . z = 35
Answer: 7

Explanation:
Given the equation
5 . z = 35
z = 35/5
z = 7

Question 22.
40 = 4y
Answer: 10

Explanation:
Given the equation
40 = 4y
40/4 = y
y = 10

Question 23.
42 = 7k
Answer: 6

Explanation:
Given the equation
42 = 7k
7k = 42
k = 42/7

Question 24.
7x = 105
Answer: 15

Explanation:
Given the equation
7x = 105
x = 105/7
x = 15

Question 25.
75 = 6 . w
Answer: 12.5

Explanation:
Given the equation
75 = 6 . w
w = 75/6
w = 12.5

Question 26.
13 = d ÷ 6
Answer: 78

Explanation:
Given the equation
13 = d ÷ 6
d = 13 × 6
d = 78

Question 27.
9 = v ÷ 5
Answer: 45

Explanation:
Given the equation
9 = v ÷ 5
v = 9 × 5
v = 45

Question 28.
\(\frac{5d}{9}\) = 10
Answer: 18

Explanation:
Given the equation
\(\frac{5d}{9}\) = 10
5d = 10 × 9
5d = 90
d = 18

Question 29.
\(\frac{3}{5}\) = 4m
Answer: 0.15

Explanation:
Given the equation
\(\frac{3}{5}\) = 4m
3 = 4m × 5
20m = 3
m = 3/20
m = 0.15

Question 30.
136 = 17b
Answer: 19.4

Explanation:
Given the equation
136 = 17b
b = 136/17
b = 19.4

Question 31.
\(\frac{2}{3}\) = \(\frac{1}{4}\)k
Answer: 2.6

Explanation:
Given the equation
\(\frac{2}{3}\) = \(\frac{1}{4}\)k
k = \(\frac{8}{3}\)
k = 2.6

Question 32.
\(\frac{2c}{15}\) = 8.8
Answer: 66

Explanation:
Given the equation
\(\frac{2c}{15}\) = 8.8
2c = 8.8 × 15
2c = 132
c = 132/2
c = 66

Question 33.
7b ÷ 12 = 4.2
Answer: 7.2

Explanation:
Given the equation
7b ÷ 12 = 4.2
7b = 4.2 × 12
b = 7.2

Question 34.
12.5 . n = 32
Answer: 2.56

Explanation:
Given the equation
12.5 . n = 32
n = 32/12.5
n = 2.56

Question 35.
3.4 m = 20.4
Answer: m = 6

Explanation:
Given the equation
3.4 m = 20.4
m = 20.4/3.4
m = 6

Question 36.
YOU BE THE TEACHER
Your friend solves the equation x ÷ 4 =28. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 6 Equations 6.3 8
Answer: Your friend is correct

Explanation:
Your friend solves the equation x ÷ 4 =28.
x ÷ 4 = 28
x = 28/4
x = 7

Question 37.
ANOTHER WAY
Show how you can solve the equation 3x = 9 by multiplying each side by the reciprocal of 3.
Answer:
3x = 9
x = 9 × 1/3
x = 3

Question 38.
MODELING REAL LIFE
Forty-five basketball players participate in a three-on-three tournament. Write and solve an equation to find the number of three-person teams in the tournament.
Big Ideas Math Answers Grade 6 Chapter 6 Equations 6.3 9
Answer: 15

Explanation:
Let the number of teams be x
x × 3 = 45
x = 45/3
x = 15

Question 39.
MODELING REAL LIFE
A theater has 1200 seats. Each row has 20 seats. Write and solve an equation to find the number of rows in the theater.
Answer:
Given,
A theater has 1200 seats. Each row has 20 seats.
Let x be the number of rows.
1200 = 20 × x
x = 1200/20
x = 60

GEOMETRY Solve for x. Check your answer.
Question 40.
Area = 45 square units
Big Ideas Math Answers Grade 6 Chapter 6 Equations 6.3 11
Answer:
l = x
w = 5
Area = 45 square units
We know that,
Area of the Rectangle = l × w
45 = x × 5
x = 45/5
x = 9 units

Question 41.
Area = 176 square units
Big Ideas Math Answers Grade 6 Chapter 6 Equations 6.3 12
Answer:
a = 16
Area = 176 square units
We know that,
Area of the square = a × a
176 = 16 × x
x = 16

Question 42.
LOGIC
Ona test, you earn 92% of the possible points by correctly answering 6 five-point questions and 8 two-point questions. How many points p is the test worth?
Answer:
Given,
Ona test, you earn 92% of the possible points by correctly answering 6 five-point questions and 8 two-point questions.
(6 × 5) + (8 × 2) = 46
92 × 1/100 = 46
0.92x = 46
x = 46/0.92
x = 50

Question 43.
MODELING REAL LIFE
You use index cards to play a homemade game. The object is to be the first to get rid of all your cards. How many cards are in your friend’s stack?
Big Ideas Math Answers Grade 6 Chapter 6 Equations 6.3 13
Answer:
The number of cards in your friend’s stack divided by the height of your’s friends stack equals the number of cards in your stack divided by the height of your stack.
x = the number of cards in your friend’s stack
x ÷ 5 = 48 ÷ 12
x ÷ 5 = 4
x = 4 × 5
x = 20

Question 44.
DIG DEEPER!
A slush drink machine fills 1440 cups in 24 hours.
a. Find the number c of cups each symbol represents.
b. To lower costs, you replace the cups with paper cones that hold 20% less. Find the number n of paper cones that the machine can fill in 24 hours.
Big Ideas Math Answers Grade 6 Chapter 6 Equations 6.3 14
Answer:
The number of symbols times the number of cups per symbol equals the total number of cups filled.
C = The number of cups filled
30 × c = 1440
c = 1440/30
c = 48

Question 45.
NUMBER SENSE
The area of the picture is 100 square inches. The length is 4 times the width. Find the length and width of the picture.
Big Ideas Math Answers Grade 6 Chapter 6 Equations 6.3 15
Answer:
Given,
The area of the picture is 100 square inches.
The length is 4 times the width.
We know that,
Area of Rectangle = l × w
100 sq. in = 4w × w
100 = 4w²
w = √25 = 5
L = 4w
L = 4 × 5
L = 20

Lesson 6.4 Writing Equations in Two Variables

EXPLORATION 1

Writing Equations in Two Variables
Work with a partner. section 3.4 Exploration 1, you used a ratio table to create a graph for an airplane traveling 300 miles per hour. Below is one possible ratio table and graph.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 1
a. Describe the relationship between the two quantities. Which quantity depends on the other quantity?

Answer: By seeing the above graph we can say that miles depend on hours (time).

b. Use variables to write an equation that represents the relationship between the time and the distance. What can you do with this equation? Provide an example.
c. Suppose the airplane is 1500 miles away from its destination. Write an equation that represents the relationship between time and distance from the destination. How can you represent this relationship using a graph?
Answer: The relationship between distance and time is distance is inversely proportional to the time.
5x = 1500

An equation in two variables represents two quantities that change in relationship to one another. A solution of an equation in two variables is an ordered pair that makes the equation true.

Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 4

Try It

Tell whether the ordered pair is a solution of the equation.
Question 1.
y = 7x, (2, 21)
Answer: No

Explanation:
Given the equation y = 7x
y = 7 × 2
y = 14
21 ≠ 14
The ordered pair is not the solution.

Question 2.
y = 5x + 1; (3, 16)
Answer: Yes

Explanation:
Given the equation y = 5x + 1
y = 5 × 3 + 1
y = 15 + 1
y = 16
The ordered pair is the solution.

Question 3.
The equation y = 10x + 25 represents the amount y(in dollars) in your savings account after x weeks. Identify the independent and dependent variables. How much is in your savings account after 8 weeks?
Answer:
Because the amount y remaining depends on the number x weeks.
Y is the dependent variable
X is the independent variable
y = 10x + 25
After 8 weeks x = 8
y = 10 (8) + 25
y = 80 + 25
y = 105

Graph the equation.
Question 4.
y = 3x
Answer:
Big Ideas Math Grade 6 Chapter 6 Equations img_2
When x = 0
y = 3(0) = 0
A(x,y) = (0,0)
When x = 1
y = 3(1) = 3
B(x,y) = (3, 1)
When x = 2
y = 3(2) = 6
C(x,y) = (6, 2)
When x = 3
y = 3(3) = 9
D(x,y) = (9, 3)

Question 5.
y = 4x + 1
Answer:
y = 4x + 1
When x = 0
y = 4(0) + 1
y = 1
When x = 1
y = 4(1) + 1
y = 5
When x = 2
y = 4(2) + 1
y = 9
Big Ideas Math Grade 6 Chapter 6 Equations img_3

Question 6.
y = \(\frac{1}{2}\)x + 2
Answer:
Given,
y = \(\frac{1}{2}\)x + 2
when x = 0
y = \(\frac{1}{2}\)0 + 2
y = 2
when x = 1
y = \(\frac{1}{2}\)1 + 2
y = 2\(\frac{1}{2}\)
y = 2.5
when x = 2
y = \(\frac{1}{2}\)2 + 2
y = 1 + 2
y = 3
Big Ideas Math Grade 6 Chapter 6 Equations img_4

Question 7.
It costs $25 to rent a kayak plus $8 for each hour. Write and graph an equation that represents the total cost (in dollars) of renting the kayak.
Answer:
Given,
It costs $25 to rent a kayak plus $8 for each hour.
y = 8x + 25

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 8.
WRITING
Describe the difference between independent variables and dependent variables.
Answer:
The independent variable is the cause. Its value is independent of other variables in your study. The dependent variable is the effect. Its value depends on changes in the independent variable.

IDENTIFYING SOLUTIONS Tell whether the ordered pair is a solution of the equation.
Question 9.
y = 3x + 8; (4, 20)
Answer:
Given the equation
y = 3x + 8
x = 4
y = 20
20 = 3(4) 8
20 = 12 + 8
20 = 20
The above equation is the solution.

Question 10.
y = 6x – 14; (7, 29)
Answer:
Given the equation
y = 6x – 14
29 = 6(7) – 14
29 = 42 – 14
29 ≠ 28
The above equation is not the solution.

Question 11.
PRECISION
Explain how to graph an equation in two variables.
Answer:

  • Find three points whose coordinates are solutions to the equation. Organize them in a table.
  • Plot the points in a rectangular coordinate system. Check that the points line up. …
  • Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.

Question 12.
WHICH ONE DOESN’T BELONG?
Which one does not belong with the other three? Explain your reasoning.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 5
Answer: n = 4n – 6 does not belong to the other three equations because we can take n has common and we can solve the equation.
Remaining there is not possible to solve the equation.

Question 13.
A sky lantern rises at an average speed of 8 feet per second. Write and graph an equation that represents the relationship between the time and the distance risen. How long does it take the lantern to rise 100 feet?
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 6
Answer:
Given,
A sky lantern rises at an average speed of 8 feet per second.
the lantern to rise 100 feet = ?
8 feets = 1 sec
100 feets = 8 × x
x = 100/8
x = 12.5 sec

Question 14.
You and a friend start biking in opposite directions from the same point. You travel 108 feet every 8 seconds. Your friend travels 63 feet every 6 seconds. How far apart are you and your friend after 15 minutes?
Answer:
Given,
You and a friend start biking in opposite directions from the same point.
You travel 108 feet every 8 seconds. Your friend travels 63 feet every 6 seconds.
Your distance Y,
Y = 108ft/8 seconds × 60 sec/min × 15 min × 1mile/5280 ft
H is determined similarly
Total distance apart in miles = Y + H
You have only 36 minutes while he travels for all 40.

Writing Equations in Two Variables Homework & Practice 6.4

Review & Refresh
Solve the equation.
Question 1.
4x = 36
Answer: 9

Explanation:
Given the equation 4x = 36
x = 36/4
x = 9

Question 2.
\(\frac{x}{8}\) = 5
Answer: 40

Explanation:
Given the equation \(\frac{x}{8}\) = 5
x = 5 × 8
x = 40

Question 3.
\(\frac{4x}{3}\) = 8
Answer: 6

Explanation:
Given the equation \(\frac{4x}{3}\) = 8
4x = 8 × 3
4x = 24
x = 24/4
x = 6

Question 4.
\(\frac{2}{5}\)x = 6
Answer: 15

Explanation:
Given the equation \(\frac{2}{5}\)x = 6
2x = 5 × 6
2x = 30
x = 30/2
x = 15

Divide. Write the answer in simplest form.
Question 5.
3\(\frac{1}{2}\) ÷ \(\frac{4}{5}\)
Answer: 4 \(\frac{3}{8}\)

Explanation:
Convert any mixed numbers to fractions.
Then your initial equation becomes:
3\(\frac{1}{2}\) = \(\frac{7}{2}\)
\(\frac{7}{2}\) × \(\frac{5}{4}\)
= \(\frac{35}{8}\)
Now convert the improper fraction to the mixed fraction.
\(\frac{35}{8}\) = 4 \(\frac{3}{8}\)

Question 6.
7 ÷ 5\(\frac{1}{4}\)
Answer: 1 \(\frac{1}{3}\)

Explanation:
Convert any mixed numbers to fractions.
Then your initial equation becomes:
7 ÷ 5\(\frac{1}{4}\)
5\(\frac{1}{4}\) = \(\frac{21}{4}\)
\(\frac{7}{1}\) ÷ \(\frac{21}{4}\) = 1 \(\frac{1}{3}\)

Question 7.
\(\frac{3}{11}\) ÷ 1\(\frac{1}{8}\)
Answer: \(\frac{8}{33}\)

Explanation:
Convert any mixed numbers to fractions.
Then your initial equation becomes:
\(\frac{3}{11}\) ÷ 1\(\frac{1}{8}\)
1\(\frac{1}{8}\) = \(\frac{9}{8}\)
\(\frac{3}{11}\) ÷ \(\frac{9}{8}\) = \(\frac{8}{33}\)

Question 8.
7\(\frac{1}{2}\) ÷ 1\(\frac{1}{3}\)
Answer: 5 \(\frac{5}{8}\)

Explanation:
Convert any mixed numbers to fractions.
Then your initial equation becomes:
7\(\frac{1}{2}\) = \(\frac{15}{2}\)
1\(\frac{1}{3}\) = \(\frac{4}{3}\)
\(\frac{15}{2}\) ÷ \(\frac{4}{3}\) = \(\frac{45}{8}\)
Now convert the improper fraction to the mixed fraction.
\(\frac{45}{8}\) = 5 \(\frac{5}{8}\)

Question 9.
Find the area of the carpet tile. Then find the area covered by120 carpet tiles.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 7
Answer:
a = 16 in
Area of the square = a × a
A = 16 in × 16 in
A = 256 sq. in
Now we have to find the area covered by120 carpet tiles.
120 × 256 = 30720

Copy and complete the statement. Round to the nearest hundredth if necessary.
Question 10.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 8
Answer: 800

Explanation:
convert from meters to centimeters
1 m = 100 cm
8 m = 8 × 100 cm = 800 cm
Thus 8m = 800cm

Question 11.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 9
Answer:
Explanation:
Convert from ounces to pounds
88 oz = 5.5 pounds

Question 12.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 10
Answer: 709 mL

Explanation:
Convert from cups to milliliters
1 cup = 236.588 mL
3 cups = 709 mL

Question 13.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 11
Answer: 9.321 mi

Explanation:
Convert from km to miles
1 km = 0.621 mi
15 km = 15 × 0.621 mi
15 km = 9.321 miles

Divide.
Question 14.
\(\sqrt [ 6 ]{ 34.8 } \)
Answer: 6th root of 34.8 is 1.806

Question 15.
\(\sqrt [ 4 ]{ 12.8 } \)
Answer: 4th root of 12.8 is 1.891

Question 16.
45.92 ÷ 2.8
Answer: 16.2

Explanation:
Multiplying two decimal numbers
45.92 ÷ 2.8 = 16.2

Question 17.
39.525 ÷ 4.25
Answer: 9.3

Explanation:
Multiplying two decimal numbers
39.525 ÷ 4.25 = 9.3

Concepts, Skills, &Problem Solving

WRITING EQUATIONS Use variables to write an equation that represents the relationship between the time and the distance. (See Exploration 1, p. 265.)
Question 18.
An eagle flies 40 miles per hour.
Answer:
y = distance, x = time, rate = 40 miles per minute
distance = rate . time
y = 40 . x

Question 19.
A person runs 175 yards per minute.
Answer:
y = distance, x = time, rate = 175 yards per minute
distance = rate . time
so y = 175 . x

IDENTIFYING SOLUTIONS Tell whether the ordered pair is a solution of the equation.
Question 20.
y = 4x; (0, 4)
Answer:
x = 0
y = 4
4 = 4(0)
4 ≠ 0
The ordered pair is not a solution of the equation.

Question 21.
y = 3x; (2, 6)
Answer:
x = 2
y = 6
y = 3x
6 = 3(2)
6 = 6
The ordered pair is a solution of the equation.

Question 22.
y = 5x – 10; (3, 5)
Answer:
x = 3
y = 5
y = 5x – 10
5 = 5(3) – 10
5 = 15 – 10
5 = 5
The ordered pair is a solution of the equation.

Question 23.
y = x + 7; (1, 6)
Answer:
x = 1
y = 6
y = x + 7
6 = 1 + 7
6 ≠ 8
The ordered pair is not a solution of the equation.

Question 24.
y = x + 4; (2, 4)
Answer:
x = 2
y = 4
4 = 2 + 4
4 ≠ 6
The ordered pair is not a solution of the equation.

Question 25.
y = x – 5; (6, 11)
Answer:
x = 6
y = 11
11 = 6 – 5
11 ≠ 1
The ordered pair is not a solution of the equation.

Question 26.
y = 6x + 1; (2, 13)
Answer:
x = 2
y = 13
13 = 6(2) + 1
13 = 12 + 1
13 = 13
The ordered pair is a solution of the equation.

Question 27.
y = 7x + 2; (2, 0)
Answer:
x = 2
y = 0
0 = 7(2) + 2
0 = 14 + 2
0 ≠ 16
The ordered pair is not a solution of the equation.

Question 28.
y = 2x – 3; (4, 5)
Answer:
x = 4
y = 5
y = 2x – 3
5 = 2(4) – 3
5 = 8 – 3
5 = 5
The ordered pair is a solution of the equation.

Question 29.
y = 3x – 3; (1, 0)
Answer:
x = 1
y = 0
y = 3x – 3
0 = 3(1) – 3
0 = 3 – 3
0 = 0
The ordered pair is a solution of the equation.

Question 30.
7 = y – 5x; (4, 28)
Answer:
x = 4
y = 28
7 = y – 5x
7 = 28 – 5(4)
7 = 28 – 20
7 ≠ 8
The ordered pair is not a solution of the equation.

Question 31.
y + 3 = 6x; (3, 15)
Answer:
x = 3
y = 15
y + 3 = 6x
15 + 3 = 6(3)
18 = 18
The ordered pair is a solution of the equation.

Question 32.
YOU BE THE TEACHER
Your friend determines whether (5, 1) is a solution of y = 3x + 2. Is your friend correct? Explain your reasoning.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 12
Answer:
x = 5
y = 1
y = 3x + 2
1 = 3(5) + 2
1 = 15 + 2
1 ≠ 17
Your friend is correct.

IDENTIFYING VARIABLES Identify the independent and dependent variables.
Question 33.
The equation A = 25w represents the area A (in square feet) of a rectangular dance floor with a width of w feet.
Answer:
The area of the dance floor (A) depends on the dance floor
A is the dependent variable
and w is the independent variable

Question 34.
The equation c= 0.09s represents the amount c(in dollars) of commission a salesperson receives for making a sale of s dollars.
Answer:
The commissioner a salesperson receives (c) depends on the sales the salesperson makes
c is dependent variable
s is independent variable

Question 35.
The equation t = 12p+ 12 represents the total cost t (in dollars) of a meal with a tip of p percent (in decimal form).
Answer:
The total cost of a meal depends on the tip of percent
the dependent variable is t
the independent variable is p

Question 36.
The equation h = 60 − 4m represents the height h(in inches) of the water in a tank m minutes after it starts to drain.
Answer:
The height of the water (h) depends on the minutes the tank has been draining
the dependent variable is h
the independent variable is m

OPEN-ENDED Complete the table by describing possible independent or dependent variables.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 24
Answer:
37. Independent variable:
The grade you receive on the test dependent variable
38. Independent variable:
The time you reach your destination dependent variable.
39. Dependent variable:
The amount of minutes used to talk independent variable.
40. Dependent variable:
The number of hours you work independent variable.

GRAPHING EQUATIONS Graph the equation.
Question 41.
y = 2x
Answer:
Given,
y = 2x
when x = 0
y = 2(0)
y = 0
when x = 1
y = 2(1)
y = 2
when x = 2
y = 2(2)
y = 4
Big Ideas Math Grade 6 Chapter 6 Equations img_5

Question 42.
y = 5x
Answer:
Given,
y = 5x
when x = 0
y = 5(0)
y = 0
when x = 1
y = 5(1)
y = 5
Big Ideas Math Grade 6 Chapter 6 Equations img_6

Question 43.
y = 6x
Answer:
Big Ideas Math Grade 6 Chapter 6 Equations img_7

Question 44.
y = x + 2
Answer:
Given
y = x + 2
when x = 0
y = 0 + 2
y = 2
(x, y) = (0,2)
when x = 1
y = 1 + 2
y = 3
(x, y) = (1,3)
when x = 2
y = 2 + 2
y = 4
(x, y) = (2,4)
Big Ideas Math Grade 6 Chapter 6 Equations img_8

Question 45.
y = x + 0.5
Answer:
Given,
y = x + 0.5
when x = 0
y = 0 + 0.5
y = 0.5
when x = 1
y = 1 + 0.5
y = 1.5
Big Ideas Math Grade 6 Chapter 6 Equations img_9

Question 46.
y = x + 4
Answer:
Big Ideas Math Grade 6 Chapter 6 Equations img_10

Question 47.
y = x + 10
Answer:
Big Ideas Math Grade 6 Chapter 6 Equations img_11

Question 48.
y = 3x + 2
Answer:
Given,
y = 3x + 2
when x = 0
y = 3(0) + 2
y = 0 + 2
y = 2
(x,y) = (0,2)
when x = 1
y = 3(1) + 2
y = 3 + 2
y = 5
(x,y) = (1,5)
when x = 2
y = 3(2) + 2
y = 6 + 2
y = 8
(x,y) = (2,8)
Big Ideas Math Grade 6 Chapter 6 Equations img_12

Question 49.
y = 2x + 4
Answer:
Given,
y = 2x + 4
when x = 0
y = 2x + 4
y = 2(0) + 4
y = 0 + 4
y = 4
when x = 1
y = 2x + 4
y = 2(1) + 4
y = 2 + 4
y = 6
when x = 2
y = 2x + 4
y = 2(2) + 4
y = 4 + 4
y = 8
Big Ideas Math Grade 6 Chapter 6 Equations img_13

Question 50.
y = \(\frac{2}{3}\)x + 8
Answer:
Big Ideas Math Grade 6 Chapter 6 Equations img_16

Question 51.
y = \(\frac{1}{4}\)x + 6
Answer:
Big Ideas Math Grade 6 Chapter 6 Equations img_15

Question 52.
y = 2.5x + 12
Answer:
Big Ideas Math Grade 6 Chapter 6 Equations img_14

Question 53.
MODELING REAL LIFE
A cheese pizza costs $5. Additional toppings cost $1.50 each. Write and graph an equation that represents the total cost (in dollars) of a pizza.
Answer:
Let x be the total cost of pizza
let x be the number of toppings
Total cost equals the cost of cheese pizza plus the cost of additional toppings times the number of toppings
The equation would be x = 5 + 1.50t
Table & Graph:
Number of toppings                 Total cost, x = 5 + 1.50 t                 Ordered pairs (t, x)

1                                                                6                                           (1, 6.50)
2                                                                9                                           (2, 8)
3                                                                9                                           (3, 9.50)
Big Ideas Math Grade 6 Chapter 6 Equations img_1

Question 54.
MODELING REAL LIFE
It costs $35 for a membership at a wholesale store. The monthly fee is $15. Write and graph an equation that represents the total cost (in dollars) of a membership.
Answer:
The equation is y = 35 + 25x
Table & Graph:
Number of months(x)                 Total cost, y = 35 + 25x                 Ordered pairs (t, x)

1                                                                60                                           (1, 60)
2                                                                85                                           (2, 85)
3                                                                110                                           (3, 110)

Question 55.
PROBLEM SOLVING
The maximum size of a text message is 160 characters. A space counts as one character.
a. Write an equation that represents the number of remaining (unused) characters in a text message as you type.
b. Identify the independent and dependent variables.
c. How many characters remain in the message shown?
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 14
Answer:
x = the number of characters used
y = the number of characters unused
The equation would be y = 160 – x
The number of unused characters (y) depends on the number of used character (x)
The dependent variable is y
The Independent variable is x
Including space and punctuation, 15 characters were used
y = 160 – x
y = 160 – 15
y = 145

Question 56.
CHOOSE TOOLS
A car averages 60 miles per hour on a road trip. Use a graph to represent the relationship between the time and the distance traveled.
Answer:
Big Ideas Math Grade 6 Chapter 6 Equations img_17

PRECISION Write and graph an equation that represents the relationship between the time and the distance traveled.
Question 57.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 15
Answer:
Big Ideas Math Grade 6 Ch 6 Answer Key img_16

Question 58.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 16
Answer:
Big Ideas Math Grade 6 Ch 6 Answer Key img_17

Question 59.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 17
Answer:
Big Ideas Math Grade 6 Ch 6 Answer Key img_18

Question 60.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 18
Answer:
Big Ideas Math Grade 6 Ch 6 Answer Key img_19

IDENTIFYING SOLUTIONS Fill in the blank so that the ordered pair is a solution of the equation.
Question 61.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 19
Answer:
y = 8x + 3
x = 1
y = 8(1) + 3
y = 11
Thus the ordered pair (1, 11)

Question 62.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 20
Answer:
y = 12x + 2
y = 14
14 = 12x + 2
14 – 2 = 12x
12 = 12x
x = 1
Thus the ordered pair (1, 14)

Question 63.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 21
Answer:
y = 9x + 4
y = 22
22 = 9x + 4
22 – 4 = 9x
9x = 18
x = 2

Question 64.
DIG DEEPER!
Can the dependent variable cause a change in the independent variable? Explain.
Answer:
Just like an independent variable, a dependent variable is exactly what it sounds like. It is something that depends on other factors.

Question 65.
OPEN-ENDED
Write an equation that has (3, 4) as a solution.
Answer:
Standard form linear equation
ax + by = c
When a, b and c are constants
We want to make two equations that
i. have that form
ii. do not have all the same solutions and
iii. (3, 4) is a solution to both
a(3) + b(4) = c
3a + 4b = c

Question 66.
MODELING REAL LIFE
You walk 5 city blocks in 12 minutes. How many city blocks can you walk in 2 hours?
Answer:
Given,
You walk 5 city blocks in 12 minutes.
12 min = 5 city
2 hours = 120 minutes
120 minutes = 300 minutes

Question 67.
GEOMETRY
How fast should the ant walk to go around the rectangle in 4 minutes?
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 22
Answer:
First find the perimeter of the rectangle
P = 2L + 2W
P = 2(16) + 2(12)
P = 32 + 24 = 56 in
r = d/t
r = 56/4
r = 14 in/min

Question 68.
MODELING REAL LIFE
To estimate how far you are from lightning (in miles), count the number of seconds between a lightning flash and the thunder that follows. Then divide the number of seconds by 5. Use two different methods to find the number of seconds between a lightning flash and the thunder that follows when a storm is 2.4 miles away.
Answer:
If you count the number of seconds between the flash of lightning and the sound of thunder, and then divide by 5, you’ll get the distance in miles to the lightning: 5 seconds = 1 mile, 15 seconds = 3 miles, 0 seconds = very close.

Question 69.
REASONING
The graph represents the cost c (in dollars) of buying n tickets to a baseball game.
a. Should the points be connected with a line to show all the solutions? Explain your reasoning.
b. Write an equation that represents the graph.
Big Ideas Math Solutions Grade 6 Chapter 6 Equations 6.4 23
Answer: y = 5x + 0.5

Equations Connecting Concepts

Using the Problem-Solving Plan
Question 1.
A tornado forms 12.25 miles from a weather station. It travels away from the station at an average speed of 440 yards per minute. How far from the station is the tornado after 30 minutes?
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations cc 1
Understand the problem.
You know the initial distance between the tornado and the station, and the average speed the tornado is traveling away from the station. You are asked to determine how far the tornado is from the station after 30 minutes.

Make a plan.
First, convert the average speed to miles per minute. Then write an equation that represents the distance d (in miles) between the tornado and the station after t minutes. Use the equation to find the value of d when t = 30.

Solve and check.
Use the plan to solve the problem. Then check your solution.
Answer:
440 × 30 = 13200

Question 2.
You buy 96 cans of soup to donate to a food bank. The store manager discounts the cost of each case for a total discount of $40. Use an equation in two variables to find the discount for each case of soup. What is the total cost when each can of soup originally costs $1.20?
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations cc 2
Answer:
Given,
You buy 96 cans of soup to donate to a food bank.
The store manager discounts the cost of each case for a total discount of $40.
1 case = 12 cans
x = 96 cans
96 = 12 × x
x = 96/12
x = 8
8 cases
8 × $40 = $320
8 × $1.20 = $9.6

Question 3.
The diagram shows the initial amount raised by an organization for cancer research. A business agrees to donate $2 for every $5 donated by the community during an additional fundraising event. Write an equation that represents the total amount raised (in dollars). How much money does the community need to donate for the organization to reach its fundraising goal?
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations cc 3

Answer: 13,000 – 8000 = 5,000

Performance Task

Planning the Climb
At the beginning of this chapter, you watched a STEAM video called “Rock Climbing.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations cc 4

Equations Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations cr 1

Graphic Organizers

You can use an Example and Non-Example Chart to list examples and non-examples of a concept. Here is an Example and Non-Example Chart for the vocabulary term equation.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations cr 2

Choose and complete a graphic organizer to help you study the concept.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations cr 3
1. inverse operations
2. solving equations using addition or subtraction
3. solving equations using multiplication or division
4. equations in two variables
5. independent variables
6. dependent variables

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations crr 1

6.1 Writing Equations in One Variable (pp. 245–250)
Learning Target: Write equations in one variable and write equations that represent real-life problems.

Write the word sentence as an equation.
Question 1.
The product of a number m and 2 is 8.
Answer: m × 2 = 8

Explanation:
We have to write the word sentence in the equation.
The phrase product indicates ‘×’
The equation would be m × 2 = 8

Question 2.
6 less than a number t is 7.
Answer: 6 – t = 7

Explanation:
We have to write the word sentence in the equation.
The phrase less than indicates ‘-‘
The equation would be 6 – t = 7

Question 3.
A number m increased by 5 is 7.
Answer: m + 5 = 7

Explanation:
We have to write the word sentence in the equation.
The phrase increased indicates ‘+’
The equation would be m + 5 = 7

Question 4.
8 is the quotient of a number g and 3.
Answer: g ÷ 3 = 8

Explanation:
We have to write the word sentence in the equation.
The phrase quotient indicates ‘÷’
The equation would be g ÷ 3 = 8

Question 5.
The height of the 50-milliliter beaker is one-third the height of the 2000-milliliter beaker. Write an equation you can use to find the height (in centimeters) of the 2000-milliliter beaker.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations crr 5
Answer: y = 3x

Explanation:
Given,
The height of the 50-milliliter beaker is one-third the height of the 2000-milliliter beaker.
Let the height of 2000 ml beaker = x
Given,
Height of 50 ml beaker, y = 1/3 of x
The equation to find the height of the 2000 ml beaker will be
y = 3x
which means the height of the 2000 ml beaker is three times the height of the 500 ml beaker.
Therefore, the equation is y = 3x.

Question 6.
There are 16 teams in a basketball tournament. After two rounds, 12 teams are eliminated. Write and solve an equation to find the number of teams remaining after two rounds.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations crr 6
Answer:
Given,
There are 16 teams in a basketball tournament. After two rounds, 12 teams are eliminated.
x = 16
y = 12
Number of teams remaining after two rounds = z
z = x – y
z = 16 – 12
z = 4

Question 7.
Write an equation that has a solution of x = 8.
Answer: 4x = 32

Question 8.
Write a word sentence for the equation y + 3 = 5.
Answer: The sum of the numbers y and 3 is 5.

6.2 Solving Equations Using Addition or Subtraction (pp. 251–258)
Learning Target: Write and solve equations using addition or subtraction.

Question 9.
Tell whether x = 7 is a solution of x + 9 = 16.
Answer: Yes

Explanation:
Given the equation
x + 9 = 16
x = 16 – 9
x = 7

Solve the equation. Check your solution.
Question 10.
x – 1 = 8
Answer: 9

Explanation:
Given the equation
x – 1 = 8
x = 8 + 1
x = 9

Question 11.
m + 7 = 11
Answer: 4

Explanation:
Given the equation
m + 7 = 11
m = 11 – 7
m = 4

Question 12.
21 = p – 12
Answer: 33

Explanation:
Given the equation
21 = p – 12
p – 12 = 21
p = 21 + 12
p = 33

Write the word sentence as an equation. Then solve the equation.
Question 13.
5 more than a number x is 9.
Answer: 5 + x = 9

Explanation:
We have to write the word sentence as an equation
The phrase more than indicates ‘+’
Thus the equation would be 5 + x = 9

Question 14.
82 is the difference of a number b and 24.
Answer: b – 24 = 82

Explanation:
We have to write the word sentence as an equation
The phrase difference indicates ‘-‘
Thus the equation would be b – 24 = 82

Question 15.
A stuntman is running on the roof of a train. His combined speed is the sum of the speed of the train and his running speed. The combined speed is 73 miles per hour, and his running speed is 15 miles per hour. Find the speed of the train.
Answer:
Given,
A stuntman is running on the roof of a train. His combined speed is the sum of the speed of the train and his running speed.
The combined speed is 73 miles per hour, and his running speed is 15 miles per hour.
Speed of the train = ?
Z =  x – y
z = 73 – 15
z = 58

Question 16.
Before swallowing a large rodent, a python weighs 152 pounds. After swallowing the rodent, the python weighs 164 pounds. Find the weight of the rodent.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations crr 16
Answer:
Given that,
Before swallowing a large rodent, a python weighs 152 pounds.
After swallowing the rodent, the python weighs 164 pounds.
164 – 152 = 12 pounds

6.3 Solving Equations Using Multiplication or Division (pp. 259–264)
Learning Target: Write and solve equations using multiplication or division.

Solve the equation. Check your solution.
Question 17.
6 . q = 54
Answer: 9

Explanation:
Given the equation
6 . q = 54
q = 54/6
q = 9

Question 18.
k ÷ 3 = 21
Answer: 63

Explanation:
Given the equation
k ÷ 3 = 21
k = 21 × 3
k = 63

Question 19.
\(\frac{5}{7}\)a = 25
Answer: 35

Explanation:
Given the equation
\(\frac{5}{7}\)a = 25
5a = 7 × 25
a = 7 × 5
a = 35

Question 20.
The weight of an object on the Moon is about 16.5% of its weight on Earth. The weight of an astronaut on the Moon is 24.75 pounds. How much does the astronaut weigh on Earth?
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations crr 20
Answer: 150 pounds

Explanation:
Given,
The weight of an object on the Moon is about 16.5% of its weight on Earth.
The weight of an astronaut on the Moon is 24.75 pounds.
Let the astronaut weight on Earth be represented by x.
Based on the information given in the question, thus can be formed into an equation as:
16.5% of x = 24.75
16.5% × x = 24.75
16.5/100 × x = 24.75
0.165x = 24.75
x = 24.75/0.165
x = 150 pounds
The astronaut weighs 150 pounds on Earth.

Question 21.
Write an equation that can be solved using multiplication and has a solution of x = 12.
Answer: 3x = 36

Question 22.
At a farmers’ market, you buy 4 pounds of tomatoes and 2 pounds of sweet potatoes. You spend 80% of the money in your wallet. How much money is in your wallet before you pay?
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations crr 22
Answer: The money in your wallet before you pay is 20 dollars

Explanation:
Cost of 1 pound of tomato = 3 dollars
Therefore,
Cost of 4 pound of tomato = 4 x 3 = 12 dollars
Cost of 4 pound of tomato = 12 dollars
Cost of 1 pound of sweet potatoes = 2 dollars
Therefore,
Cost of 2 pound of sweet potatoes = 2 x 2 = 4 dollars
Cost of 2 pounds of sweet potatoes = 4 dollars
The combined cost spend at the market is:
cost spend at market = Cost of 4 pound of tomato + Cost of 2 pound of sweet potatoes
cost spend at market = 12 + 4 = 16 dollars
You spent 80% of the money in your wallet
Therefore, 80% of the money in your wallet is equal to 16 dollars
Let x be the money in your wallet
Then, we get
80 % of x = 16
80/100 × x = 16
0.8 x = 16
x = 16/0.8
x = 20
Thus money in your wallet before you pay is $20.

6.4 Writing Equations in Two Variables (pp. 265-272)
Learning Target: Write equations in two variables and analyze the relationship between the two quantities.

Tell whether the ordered pair is a solution of the equation.
Question 23.
y = 3x + 1; (2, 7)
Answer:
Given the equation
x = 2
y = 7
7 = 3(2) + 1
7 = 6 + 1
7 = 7
Yes, it is the solution of the equation.

Question 24.
y = 7x – 4; (4, 22)
Answer:
Given the equation
x = 4
y = 22
22 = 7(4) – 4
22 = 28 – 4
22 ≠ 24
No, it is not the solution of the equation.

Question 25.
The equation E = 360m represents the kinetic energy E (in joules) of a roller-coaster car with a mass of m kilograms. Identify the independent and dependent variables.
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations crr 25
Answer: E is the dependent variable
m is the independent variable

Graph the equation.
Question 26.
y = x + 1
Answer:
Given,
y = x + 1
when x = 0
y = 0 + 1
y = 1
(x, y)  = (0, 1)
when x = 1
y = 1 + 1
y = 2
(x, y)  = (1, 2)
when x = 2
y = 2 + 1
y = 3
(x, y)  = (2, 3)
Big Ideas Math Grade 6 Chapter 6 Equations img_19

Question 27.
y = 7x
Answer:
Big Ideas Math Grade 6 Chapter 6 Equations img_20

Question 28.
y = 4x + 3
Answer:

Question 29.
y = \(\frac{1}{2}\)x + 5
Answer:
Given,
y = \(\frac{1}{2}\)x + 5
when x = 0
y = \(\frac{1}{2}\)0+ 5
y = 5
when x = 0
y = \(\frac{1}{2}\)1+ 5
y = 5\(\frac{1}{2}\)
Big Ideas Math Grade 6 Chapter 6 Equations img_21

Question 30.
A taxi ride costs $3 plus $2.50 per mile. Write and graph an equation that represents the total cost (in dollars) of a taxi ride. What is the total cost of a five-mile taxi ride?
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations crr 30
Answer:
Given,
A taxi ride costs $3 plus $2.50 per mile.
5 × 2.50 = $25.50

Question 31.
Write and graph an equation that represents the total cost (in dollars) of renting the bounce house. How much does it cost to rent the bounce house for 6 hours?
Big Ideas Math Answer Key Grade 6 Chapter 6 Equations crr 31
Answer:
C= $25 x 6 = $150
25×6=150+100=250

Question 32.
A car averages 50 miles per hour on a trip. Write and graph an equation that represents the relationship between the time and the distance traveled. How long does it take the car to travel 525 miles?
Answer:
Given,
A car averages 50 miles per hour on a trip.
50 miles – 1 hour
525 miles – x
50 × x = 525
x = 525/50
x = 10.50 hours
Thus it takes 10.5 hours to travel 525 miles.

Equations Practice Test

Question 1.
Write “7 times a number is 84” as an equation.
Answer: 7 × n = 84

Explanation:
We have to write the word sentence into the equation
The phrase times indicates ‘×’
Thus the equation would be 7 × n = 84

Solve the equation. Check your solution.
Question 2.
15 = 7 + b
Answer: b = 8

Explanation:
Given,
15 = 7 + b
b = 15 – 7
b = 8

Question 3.
v – 6 = 16
Answer: v = 22

Explanation:
Given,
v – 6 = 16
v = 16 + 6
v = 22

Question 4.
5x = 70
Answer: x = 14

Explanation:
Given,
5x = 70
x = 70/5
x = 14

Question 5.
\(\frac{6m}{7}\) = 30
Answer:
Given,
\(\frac{6m}{7}\) = 30
6m = 30 × 7
6m = 210
m = 210 ÷ 6
m = 35

Question 6.
Tell whether (3, 27) is a solution of y = 9x
Answer: solution

Explanation:
Given,
y = 9x
x = 3
y = 27
27 = 9(3)
27 = 27
Thus the ordered pair is a solution.

Question 7.
Tell whether (8, 36) is a solution of y = 4x + 2.
Answer: not a solution

Explanation:
Given,
y = 4x + 2.
x = 8
y = 36
36 = 4(8) + 2
36 = 32 + 2
36 ≠ 34

Question 8.
The drawbridge shown consists of two identical sections that open to allow boats to pass. Write s an equation you can use to find the length (in feet) of each section of the drawbridge.
Big Ideas Math Answers 6th Grade Chapter 6 Equations pt 8
Answer: 2s = 366ft

Question 9.
Each ticket to a school dance is $4. The total amount collected in ticket sales is $332. Find the number of students attending the dance.
Answer:
Given,
Each ticket to a school dance is $4.
The total amount collected in ticket sales is $332.
The equation would be
4s = 332
s = 83

Question 10.
A soccer team sells T-shirts for a fundraiser. The company that makes the T-shirts charges $10 per shirt plus a $20 shipping fee per order.
a. Write and graph an equation that represents the total cost (in dollars) of ordering the shirts.

Answer:
For this case, the first thing we must do is define variables:
c = total cost
x = x number of shirts.
The equation that adapts to the problem is:
c (x) = 10x + 20

b. Choose an ordered pair that lies on your graph in part(a). Interpret it in the context of the problem.
Big Ideas Math Answers 6th Grade Chapter 6 Equations pt 10
Answer:
Let’s choose the next ordered pair:
(x, c (x)) = (0, 20)
We verify that it is in the graph:
c (20) = 10 (0) + 20
c (20) = 20 (yes, it belongs to the graph).
In the context of the problem, this point means that the cost per shipment is $ 20

Question 11.
You hand in 2 homework pages to your teacher. Your teacher now has 32 homework pages to grade. Find the number of homework pages that your teacher originally had to grade.
Answer:
Given that,
You hand in 2 homework pages to your teacher.
Your teacher now has 32 homework pages to grade.
32 – 2 = 30

Question 12.
Write an equation that represents the total cost (in dollars) of the meal shown with a tip that is a percent of the check total. What is the total cost of the meal when the tip is 15%?
Big Ideas Math Answers 6th Grade Chapter 6 Equations pt 12
Answer: $41.40

Equations Cumulative Practice

Big Ideas Math Answers 6th Grade Chapter 6 Equations cp 1
Question 1.
You buy roses at a flower shop for $3 each. How many roses can you buy with $27?
A. 9
B. 10
C. 24
D. 81
Answer: 9

Explanation:
given,
You buy roses at a flower shop for $3 each.
27/3 = 9
Thus you can buy 9 roses with $27.
Thus the correct answer is option A.

Question 2.
You are making identical fruit baskets using 16 apples, 24 pears, and 32 bananas. What is the greatest number of baskets you can make using all of the fruit?
F. 2
G. 4
H. 8
I. 16
Answer: 8

Explanation:
Given,
You are making identical fruit baskets using 16 apples, 24 pears, and 32 bananas.
8 baskets
MULTIPLES OF 16, 24, and, 32
16: 1, 2, 4, 8, 16
24: 1, 2, 3, 4, 6, 8, 12, 24
32: 1, 2, 4, 8, 16, 32

Question 3.
Which equation represents the word sentence?
Big Ideas Math Answers 6th Grade Chapter 6 Equations cp 3
A. 18 – 5 = 9 – y
B. 18 + 5 = 9 – y
C. 18 + 5 = y – 9
D. 18 – 5 = y – 9
Answer: 18 + 5 = 9 – y

Explanation:
The suitable equation for the above word sentence is
18 + 5 = 9 – y
Thus the correct answer is option B.

Question 4.
The tape diagram shows the ratio of tickets sold by you and your friend. How many more tickets did you sell than your friend?
Big Ideas Math Answers 6th Grade Chapter 6 Equations cp 4
F. 6
G. 12
H. 18
I. 30
Answer: 6

Explanation:
Each rectangle = 6
6 × 5 = 30
6 × 2 = 12
30 + 12 = 42
Thus the correct answer is option A.

Question 5.
What is the value of x that makes the equation true?
Big Ideas Math Answers 6th Grade Chapter 6 Equations cp 5
59 + x = 112
Answer:
Given the equation
59 + x = 112
x = 112 – 59
x = 53
Thus the value of x that makes the equation true is 53

Question 6.
The steps your friend took to divide two mixed numbers are shown.
Big Ideas Math Answers 6th Grade Chapter 6 Equations cp 6
What should your friend change in order to divide the two mixed numbers correctly?
A. Find a common denominator of 5 and 2.
B. Multiply by the reciprocal of \(\frac{18}{5}\).
C. Multiply by the reciprocal of \(\frac{3}{2}\).
D. Rename 3\(\frac{3}{5}\) as 2\(\frac{8}{5}\).
Answer:  Multiply by the reciprocal of \(\frac{3}{2}\).

Question 7.
A company ordering parts receives a charge of $25 for shipping and handling plus cp$20 per part. Which equation represents the cost (in dollars) of ordering parts?
F. c = 25 + 20p
G. c = 20 + 25p
H. p = 25 + 20c
I. p = 20 + 25c
Answer: c = 25 + 20p

Question 8.
Which property is illustrated by the statement?
5(a + 6) = 5(a) + 5(6)
A. Associative Property of Multiplication
B. Commutative Property of Multiplication
C. Commutative Property of Addition
D. Distributive Property
Answer: Distributive Property

Question 9.
What is the value of the expression?
Big Ideas Math Answers 6th Grade Chapter 6 Equations cp 5
46.8 ÷ 0.156
Answer:
Divide the two decimal numbers
we get the answer
300

Question 10.
In the mural below, the squares that are painted red are marked with the letter R.
Big Ideas Math Answers 6th Grade Chapter 6 Equations cp 10
What percent of the mural is painted red?
F. 24%
G. 25%
H. 48%
I. 50%
Answer: 48%

Question 11.
Which expression is equivalent to 28x + 70?
A. 14 (2x + 5)
B. 14 (5x + 2)
C. 2 (14x + 5)
D. 14 (7x)
Answer: 14 (2x + 5)

Explanation:
28x + 70
Taking 14 as the common factor
14(2x + 5)
Thus the correct answer is option A.

Question 12.
What is the first step in evaluating the expression?
3 . (5 + 2)2 ÷ 7
F. Multiply 3 and 5.
G. Add 5 and 2
H. Evaluate 52.
I. Evaluate 22.
Answer: G. Add 5 and 2

Question 13.
Jeff wants to save $4000 to buy a used car. He has already saved $850. He plans to save an additional $150 each week.
Big Ideas Math Answers 6th Grade Chapter 6 Equations cp 13
Part A Write and solve an equation to represent the number of weeks remaining until he can afford the car.
Jeff saves $150 per week by saving \(\frac{3}{4}\) of what he earns at his job each week.
He works 20 hours per week.
Part B Write an equation to represent the amount per hour that Jeff must earn to save $150 per week. Explain your reasoning.
Part C What is the amount per hour that Jeff must earn? Show your work and explain your reasoning.
Answer: 21 weeks

Explanation:
150 × 21 = 3150
3150 + 850 = 4000

Conclusion:

We put all our efforts and experience to prepare Big Ideas Math Book 6th Grade Answer Key Chapter 6 Equations pdf. Hope you are satisfied with the given data and information regarding Big Ideas Math Book 6th Grade Answer Key Chapter 6 Equations. If you have any kind of doubts, clarify with us here. We wish all the very best for all the aspirants preparing for the exam.

Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume

Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume

High School Students who are seeking homework help can refer to this Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume Ultimate Guide. By using this ultimate preparation material, you all can easily understand and learn the concepts covered in chapter 11 Circumference, Area, and Volume. Enhance your subject knowledge and score high in various exams after referring and practicing the questions and solutions provided in the BIM Geometry Ch 11 Circumference, Area, and volume Textbook Answer Key Pdf. In this article, you will find the aligned Lesson-wise Big Ideas Geometry Answers Ch 11 in pdf format to download & access offline too for free of cost.

Big Ideas Math Book Geometry Answer Key Chapter 11 Circumference, Area, and Volume

Access the Lesson-wise Big Ideas Math Geometry Chapter 11 Solution Key and solve all easy & complex questions of Circumference, Area, and Volume with ease. Also, you can excel in math concepts by practicing the BIM Geometry Textbook Answer Key. Ace up your preparation with BIM Geometry Ch 11 Circumference, Area, and Volume Answer Key and check your knowledge accordingly for better scores and math skills.

Circumference, Area, and Volume Maintaining Mathematical Proficiency

Find the surface area of the prism.

Question 1.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 1

Answer:
The surface area of the prism = 158.

Explanation:
In the above-given question,
given that,
l = 5 ft, w = 8 ft, and h = 3 ft.
where l = length, w = width, and h = height.
The surface area of the rectanguler prism = 2(lw + lh + wh).
surface area = 2(5×8 + 5×3 + 8×3).
surface area = 2(40 + 15 + 24).
surface area = 2(79).
surface area = 158.

Question 2.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 2

Answer:
The surface area of the triangular prism = 68m.

Explanation:
In the above-given question,
given that,
l = 10 m, p = 4 m, and h = 10.
the surface area of the triangular prism = 2B + ph.
b = base, p = perimeter, and h = height.
surface area = 2(6 + 8) + 4(10).
surface area = 2(14) + 40.
surface area = 28 + 40.
surface area = 68m.

Question 3.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 3

Answer:
The surface area of the triangular prism = 42m.

Explanation:
In the above-given question,
given that,
w = 10 cm, p = 4 cm, and h = 5 cm, l = 6 cm.
the surface area of the triangular prism = 2B + ph.
b = base, p = perimeter, and h = height.
surface area = 2(6 + 5) + 4(5).
surface area = 2(11) + 20.
surface area = 22 + 20.
surface area = 42cm.

Find the missing dimension.

Question 4.
A rectangle has a perimeter 0f 28 inches and a width of 5 inches. What is the length of the rectangle?

Answer:
The length of the rectangle = 9 in.

Explanation:
In the above-given question,
given that,
A rectangle has a perimeter of 28 inches and a width of 5 inches.
length of the rectangle = p/2 – w.
length = 28/2 – 5.
where perimeter = 28 in, and w = 5 in.
length = 14 – 5.
length = 9.
so the length of the rectangle = 9 in.

Question 5.
A triangle has an area of 12 square centimeters and a height of 12 centimeters. What is the base of the triangle?

Answer:
The base of the triangle = 2 cm.

Explanation:
In the above-given question,
given that,
A triangle has an area of 12 sq cm and a height of 12 cm.
The base of the triangle = 2(A)/h.
base = 2(12)/12.
base = 24/12.
base = 2cm.
so the base of the triangle = 2 cm.

Question 6.
A rectangle has an area of 84 square feet and a width of 7 feet. What is the length of the rectangle?

Answer:
The length of the rectangle = 12 ft.

Explanation:
In the above-given question,
given that,
A rectangle has an area of 84 sq ft and a width of 7 feet.
area of the rectangle = l x b.
84 = l x 7.
l = 84/7.
l = 12.
so the length of the rectangle = 12 ft.

Question 7.
ABSTRACT REASONING
Write an equation for the surface area of a Prism with a length, width, and height of x inches. What solid figure does the prism represent?

Answer:
The surface area of a prism = 2(lw + wh + lh).

Explanation:
In the above-given question,
given that,
length = l, width = w, and height = x inches.
the surface area of the prism = 2(lw + wh + lh).
the solid figure does the prism represent the rectangular prism.

Circumference, Area, and Volume Monitoring Progress

Draw a net of the three-dimensional figure. Label the dimensions.

Question 1.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 4

Answer:
The surface area of the prism = 64 cm.

Explanation:
In the above-given question,
given that,
l = 2 ft, w = 4 ft, and h = 4 ft.
where l = length, w = width, and h = height.
The surface area of the rectanguler prism = 2(lw + lh + wh).
surface area = 2(2×4 + 4×4 + 4×2).
surface area = 2(8 + 16 + 8).
surface area = 2(32).
surface area = 64.

Question 2.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 5

Answer:
The surface area of the prism = 392 m.

Explanation:
In the above-given question,
given that,
l =8 m, w = 12 m, and h = 5 m.
where l = length, w = width, and h = height.
The surface area of the rectanguler prism = 2(lw + lh + wh).
surface area = 2(8×12 + 12×5 + 5×8).
surface area = 2(96 + 60 + 40).
surface area = 2(196).
surface area = 392.

Question 3.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 6

Answer:
The surface area of the triangular prism = 170 in.

Explanation:
In the above-given question,
given that,
B = 10 in, p = 15 in, and h = 15 in, l = 10 in.
the surface area of the triangular prism = 2B + ph.
b = base, p = perimeter, and h = height.
surface area = 2(10) + 15(10).
surface area = 2(10) + 150.
surface area = 20 + 150.
surface area = 170 in.

11.1 Circumference and Arc Length

Exploration 1

Finding the Length of a Circular Arc

Work with a partner: Find the length of each red circular arc.

a. entire circle
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 7

Answer:

b. one-fourth of a circle
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 8
Answer:

c. one-third of a circle
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 9
Answer:

d. five-eights of a circle
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 10
Answer:

Exploration 2

Using Arc Length

Work with a partner: The rider is attempting to stop with the front tire of the motorcycle in the painted rectangular box for a skills test. The front tire makes exactly one-half additional revolution before stopping. The diameter of the tire is 25 inches. Is the front tire still in contact with the painted box? Explain.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11

Answer:

Communicate Your Answer

Question 3.
How can you find the length of a circular arc?
Answer:
The length of a circular arc = 2
LOOKING FOR REGULARITY IN REPEATED REASONING
To be proficient in math, you need to notice if calculations are repeated and look both for general methods and for shortcuts.
Answer:

Question 4.
A motorcycle tire has a diameter of 24 inches. Approximately how many inches does the motorcycle travel when its front tire makes three-fourths of a revolution?
Answer:

Lesson 11.1 Circumference and Arc Length

Monitoring Progress

Question 1.
Find the circumference of a circle with a diameter of 5 inches.

Answer:
Circumference C = πd
C = 3.14x 5 = 15.7 in

Question 2.
Find the diameter of a circle with a circumference of 17 feet.

Answer:
Diameter d = C/π
d = 17/π = 5.41 ft

Find the indicated measure.

Question 3.
arc length of \(\widehat{P Q}\)
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 12

Answer:
arc length of \(\widehat{P Q}\) is 5.887

Explanation:
\(\widehat{P Q}\) = \(\frac { 75 }{ 360 } \) . π(9)
= 5.887

Question 4.
circumference of ⊙N
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 13

Answer:
arc length of LM/C = LM/360
61.26/C = 270/360
C = 81.68

Question 5.
radius of ⊙G
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 14

Answer:
arc length of EF = \(\frac { 60 }{ 360 } \) • 2πr
10.5 = \(\frac { 1 }{ 6 } \) • 2πr
r = 10.02

Question 6.
A car tire has a diameter of 28 inches. How many revolutions does the tire make while traveling 500 feet?

Answer:
The car tire have to make 69 revolutions to travel 500 ft.

Explanation:
Circumference C = 2πr = πd
C = 28π
Distance travelled = number of revolutions x C
500 x 12 = number of revolutions x 28π
number of revolutions = 68.2

Question 7.
In Example 4. the radius of the arc for a runner on the blue path is 44.02 meters, as shown in the diagram. About how far does this runner travel to go once around the track? Round to the nearest tenth of a meter.

Answer:

Question 8.
Convert 15° to radians.

Answer:
15° = 15 . \(\frac { π radians }{ 180° } \) = \(\frac { π }{ 12 } \) radians

Question 9.
Convert \(\frac{4 \pi}{3}\) radians to degrees.

Answer:
\(\frac{4 \pi}{3}\) radians = \(\frac{4 \pi}{3}\) radians . \(\frac { 180° }{ π radians } \) = 240 degrees

Exercise 11.1 Circumference and Arc Length

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The circumference of a circle with diameter d is C = _______ .
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 1

Question 2.
WRITING
Describe the difference between an arc measure and an arc length.

Answer:
An arc measure is measured in degrees while an arc length is the distance along an arc measured in linear units.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 10, find the indicated measure.

Question 3.
circumference of a circle with a radius of 6 inches
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 3

Question 4.
diameter of a circle with a circumference of 63 feet

Answer:
C = 63 ft
πd = 63
d = 20.05

Question 5.
radius of a circle with a circumference of 28π
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 5

Question 6.
exact circumference of a circle with a diameter of 5 inches

Answer:
C = πd
C = 5π = 15.707

Question 7.
arc length of \(\widehat{A B}\)
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 15
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 7

Question 8.
m\(\widehat{D E}\)
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 16

Answer:
\(\frac { arc length of DE }{ 2πr } \) = \(\frac { DE }{ 360 } \)
\(\frac { 8.73 }{ 2π(10) } \) = \(\frac { DE }{ 360 } \)
DE = 50.01°

Question 9.
circumference of ⊙C
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 17
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 9

Question 10.
radius of ⊙R
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 18
Answer:
\(\frac { arc length of LM }{ 2πr } \) = \(\frac { LM }{ 360 } \)
\(\frac { 38.95 }{ 2πr } \) = \(\frac { 260 }{ 360 } \)
r = 8.583

Question 11.
ERROR ANALYSIS
Describe and correct the error in finding the circumference of ⊙C.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 19
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 11

Question 12.
ERROR ANALYSIS
Describe and correct the error in finding the length of \(\widehat{G H}\).
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 20

Answer:
\(\frac { arc length of GH }{ 2πr } \) = \(\frac { m GH }{ 360 } \)
\(\widehat{G H}\). = \(\frac { 5 }{ 24 } \) . 2π(10)
= 13.08

Question 13.
PROBLEM SOLVING
A measuring wheel is used to calculate the length of a path. The diameter of the wheel is 8 inches. The wheel makes 87 complete revolutions along the length of the path. To the nearest foot, how long is the path?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 21
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 13

Question 14.
PROBLEM SOLVING
You ride your bicycle 40 meters. How many complete revolutions does the front wheel make?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 22

Answer:
Circumference of the front wheel = 2π(32.5)
= 65π cm
Distance covered = 40 m = 40 x 100 = 4000 cm
Number of revolutions = \(\frac { 4000 }{ 65π } \) = 19.58

In Exercises 15-18 find the perimeter of the shaded region.

Question 15.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 23
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 15

Question 16.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 24
Answer:
Two horizontal edges are 2 . 3 = 6
Circumference of circle = 2π(3) = 6π
The perimeter of the shaded region = 6 + 6π

Question 17.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 25
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 17

Question 18.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 26
Answer:

In Exercises 19 – 22, convert the angle measure.

Question 19.
Convert 70° to radians.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 19

Question 20.
Convert 300° to radians.

Answer:
300 • (\(\frac { π }{ 180 } \)) = \(\frac { 5π }{ 3 } \) radian

Question 21.
Convert \(\frac{11 \pi}{12}\) radians to degrees.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 21

Question 22.
Convert \(\frac{\pi}{8}\) radian to degrees.

Answer:
\(\frac { π }{ 8 } \) • \(\frac { 180 }{ π } \)) = 22.5°

Question 23.
PROBLEM SOLVING
The London Eye is a Ferris wheel in London, England, that travels at a speed of 0.26 meter per second. How many minutes does it take the London Eye to complete one full revolution?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 27
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 23

Question 24.
PROBLEM SOLVING
You are planning to plant a circular garden adjacent to one of the corners of a building, as shown. You can use up to 38 feet of fence to make a border around the garden. What radius (in feet) can the garden have? Choose all that apply. Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 28
(A) 7
(B) 8
(C) 9
(D) 10

Answer:
C = 38 ft
2πr = 38
r = 6.04

In Exercises 25 and 26, find the circumference of the circle with the given equation. Write the circumference in terms of π

Question 25.
x2 + y2 = 16
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 25

Question 26.
(x + 2)2 + (y – 3)2 = 9

Answer:
The radius of circle (x + 2)² + (y – 3)² = 9 is 3
C = 2πr = 2π(3) = 6π
The circumference of the circle is 6π units.

Question 27.
USING STRUCTURE
A semicircle has endpoints (- 2, 5) and (2, 8). Find the arc length of the semicircle.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 27

Question 28.
REASONING
\(\widehat{E F}\) is an arc on a circle with radius r. Let x° be the measure of \(\widehat{E F}\). Describe the effect on the length of \(\widehat{E F}\) if you (a) double the radius of the circle, and (b) double the measure of \(\widehat{E F}\).
Answer:

Question 29.
MAKING AN ARGUMENT
Your friend claims that it is possible for two arcs with the same measure to have different arc lengths. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 29

Question 30.
PROBLEM SOLVING
Over 2000 years ago, the Greek scholar Eratosthenes estimated Earth’s circumference by assuming that the Sun’s rays were Parallel. He chose a day when the Sun shone straight down into a well in the city of Syene. At noon, he measured the angle the Sun’s rays made with a vertical stick in the city of Alexandria. Eratosthenes assumed that the distance from Syene to Alexandria was equal to about 575 miles. Explain how Eratosthenes was able to use this information to estimate Earth’s circumference. Then estimate Earth’s circumference.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 29
Answer:

Question 31.
ANALYZING RELATIONSHIPS
In ⊙C the ratio of the length of \(\widehat{P Q}\) to the length of \(\widehat{R S}\) is 2 to 1. What is the ratio of m∠PCQ to m∠RCS?
(A) 4 to 1
(B) 2 to 1
(C) 1 to 4
(D) 1 to 2
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 31

Question 32.
ANALYZING RELATIONSHIPS
A 45° arc in ⊙C and a 30° arc in ⊙P have the same length. What is the ratio of the radius r1 of ⊙C to the radius r2 of ⊙P? Explain your reasoning.
Answer:

Question 33.
PROBLEM SOLVING
How many revolutions does the smaller gear complete during a single revolution of the larger gear?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 30
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 33

Question 34.
USING STRUCTURE
Find the circumference of each circle.
a. a circle circumscribed about a right triangle whose legs are 12 inches and 16 inches long
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 1
c² = a² + b²
c² = 12²+ 16²= 400
c = 20 in
circumference c = dπ
= 20π = 62.83 in

b. a circle circumscribed about a square with a side length of 6 centimeters
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 2
d² = 6²+ 6² = 72
d = 8.49 cm
C = dπ
C = 8.49π
C = 26.67 cm

c. a circle inscribed in an equilateral triangle with a side length of 9 inches
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 3
r = \(\frac { a√3 }{ 3 } \)
r = \(\frac { 9√3 }{ 3 } \)
r = 3√3 = 5.2
C = 2πr
C = 2π (5.2) = 32.67 in

Question 35.
REWRITING A FORMULA
Write a formula in terms of the measure θ (theta) of the central angle in radians) that can he used to find the length of an arc of a circle. Then use this formula to find the length of an arc of a circle with a radius of 4 inches and a central angle of \(\frac{3 \pi}{4}\) radians.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 35

Question 36.
HOW DO YOU SEE IT?
Compare the circumference of ⊙P to the length of \(\widehat{D E}\). Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 31
Answer:

Question 37.
MAKING AN ARGUMENT
In the diagram. the measure of the red shaded angle is 30°. The arc length a is 2. Your classmate claims that it is possible to find the circumference of the blue circle without finding the radius of either circle. Is your classmate correct? Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 32
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 37

Question 38.
MODELING WITH MATHEMATICS
What is the measure (in radians) of the angle formed by the hands of a clock at each time? Explain your reasoning.
a. 1 : 30 P.M.

Answer:
3π/4

b. 3:15 P.M.

Answer:
π/24

Question 39.
MATHEMATICAL CONNECTIONS
The sum of the circumferences of circles A, B, and C is 63π. Find AC.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 33
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 39

Question 40.
THOUGHT PROVOKING
Is π a rational number? Compare the rational number \(\frac{355}{113}\) to π. Find a different rational number that is even closer π.

Answer:
π is not a rational number as it can not be represented as an equivalent fraction. π = 3.14 and 355/113 = 3.14. This fraction resembles that value of π. Therefore a more accurate fraction will be starting by the value of 7 decimla places of π, therefore 3.1415926 x x = a.

Question 41.
PROOF
The circles in the diagram are concentric and \(\overrightarrow{F G}\) ≅ \(\overrightarrow{G H}\) Prove that \(\widehat{J K}\) and \(\widehat{N G}\) have the same length.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 34
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 41.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 41.2

Question 42.
REPEATED REASONING
\(\overline{A B}\) is divided into four congruent segments, and semicircles with radius r are drawn.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 35
a. What is the sum of the four arc lengths?
Answer:

b. What would the sum of the arc lengths be if \(\overline{A B}\) was divided into 8 congruent segments? 16 congruent segments? n congruent segments? Explain your reasoning.
Answer:

Maintaining Mathematical Proficiency

Find the area of the polygon with the given vertices.

Question 43.
X(2, 4), Y(8, – 1), Z(2, – 1)
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 43.1

Question 44.
L(- 3, 1), M(4, 1), N(4, – 5), P(- 3, – 5)

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 4
LP = √(-3 + 3)² + (-5 – 1)² = 6
PN = √(4 + 3)² + (-5 + 5)² = 7
MN = √(4 – 4)² + (-5 – 1)² = 6
LM = √(4 + 3)² + (1 – 1)²= 7
Area = 6 x 7 = 42 units

11.2 Areas of Circles and Sectors

Exploration 1

Finding the Area of a Sector of a Circle

Work with a partner: A sector of a circle is the region bounded by two radii of the circle and their intercepted arc. Find the area of each shaded circle or sector of a circle.

a. entire circle
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 36
Answer:

b. one – fourth of a circle
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 37
Answer:

c. seven – eighths of a circle.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 38
Answer:

d. two – thirds of a circle
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 39
Answer:

Exploration 2

Finding the Area of a Circular Sector

Work with a partner: A center pivot irrigation system consists of 400 meters of sprinkler equipment that rotates around a central pivot point at a rate of once every 3 days to irrigate a circular region with a diameter of 800 meters. Find the area of the sector that is irrigated by this system in one day.
REASONING ABSTRACTLY
To be proficient in math, you need to explain to yourself the meaning of a problem and look for entry points to its solution.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 40
Answer:

Communicate Your Answer

Question 3.
How can you find the area of a sector of a circle?
Answer:

Question 4.
In Exploration 2, find the area of the sector that is irrigated in 2 hours.
Answer:

Lesson 11.2 Areas of Circles and Sectors

Monitoring progress

Question 1.
Find the area of a circle with a radius of 4.5 meters.

Answer:
Circle area = πr²
A = π(4.5)² = 20.25π

Question 2.
Find the radius of a circle with an area of 176.7 square feet.

Answer:
Circle area = πr²
176.7 = πr²
r² = 56.24
r = 7.499

Question 3.
About 58,000 people live in a region with a 2-mile radius. Find the population density in people per square mile.

Answer:
The population density is about 4615.49 people per square mile.

Explanation:
A = πr² = π • 2² = 4π
Population density = \(\frac { number of people }{ area of land } \)
= \(\frac { 58000 }{ 4π } \) = 4615.49

Question 4.
A region with a 3-mile radius has a population density of about 1000 people per square mile. Find the number of people who live in the region.

Answer:
The number of people who live in the region are 28274.

Explanation:
A = πr² = π • 3² = 9π
Population density = \(\frac { number of people }{ area of land } \)
Number of people = 1000 x 9π = 28274

Find the indicated measure

Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 41

Question 5.
area of red sector

Answer:
The area of red sector = 205.25

Explanation:
m∠FDE = 120°, FE = 120° and FGE = 360° – 120° = 240°
Area of red sector = \(\frac { FE }{ 360° } \) • πr²
= \(\frac { 120 }{ 360° } \) • π(14²)
= 205.25

Question 6.
area of blue sector

Answer:
Area of blue sector = 410.5

Explanation:
Area of blue sector = \(\frac { FGE }{ 360° } \) • πr²
= \(\frac { 240 }{ 360° } \) • π(14²) = 410.5

Question 7.
Find the area of ⊙H.

Answer:
Area of ⊙H = 907.92 sq cm

Explanation:
Area of sector FHG =\(\frac { FG }{ 360° } \) • Area of ⊙H
214.37 = \(\frac { 85 }{ 360° } \) • Area of ⊙H
Area of ⊙H = 907.92 sq cm

Question 8.
Find the area of the figure.

Answer:
Area of triangle = \(\frac { 1 }{ 2 } \) • 7 • 7
= 24.5 sq m
Area of semi circle = πr²/2
= π(3.5)²/2
= 19.242255
Area of the figure = 24.5 + 19.24 = 43.74 sq m

Question 9.
If you know the area and radius of a sector of a circle, can you find the measure of the intercepted arc? Explain.

Answer:

Exercise 11.2 Areas of Circles and Sectors

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
A(n) ____________ of a circle is the region bounded by two radii of the circle and their intercepted arc.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 1

Question 2.
WRITING
The arc measure of a sector in a given circle is doubled. will the area of the sector also be doubled? Explain our reasoning.

Answer:
Yes

Explanation:
Area of sector with arc measure x and radius r is s = π/180(xr)
If x becomes doube, then s1 = π/180(2xr) = 2s
This means that if the arc measure doubles, area of the sector also doubles.

Monitoring Progress and Modeling with Mathematics

In Exercise 3 – 10, find the indicated measure,

Question 3.
area of ⊙C
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 42
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 3

Question 4.
area of ⊙C
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 43

Answer:
Area A = πr²
A = π(10)² = 100π sq in

Question 5.
area of a circle with a radius of 5 inches
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 5

Question 6.
area of a circle with a diameter of 16 feet

Answer:
d = 2r
Circle area = πr² = (π/4)d²
= (π/4)16² = 64π

Question 7.
radius of a circle with an area of 89 square feet
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 7

Question 8.
radius of a circle with an area of 380 square inches

Answer:
A = πr²
380 = πr²
r = 10.99

Question 9.
diameter of a circle with an area of 12.6 square inches
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 9

Question 10.
diameter of a circle with an area of 676π square centimeters

Answer:
Area A = 676π square centimeters
(π/4)d² = 676π
d² = 2704
d = 52

In Exercises 11 – 14, find the indicated measure.

Question 11.
About 210,000 people live in a region with a 12-mile radius. Find the population density in people per square mile.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 11

Question 12.
About 650,000 people live in a region with a 6-mile radius. Find the population density in people per square mile.

Answer:
The population density is about 5747 people per square mile.

Explanation:
Area of region = π(6)² = 36π
Population density = \(\frac { Number of people }{ area of land } \)
= \(\frac { 650,000 }{ 36π } \) = 5747.2

Question 13.
A region with a 4-mile radius has a population density of about 6366 people per square mile. Find the number of people who live in the region.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 13

Question 14.
About 79,000 people live in a circular region with a population density of about 513 people per square mile. Find the radius of the region.

Answer:
The radius of the region is 7

Explanation:
Population density = \(\frac { Number of people }{ area of land } \)
513 = \(\frac { 79,000 }{ πr² } \)
πr² = 153.99
r = 7

In Exercises 15-18 find the areas of the sectors formed by∠DFE.

Question 15.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 44
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 15

Question 16.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 45

Answer:
Area of sector = \(\frac { 104° }{ 360° } \) • π(14)²
= 177.88
Area of red region is 177.88 sq cm
Area of blue region = \(\frac { 256° }{ 360° } \) • π(14)²
= 437.86 sq cm

Question 17.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 46
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 17

Question 18.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 47

Answer:
Area of red region is 10.471 sq ft
Area of the blue region is 39.79 sq ft

Explanation:
Area of sector = \(\frac { 75° }{ 360° } \) • π(4)²
= 10.471
Area of red region is 10.471 sq ft
Area of blue region = \(\frac { 285° }{ 360° } \) • π(4)²
= 39.79 sq ft

Question 19.
ERROR ANALYSIS
Describe and correct the error in finding. the area of the circle.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 48
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 19

Question 20.
ERROR ANALYSIS
Describe and correct the error in finding the area of sector XZY when the area of ⊙Z is 255 square feet.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 49

Answer:
Area of ⊙Z is 255 square feet
πr² = 255
r = 9
Area of sector XZY = \(\frac { 115 }{ 360 } \) • 255
n = 81.458 sq ft

In Exercises 21 and 22, the area of the shaded sector is show. Find the indicated measure.

Question 21.
area of ⊙M
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 50
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 21

Question 22.
radius of ⊙M
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 51

Answer:
radius of ⊙M = 3.98

Explanation:
Area of region = \(\frac { 89 }{ 360 } \) . Area of ⊙M
12.36 = \(\frac { 89 }{ 360 } \) . Area of ⊙M
Area of ⊙M = 49.99
πr² = 49.99
r = 3.98

In Exercises 23 – 28, find the area of the shaded region.

Question 23.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 52
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 23

Question 24.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 53

Answer:
The area of the shaded region is 85.840 sq in.

Explanation:
Area of square = 20² = 400
Diameter of one circle = 10
radius of one circle = 5 in
Area of one circle = π(5)² = 78.53
Areas of four circle = 314.159
Area of shaded region = 400 – 314.159 = 85.840

Question 25.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 54
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 25.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 25.2

Question 26.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 55

Answer:
The area of shaded region is 301.59

Explanation:
The radius of smaller circle is 8 cm
The radius of bigger circle is 16 cm
Area of smaller semicircle = \(\frac { 1 }{ 2 } \)(π(8)²) = 100.53
Area of lager semicircle = \(\frac { 1 }{ 2 } \)(π(16)²) = 402.123
Area of shaded region = 402.123 – 100.53 = 301.59

Question 27.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 56
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 27.1

Question 28.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 57
Answer:
Area of shaded region = 7.63

Explanation:
c² = 3² + 4² = 25
c = 5
Radius = 2.5
Circle area = π(2.5)² = 19.63
Area of triangle = (3 x 4)/2 = 6
Area of shaded region = 19.63 – 12 = 7.63

Question 29.
PROBLEM SOLVING
The diagram shows the shape of a putting green at a miniature golf course. One part of the green is a sector of a circle. Find the area of the putting green.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 58
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 29.1

Question 30.
MAKING AN ARGUMENT
Your friend claims that if the radius of a circle is doubled, then its area doubles. Is your friend correct? Explain your reasoning.

Answer:
The friend is not correct. doubling the radius quadruples the area.

Explanation:
Area of circle with radius r = πr²
Area of circle with radius 2r = π(2r)² = 4πr²

Question 31.
MODELING WITH MATHEMATICS
The diagram shows the area of a lawn covered by a water sprinkler.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 59
a. What is the area of the lawn that is covered by the sprinkler?
b. The water pressure is weakened so that the radius is 12 feet. What is the area of the lawn that will be covered?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 31

Question 32.
MODELING WITH MATHEMATICS
The diagram shows a projected beam of light from a lighthouse.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 60
a. What is the area of water that can be covered by the light from the lighthouse?
Answer:
Area = \(\frac { 115 }{ 360 } \) x π(18)²
= 325.15 sq mi

b. What is the area of land that can be covered by the light from the lighthouse?
Answer:
Area = \(\frac { 245 }{ 360 } \) x π(18)²
= 692.72 sq mi

Question 33.
ANALYZING RELATIONSHIPS
Look back at the Perimeters of Similar Polygons Theorem (Theorem 8.1) and the Areas of Similar PoIyons Theorem (Theorem 8.2) in Section 8.1. How would you rewrite these theorems to apply to circles? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 33

Question 34.
ANALYZING RELATIONSHIPS
A square is inscribed in a circle. The same square is also circumscribed about a smaller circle. Draw a diagram that represents this situation. Then find the ratio of the area of the larger circle to the area of the smaller circle.

Answer:
We start by assigning a variable to the radius of the inner circle. It is r, therefore the area of the circle is πr²
It can be seen that the side length of square is twice this radius. Therefore it can be said that the side length of this square is 2r.
Next, it can be seen that the diagonal of the square is diameter of outer circle. Therefore, length of the diagonal of the circle d = 2r√2. outer circle radius = r√2
Area of outer circle 2πr²
The ratio of the area of larger circle to the smaller circle = 2.

Question 35.
CONSTRUCTION
The table shows how students get to school.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 61
a. Explain why a circle graph is appropriate for the data.
b. You will represent each method by a sector of a circle graph. Find the central angle to use for each sector. Then construct the graph using a radius of 2 inches.
c. Find the area of each sector in your graph.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 35.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 35.2

Question 36.
HOW DO YOU SEE IT?
The outermost edges of the pattern shown form a square. If you know the dimensions of the other square, is it possible to compute the total colored area? Explain.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 62
Answer:

Question 37.
ABSTRACT REASONING
A circular pizza with a 12-inch diameter is enough for you and 2 friends. You want to buy pizzas for yourself and 7 friends. A 10-inch diameter pizza with one topping Costs $6.99 and a 14-inch diameter pizza with one topping Costs $12.99. How many 10-inch and 14-inch pizzas should you buy in each situation? Explain.
a. You want to spend as little money as possible.
b. You want to have three pizzas. each with a different topping, and spend as little money as possible.
C. You want to have as much of the thick outer crust as possible.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 37.1

Question 38.
THOUGHT PROVOKING
You know that the area of a circle is πr2. Find the formula for the area of an ellipse, shown below.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 63

Answer:
Ellipse area = πab

Question 39.
MULTIPLE REPRESENTATIONS
Consider a circle with a radius of 3 inches.
a. Complete the table, where x is the measure of the arc and is the area of the corresponding sector. Round your answers to the nearest tenth.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 64
b. Graph the data in the table.
c. Is the relationship between x and y linear? Explain.
d. If parts (a) – (c) were repeated using a circle with a radius of 5 inches, would the areas in the table change? Would your answer to part (c) change? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 39.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 39.2
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 39.3

Question 40.
CRITICAL THINKING
Find the area between the three congruent tangent circles. The radius of each circle is 6 inches.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 65
Answer:

Question 41.
PROOF
Semicircles with diameters equal to three sides of a right triangle are drawn, as shown. Prove that the sum of the areas of the two shaded crescents equals the area of the triangle.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 66
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 41

Maintaining Mathematical proficiency

Find the area of the figure.

Question 42.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 67

Answer:
Area = \(\frac { 1 }{ 2 } \)(base x height)
Area = \(\frac { 1 }{ 2 } \)(18 x 6) = 54 sq in

Question 43.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 68
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 43

Question 44.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 69
Answer:
Area = \(\frac { 1 }{ 2 } \)(base x height)
A = \(\frac { 1 }{ 2 } \)(13 x 9) = 58.5 sq in

Question 45.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 70
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 45

11.3 Areas of Polygons

Exploration 1

Finding the Area of a Regular Polygon

Work with a partner: Use dynamic geometry software to construct each regular polygon with side lengths of 4, as shown. Find the apothem and use it to find the area of the polygon. Describe the steps that you used.
a.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 71
Answer:

b.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 72
Answer:

c.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 73
Answer:

d.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 74
Answer:

Exploration 2

Writing a Formula for Area

Work with a partner: Generalize the steps you used in Exploration 1 to develop a formula for the area of a regular polygon.
REASONING ABSTRACTLY
To be proficient in math, you need to know and flexibly use different properties of operations and objects.
Answer:

Communicate Your Answer

Question 3.
How can you find the area of a regular polygon?
Answer:

Question 4.
Regular pentagon ABCDE has side lengths of 6 meters and an apothem of approximately 4.13 meters. Find the area of ABCDE.
Answer:

Lesson 11.3 Areas of Polygons

Monitoring Progress

Question 1.
Find the area of a rhombus with diagonals d1 = 4 feet and d2 = 5 feet.

Answer:
Area of rhombus = \(\frac { 1 }{ 4 } \)(d₁d₂)
= \(\frac { 1 }{ 4 } \)(4 x 5) = 5 sq ft

Question 2.
Find the area of a kite with diagonals d1 = 12 inches and d1 = 9 inches.

Answer:
Area of kite = \(\frac { 1 }{ 4 } \)(d₁d₂)
= \(\frac { 1 }{ 4 } \)(12 x 9) = 27 sq in

In the diagram. WXYZ is a square inscribed in ⊙P.

Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 75

Question 3.
Identify the center, a radius, an apothem, and a central angle of the polygon.

Answer:
P is the center, PY or PX is the radius, PQ is apothem, ∠XPY is the central angle.

Question 4.
Find m∠XPY, m∠XPQ, and m∠PXQ.

Answer:
m∠XPY = \(\frac { 360 }{ 4 } \) = 90
m∠XPQ = 90/2 = 45
m∠PXQ = 180 – (90 + 45) = 45

Find the area of regular polygon

Question 5.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 76

Answer:
c = √(8² + 6.5²) = 10.3
a = 20.61
Area = 0.25(√5(5+2√5) a²
Area = 0.25(√5(5+2√5) 20.61² = 730.8

Question 6.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 77
Answer:
Area = \(\frac { 5a² }{ 2 } \)√(5+2√5)
= \(\frac { 5(7²) }{ 2 } \)√(5+2√5)
Area = 55377

Exercise 11.3 Areas of Polygons

Vocabulary and Core Concept Check

Question 1.
WRITING
Explain how to find the measure of a central angle of a regular polygon.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 78
Find the radius of ⊙F. Answer:

Answer:
EF = radius = 6.8

Find the apothem of polygon ABCDE.

Answer:
GF = apothem = 5.5

Find AF.

Answer:
AF = √4² + 5.5²
AF = 6.8

Find the radius of polygon ABCDE.
Answer:
AF = radius = 6.8

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the area of the kite or rhombus.

Question 3.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 79
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 3

Question 4.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 80

Answer:
d₁ = 6 + 6 = 12
d₂ = 2 + 10 = 12
area A = \(\frac { 1 }{ 4 } \)(d₁d₂)
= \(\frac { 1 }{ 4 } \)(12 x 12)
= 36

Question 5.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 81
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 5

Question 6.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 82
Answer:
area A = \(\frac { 1 }{ 4 } \)(d₁d₂)
A = \(\frac { 1 }{ 4 } \)(5 x 6) = 7.5

In Exercises 7 – 10, use the diagram

Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 83

Question 7.
Identify the center of polygon JKLMN?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 7

Question 8.
Identify a central angle of polygon JKLMN.

Answer:
∠NPM is the central angle of polygon JKLMN

Question 9.
What is the radius of polygon JKLMN?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 9

Question 10.
What is the apothem of polygon JKLMN?
Answer:
QP is the apothem of polygon JKLMN

In Exercises 11 – 14, find the measure of a central angle of a regular polygon with the given number of sides. Round answers to the nearest tenth of a degree, if necessary.

Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 84

Question 11.
10 sides
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 11

Question 12.
18 sides

Answer:
The measure of central angle = \(\frac { 360 }{ 18 } \) = 20

Question 13.
24 sides
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 13

Question 14.
7 sides

Answer:
The measure of central angle = \(\frac { 360 }{ 7 } \) = 51.42

In Exercises 15 – 18, find the given angle measure for regular octagon ABCDEFGH.

Question 15.
m∠GJH
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 15

Question 16.
m∠GJK

Answer:
m∠GJK = m∠GJH/2
m∠GJK = 22.5

Question 17.
m∠KGJ
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 17

Question 18.
m∠EJH
Answer:
m∠EJH = 3(45) = 135

In Exercises 19 – 24, find the area of the regular polygon.

Question 19.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 85
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 19

Question 20.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 86
Answer:
Area = \(\frac { 5a² }{ 2 } \)√(5+2√5)
Area = \(\frac { 5(6.84)² }{ 2 } \)√(5+2√5)
A = 359.9784

Question 21.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 87
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 21

Question 22.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 88
Answer:
Area = \(\frac { 3√3 a²}{ 2 } \)
A = \(\frac { 3√3 (7)²}{ 2 } \)
A = 127.30

Question 23.
an octagon with a radius of 11 units
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 23

Question 24.
a pentagon with an apothem of 5 units

Answer:
A = 90.75

Explanation:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 6

We know apothem a = and it divides pentagon into triangles, the central angle is divided into 360/5 = 72
After that, we halved this angle and got 2 right triangles with x = 44 and y = 36. Since we know one side and all three angles of the triangle, we can calculate p with the tangent function.
tan y = p/a
tan 36 = p/5
p = 3.63
Since p is just half of the length of the side, we have to multiply it by 2
2 . p = 2 . 3.63 = 7.26 = s
Area = \(\frac { a . s. n }{ 2 } \)
A = \(\frac { 15 x 7.26 x 5 }{ 2 } \)
A = 90.75

Question 25.
ERROR ANALYSIS
Describe and correct the error in finding the area of the kite.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 89
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 25

Question 26.
ERROR ANALYSIS
Describe and correct the error in finding. the area of the regular hexagon.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 90

Answer:
s = √15² – 13² = 7.48
Area = \(\frac { 1 }{ 2 } \)(a . ns)
A = \(\frac { 1 }{ 2 } \)(13 x 6 x 7.48)
A = 291.72

In Exercises 27 – 30, find the area of the shaded region.

Question 27.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 91
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 27

Question 28.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 92

Answer:
Area of the shaded region = 223.75

Explanation:
Square side = diagonal/√2
= 28/√2 = 19.79
Area of square = 19.79² = 392
Circle area = π(14)² = 615.75
Area of the shaded region = 615.75 – 392 = 223.75

Question 29.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 93
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 29

Question 30.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 94
Answer:

Question 31.
MODELING WITH MATHEMATICS
Basaltic columns arc geological formations that result from rapidly cooling lava. Giant’s Causeway in Ireland contains many hexagonal basaltic columns. Suppose the top of one of the columns is in the shape of a regular hexagon with a radius of 8 inches. Find the area of the top of the column to the nearest square inch.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 95
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 31

Question 32.
MODELING WITH MATHEMATICS
A watch has a circular surface on a background that is a regular octagon. Find the area of the octagon. Then find the area of the silver border around the circular face.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 96
Answer:

CRITICAL THINKING
In Exercises 33 – 35, tell whether the statement is true or false. Explain your reasoning

Question 33.
The area of a regular n-gon of a fixed radius r increases as n increases.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 33

Question 34.
The apothem of a regular polygon is always less than the radius.

Answer:
true, the radius always reaches the end of the circle but the apothem never does

Question 35.
The radius of a regular polygon is always less than the side length.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 35

Question 36.
REASONING
Predict which figure has the greatest area and which has the least area. Explain your reasoning. Check by finding the area of each figure.
(A) Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 97
(B) Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 98
(C) Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 99
Answer:
(B) has the highest area and C has the lowest area.

Explanation:
(A) area = π(6.5)² = 132.73
(B) area = 139.25
(C) area = \(\frac { 1 }{ 2 } \)(18 x 15) = 135

Question 37.
USING EQUATIONS
Find the area of a regular
pentagon inscribed in a circle whose equation is given by (x – 4)2 + y + 2)2 = 25.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 37.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 37.2

Question 38.
REASONING
What happens to the area of a kite if you double the length of one of the diagonals? if you double the length of both diagonals? Justify your answer.

Answer:
Area of a kite = \(\frac { 1 }{ 4 } \)(d₁d₂)
If you double the length of one diagonal, then d₁ = 2d₁
Area of kite = \(\frac { 1 }{ 2 } \)(d₁d₂)
If you double length of both diagonals
Area = \(\frac { 1 }{ 4 } \)(2d₁2d₂) = d₁d₂
If you double the length of one diagonal, then the area becomes halve. If you double length of both diagonals, then area becomes 4 times.

MATHEMATICAL CONNECTIONS
In Exercises 39 and 40, write and solve an equation to find the indicated lengths. Round decimal answers to the nearest tenth.

Question 39.
The area of a kite is 324 square inches. One diagonal is twice as long as the other diagonal. Find the length of each diagonal.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 39

Question 40.
One diagonal of a rhombus is four times the length of the other diagonal. The area of the rhombus is 98 square feet. Find the length of each diagonal.

Answer:
The length of each diagonal is 9.89, 2.47.

Explanation:
One diagonal of a rhombus is four times the length of the other diagonal.
d₁ = 4d₂
Area = \(\frac { 1 }{ 4 } \)(d₁d₂)
98 = \(\frac { 1 }{ 4 } \)(d₁(4d₁))
d₁ = 9.89
d₂ = 2.47

Question 41.
REASONING
The perimeter of a regular nonagon. or 9-gon, is 18 inches. Is this enough information to find the area? If so, find the area and explain your reasoning. If not, explain why not.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 41

Question 42.
MAKING AN ARGUMENT
Your friend claims that it is possible to find the area of any rhombus if you only know the perimeter of the rhombus. Is your friend correct? Explain your reasoning.

Answer:
No; A rhombus is not a regular polygon.

Question 43.
PROOF
Prove that the area of any quadrilateral with perpendicular diagonals is A = \(\frac{1}{2}\)d1d2, where d1 and d2 are the lengths of the diagonals.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 100
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 43.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 43.2

Question 44.
HOW DO YOU SEE IT?
Explain how to find the area of the regular hexagon by dividing the hexagon into equilateral triangles.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 101

Answer:

Question 45.
REWRITING A FORMULA
Rewrite the formula for the area of a rhombus for the special case of a square with side length s. Show that this is the same as the formula for the area of a square, A = s2.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 45

Question 46.
REWRITING A FORMULA
Use the formula for the area of a regular polygon to show that the area of an equilateral triangle can be found by using the formula A = \(\frac{1}{4}\)s2√3 where s is the side length.
Answer:

Question 47.
CRITICAL THINKING
The area of a regular pentagon is 72 square centimeters. Find the length of one side.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 47

Question 48.
CRITICAL THINKING
The area of a dodecagon, or 12-gon, is 140 square inches. Find the apothem of the polygon.

Answer:
Let the side length of dodecagon be 2x. The measure of each interior angle of a regular decagon is 150. This implies that the base angle C and A of the resulting isosceles triangle formed by the red sides is equal to 150/2 = 75. The adjacent to this angle is the length 2x/2 = x inches, while the opposite to it is the blue apothem in the right triangle BDC formed. Therefore a = x tan 75. Therefore, area of dodecagon is
140 = 1/2 (x tan75)(12 . 2x)
140 = 44.785 x²
x² = 3.126
x = 1.768

Question 49.
USING STRUCTURE
In the figure, an equilateral triangle lies inside a square inside a regular pentagon inside a regular hexagon. Find the approximate area of the entire shaded region to the nearest whole number.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 102
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 49.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 49.2

Question 50.
THOUGHT PROVOKING
The area of a regular n-gon is given by A = \(\frac{1}{2}\)aP. As n approaches infinity, what does the n-gon approach? What does P approach? What does a approach? What can you conclude from your three answers? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 103
Answer:

Question 51.
COMPARING METHODS
Find the area of regular pentagon ABCDE by using the formula A = \(\frac{1}{2}\)aP, or A = \(\frac{1}{2}\)a • ns. Then find the area by adding the areas of smaller polygons. Check that both methods yield the same area. Which method do you prefer? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 104
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 51.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 51.2
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 51.3

Question 52.
USING STRUCTURE
Two regular polygons both have n sides. One of the polygons is inscribed in, and the other is circumscribed about, a circle of radius r. Find the area between the two polygons in terms of n and r.

Answer:
The radius of the smaller polygon is equal to the apothem of the larger polygon. The central angle is 360/n, therefore the apothem makes an angle of 180/n. Use sine and cosine to find the apothem and side length of the smaller polygon.
asmall = r sin\(\frac { 180 }{ n } \)
ssmall = 2r cos\(\frac { 180 }{ n } \)
Use tangent to find the side length of the large polygon.
Slarge = 2r tan\(\frac { 180 }{ n } \)
Use the formula to find the area of the smaller polygon.
Asmall = 1/2 . asmall . n . ssmall
Asmall = 1/2 . r sin\(\frac { 180 }{ n } \) . n . 2r cos\(\frac { 180 }{ n } \)
Asmall = nr² sin \(\frac { 180 }{ n } \) cos\(\frac { 180 }{ n } \)
Use the formula to find the area of the larger polygon.
ALarge = 1/2 . alarge . n . slarge
= nr² tan\(\frac { 180 }{ n } \)
The area between the polygons is equal to the area of the larger polygon minus the area of the smaller polygon. Use some trig identities to simplify the expression.
A = Alarge – Asmall
A = nr² tan\(\frac { 180 }{ n } sin²[latex]\frac { 180 }{ n }

Maintaining Mathematical Proficiency

Determine whether the figure has line symmetry, rotational symmetry, both, or neither. If the
figure has line symmetry. determine the number of lines of symmetry. It the figure has rotational
symmetry, describe any rotations that map the figure onto itself.

Question 53.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 105
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 53

Question 54.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 106
Answer:
The figure has rotational symmetry.

Question 55.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 107
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 55

Question 56.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 108
Answer:
The figure has one line symmetry.

11.4 Three-Dimensional Figures

Exploration 1

Analyzing a Property of Polyhedra

Work with a partner: The five Platonic solids are shown below. Each of these solids has congruent regular polygons as faces. Complete the table by listing the numbers of vertices, edges, and faces of each Platonic solid.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 109
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 110
Answer:

Communicate Your Answer

Question 2.
What is the relationship between the numbers of vertices V, edges E, and faces F of a polyhedron? (Note: Swiss mathematician Leonhard Euler (1707 – 1783) discovered a formula that relates these quantities.)
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to reason inductively about data.
Answer:

Question 3.
Draw three polyhedra that are different from the Platonic solids given in Exploration 1. Count the numbers of vertices, edges, and faces of each polyhedron Then verify that the relationship you found in Question 2 is valid for each polyhedron.
Answer:

Lesson 11.4 Three-Dimensional Figures

Monitoring progress

Tell whether the solid is a polyhedron. If it is, name the polyhedron.

Question 1.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 111

Answer:
The solid is formed by polygons, so it is a polyhedron. The base is a square, it is a square pyramid.

Question 2.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 112
Answer:
The solid have curved faces. So it is not a polyhedron.

Question 3.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 113
Answer:
The solid is formed by polygons, so it is a polyhedron. iT has two triangles, one rectangle and two squares.

Describe the shape formed by the intersection of the plane and the solid.

Question 4.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 114
Answer:
The cross-section is a pentagon.

Question 5.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 115
Answer:
The cross-section is a hexagon.

Question 6.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 116
Answer:
The cross-section is a circle.

Sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.

Question 7.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 117
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 8

Question 8.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 118
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 9

Question 9.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 119
Answer:

Exercise 11.4 Three-Dimensional Figures

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
A(n) ___________ is a solid that is bounded by polygons.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which solid does not belong with the other three? Explain your reasoning.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 120

Answer:
Cone does not belong with the other three as it has a curved surface and others not.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, match the polyhedron with its name.

3. Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 121A. triangular Prism
4. Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 122B. rectangular pyramid
5. Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 123C. hexagonal pyramid
6. Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 124D. Pentagonal prism

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 5

In Exercises 7 – 10, tell whether the solid is a polyhedron. If it is, name the polyhedron.

Question 7.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 125
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 7

Question 8.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 126
Answer:
Yes, it is a polyhedron. It is a hexagonal prism.

Question 9.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 127
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 9

Question 10.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 128
Answer:
Yes, it is a polyhedron, truncated square pyramid.

In Exercises 11 – 14, describe the cross section formed by the intersection of the plane and the solid.

Question 11.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 129
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 11

Question 12.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 130
Answer:
The cross-section is a square.

Question 13.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 131
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 13

Question 14.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 132
Answer:
The cross-section is a hexagon

In Exercises 15 – 18, sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.

Question 15.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 133
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 15

Question 16.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 134
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume

Question 17.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 135
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 17

Question 18.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 136
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 9

Question 19.
ERROR ANALYSIS
Describe and correct the error in identifying the solid.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 137
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 19

Question 20.
HOW DO YOU SEE IT?
Is the swimming pool shown a polyhedron? If it is, name the polyhedron. If not, explain why not.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 138
Answer:
It is an octagonal polyhedron.

In Exercises 21 – 26, sketch the polyhedron.

Question 21.
triangular prism
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 21

Question 22.
rectangular prism

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11

Question 23.
pentagonal prism
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 23

Question 24.
hexagonal prism
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 12

Question 25.
square pyramid
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 25

Question 26.
pentagonal pyramid

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 13

Question 27.
MAKING AN ARGUMENT
Your Friend says that the polyhedron shown is a triangular prism. Your cousin says that it is a triangular pyramid. Who is correct? Explain your reasoning.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 139
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 27

Question 28.
ATTENDING TO PRECISION
The figure shows a plane intersection a cube through four of its vertices. The edge length of the cube is 6 inches.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 140
a. Describe the shape of the cross section.
Answer:
The cross-section is a rectangle.

b. What is the perimeter of the cross section?
Answer:
The perimeter is 2(l + b)

c. What is the area of the cross section?
Answer:
Area is lb.

REASONING
In Exercises 29 – 34, tell whether it is possible for a cross section of a cube to have the given shape. If it is, describe or sketch how the plane could intersect the cube.

Question 29.
circle
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 29

Question 30.
pentagon
Answer:
yes, cross-section of the cube can be a pentagon.

Question 31.
rhombus
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 31

Question 32.
isosceles triangle
Answer:
Yes, the cross-section can be an isosceles triangle.

Question 33.
hexagon
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 33

Question 34.
scalene triangle
Answer:
Yes, the cross-section can be scalene triangle.

Question 35.
REASONING
Sketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.
a.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 141
b.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 142
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 35

Question 36.
THOUGHT PROVOKING
Describe how Plato might have argued that there are precisely five Platonic Solids (see page 617). (Hint: Consider the angles that meet at a vertex.)
Answer:

Maintaining Mathematical proficiency

Decide whether enough information is given to prove that the triangles are congruent. It so, state the theorem you would use.

Question 37.
∆ABD, ∆ CDB
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 143
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 37

Question 38.
∆JLK, ∆JLM
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 144

Answer:
∆JLK ≅ ∆JLM by SAS congruence theorem.

Question 39.
∆RQP, ∆RTS
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 145
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 39

11.1 – 11.4 Quiz

Find the indicated measure.

Question 1.
m[latex]\widehat{E F}\)
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 146
Answer:
13.7 = \(\frac { m[latex]\widehat{E F}\) }{ 360 } [/latex] • 2π(7)
m\(\widehat{E F}\) = 112.13

Question 2.
arc length of \(\widehat{Q S}\)
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 147
Answer:
arc length of \(\widehat{Q S}\) = \(\frac { 83 }{ 360 } \) • 2π(4)
= 5.79

Question 3.
circumference of ⊙N
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 148
Answer:
8 = \(\frac { 48 }{ 360 } \) • 2πr
C = 60

Question 4.
Convert 26° to radians and \(\frac{5 \pi}{9}\) radians to degrees.

Answer:
26° = 26 . \(\frac { π }{ 180 } \) = \(\frac { 13π }{ 90 } \) radians
\(\frac{5 \pi}{9}\) = \(\frac{5 \pi}{9}\) . \(\frac { 180 }{ π } \) = 100°

Use the figure to find the indicated measure.

Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 149

Question 5.
area of red sector

Answer:
area of red sector = \(\frac { 100 }{ 360 } \) . π(12)²
= 125.66

Question 6.
area of blue sector

Answer:
area of blue sector = \(\frac { 260 }{ 360 } \) . π(12)²
= 326.72

In the diagram, RSTUVWXY is a reuIar octagon inscribed in ⊙C.

Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 150

Question 7.
Identify the center, a radius, an apothem, and a central angle of the polygon.

Answer:
C is center, CY is radius, CZ is apothem, ∠YCR is central angle of the polygon

Question 8.
Find m∠RCY, m∠RCZ, and m∠ZRC.

Answer:
m∠RCY = 360/8 = 45
m∠RCZ = 45/2 = 22.5
m∠ZRC = 180 – (22.5 + 90) = 67.5

Question 9.
The radius of the circle is 8 units. Find the area of the octagon.

Answer:
Area of octagon = 0.5 x 8 x 8 sin 45 = 22.62

Tell whether the solid is a polyhedron. If it is, name the polyhedron.

Question 10.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 151
Answer:
It is not a polyhedron.

Question 11.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 152
Answer:
The solid is a polyhedron. It is an octagonal pyramid.

Question 12.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 153
Answer:
The solid is a polyhedron. It is a pentagonal prism.

Question 13.
Sketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 154

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 14

Question 14.
The two white congruent circles just fit into the blue circle. What is the area of the blue region?
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 155
Answer:
White circle diameter = radius of the blue circle.
6 = radius of the blue circle.
Area of blue circle = π6² = 113.09
Area of white circle = π3² = 28.27
Area of blue region = 113.09 – 2(28.27) = 56.541

Question 15.
Find the area of each rhombus tile. Then find the area of the pattern.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 156

Answer:
Area of yellow tile = \(\frac { 1 }{ 4 } \)(15.7 x 11.4) = 44.745
area of red tile = \(\frac { 1 }{ 4 } \)(18.5 x 6) = 27.75
Area of pattern = 32(44.745) + 23(27.75) = 2070.09 sq mm

11.5 Volumes of Prisms and Cylinders

Exploration 1

Finding volume

Work with a partner: Consider a stack of square papers that is in the form of a right prism.

Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 157

a. What is the volume the prism?
Answer:

b. When you twist the slack of papers, as shown at the right, do you change the volume? Explain your reasoning.
Answer:

c. Write a carefully worded conjecture that describes the conclusion you reached in part (b).
ATTENDING TO PRECISION
To be proficient in math, you need to communicate precisely to others.
Answer:

d. Use your conjecture to find the volume of the twisted stack of papers.
Answer:

Exploration 2

Finding volume

Work with a partner: Use the conjecture you wrote in Exploration I to find the volume of the cylinder.

a.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 158
Answer:

b.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 159
Answer:

Communicate Your Answer

Question 3.
How can you find the volume of a prism or cylinder that is not a right prism or right cylinder?
Answer:

Question 4.
In Exploration 1, would the conjecture you wrote change if the papers in each stack were not squares? Explain your reasoning.
Answer:

Lesson 11.5 Volumes of Prisms and Cylinders

Monitoring Progress

Find the volume of the solid.

Question 1.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 160
Answer:
Volume = Area * height
Area = \(\frac { 1 }{ 2 } \)(5 x 9) = 22.5
Volume = 22. 5 x 8 = 180 cubic m.

Question 2.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 161

Answer:
Area of circle = πr² = π(8)² = 64π
Volume = 64π x 14 = 2814.86 cubic ft

Question 3.
The diagram shows the dimensions of a concrete cylinder. Concrete has a density of 2.3 grams per cubic centimeter. Find the mass of the concrete cylinder to the nearest gram.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 162
Answer:
Mass of the concrete cylinder = 32 x π(24)² = 18432π cubic in

Question 4.
WHAT IF?
In Example 4, you want the length to be 5 meters, the width to be 3 meters. and the volume to be 60 cubic meters. What should the height be?
Answer:
volume = lbh
60 = 5 x 3 x h
h = 4 m

Question 5.
WHAT IF?
In Example 5, you want the height to be 5 meters and the volume to be 75 cubic meters. What should the area of the base be? Give a possible length and width.
Answer:
volume V = base x height
75 = base x 5
Base = 15 sq m

Question 6.
Prism C and prism D are similar. Find the volume of prism D.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 163
Answer:
\(\frac { 12 }{ 3 } \) = \(\frac { 1536 }{ v } \)
v = 384
Volume of prism D = 384 cubic m

Question 7.
Find the volume of the composite solid.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 164
Answer:
Volume = area x height
Volume = \(\frac { 1 }{ 2 } \)(10 x 3 x 6)
= 90 cubic ft

Exercise 11.5 Volumes of Prisms and Cylinders

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
In what type of units is the volume of a solid measured?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 1

Question 2.
COMPLETE THE SENTENCE
Density is the amount of _______ that an object has in a given unit of __________ .

Answer:
Density is the mass of the object divided by its volume.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the volume of the prism.

Question 3.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 165
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 3

Question 4.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 166
Answer:
Volume V = lbh
V = 1.5 x 2 x 4 = 12 m³

Question 5.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 167
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 5

Question 6.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 168
Answer:
Volume = 6 x 11 x 14
V = 924 m³

In Exercises 7 – 10. find the volume of the cylinder.

Question 7.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 169
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 7

Question 8.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 170
Answer:
Volume = πr²h
V = π(13.4)² x 9.8
V = 1759.6π cm³

Question 9.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 171
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 9

Question 10.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 172
Answer:
Volume V = πr²h
Shoter side = 18/2 = 9
Height h = √18² – 9² = 15.588
V = π6² x 15.588
= 1763 m³

In Exercises 11 and 12. make a sketch of the solid and find its volume. Round your answer to the nearest hundredth.

Question 11.
A prism has a height of 11.2 centimeters and an equilateral triangle for a base, where each base edge is 8 centimeters.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 11

Question 12.
A pentagonal prism has a height of 9 feet and each base edge is 3 feet.

Answer:
volume is 139.32 ft³

explanation:
Pentagon area = 15.48
Height h = 9 ft
Volume V = area x height
= 15.48 x 9 = 139.32

Question 13.
PROBLEM SOLVING
A piece of copper with a volume of 8.25 cubic centimeters has a mass of 73.92 grams. A piece of iron with a volume of 5 cubic centimeters has a mass of 39.35 grams. Which metal has the greater density?
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 173
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 13

Question 14.
PROBLEM SOLVING
The United States has minted one-dollar silver coins called the American Eagle Silver Bullion Coin since 1986. Each coin has a diameter of 40.6 millimeters and is 2.98 millimeters thick. The density of silver is 10.5 grams per cubic centimeter. What is the mass of an American Eagle Silver Bullion Coin to the nearest grain?
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 174
Answer:
V = πr²h
V =π(20.3)²  x (2.98)
V = 3856 mm³

Question 15.
ERROR ANALYSIS
Describe and correct the error in finding the volume of the cylinder.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 175
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 15

Question 16.
ERROR ANALYSIS
Describe and correct the error in finding the density of an object that has a mass of 24 grams and a volume of 28.3 cubic centimeters.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 176

Answer:
Density = mass / volume
Density = \(\frac { 24 }{ 28.3 } \)
Density = 0.8480

In Exercises 17 – 22, find the missing dimension of the prism or cylinder.

Question 17.
Volume = 500 ft3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 177
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 17

Question 18.
Volume = 2700 yd3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 178

Answer:
Volume = 2700 yd³
12 x 5 x v = 2700
v = 15 yd

Question 19.
Volume = 80 cm3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 179
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 19

Question 20.
Volume = 72.66 in.3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 180
Answer:
Volume = 72.66 in.3
Area . x = 72.66
10.39 x = 72.66
x = 6.9 in

Question 21.
Volume = 3000 ft3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 181
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 21

Question 22.
Volume = 1696.5 m3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 182
Answer:
Volume = 1696.5
πr²h = 1696.5
πz² x 15 = 1696.5
z² = 36.00
z = 6

In Exercises 23 and 24, find the area of the base of the rectangular prism with the given volume and height. Then give a possible length and width.

Question 23.
V= 154 in.3, h = 11 in.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 23

Question 24.
V = 27 m3,h = 3m

Answer:
V = Bh
27 = B x 3
B = 9

In Exercises 25 and 26, the solids are similar. Find the volume of solid B.

Question 25.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 183
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 25

Question 26.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 184
Answer:
\(\frac { 12 }{ 15 } \) = \(\frac { 4608π }{ V } \)
V = 5760π
Volume of cylinder B = 5760π

In Exercises 27 and 28, the solids are similar. Find the indicated measure.

Question 27.
height x of the base of prism A
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 185
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 27

Question 28.
height h of cylinder B
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 186
Answer:
height h of cylinder B is 40 ft

Explanation:
\(\frac { 7π }{ 5 } \) = \(\frac { 56π }{ h } \)
h = 40

In Exercises 29 – 32. find the volume of the composite solid.

Question 29.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 187
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 29

Question 30.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 188
Answer:
Volume V = 89.32

Explanation:
Volume of square = 4³ = 64
Volume of semicircle = π(2)² x 4 = 8π

Question 31.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 189
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 31

Question 32.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 190

Answer:
The volume of composite solid is 35 cubic ft

Explanation:
Volume of larger prism = 4 x 2 x 5 = 40
Volume of the smaller prism = 1 x 1 x 5 = 5
Volume of larger prism – volume of the smaller prism = 40 – 5 = 35 cubic ft

Question 33.
MODELING WITH MATHEMATICS
The Great Blue Hole is a cylindrical trench located off the coast of Belize. It is approximately 1000 feet wide and 400 feet deep. About how many gallons of water does the Great Blue Hole contain? (1 ft3 ≈ 7.48 gallons)
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 191
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 33

Question 34.
COMPARING METHODS
The Volume Addition Postulate states that the volume of a solid is the sum of the volumes of all its non overlapping parts. Use this postulate to find the volume of the block of concrete in Example 7 by subtracting the volume of each hole from the volume of the large rectangular prism. Which method do you prefer? Explain your reasoning.
Answer:

REASONING
In Exercises 35 and 36, you are melting a rectangular block of wax to make candles. how many candles of the given shape can be made using a block that measures 10 centimeters by 9 centimeters by 20 centimeters?

Question 35.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 192
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 35

Question 36.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 193

Answer:
7 triangular prism candles with the given measures can be made.

Explanation:
Volume of block = 1800
The volume of triangular prism = 4 x 6 x 10 = 240
1800/240 = 7.5

Question 37.
PROBLEM SOLVING
An aquarium shaped like a rectangular prism has a length of 30 inches, a width of 10 inches, and a height of 20 inches. You fill the aquarium \(\frac{3}{4}\) fill with water. When you submerge a rock in the aquarium, the water level rises 0.25 inch.
a. Find the volume of the rock.
b. How many rocks of this size can you place in the aquarium before water spills out?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 37

Question 38.
PROBLEM SOLVING
You drop an irregular piece of metal into a container partially filled with water and measure that the waler level rises 4.8 centimeters. The square base of the container has a side length of 8 centimeters. You measure the mass of the metal to be 450 grams. What is the density of the metal?

Answer:
The density of metal is 1.4648

Explanation:
Density = \(\frac { Mass }{ Volume } \)
Volume V = 4.8 x 64 = 307.2
Density = \(\frac { 450 }{ 307.2 } \) = 1.4648

Question 39.
WRITING
Both of the figures shown arc made up of the same number of congruent rectangles. Explain how Cavalieri’s Principle can be adapted to compare the areas of these figures.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 194
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 39

Question 40.
HOW DO YOU SEE IT?
Each stack of memo papers contains 500 equally-sized sheets of paper. Compare their volumes. Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 195
Answer:

Question 41.
USING STRUCTURE
Sketch the solid formed by the net. Then find the volume of the solid.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 196
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 41

Question 42.
USING STRUCTURE
Sketch the solid with the given views. Then find the volume of the solid.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 197

Answer:
Volume = 2.5 x 3.5 x 6
Volume = 52.5

Question 43.
OPEN-ENDED
Sketch two rectangular prisms that have volumes of 1000 cubic inches hut different surface areas. Include dimensions in your sketches.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 43

Question 44.
MODELING WITH MATHEMATICS
Which box gives you more cereal for your money? Explain.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 199

Answer:
First one gives more cerel for your money.

Explanation:
Bigger one volume = 16 x 4 x 10 = 640
Smaller one volume = 2 x 8 x 10 = 160
6 – 640 means 1 – 106.66
2 – 160 means 1 – 80

Question 45.
CRITICAL THINKING
A 3-inch by 5-inch index card is rotated around a horizontal line and a vertical line to produce two different solids. Which solid has a greater volume? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 199
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 45

Question 46.
CRITICAL THINKING
The height of cylinder X is twice the height of cylinder Y. The radius of cylinder X is half the radius of cylinder Y. Compare the volumes of cylinder X and cylinder Y. Justify your answer.

Answer:
Let the height of cylinder X be h, radius be r and its volume is πr²h
So, the height of cylinder Y is h/2 and radius is 2r, then the volume is 2πr²h
From both expressions, it can be seen that the volume of cylinder y is twice that of cylinder X.

Question 47.
USING STRUCTURE
Find the volume of the solid shown. The bases of the solid are sectors of circles.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 200
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 47

Question 48.
MATHEMATICAL CONNECTIONS
You drill a circular hole of radius r through the base of a cylinder of radius R. Assume the hole is drilled completely through to the other base. You want the volume of the hole to be half the volume of the cylinder. Express r as a function of R.

Answer:
r = √R²/2

Explanation:
The radius of a solid cylinder without a hole is R. So its volume is πR²h
As per the given condition, the volume of the hole must be half of that of the solid cylinder, hole volume is πR²h/2
Volume of cylinder V = πr²h
πR²h/2 = πr²h
R²/2 = r²
r = √R²/2
r = \(\frac { R√2 }{ 2 } \)

Question 49.
ANALYZING RELATIONSHIPS
How can you change the height of a cylinder so that the volume is increased by 25% but the radius remains the same?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 49

Question 50.
ANALYZING RELATIONSHIPS
How can you change the edge length of a cube so that the volume is reduced by 40%?

Answer:
Write the equation of volume of rectangular prism which can be used to evaluate the cube volume
Volume = s x s x s
The above equation shows that the volume of a cube is directly proportional to one of its side length, therefore, if the volume is to be reduced by 40%, then its the length of one of its side must be reduced by 40%, without changing the 2 other of its sides.

Question 51.
MAKING AN ARGUMENT
You have two objects of equal V0lume. Your friend says you can compare the densities of the objects by comparing their mass, because the heavier object will have a greater density. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 51

Question 52.
THOUGHT PROVOKING
Cavalieri’s Principle states that the two solids shown below have the same volume. Do they also have the same surface area? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 201
Answer:

Question 53.
PROBLEM SOLVING
A barn is in the shape of a pentagonal prism with the dimensions shown, The volume of the barn is 9072 cubic feel. Find the dimensions of each half of the root.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 202
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 53

Question 54.
PROBLEM SOLVING
A wooden box is in the shape of a regular pentagonal prism. The sides, top, and bottom of the box are 1 centimeter thick. Approximate the volume of wood used to construct the box. Round your answer to the nearest tenth.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 203
Answer:
Pentagon area = \(\frac { 5 }{ 2 } \)(4)² sin 72
A = 38
Volume = Area x height
Volume = 38 x 6 = 228

Maintaining Mathematical Proficiency

Find the surface area of the regular pyramid.

Question 55.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 204
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 55

Question 56.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 205
Answer:
Surface area = base area + 3bs
Surface area = 166.3 + 3(8)(10)
= 406.3

Question 57.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 206
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 57

11.6 Volumes of Pyramids

Exploration 1

Finding the Volume of a Pyramid

Work with a partner: The pyramid and the prism have the same height and the same square base.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 207
When the pyramid is filled with sand and poured into the prism, it takes three pyramids to fill the prism.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 208
Use this information to write a formula for the volume V of a pyramid.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.
Answer:

Exploration 2

Finding the Volume of a Pyramid
Work with a partner: Use the formula you wrote in Exploration 1 to find the volume of the hexagonal pyramid.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 209
Answer:

Communicate Your Answer

Question 3.
How can you find the volume of a pyramid?
Answer:

Question 4.
In Section 11 .7, you will study volumes of cones. How do you think you could use a method similar to the one presented in Exploration 1 to write a formula for the volume of a cone? Explain your reasoning.
Answer:

Lesson 11.6 Volumes of Pyramids

Monitoring Progress

Find the volume of the pyramid.

Question 1.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 210
Answer:
The volume of the pyramid is 400 cm³

Explanation:
Volume V = \(\frac { 1 }{ 3 } \)Bh
V = \(\frac { 1 }{ 3 } \)(10 x 10 x 12)
V = 400

Question 2.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 211

Answer:
The volume of the pyramid is 2494.13 cm³

Explanation:
Volume V = \(\frac { 1 }{ 3 } \)Bh
V = \(\frac { 1 }{ 3 } \)(374.12 x 20)
V = 2494.13

Question 3.
The volume of a square pyramid is 75 cubic meters and the height is 9 meters. Find the side length of the square base.

Answer:
The side length of the square base is 5 m

Explanation:
Volume V = \(\frac { 1 }{ 3 } \)Bh = 75
\(\frac { 1 }{ 3 } \)B(9) = 75
B = 25
s = 5

Question 4.
Find the height of the triangular pyramid at the left.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 212

Answer:
The height of the triangular pyramid is 8 m

Explanation:
V = 24
\(\frac { 1 }{ 3 } \)Bh = 24
B = \(\frac { 1 }{ 2 } \)(3 x 6) = 9
\(\frac { 1 }{ 3 } \)(9)h = 24
h = 8

Question 5.
Pyramid C and pyramid D are similar. Find the volume of pyramid D.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 213

Answer:
The volume of pyramid D is 12 m³

Explanation:
\(\frac { volume of pyramid C }{ volume of pyramid D } \) = (\(\frac { pyramid C base }{ pyramid D base } \))³
\(\frac { 324 }{ V } \) = (\(\frac { 9 }{ 3 } \))³
V = 12

Question 6.
Find the volume of the composite solid.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 214

Answer:
the volume of solid = 96

Explanation:
Volume of prism = Bh
B = 8 x 2 = 16
V = 16 x 5 = 80
\(\frac { 1 }{ 3 } \)Bh
= \(\frac { 1 }{ 3 } \)(16 x 3) = 16
the volume of solid = 16 + 80 = 96

Exercise 11.6 Volumes of Pyramids

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Explain the difference between a triangular prism and a triangular pyramid.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 1

Question 2.
REASONING
A square pyramid and a cube have the same base and height. Compare the volume of the square pyramid to the volume of the cube.

Answer:
Square pyramid = 1/3 Bh
Cube = BH
So, the volume of the square pyramid is 1/3 of the volume of the cube.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, find the volume of the pyramid.

Question 3.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 215
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 3

Question 4.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 216

Answer:
V = 6 in³

Explanation:
V =  \(\frac { 1 }{ 3 } \)Bh
B = 2 x 3 = 6
V = \(\frac { 1 }{ 3 } \)(6 x 3)

In Exercises 5 – 8, find the indicated measure.

Question 5.
A pyramid with a square base has a volume of 120 cubic meters and a height of 10 meters. Find the side length of the square base.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 5

Question 6.
A pyramid with a square base has a volume of 912 cubic feet and a height of 19 feet. Find the side length of the square base.

Answer:
The side length of the square base is 12 ft

Explanation:
A pyramid with a square base has a volume of 912 cubic feet
h = 19
\(\frac { 1 }{ 3 } \)Bh = 912
\(\frac { 1 }{ 3 } \)B(19) = 912
B = 144
s = 12

Question 7.
A pyramid with a rectangular base has a volume of 480 cubic inches and a height of 10 inches. The width of the rectangular base is 9 inches. Find the length of the rectangular base.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 7

Question 8.
A pyramid with a rectangular base has a volume of 105 cubic centimeters and a height of 15 centimeters. The length of the rectangular base is 7 centimeters. Find the width of the rectangular base.

Answer:
The width of the rectangular base is 3 cm

Explanation:
A pyramid with a rectangular base has a volume of 105 cubic centimeters
h = 15
l = 7
\(\frac { 1 }{ 3 } \)Bh = 105
\(\frac { 1 }{ 3 } \)lbh = 105
\(\frac { 1 }{ 3 } \)(7 x 15 x b) = 105
b = 3

Question 9.
ERROR ANALYSIS
Describe and correct the error in finding the volume of the pyramid.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 217
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 9

Question 10.
OPEN-ENDED
Give an example of a pyramid and a prism that have the same base and the same volume. Explain your reasoning.
Answer:
Let the rectangular prism have the base dimensions 4 x 2 nad a height of 5 so its volume is 4 x 2 x 5 = 40 cubic units
Therefore the base of the rectangular prism also have the dimensions of 4 x 2 and a height of 5 x 3 = 15 units so its volume V = 1/3 x 4 x 2 x 15 = 40 cubic units

In Exercises 11 – 14, find the height of the pyramid.

Question 11.
Volume = 15 ft3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 218
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 11

Question 12.
Volume = 224 in.3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 219

Answer:
The height of the pyramid is 10.5 in

Explanation:
Volume = 224
\(\frac { 1 }{ 3 } \)Bh = 224
B = 8² = 64
\(\frac { 1 }{ 3 } \)(64)h = 224
h = 10.5

Question 13.
Volume = 198 yd3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 220
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 13

Question 14.
Volume = 392 cm3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 221

Answer:
The height of the pyramid is 12 cm

Explanation:
Volume = 392
\(\frac { 1 }{ 3 } \)Bh = 392
B = 14 x 7 = 98
\(\frac { 1 }{ 3 } \)(98)h = 392
h = 12

In Exercises 15 and 16, the pyramids are similar. Find the volume of pyramid B.

Question 15.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 222
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 15

Question 16.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 223

Answer:
Volume of A = 80

Explanation:
\(\frac { Volume of B }{ Volume of A } \) = (\(\frac { Side of B }{ side of A } \))³
\(\frac { V }{ 10 } \) = (\(\frac { 6 }{ 3 } \))³
V = 8 x 10
Volume of A = 80

In Exercises 17 – 20, find the volume of the composite solid.

Question 17.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 224
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 17

Question 18.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 225

Answer:
Composite solid volume = 306

Explanation:
Base area = \(\frac { 1 }{ 2 } \)bh = \(\frac { 1 }{ 2 } \)(12 x 9) = 54
Bottom solid volume V = \(\frac { 1 }{ 3 } \)Bh = \(\frac { 1 }{ 3 } \)(54 x 10)
V = 180
Top solid volume v = \(\frac { 1 }{ 3 } \)(54 x 7) = 126
Composite solid volume = 180 + 126 = 306

Question 19.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 226
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 19

Question 20.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 227

Answer:
Composite solid volume = 1152

Explanation:
Volume of Box = 12 x 12 x 12 = 1728
Square pyramid volume = \(\frac { 1 }{ 3 } \)Bh = \(\frac { 1 }{ 3 } \)(144 x 12)
= 576
Composite solid volume = 1728 – 576 = 1152

Question 21.
ABSTRACT REASONING
A pyramid has a height of 8 feet and a square base with a side length of 6 feet.

a. How does the volume of the pyramid change when the base slays the same and the height is doubled?
b. How does the volume of the pyramid change when the height stays the same and the side length of the base is doubled?
C. Are your answers 10 parts (a) and (b) true for any square pyramid? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 21.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 21.2

Question 22.
HOW DO YOU SEE IT?
The cube shown is formed by three pyramids. each with the same square base and the same height. How could you use this to verify the formula for the volume of a pyramid?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 228
Answer:

Question 23.
CRITICAL THINKING
Find the volume of the regular pentagonal pyramid. Round your answer to the nearest hundredth. In the diagram. m∠ABC = 35°
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 229
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 23.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 23.2

Question 24.
THOUGHT PROVOKING
A frustum of a pyramid is the part of the pyramid that lies between the base and a plane parallel to the base, as shown. Write a formula for the volume of the frustum of a square pyramid in terms of a, b, and h. (Hint: Consider the “missing” top of the pyramid and use similar triangles.)
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 230
Answer:

Question 25.
MODELING WITH MATHEMATICS
Nautical deck prisms were used as a safe way to illuminate decks on ships. The deck prism shown here is composed of the following three solids: a regular hexagonal prism with an edge length of 3.5 inches and a height of 1.5 inches, a regular hexagonal prism with an edge length of 3.25 inches arid a height of 0.25 inch, and a regular hexagonal pyramid with an edge length of 3 inches and a height of 3 inches. Find the volume of the deck prism.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 231
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 25.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 25.2

Maintaining Mathematical Proficiency

Find the value of X. Round your answer to the nearest tenth.

Question 26.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 232
Answer:
tan 35 = \(\frac { 9 }{ x } \)
0.7 = \(\frac { 9 }{ x } \)
x = 12.8

Question 27.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 233
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 27

Question 28.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 234
Answer:
tan 30 = \(\frac { x }{ 10 } \)
0.577 = \(\frac { x }{ 10 } \)
x = 5.77

Question 29.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 235
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 29

11.7 Surface Areas and Volumes of Cones

Exploration 1

Finding the Surface Area of a Cone

Work with a partner: Construct a circle with a radius of 3 inches. Mark the circumference of the circle into six equal parts, and label the length of each part. Then cut out one sector of the circle and make a cone.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 236
a. Explain why the base of the cone is a circle. What are the circumference and radius
of the base?
Answer:

b. What is the area of the original circle? What is the area with one sector missing?
Answer:

c. Describe the surface area of the cone, including the base. Use your description to find the surface area.
Answer:

Exploration 2

Finding the Volume of a Cone

Work with a partner: The cone and the cylinder have the same height and the same circular base.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 237
When the cone is filled with sand and poured into the cylinder. it takes three cones to fill the cylinder.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 238
Use this information to write a formula for the volume V of a cone.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results in constructing arguments.
Answer:

Communicate Your Answer

Question 3.
How can you find the surface area and the volume of a cone?
Answer:

Question 4.
In Exploration 1, cut another sector from the circle and make a cone. Find the radius of the base and the surface area of the cone. Repeat this three times, recording your results in a table. Describe the pattern.
Answer:

Lesson 11.7 Surface Areas and Volumes of Cones

Monitoring progress

Question 1.
Find the surface area of the right cone.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 239

Answer:
The surface area of the right cone is 436.17 m²

Explanation:
r = 7.8
l = 10
S = πr² + πrl
S = π(7.8)² + π(7.8 x 10)
S = 436.17

Find the volume of the cone.

Question 2.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 240

Answer:
The volume of the cone is 2206.44 in³

Explanation:
r = 7, h = 13
l = √13² – 7²= 10.95
S = πr² + πrl
S = π7² + π(7 x 10.95)
S = 394.74
Volume V = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 7² x 13)
V = 2206.44

Question 3.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 241

Answer:
The volume of the cone is 163.4 m³

Explanation:
h = √8² – 5² = 6.24
Volume V = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 5² x 6.24)
V = 163.4

Question 4.
Cone C and cone D are similar. Find the volume of cone D.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 242

Answer:
Volume of cone D = 18.84 cm³

Explanation:
\(\frac { Volumeof cone C }{ Volume of cone D } \) = (\(\frac { height of C }{ height of D } \))³
\(\frac { 384π }{ Volume of cone D } \)= (\(\frac { 8 }{ 2 } \))³
Volume of cone D = 18.84

Question 5.
Find the volume of the composite solid.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 243

Answer:
Composite solid volume = 329.86 cm³

Explanation:
Volume of cylinder = πr²h = π(3)² x 10 = 90π
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
= \(\frac { 1 }{ 3 } \)(π x 3² x 5)
= 15π
Composite solid volume = 15π + 90π = 105π

Exercise 11.7 Surface Areas and Volumes of Cones

Vocabulary and Core Concept Check

Question 1.
WRITING
Describe the differences between pyramids and cones. Describe their similarities.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 1

Question 2.
COMPLETE THE SENTENCE
The volume of a cone with radius r and height h is \(\frac{1}{3}\) the volume of a(n) __________ with radius r and height h.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the surface area of the right cone.

Question 3.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 244
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 3

Question 4.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 245

Answer:
The surface area of cone is 219.44 sq cm.

Explanation:
S = πr² + πrl
S = π(5.5)² + π(5.5 x 7.2)
S = 219.44

Question 5.
A right cone has a radius of 9 inches and a height of 12 inches.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 5

Question 6.
A right cone has a diameter of 11.2 feet and a height of 19.2 feet.

Answer:
The surface area is 421.52 sq ft.

Explanation:
r = 5.6
h = 19.2
l = √19.2² – 5.6² = 18.36
Surface area S = πr² + πrl
S = π(5.6)² + π(5.6 x 18.36)
S = 421.52

In Exercises 7 – 10, find the volume of the cone.

Question 7.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 246Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 246
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 7

Question 8.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 247

Answer:
The volume is 2.09 cubic meter

Explanation:
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π(1)² x 2)
V = 2.09

Question 9.
A cone has a diameter of 11.5 inches and a height of 15.2 inches.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 9

Question 10.
A right cone has a radius of 3 feet and a slant height of 6 feet.

Answer:
The volume is 56.54 cubic ft

Explanation:
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 3² x 6)
V = 56.54

In Exercises 11 and 12, find the missing dimension(s).

Question 11.
Surface area = 75.4 cm2
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 11

Question 12.
Volume = 216π in.3

Answer:
The radius is 6.13 in

Explanation:
Volume = 216π in.3
\(\frac { 1 }{ 3 } \)(πr²h) = 216
\(\frac { 1 }{ 3 } \)(πr² x 18) = 216
r = 6.13

In Exercises 13 and 14, the cones are similar. Find the volume of cone B.

Question 13.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 248
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 13

Question 14.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 249

Answer:
Volume of cone B = 24.127

Explanation:
\(\frac { Volumeof cone A }{ Volume of cone B } \) = (\(\frac { height of A }{ height of B } \))³
\(\frac { 120π }{ Volume of cone B } \) = (\(\frac { 10 }{ 4 } \))³
Volume of cone B = 24.127

In Exercises 15 and 16, find the volume of the composite solid.

Question 15.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 250
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 15

Question 16.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 251

Answer:
Volume of the composite solid = 97.93 cubic m

Explanation:
Volume of box = lbh
V = 51 x 5.1 x 5.1 = 132.651
Cone volume = \(\frac { 1 }{ 3 } \)(πr²h)
v = \(\frac { 1 }{ 3 } \)(π x 2.55² x 5.1)
v = 34.72
Volume of the composite solid = 132.651 – 34.72 = 97.93

Question 17.
ANALYZING RELATIONSHIPS
A cone has height h and a base with radius r. You warn to change the cone so its volume is doubled. What is the new height if you change only the height? What is the new radius if you change only the radius? Explain.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 17

Question 18.
HOW DO YOU SEE IT
A snack stand serves a small order of popcorn in a cone-shaped container and a large order of popcorn in a cylindrical container. Do not perform any calculations.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 252
a. How many small containers of popcorn do you have to buy to equal the amount of popcorn in a large container? Explain.

Answer:
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 3² x 8) = 75.39
Volume of cylinder = πr²h = π x 3² x 8 = 226.19
Volume of cylinder / Volume of cone = \(\frac { 226.19 }{ 75.39 } \) = 3
You have to buy 3 small containers of popcorn to equal the amount of popcorn in a large containe.

b. Which container gives you more popcorn for your money? Explain.
Answer:
$1.25 -> 75.39 i.e $1 = 60.312
$2.50 -> 226.19 i.e $1 = 90.47
So, large containers gives you more popcorn for your money

In Exercises 19 and 20. find the volume of the right cone.

Question 19.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 253
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 19

Question 20.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 254

Answer:
Volume of cone is 575.62 cubic yd

Explanation:
tan 32 = \(\frac { 7 }{ h } \)
h = 11.21
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 7² x 11.21)
V = 575.62

Question 21.
MODELING WITH MATHEMATICS
A cat eats hail a cup of food, twice per day. Will the automatic pet feeder hold enough food for 10 days? Explain your reasoning. (1 cup ≈ 14.4 in.3)
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 255
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 21

Question 22.
MODELING WITH MATHEMATICS
During a chemistry lab, you use a funnel to pour a solvent into a flask. The radius of the funnel is 5 centimeters and its height is 10 centimeters. You pour the solvent into the funnel at a rate of 80 milliliters per second and the solvent flows out of the funnel at a rate of 65 milliliters per second. How long will it be before the funnel overflows? (1 mL = 1 cm3)
Answer:
17.45 seconds

Explanation:
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 5² x 10)
V = 261.8
\(\frac { 261.8 }{ 15 } \) = 17.45

Question 23.
REASONING
To make a paper drinking cup, start with a circular piece of paper that has a 3-inch radius, then follow the given steps. How does the surface area of the cup compare to the original paper circle? Find m∠ABC.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 256
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 23

Question 24.
THOUGHT PROVOKING
A frustum of a cone is the part of the cone that lies between the base and a plane parallel to the base, as shown. Write a formula for the volume of the frustum of a cone in terms of a, b, and h. (Hint: Consider the “missing” top of the cone and use similar triangles.)
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 257
Answer:
Volume V  = (1/3) * π * h * (r1² + r2² + (r1 * r2))

Question 25.
MAKING AN ARGUMENT
In the figure, the two cylinders are congruent The combined height of the two smaller cones equals the height of the larger cone. Your friend claims that this means the total volume of the two smaller cones is equal to the volume of the larger cone. Is your friend correct? Justify your answer.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 258
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 25

Question 26.
CRITICAL THINKING
When the given triangle is rotated around each of its sides. solids of revolution are formed. Describe the three solids and find their volumes. Give your answers in terms of π.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 259
Answer:

Maintaining Mathematical Proficiency

Find the indicated measure.

Question 27.
area of a circle with a radius of 7 feet
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 27

Question 28.
area of a circle with a diameter of 22 centimeters

Answer:
d = 11
A = πr²
A = 121π

Question 29.
diameter of a circle with an area of 256 square meters
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 29

Question 30.
radius of a circle with an area of 529 π square inches
Answer:
A = πr²
529π = πr²
r = 23

11.8 Surface Areas and Volumes of Spheres

Exploration 1

Finding the Surface Area of a Sphere

Work with a partner: Remove the covering from a baseball or softball.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 260
You will end up with two “figure 8” pieces of material, as shown above. From the amount of material it takes to cover the ball, what would you estimate the surface area S of the ball to be? Express your answer in terms of the radius r of the ball.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 261
Use the Internet or some other resource to confirm that the formula you wrote for the surface area of a sphere is correct.
USING TOOLS STRATEGICALLY
To be proficient in math, you need to identify relevant external mathematical resources, such as content located on a website.
Answer:

Exploration 2

Finding the volume of a sphere

Work with a partner: A cylinder is circumscribed about a sphere, as shown. Write a formula for the volume V of the cylinder in terms of the radius r.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 262
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 263
When half of the sphere (a hemisphere) is filled with sand and poured into the cylinder, it takes three hemispheres to till the cylinder. Use this information to write a formula for the volume V of a sphere in terms of the radius r.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 264
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 265
Answer:

Communicate Your Answer

Question 3.
How can you find the surface area and the volume of a sphere?
Answer:

Question 4.
Use the results of Explorations 1 and 2 to find the surface area and the volume of a sphere with a radius of(a) 3 inches and (b) 2 centimeters.
Answer:

Lesson 11.8 Surface Areas and Volumes of Spheres

Monitoring Progress

Find the surface area of the sphere.

Question 1.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 266

Answer:
The surface area of the sphere is 5026.54 ft²

Explanation:
D = 40
r = 20
The surface area of the sphere = 4πr²
S = 4 x π x (20)²
S = 5026.54 ft²

Question 2.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 267

Answer:
The surface area of the sphere is 113.09 ft²

Explanation:
Circumference C = 6π
2πr = 6π
r = 3
The surface area of the sphere = 4πr²
S = 4π x 3²
S = 113.09

Question 3.
Find the radius of the sphere.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 268

Answer:
The radius of the sphere is 2.73 m

Explanation:
The surface area of the sphere = 4πr²
30π = 4πr²
r = 2.73

Question 4.
The radius of a sphere is 5 yards. Find the volume of the sphere.

Answer:
The volume of the sphere is 523.59 yards³

Explanation:
r = 5
The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 5³
V = 523.59 yards³

Question 5.
The diameter of a sphere is 36 inches. Find the volume of the sphere.

Answer:
The volume of the sphere is 24429.02 in³

Explanation:
D = 36
r = 18
The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 18³
V = 24429.02

Question 6.
The surface area of a sphere is 576π square centimeters. Find the volume of the sphere.

Answer:
The volume of the sphere is 2304π cm³

Explanation:
The surface area of the sphere = 4πr²
576π = 4πr²
r = 12
The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 12³
V = 2304π

Question 7.
Find the volume of the composite solid at the left.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 269

Answer:
The volume of the composite solid is 7.324 m³

Explanation:
The volume of cone = πr²\(\frac { h }{ 3 } \)
= π x 1² x \(\frac { 5 }{ 3 } \) = 5.23
The volume of sphere = \(\frac { 4 }{ 3 } \)πr³
= \(\frac { 4 }{ 3 } \)π x 1³ = 4.188
The volume of the composite solid = The volume of cone + The volume of sphere/2
= 5.23 + 4.188/2
= 7.324 m³

Exercise 11.8 Surface Areas and Volumes of Spheres

Question 1.
VOCABULARY
When a plane intersects a sphere. what must be true for the intersection to be a great circle?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 1

Question 2.
WRITING
Explain the difference between a sphere and a hemisphere.

Answer:
Hemisphere is a related term of the sphere. Sphere and hemisphere are three-dimensional solids. The volume of sphere is \(\frac { 4 }{ 3 } \)πr³ and hemisphere volume is \(\frac { 2 }{ 3 } \)πr³. The surface area of the sphere is 4πr² and hemisphere surface area is 3πr².

Monitoring progress and Modeling with Mathematics

In Exercises 3 – 6, find the surface area of the sphere.

Question 3.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 270
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 3

Question 4.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 271

Answer:
The surface area of the sphere is 225π cm²

Explanation:
The surface area of the sphere = 4πr²
S = 4π x 7.5²
S = 225π

Question 5.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 272
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 5

Question 6.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 273

Answer:
The surface area of the sphere is 8π ft²

Explanation:
C = 4π
2πr = 4π
r = 2
The surface area of the sphere = 4πr²
S = 4π x 2²
S = 8π

In Exercises 7 – 10. find the indicated measure.

Question 7.
Find the radius of a sphere with a surface area of 4π square feet.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 7

Question 8.
Find the radius of a sphere with a surface area of 1024π square inches.

Answer:
The radius of a sphere is 16 in

Explanation:
The surface area of the sphere = 1024π
4πr² = 1024π
r = 16

Question 9.
Find the diameter of a sphere with a surface area of 900π square meters.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 9

Question 10.
Find the diameter of a sphere with a surface area of 196π square centimeters.

Answer:
The diameter of a sphere is 14 cm

Explanation:
The surface area of the sphere = 196π
4πr² = 196π
r = 7
D = 2(7) = 14

In Exercises 11 and 12, find the surface area of the hemisphere.

Question 11.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 274
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 11

Question 12.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 275

Answer:
The surface area of the hemisphere is 108π in²

Explanation:
D = 12, r = 6
The surface area of the sphere = 3πr²
S = 3π x 6²
S = 108π

In Exercises 13 – 18. find the volume of the sphere.

Question 13.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 276
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 13

Question 14.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 277

Answer:
The volume of the sphere is 268.08 ft³

Explanation:
r = 4 ft
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 4³
V = 268.08 ft

Question 15.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 278
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 15

Question 16.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 279

Answer:
The volume of the sphere is 1436.75 ft³

Explanation:
D = 14 ft
r = 7 ft
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 7³
V = 1436.75 ft

Question 17.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 282
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 17

Question 18.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 280

Answer:
The volume of the sphere is 179.89 in³

Explanation:
C = 7π
2πr = 7π
r = 3.5
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 3.5³
V = 179.89 in

In Exercises 19 and 20, find the volume of the sphere with the given surface area.

Question 19.
Surface area = 16π ft2
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 19

Question 20.
Surface area = 484π cm2

Answer:
The volume of the sphere is 5575.27 cm³

Explanation:
Surface area = 484π
4πr² = 484π
r = 11
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 11³
V = 5575.27

Question 21.
ERROR ANALYSIS
Describe and correct the error in finding the volume of the sphere.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 281
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 21

Question 22.
ERROR ANALYSIS
Describe and correct the error in finding the volume of the sphere.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 283

Answer:
Diameter = 3
radius = 1.5
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x (1.5)³
V = 14.137 cubic in

In Exercises 23 – 26, find the volume of the composite solid.

Question 23.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 284
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 23

Question 24.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 285

Answer:
Volume is 288π ft³

Explanation:
Volume of hemipshere = \(\frac { 2 }{ 3 } \)πr³
= \(\frac { 2 }{ 3 } \)π x 6³ = 144π
volume of the cone = πr²\(\frac { h }{ 3 } \)
= π x 6² x \(\frac { 12 }{ 3 } \) = 144π
Area of circle = πr² = π x 6² = 36π
Volume of hemipshere + volume of the cone = 144π + 144π = 288π

Question 25.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 286
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 25

Question 26.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 287

Answer:
The volume of solid is 296π m³

Explanation:
Volume of hemipshere = \(\frac { 2 }{ 3 } \)πr³
= \(\frac { 2 }{ 3 } \)π x 6³ = 144π
Volume of cylinder = πr²h
= π x 6² x 14 = 504π
Volume of solid = 504π – 2(144π) = 296π

In Exercises 27 – 32, find the surface area and volume of the ball.

Question 27.
bowling ball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 289
d = 8.5 in.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 27

Question 28.
basketball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 289
C = 29.5 in.

Answer:
The surface area is 277 in², volume is 43212.27 in³

Explanation:
C = 29.5
2πr = 29.5
r = 4.69
Surface area = 4πr²
S = 4π x 4.69² = 277
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 4.69³
V = 43212.27

Question 29.
softball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 290
C = 12 in.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 29

Question 30.
golf ball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 291
d = 1.7 in.

Answer:
The surface area is 9.07 in², volume is 2.57 in³

Explanation:
d = 1.7
r = 0.85
Surface area = 4πr²
S = 4π x 0.85² = 9.07
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 0.85³
V = 2.57

Question 31.
volleyball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 292
C = 26 in.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 31

Question 32.
baseball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 293
C = 9 in.

Answer:
The surface area is 25.78 in², volume is 12.24 in³

Explanation:
C = 9
2πr = 9
r = 1.43
Surface area = 4πr²
S = 4π x 1.43²
S = 25.78
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 1.43³
V = 12.24

Question 33.
MAKING AN ARGUMENT
You friend claims that if the radius of a sphere is doubled, then the surface area of the sphere will also be doubled. Is our friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 33

Question 34.
REASONING
A semicircle with a diameter of 18 inches is rotated about its diameter. Find the surface area and the volume of the solid formed.

Answer:
The surface area is 1018 in², volume is 3054.02 in³

Explanation:
Diameter = 18
radius r = 9
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 9³
V = 3054.02
Surface area = 4πr²
S = 4π x 9²
S = 1018

Question 35.
MODELING WITH MATHEMATICS
A silo has the dimensions shown. The top of the silo is a hemispherical shape. Find the volume of the silo.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 294
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 35

Question 36.
MODELING WITH MATHEMATICS
Three tennis balls are stored in a cylindrical container with a height of 8 inches and a radius of 1.43 inches. The circumference of a tennis ball is 8 inches.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 295
a. Find the volume of a tennis ball.

Answer:
C = 8 in
2πr = 8
r = 1.27
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 1.27³ = 8.64
The volume of tennis ball = 8.64 in³

b. Find the amount of space within the cylinder not taken up by the tennis balls.
Answer:
The surface area of tennis ball S = 4πr²
S = 4π x 1.27² = 20.26
Area of cylinder s = 2πrh+2πr²
s = 2π x 1.43 x 8+2π x 1.43²
s = 84.72
Remaining space = 84.72 – 20.26 = 64.46 in²

Question 37.
ANALYZING RELATIONSHIPS
Use the table shown for a sphere.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 296
a. Copy and complete the table. Leave your answers in terms of π.
b. What happens to the surface area of the sphere when the radius is doubled? tripled? quadrupled?
c. What happens to the volume of the sphere when the radius is doubled? tripled? quadrupled?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 37

Question 38.
MATHEMATICAL CONNECTIONS
A sphere has a diameter of 4(x + 3) centimeters and a surface area of 784 π square centimeters. Find the value of x.

Answer:
x =11

Explanation:
Surface area = 4πr²
784π = πr²
r = 28
2r = diameter = 4(x + 3)
r = 2(x + 3)
28 = 2(x + 3)
x = 11

Question 39.
MODELING WITH MATHEMATICS
The radius of Earth is about 3960 miles. The radius of the moon is about 1080 miles.
a. Find the surface area of Earth and the moon.
b. Compare the surface areas of Earth and the moon.
c. About 70% of the surface of Earth is water. How many square miles of water are on Earth’s surface?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 39

Question 40.
MODELING WITH MATHEMATICS
The Torrid Zone on Earth is the area between the Tropic of Cancer and the Tropic of Capricorn. The distance between these two tropics is about 3250 miles. You can estimate the distance as the height of a cylindrical belt around the Earth at the equator.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 297
a. Estimate the surface area of the Torrid Zone. (The radius of Earth is about 3960 miles.)
Answer:
Surface area of cylinder = 2πrh
S = 2π x 3960 x 3250 = 80875080
surface area of earth = 4πr²
= 4π x 3960² = 197086348.8

b. A meteorite is equally likely to hit anywhere on Earth. Estimate the probability that a meteorite will land in the Torrid Zone.
Answer:
Probability of meteorites hitting the torrid zone = 80875080/197086348.8 = 0.4104

Question 41.
ABSTRACT REASONING
A sphere is inscribed in a cube with a volume of 64 cubic inches. What is the surface area of the sphere? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 41

Question 42.
HOW DO YOU SEE IT?
The formula for the volume of a hemisphere and a Cone are shown. If each solid has the same radius and r = h, which solid will have a greater volume? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 298
Answer:
The hemisphere has the highest volume.

Explanation:
Volume of hemisphere v = \(\frac { 2 }{ 3 } \)πr³
Volume of cone V = \(\frac { 1 }{ 3 } \)πr²h
If r = h
Volume of cone V = \(\frac { 1 }{ 3 } \)πr² x r = \(\frac { 1 }{ 3 } \)πr³
So, the hemisphere has the highest volume.

Question 43.
CRITICAL THINKING
Let V be the volume of a sphere. S be the surface area of the sphere, and r be the radius of the sphere. Write an equation for V in terms of r and S. (Hint: Start with the ratio \(\frac{V}{S}\).)
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 43

Question 44.
THOUGHT PROVOKING
A spherical lune is the region between two great circles of a sphere. Find the formula for the area of a lune.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 299
Answer:

Question 45.
CRITICAL THINKING
The volume of a right cylinder is the same as the volume of a sphere. The radius of the sphere is 1 inch. Give three possibilities for the dimensions of the cylinder.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 45.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 45.2

Question 46.
PROBLEM SOLVING
A spherical cap is a portion of a sphere cut off by a plane. The formula for the volume of a spherical cap is V = \(\frac{\pi h}{6}\) (3a2 + h2), where a is the radius of the base of the cap and h is the height of the cap. Use the diagram and given information to find the volume of each spherical cap.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 300
a. r = 5ft, a = 4ft
Answer:

b. r = 34 cm, a = 30 cm
Answer:

c. r = 13 m, h = 8 m
Answer:

d. r=75 in., h = 54in.
Answer:

Question 47.
CRITICAL THINKING
A sphere with a radius of 2 inches is inscribed in a right cone with a height of 6 inches. Find the surface area and the volume of the cone.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 47.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 47.2
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 47.3

Maintaining Mathematical Proficiency

Solve the triangle. Round decimal answers to the nearest tenth.

Question 48.
A = 26°, C = 35°, b = 13

Answer:
B = 119°, a = 7.16, c = 9.5

Explanation:
B = 180 – (26 + 35) = 119
\(\frac { sin A }{ a } \) = \(\frac { sin B }{ b } \)
\(\frac { sin 26 }{ a } \) = \(\frac { sin 119 }{ 13 } \)
a = 7.16
\(\frac { sin C }{ c } \) = \(\frac { sin B }{ b } \)
\(\frac { sin 35 }{ c } \) = \(\frac { sin 119 }{ 13 } \)
c = 9.5

Question 49.
B = 102°, C = 43°, b = 21
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 49

Question 50.
a = 23, b = 24, c = 20

Answer:
A = 62.2, B = 65.5, C = 49.4

Explanation:
a² = b² + c² – 2bc cos A
23² = 24²+ 20² – 2(24 x 20) cos A
A = 62.2
\(\frac { sin 62.2 }{ 23 } \) = \(\frac { sin B }{ 24 } \)
B = 65.5
\(\frac { sin 62.2 }{ 23 } \) = \(\frac { sin C }{ 20 } \)
C = 49.4

Question 51.
A = 103°, b = 15, c = 24
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 51

Circumference, Area, and Volume Review

11.1 Circumference and Arc Length

Find the indicated measure.

Question 1.
diameter of ⊙P
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 301

Answer:
diameter of ⊙P is 29.99

Explanation:
Circumference = 94.24
πd = 94.24
d = 29.99

Question 2.
circumference of ⊙F
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 302

Answer:
circumference of ⊙F = 56.57

Explanation:
5.5 = \(\frac { 35 }{ 360 } \) . C
C = 56.57

Question 3.
arc length of \(\widehat{A B}\)
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 303

Answer:
arc length of \(\widehat{A B}\) = 26.09

Explanation:
arc length of \(\widehat{A B}\) = \(\frac { 115 }{ 360 } \) . 2π(13)
= 26.09

Question 4.
A mountain bike tire has a diameter of 26 inches. To the nearest foot, how far does the tire travel when it makes 32 revolutions?

Answer:
The tire travels 2613.80 inches.

Explanation:
D = 26 in
r = 13 in
Circumference C = 2π(13) = 81.68
32 revolutions = 32 x 81.68 = 2613.80

11.2 Areas of Circles and Sectors

Find the area of the blue shaded region.

Question 5.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 304

Answer:
Area = \(\frac { 240 }{ 360 } \) . π(9)²
= 169.64

Question 6.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 305

Answer:
Area of shaded region = 11.43

Explanation:
Area of rectangle = 6 x 4 = 24
Area of semicircle = π(2)² = 4π
Area of shaded region = 24 – 4π = 11.43

Question 7.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 306

Answer:
Area of shaded region = 173.13

Explanation:
Area of small region = 27.93 = \(\frac { 50 }{ 360 } \) . πr²
πr² = 201.096
r = 8
Area of shaded region = \(\frac { 310 }{ 360 } \) . π(8)²
= 173.13

11.3 Areas of Polygons

Find the area of the kite or rhombus.

Question 8.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 307
Answer:
Area = 65

Explanation:
Area = \(\frac { 1 }{ 4 } \)(d₁d₂)
A = \(\frac { 1 }{ 4 } \)(13 x 20)
A = 65

Question 9.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 308
Answer:
Area = 48

Explanation:
Area = \(\frac { 1 }{ 4 } \)(d₁d₂)
A = \(\frac { 1 }{ 4 } \)(16 x 12)
A = 48

Question 10.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 309
Answer:
Area = 52.5

Explanation:
Area = \(\frac { 1 }{ 4 } \)(d₁d₂)
A = \(\frac { 1 }{ 4 } \)(14 x 15)
A = 52.5

Find the area of the regular polygon.

Question 11.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 310
Answer:
Area = \(\frac { 3√3 }{ 2 } \)a²
A = \(\frac { 3√3 }{ 2 } \)(8.8)²
A = 201.195

Question 12.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 311
Answer:
Area = \(\frac { 1 }{ 2 } \)(n . a. s)
A = \(\frac { 1 }{ 2 } \)(9 . 5.2 . 7.6)
A = 117.84

Question 13.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 312
Answer:
Area = \(\frac { 1 }{ 2 } \)(n . a. s)
A = \(\frac { 1 }{ 2 } \)(5 . 4 .  3.3)
A = 33

Question 14.
A platter is in the shape of a regular octagon with an apothem of 6 inches. Find the area of the platter.

Answer:
Area = \(\frac { 1 }{ 2 } \)(n . a. s)
A = \(\frac { 1 }{ 2 } \)(8 . 6 . sin 45)
A = 16.97

11.4 Three-Dimensional Figures

Sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.

Question 15.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 313
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume

Question 16.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 314
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 15

Question 17.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 315
Answer:

Describe the cross section formed by the intersection of the plane and the solid.

Question 18.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 316
Answer:
The cross section is a rectangle.

Question 19.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 317
Answer:
The cross-section is a square.

Question 20.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 318
Answer:
the cross-section is a triangle.

11.5 Volumes of Prisms and Cylinders

Find the volume of the solid.

Question 21.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 319

Answer:
Volume = lbh
V = 3.6 x 2.1 x 1.5 = 113.4 m³

Question 22.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 320

Answer:
Volume = πr²h
V = π(2)² x 8 = 100.53 mm³

Question 23.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 321

Answer:
Pentagon area = 6.88
Volume = Area x height
V = 6.88 x 4 = 27.52 yd³

11.6 Volumes of Pyramids

Find the volume of the pyramid.

Question 24.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 322

Answer:
Volume V = Base area x height/3
Base Area = 9² = 81
V = 81 x 7/3 = 189 ft³

Question 25.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 323
Answer:
Volume V = Base area x height/3
Base Area = 4 x 15 = 60
V = 60x 20/3 = 400 yd³

Question 26.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 324

Answer:
Volume V = Base area x height/3
base area = 18 x 10 = 180
V = 180 x 5/3 = 300 m³

Question 27.
The volume of a square pyramid is 60 cubic inches and the height is 15 inches. Find the side length of the square base.

Answer:
The side length of the square base is 3.46 in

Explanation:
The volume of a square pyramid is 60 cubic inches
V = 60
s²h/3 = 60
s² x 15/3 = 60
s² = 12
s = 3.46

Question 28.
The volume of a square pyramid is 1024 cubic inches. The base has a side length of 16 inches. Find the height of the pyramid
Answer:
The volume of a square pyramid is 1024 cubic inches
s²h/3 = 1024
16²h = 3072
h = 12

11.7 Surface Areas and Volumes of Cones

Find the surface area and the volume of the cone.

Question 29.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 325
Answer:
Surface area is 678.58 cm²
volume is 1017.87 cm³

Explanation:
Surface area of cone S = πr² + πrl
S = π x 9² + π x 9 x 15
S = 678.58
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 9² x 12)
V = 1017.87

Question 30.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 326

Answer:
Surface area is 2513.27 cm²
volume is 8042.47 cm³

Explanation:
Surface area of cone S = πr² + πrl
S = π x 16² + π x 16 x 34
S = 2513.27
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 16² x 30)
V = 8042.47

Question 31.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 327

Answer:
Surface area is 439.82 m²
volume is 562.102 m³

Explanation:
Surface area of cone S = πr² + πrl
S = π x 7² + π x 7 x 13
S = 439.82
h = √13² – 7² = 10.95
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 7² x 10.95)
V = 562.102

Question 32.
A cone with a diameter of 16 centimeters has a volume of 320π cubic centimeters. Find the height of the cone.

Answer:
The height of the cone = 15 cm.

Explanation:
r = 8
Volume V = 320π
\(\frac { 1 }{ 3 } \)(πr²h) = 320π
\(\frac { 1 }{ 3 } \)(π x 8² x h) = 320π
h = 15

11.8 Surface Areas and Volumes of Spheres

Find the surface area and the volume of the sphere.

Question 33.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 328

Answer:
The surface area is 615.75 in², volume is 1436.75 in³

Explanation:
Surface area S = 4πr²
S = 4π x 7²
S = 615.75
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 7³
V = 1436.75

Question 34.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 329

Answer:
The surface area is 907.92 ft², volume is 2572.44 ft³

Explanation:
d = 17
r = 8.5
Surface area S = 4πr²
S = 4π x 8.5²
S = 907.92
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 7³
V = 2572.44

Question 35.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 330

Answer:
The surface area is 2827.43 ft², volume is 14137.16 ft³

Explanation:
C = 30π
2πr = 30π
r = 15
Surface area S = 4πr²
S = 4π x 15²
S = 2827.43
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 15³
V = 14137.16

Question 36.
The shape of Mercury can be approximated by a sphere with a diameter of 4880 kilometers. Find the surface area and the volume of Mercury.

Answer:
The surface area and the volume of Mercury is 23814400π, 19369045330π

Explanation:
d = 4880
r = 2440
Surface area S = 4πr²
S = 4π x 2440² = 23814400π
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 2440³
V = 19369045330π

Question 37.
A solid is composed of a cube with a side length of 6 meters and a hemisphere with a diameter of 6 meters. Find the volume of the composite solid.

Answer:
Volume of the composite solid = 272.52

Explanation:
Volume of cube = a³
= 6³ = 216
Volume of hemisphere = \(\frac { 4 }{ 6 } \)πr³
= \(\frac { 4 }{ 6 } \)π x 3³ = 18π
Volume of the composite solid = 216 + 18π = 272.52

Circumference, Area, and Volume Test

Find the volume of the solid.

Question 1.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 331

Answer:
Volume = 2577.29 m³

Explanation:
Volume = \(\frac { 3√3 }{ 2 } \)a²h
= \(\frac { 3√3 }{ 2 } \) x 8² x 15.5
= 2577.29

Question 2.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 332

Answer:
Volume is 17.157 ft³

Explanation:
d = 3.2
r = 1.6
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 1.6³
V = 17.157

Question 3.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 333

Answer:
Volume of sloid = 402.11 m³

Explanation:
Volume of cone = \(\frac { 1 }{ 3 } \)πr²h
= \(\frac { 1 }{ 3 } \)π x 4² x 3
= 50.26
Volume of cylinder = πr²h
= π x 4² x 6 = 301.59
Volume of sloid = 2(50.26) + 301.59 = 402.11

Question 4.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 334

Answer:
Volume of solid = 106.66

Explanation:
Volume of rectangular box = 5 x 2 x 8 = 80
Volume of pyramid = 80/3 = 26.66
Volume of solid = 80 + 26.66 = 106.66

Find the indicated measure.

Question 5.
circumference of ⊙F
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 335

Answer:
circumference of ⊙F is 109.71 in

Explanation:
64 = \(\frac { 210 }{ 360 } \) • C
C = 109.7

Question 6.
m\(\widehat{G H}\)
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 336

Answer:
m\(\widehat{G H}\) = 74.27

Explanation:
35 = \(\frac { x }{ 360 } \) • 2π x 27
x = 74.27

Question 7.
area of shaded sector
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 337

Answer:
Area is 142.41 in²

Explanation:
Area = \(\frac { 360 – 105 }{ 360 } \) • π x 8²
Area = 142.41

Question 8.
Sketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 338
Answer:

Question 9.
Find the surface area of a right cone with a diameter of 10 feet and a height of 12 feet.

Answer:
The surface area is 486.7 sq ft

Explanation:
l² = r² + h²
l² = 5² + 12²
l = 13
Surface area S = πr² + 2πrl
S = π x 5² + 2π x 5 x 13
S = 486.7

Question 10.
You have a funnel with the dimensions shown.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 339
a. Find the approximate volume of the funnel.

Answer:
Volume = \(\frac { 1 }{ 3 } \)πr²h
= \(\frac { 1 }{ 3 } \)π x 6² x 10
= 376.99

b. You use the funnel to put oil in a ear. Oil flows out of the funnel at a rate of 45 milliliters per second. How long will it take to empty the funnel when it is full of oil? (1 mL = 1 cm3)
Answer:

c. How long would it take to empty a funnel with a radius of 10 centimeters and a height of 6 centimeters if oil flows out of the funnel at a rate of 45 milliliters per second?
Answer:

d. Explain why you can claim that the time calculated in part (c) is greater than the time calculated in part (b) without doing any calculations.
Answer:

Question 11.
A water bottle in the shape of a cylinder has a volume of 500 cubic centimeters. The diameter of a base is 7.5 centimeters. What is the height of the bottle? Justify your answer.

Answer:
The height of the bottle is 11.3 cm

Explanation:
Volume of cylinder = 500
πr²h = 500
π(3.75)²h = 500
h = 11.3 cm

Question 12.
Find the area of a dodecagon (12 sides) with a side length of 9 inches.

Answer:
Area is 237.31

Explanation:
Area = \(\frac { 1 }{ 4 } \)πa²cot(π/n)
= \(\frac { 1 }{ 4 } \)π x 9² x cot(π/12)
= 237.31

Question 13.
In general, a cardboard fan with a greater area does a better job of moving air and cooling you. The fan shown is a sector of a cardboard circle. Another fan has a radius of 6 centimeters and an intercepted are of 150°. Which fan does a better job of cooling you?
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 340

Answer:

Explanation:

Circumference, Area, and Volume Cumulative Assessment

Question 1.
Identify the shape of the cross section formed by the intersection of the plane and the
solid.

a.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 341
Answer:
The cross-section is a trapezoid.

b.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 342
Answer:
The cross-section is a pentagon.

c.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 343
Answer:
The cross-section is a rectangle.

Question 2.
In the diagram, Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 344 is tangent to ⊙P at Q and \(\overline{P Q}\) is a radius of ⊙P? What must be true about Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 344 and \(\overline{P Q}\)? Select all that apply.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 345
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 346
Answer:
PQ is perpendicular to RS.

Question 3.
A crayon can be approximated by a composite solid made from a cylinder and a cone.
A crayon box is a rectangular prism. The dimensions of a crayon and a crayon box containing 24 crayons are shown.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 348
a. Find the volume of a crayon.
Answer:
The volume of a crayon = πr²h + \(\frac { 1 }{ 3 } \)πr²h
= π x 4.25² x 80 + \(\frac { 1 }{ 3 } \)π x 3.25² x 10
= 4539.6 + 110.61
= 4650.21 mm³

b. Find the amount of space within the crayon box not taken up by the crayons.
Answer:
Volume of box = 94 x 28 x 71 = 186872
The volume of a crayon = 4650.21
Remaining space = 186872 – 24 x 4650.21
= 75266.96

Question 4.
What is the equation ol the line passing through the point (2, 5) that is parallel to the line x + \(\frac{1}{2}\)y = – 1?
(A) y = – 2x + 9
(B) y = 2x + 1
(C) y = \(\frac{1}{2}\)x + 4
(D) y = –\(\frac{1}{2}\)x + 6
Answer:
(A) y = – 2x + 9

Explanation:
x + \(\frac{1}{2}\)y = – 1
y = -2 – 2x
The slope of the line is -2
The euation of line is y – 5 = -2(x – 2)
y – 5 = -2x + 4
y = -2x + 9

Question 5.
The top of the Washington Monument in Washington, D.C., is a square pyramid, called a pyramidion. What is the volume of the pyramidion?
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 347
(A) 22,019.63 ft3
(B) 172,006.91 ft3
(C) 66,058.88 ft3
(D) 207,530.08 ft3

Answer:
(A) 22,019.63 ft3

Explanation:
Volume = a²\(\frac { h }{ 3 } \)
= 34.5² x \(\frac { 55.5 }{ 3 } \)
= 22019.62

Question 6.
Prove or disprove that the point (1, √3 ) lies on the circle centered at the origin and containing the point (0, 2).
Answer:
We consider the circle centered at the origin and containing the point (0, 2).
Therefore, we canconclude that rdaius is 2 and points be (0, 0), (1, √3)
distance = √(1 – 0)² + (√3 – 0)² = 2
As radius and distance are same. The point B(1, √3) lies on the circle.

Question 7.
Your friend claims that the house shown can be described as a composite solid made from a rectangular prism and a triangular prism. Do you support your friend’s claim? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 349
Answer:
Yes

Question 8.
The diagram shows a square pyramid and a cone. Both solids have the same height, h, and the base of the cone has radius r. According to Cavalieri’s Principle, the solids will have the same volume if the square base has sides of length ______ .
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 350
Answer:
2r/√2

Explanation:
Volume of square pyramid = a²\(\frac { h }{ 3 } \)
Square diagonal = √2a
radius = √2a/2
a = 2r/√2
Volume of cone = \(\frac { 1 }{ 3 } \)πr²h

Question 9.
About 19,400 people live in a region with a 5-mile radius. Find the population density in people per square mile.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 351

Answer:
The number of people per square mile is 247

Explanation:
S = πr²
= π x 5² = 78.5
Number of people per square mile = 19400/78.5 = 247

Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations

Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations

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Big Ideas Math Book Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations

Here is the complete list of topic-wise Big Ideas Math Book Algebra 1 Chapter 5 Solving Systems of Linear Equations Solution Key which covers the questions from the BIM Textbooks based on the latest Common Core Curriculum. Ch 5 Big Ideas Math Textbook Algebra 1 Answers material given here offers Questions from Exercises(5.1 to 5.7), Chapter Tests, Review Tests, Quiz, Assessment Tests, Cumulative Assessments, etc. Practice thoroughly and gain more subject knowledge on the concepts of BIM Math Book Algebra 1 Chapter 5 Solutions.

Solving Systems of Linear Equations Maintaining Mathematical Proficiency

Graph the equation.
Question 1.
y + 4 = x
Answer:

Question 2.
6x – y = -1
Answer:

Question 3.
4x + 5y = 20
Answer:

Question 4.
-2y + 12 = -3x
Answer:

Solve the inequality. Graph the solution.
Question 5.
m + 4 > 9
Answer:

Question 6.
24 ≤ -6t
Answer:

Question 7.
2a – 5 ≤ 13
Answer:

Question 8.
-5z + 1 < -14
Answer:

Question 9.
4k – 16 < k + 2
Answer:

Question 10.
7w + 12 ≥ 2w – 3
Answer:

Question 11.
ABSTRACT REASONING
The graphs of the linear functions g and h have different slopes. The value of both functions at x = a is b. When g and h are graphed in the same coordinate plane, what happens at the point (a, b)?
Answer:

Solving Systems of Linear Equations Mathematical Practices

Mathematically proficient students use technological tools to explore concepts.

Monitoring Progress

Use a graphing calculator to find the point of intersection of the graphs of the two linear equations.
Question 1.
y = -2x – 3
y = \(\frac{1}{2}\)x – 3
Answer:

Question 2.
y = -x + 1
y = x- 2
Answer:

Question 3.
3x – 2y = 2
2x – y = 2
Answer:

Lesson 5.1 Solving Systems of Linear Equations by Graphing

Essential Question How can you solve a system of linear equations?

EXPLORATION 1

Writing a System of Linear EquationsWork with a partner. Your family opens a bed-and-breakfast. They spend $600 preparing a bedroom to rent. The cost to your family for food and utilities is $15 per night. They charge $75 per night to rent the bedroom.
a. Write an equation that represents the costs.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 1
b. Write an equation that represents the revenue (income).
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 2
c. A set of two (or more) linear equations is called a system of linear equations. Write the system of linear equations for this problem.

EXPLORATION 2

Using a Table or Graph to Solve a System
Work with a partner. Use the cost and revenue equations from Exploration 1 to determine how many nights your family needs to rent the bedroom before recovering the cost of preparing the bedroom. This is the break-even point.
a. Copy and complete the table.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 3
b. How many nights does your family need to rent the bedroom before breaking even?
c. In the same coordinate plane, graph the cost equation and the revenue equation from Exploration 1.
d. Find the point of intersection of the two graphs. What does this point represent? How does this compare to the break-even point in part (b)? Explain.

Communicate Your Answer

Question 3.
How can you solve a system of linear equations? How can you check your solution?
Answer:

Question 4.
Solve each system by using a table or sketching a graph. Explain why you chose each method. Use a graphing calculator to check each solution.
a. y = -4.3x – 1.3
y = 1.7x + 4.7

b. y = x
y = -3x + 8

c. y = -x – 1
y = 3x + 5
Answer:

Monitoring Progress

Tell whether the ordered pair is a solution of the system of linear equations.
Question 1.
(1, -2); 2x + y = 0
-x + 2y = 5
Answer:

Question 2.
(1, 4); y = 3x + 1
y = -x + 5
Answer:

Solve the system of linear equations by graphing.
Question 3.
y = x – 2
y = -x + 4
Answer:

Question 4.
y = \(\frac{1}{2}\)x + 3
y = –\(\frac{3}{2}\)x – 5
Answer:

Question 5.
2x + y = 5
3x – 2y = 4
Answer:

Question 6.
You have a total of 18 math and science exercises for homework. You have six more math exercises than science exercises. How many exercises do you have in each subject?
Answer:

Solving Systems of Linear Equations by Graphing 5.1 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
Do the equations 5y – 2x = 18 and 6x = -4y – 10 form a system of linear equations? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Consider the system of linear equations -4x + 2y = 4 and 4x – y = -6. Which is different? Find “both” answers.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 4
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, tell whether the ordered pair is a solution of the system of linear equations.
Question 3.
(2, 6); x + y = 8
3x – y = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 3

Question 4.
(8, 2); x – y = 6
2x – 10y = 4
Answer:

Question 5.
(-1, 3); y = -7x – 4
y = 8x + 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 5

Question 6.
(-4, -2); y = 2x + 6
y = -3x – 14
Answer:

Question 7.
(-2, 1); 6x + 5y = -7
2x – 4y = -8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 7

Question 8.
(5, -6); 6x + 3y = 12
4x + y = 14
Answer:

In Exercises 9–12, use the graph to solve the system of linear equations. Check your solution.
Question 9.
x – y = 4
4x + y = 1
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 9

Question 10.
x + y = 5
y – 2x = -4
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 6
Answer:

Question 11.
6y + 3x = 18
-x + 2y = 24
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 11

Question 12.
2x – y = -2
2x + 4y = 8
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 8
Answer:

In Exercises 13–20, solve the system of linear equations by graphing.
Question 13.
y = -x + 7
y = x + 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 13

Question 14.
y = -x + 4
y = 2x – 8
Answer:

Question 15.
y = \(\frac{1}{3}\)x + 2
y = \(\frac{2}{3}\)x + 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 15

Question 16.
y = \(\frac{3}{4}\)x – 4
y = –\(\frac{1}{2}\)x + 11
Answer:

Question 17.
9x + 3y = -3
2x – y = -4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 17.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 17.2

Question 18.
4x – 4y = 20
y = -5
Answer:

Question 19.
x – 4y = -4
-3x – 4y = 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 19.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 19.2

Question 20.
3y + 4x = 3
x + 3y = -6
Answer:

ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in solving the system of linear equations.
Question 21.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 21.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 21.2

Question 22.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 10
Answer:

USING TOOLS In Exercises 23–26, use a graphing calculator to solve the system of linear equations.
Question 23.
0.2x + 0.4y = 4
-0.6x + 0.6y = -3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 23

Question 24.
-1.6x – 3.2y = -24
2.6x + 2.6y = 26
Answer:

Question 25.
-7x + 6y = 0
0.5x + y = 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 25

Question 26.
4x – y = 1.5
2x + y = 1.5
Answer:

Question 27.
MODELING WITH MATHEMATICS
You have 40 minutes to exercise at the gym, and you want to burn 300 calories total using both machines. How much time should you spend on each machine?
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 11
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 27.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 27.2
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 27.3

Question 28.
MODELING WITH MATHEMATICS
You sell small and large candles at a craft fair. You collect $144 selling a total of 28 candles. How many of each type of candle did you sell?
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 12
Answer:

Question 29.
MATHEMATICAL CONNECTIONS
Write a linear equation that represents the area and a linear equation that represents the perimeter of the rectangle. Solve the system of linear equations by graphing. Interpret your solution.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 13
Answer:

Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 29.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 29.2

Question 30.
THOUGHT PROVOKING
Your friend’s bank account balance (in dollars) is represented by the equation y = 25x + 250, where x is the number of months. Graph this equation. After 6 months, you want to have the same account balance as your friend. Write a linear equation that represents your account balance. Interpret the slope and y-intercept of the line that represents your account balance.
Answer:

Question 31.
COMPARING METHODS
Consider the equation x + 2 = 3x – 4.
a. Solve the equation using algebra.
b. Solve the system of linear equations y = x + 2 and y = 3x – 4 by graphing.
c. How is the linear system and the solution in part (b) related to the original equation and the solution in part (a)?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 31.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 31.2

Question 32.
HOW DO YOU SEE IT?
A teacher is purchasing binders for students. The graph shows the total costs of ordering x binders from three different companies.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 14
a. For what numbers of binders are the costs the same at two different companies? Explain.
b. How do your answers in part (a) relate to systems of linear equations?
Answer:

Question 33.
MAKING AN ARGUMENT
You and a friend are going hiking but start at different locations. You start at the trailhead and walk 5 miles per hour. Your friend starts 3 miles from the trailhead and walks 3 miles per hour.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 15
a. Write and graph a system of linear equations that represents this situation.
b. Your friend says that after an hour of hiking you will both be at the same location on the trail. Is your friend correct? Use the graph from part (a) to explain your answer.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 33.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 33.2

Maintaining Mathematical Proficiency

Solve the literal equation for y.
Question 34.
10x + 5y = 5x + 20
Answer:

Question 35.
9x + 18 = 6y – 3x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 35

Question 36.
\(\frac{3}{4}\) x + \(\frac{1}{4}\) y = 5
Answer:

Lesson 5.2 Solving Systems of Linear Equations by Substitution

Essential Question How can you use substitution to solve a system of linear equations?

EXPLORATION 1

Using Substitution to Solve Systems
Work with a partner. Solve each system of linear equations using two methods.
Method 1 Solve for x first.
Solve for x in one of the equations. Substitute the expression for x into the other equation to find y. Then substitute the value of y into one of the original equations to find x.

Method 2 Solve for y first.
Solve for y in one of the equations. Substitute the expression for y into the other equation to find x. Then substitute the value of x into one of the original equations to find y.
Is the solution the same using both methods? Explain which method you would prefer to use for each system
a. x + y = -7
-5x + y = 5

b. x – 6y = -11
3x + 2y = 7

c. 4x + y = -1
3x – 5y = -18

EXPLORATION 2

Writing and Solving a System of Equations
Work with a partner.
a. Write a random ordered pair with integer coordinates. One way to do this is to use a graphing calculator. The ordered pair generated at the right is (-2, -3).
b. Write a system of linear equations that has your ordered pair as its solution.
c. Exchange systems with your partner and use one of the methods from Exploration 1 to solve the system. Explain your choice of method.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 1

Communicate Your Answer

Question 3.
How can you use substitution to solve a system of linear equations?
Answer:

Question 4.
Use one of the methods from Exploration 1 to solve each system of linear equations. Explain your choice of method. Check your solutions.
a. x + 2y = -7
2x – y = -9

b.x – 2y = -6
2x + y = -2

c.-3x + 2y = -10
-2x + y = -6

d. 3x + 2y = 13
x – 3y = -3

e. 3x – 2y = 9
-x – 3y = 8

f. 3x – y = -6
4x + 5y = 11
Answer:

Monitoring Progress

Solve the system of linear equations by substitution. Check your solution.
Question 1.
y = 3x + 14
y = -4x
Answer:

Question 2.
3x + 2y = 0
y = \(\frac{1}{2}\)x – 1
Answer:

Question 3.
x = 6y – 7
4x + y = -3
Answer:

Solve the system of linear equations by substitution. Check your solution.
Question 4.
x + y = -2
-3x + y = 6
Answer:

Question 5.
-x + y = -4
4x – y = 10
Answer:

Question 6.
2x – y = -5
3x – y = 1
Answer:

Question 7.
x – 2y = 7
3x – 2y = 3
Answer:

Question 8.
There are a total of 64 students in a drama club and a yearbook club. The drama club has 10 more students than the yearbook club. Write a system of linear equations that represents this situation. How many students are in each club?
Answer:

Solving Systems of Linear Equations by Substitution 5.2 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Describe how to solve a system of linear equations by substitution.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 1

Question 2.
NUMBER SENSE
When solving a system of linear equations by substitution, how do you decide which variable to solve for in Step 1?
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3−8, tell which equation you would choose to solve for one of the variables. Explain.
Question 3.
x + 4y = 30
x – 2y = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 3

Question 4.
3x – y = 0
2x + y = -10
Answer:

Question 5.
5x + 3y = 11
5x – y = 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 5

Question 6.
3x – 2y = 19
x + y = 8
Answer:

Question 7.
x – y = -3
4x + 3y = -5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 7

Question 8.
3x + 5y = 25
x – 2y = -6
Answer:

In Exercises 9–16, solve the system of linear equations by substitution. Check your solution.
Question 9.
x = 17 – 4y
y = x – 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 9.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 9.2

Question 10.
6x – 9 = y
y = -3x
Answer:

Question 11.
x = 16 – 4y
3x + 4y = 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 11.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 11.2

Question 12.
-5x + 3y = 51
y = 10x – 8
Answer:

Question 13.
2x = 12
x – 5y = -29
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 13

Question 14.
2x – y = 23
x – 9 = -1
Answer:

Question 15.
5x + 2y = 9
x + y = -3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 15.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 15.2

Question 16.
11x – 7y = -14
x – 2y = -4
Answer:

Question 17.
ERROR ANALYSIS
Describe and correct the error in solving for one of the variables in the linear system 8x + 2y = -12 and 5x – y = 4.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 17.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 17.2

Question 18.
ERROR ANALYSIS
Describe and correct the error in solving for one of the variables in the linear system 4x + 2y = 6 and 3x + y = 9.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 3
Answer:

Question 19.
MODELING WITH MATHEMATICS
A farmer plants corn and wheat on a 180-acre farm. The farmer wants to plant three times as many acres of corn as wheat. Write a system of linear equations that represents this situation. How many acres of each crop should the farmer plant?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 19.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 19.2

Question 20.
MODELING WITH MATHEMATICS
A company that offers tubing trips down a river rents tubes for a person to use and “cooler” tubes to carry food and water. A group spends $270 to rent a total of 15 tubes. Write a system of linear equations that represents this situation. How many of each type of tube does the group rent?
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 4
Answer:

In Exercises 21–24, write a system of linear equations that has the ordered pair as its solution.
Question 21.
(3, 5)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 21

Question 22.
(-2, 8)
Answer:

Question 23.
(-4, -12)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 23

Question 24.
(15, -25)
Answer:

Question 25.
PROBLEM SOLVING
A math test is worth 100 points and has 38 problems. Each problem is worth either 5 points or 2 points. How many problems of each point value are on the test?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 25.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 25.2

Question 26.
PROBLEM SOLVING
An investor owns shares of Stock A and Stock B. The investor owns a total of 200 shares with a total value of $4000. How many shares of each stock does the investor own?
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 5
Answer:

MATHEMATICAL CONNECTIONS In Exercises 27 and 28, (a) write an equation that represents the sum of the angle measures of the triangle and (b) use your equation and the equation shown to find the values of x and y.
Question 27.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 27.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 27.2

Question 28.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 7
Answer:

Question 29.
REASONING
Find the values of a and b so that the solution of the linear system is (-9, 1).
ax + by = -31 Equation 1
ax – by = -41 Equation 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 29.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 29.2

Question 30.
MAKING AN ARGUMENT
Your friend says that given a linear system with an equation of a horizontal line and an equation of a vertical line, you cannot solve the system by substitution. Is your friend correct? Explain.
Answer:

Question 31.
OPEN-ENDED
Write a system of linear equations in which (3, -5) is a solution of Equation 1 but not a solution of Equation 2, and (-1, 7) is a solution of the system.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 31.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 31.2

Question 32.
HOW DO YOU SEE IT?
The graphs of two linear equations are shown.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 8
a. At what point do the lines appear to intersect?
b. Could you solve a system of linear equations by substitution to check your answer in part (a)? Explain.
Answer:

Question 33.
REPEATED REASONING
A radio station plays a total of 272 pop, rock, and hip-hop songs during a day. The number of pop songs is 3 times the number of rock songs. The number of hip-hop songs is 32 more than the number of rock songs. How many of each type of song does the radio station play?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 33.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 33.2

Question 34.
THOUGHT PROVOKING
You have $2.65 in coins. Write a system of equations that represents this situation. Use variables to represent the number of each type of coin.
Answer:

Question 35.
NUMBER SENSE
The sum of the digits of a two-digit number is 11. When the digits are reversed, the number increases by 27. Find the original number
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 35.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 35.2

Maintaining Mathematical Proficiency

Find the sum or difference.
Question 36.
(x – 4) + (2x – 7)
Answer:

Question 37.
(5y – 12) + (-5y – 1)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 37.1

Question 38.
(t – 8) – (t + 15)
Answer:

Question 39.
(6d + 2) – (3d – 3)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 39

Question 40.
4(m + 2) + 3(6m – 4)
Answer:

Question 41.
2(5v + 6) – 6(-9v + 2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 41

Lesson 5.3 Solving Systems of Linear Equations by Elimination

Essential Question How can you use elimination to solve a system of linear equations?

EXPLORATION 1

Writing and Solving a System of Equations
Work with a partner. You purchase a drink and a sandwich for $4.50. Your friend purchases a drink and five sandwiches for $16.50. You want to determine the price of a drink and the price of a sandwich.
a. Let x represent the price (in dollars) of one drink. Let y represent the price (in dollars) of one sandwich. Write a system of equations for the situation. Use the following verbal model.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 1
Label one of the equations Equation 1 and the other equation Equation 2.
b. Subtract Equation 1 from Equation 2. Explain how you can use the result to solve the system of equations. Then find and interpret the solution.

EXPLORATION 2

Using Elimination to Solve Systems
Work with a partner. Solve each system of linear equations using two methods.
Method 1 Subtract. Subtract Equation 2 from Equation 1. Then use the result to solve the system.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 2
Method 2 Add. Add the two equations. Then use the result to solve the system.
Is the solution the same using both methods? Which method do you prefer?
a. 3x – y = 6
3x + y = 0

b. 2x + y = 6
2x – y = 2

c. x – 2y = -7
x + 2y = 5

EXPLORATION 3

Using Elimination to Solve a System
Work with a partner.
2x + y = 7 Equation 1
x + 5y = 17 Equation 2
a. Can you eliminate a variable by adding or subtracting the equations as they are? If not, what do you need to do to one or both equations so that you can?
b. Solve the system individually. Then exchange solutions with your partner and compare and check the solutions.

Communicate Your Answer

Question 4.
How can you use elimination to solve a system of linear equations?
Answer:

Question 5.
When can you add or subtract the equations in a system to solve the system? When do you have to multiply first? Justify your answers with examples.
Answer:

Question 6.
In Exploration 3, why can you multiply an equation in the system by a constant and not change the solution of the system? Explain your reasoning.
Answer:

Monitoring Progress

Solve the system of linear equations by elimination. Check your solution.
Question 1.
3x + 2y = 7
-3x + 4y = 5
Answer:

Question 2.
x – 3y = 24
3x + y = 12
Answer:

Question 3.
x + 4y = 22
4x + y = 13
Answer:

Question 4.
Solve the system in Example 3 by eliminating x.
Answer:

Solving Systems of Linear Equations by Elimination 5.3 Exercises

Vocabulary and Core Concept Check
Question 1.
OPEN-ENDED
Give an example of a system of linear equations that can be solved by first adding the equations to eliminate one variable.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 1

Question 2.
WRITING
Explain how to solve the system of linear equations by elimination.
2x – 3y = -4 Equation 1
-5x + 9y = 7 Equation 2
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3−10, solve the system of linear equations by elimination. Check your solution.
Question 3.
x + 2y = 13
-x + y = 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 3.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 3.2

Question 4.
9x + y = 2
-4x – y = -17
Answer:

Question 5.
5x + 6y = 50
x – 6y = -26
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 5.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 5.2

Question 6.
-x + y = 4
x + 3y = 4
Answer:

Question 7.
-3x – 5y = -7
-4x + 5y = 14
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 7.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 7.2

Question 8.
4x – 9y = -21
-4x – 3y = 9
Answer:

Question 9.
-y – 10 = 6x
5x + y = -10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 9.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 9.2

Question 10.
3x – 30 = y
7y – 6 = 3x
Answer:

In Exercises 11–18, solve the system of linear equations by elimination. Check your solution.
Question 11.
x + y = 2
2x + 7y = 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 11.1

Question 12.
8x – 5y = 11
4x – 3y = 5
Answer:

Question 13.
11x – 20y = 28
3x + 4y = 36
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 13.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 13.2

Question 14.
10x – 9y = 46
-2x + 3y = 10
Answer:

Question 15.
4x – 3y = 8
5x – 2y = -11
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 15.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 15.2

Question 16.
-2x – 5y = 9
3x + 11y = 4
Answer:

Question 17.
9x + 2y = 39
6x + 13y = -9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 17.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 17.2

Question 18.
12x – 7y = -2
8x + 11y = 30
Answer:

Question 19.
ERROR ANALYSIS
Describe and correct the error in solving for one of the variables in the linear system 5x – 7y = 16 and x + 7y = 8.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 19

Question 20.
ERROR ANALYSIS
Describe and correct the error in solving for one of the variables in the linear system 4x + 3y = 8 and x – 2y = -13.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 4
Answer:

Question 21.
MODELING WITH MATHEMATICS
A service center charges a fee of x dollars for an oil change plus y dollars per quart of oil used. A sample of its sales record is shown. Write a system of linear equations that represents this situation. Find the fee and cost per quart of oil.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 21.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 21.2

Question 22.
MODELING WITH MATHEMATICS
A music website charges x dollars for individual songs and y dollars for entire albums. Person A pays $25.92to download 6 individual songs and 2 albums. Person B pays $33.93 to download 4 individual songs and 3 albums. Write a system of linear equations that represents this situation. How much does the website charge to download a song? an entire album?
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 6
Answer:

In Exercises 23–26, solve the system of linear equations using any method. Explain why you chose the method.
Question 23.
3x + 2y = 4
2y = 8 – 5x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 23.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 23.2

Question 24.
-6y + 2 = -4x
y – 2 = x
Answer:

Question 25.
y – x = 2
y = – \(\frac{1}{4}\) x + 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 25

Question 26.
3x + y = \(\frac{1}{3}\)
2x – 3y = \(\frac{8}{3}\)
Answer:

Question 27.
WRITING
For what values of a can you solve the linear system ax + 3y = 2 and 4x + 5y = 6 by elimination without multiplying first? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 27

Question 28.
HOW DO YOU SEE IT?
The circle graph shows the results of a survey in which 50 students were asked about their favorite meal.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 7
a. Estimate the numbers of students who chose breakfast and lunch.
b. The number of students who chose lunch was 5 more than the number of students who chose breakfast. Write a system of linear equations that represents the numbers of students who chose breakfast and lunch.
c. Explain how you can solve the linear system in part (b) to check your answers in part (a).
Answer:

Question 29.
MAKING AN ARGUMENT
Your friend says that any system of equations that can be solved by elimination can be solved by substitution in an equal or fewer number of steps. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 29

Question 30.
THOUGHT PROVOKING
Write a system of linear equations that can be added to eliminate a variable or subtracted to eliminate a variable.
Answer:

Question 31.
MATHEMATICAL CONNECTIONS
A rectangle has a perimeter of 18 inches. A new rectangle is formed by doubling the width w and tripling the length ℓ, as shown. The new rectangle has a perimeter P of 46 inches.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 8
a. Write and solve a system of linear equations to find the length and width of the original rectangle.
b. Find the length and width of the new rectangle.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 31.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 31.2

Question 32.
CRITICAL THINKING
Refer to the discussion of System 1 and System 2 on page 248. Without solving, explain why the two systems shown have the same solution.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 9
Answer:

Question 33.
PROBLEM SOLVING
You are making 6 quarts of fruit punch for a party. You have bottles of 100% fruit juice and 20% fruit juice. How many quarts of each type of juice should you mix to make 6 quarts of 80% fruit juice?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 33.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 33.2

Question 34.
PROBLEM SOLVING
A motorboat takes 40 minutes to travel 20 miles downstream. The return trip takes 60 minutes. What is the speed of the current?
Answer:

Question 35.
CRITICAL THINKING
Solve for x, y, and z in the system of equations. Explain your steps.
x + 7y + 3z = 29 Equation 1
3z + x – 2y = -7 Equation 2
5y = 10 – 2x Equation 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 35.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 35.2
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 35.3

Maintaining Mathematical Proficiency

Solve the equation. Determine whether the equation has one solution, no solution, or infinitely many solutions.
Question 36.
5d – 8 = 1 + 5d
Answer:

Question 37.
9 + 4t = 12 – 4t
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 37

Question 38.
3n + 2 = 2(n – 3)
Answer:

Question 39.
-3(4 – 2v) = 6v – 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 39

Write an equation of the line that passes through the given point and is parallel to the given line.
Question 40.
(4, -1); y = -2x + 7
Answer:

Question 41.
(0, 6); y = 5x – 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 41

Question 42.
(-5, -2); y = \(\frac{2}{3}\)x + 1
Answer:

Lesson 5.4 Solving Special Systems of Linear Equations

Essential Question Can a system of linear equations have no solution or infinitely many solutions?

EXPLORATION 1

Using a Table to Solve a System
Work with a partner. You invest $450 for equipment to make skateboards. The materials for each skateboard cost $20. You sell each skateboard for $20.
a. Write the cost and revenue equations. Then copy and complete the table for your cost C and your revenue R.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 1
b. When will your company break even? What is wrong?
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 2

EXPLORATION 2

Writing and Analyzing a System
Work with a partner. A necklace and matching bracelet have two types of beads. The necklace has 40 small beads and 6 large beads and weighs 10 grams. The bracelet has 20 small beads and 3 large beads and weighs 5 grams. The threads holding the beads have no significant weight.
a. Write a system of linear equations that represents the situation. Let x be the weight (in grams) of a small bead and let y be the weight (in grams) of a large bead.
b. Graph the system in the coordinate plane shown. What do you notice about the two lines?
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 3
c. Can you find the weight of each type of bead? Explain your reasoning.

Communicate Your Answer

Question 3.
Can a system of linear equations have no solution or infinitely many solutions? Give examples to support your answers.
Answer:

Question 4.
Does the system of linear equations represented by each graph have no solution, one solution, or infinitely many solutions? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 4
Answer:

Monitoring Progress

Solve the system of linear equations.
Question 1.
x + y = 3
2x + 2y = 6
Answer:

Question 2.
y = -x + 3
2x + 2y = 4
Answer:

Question 3.
x + y = 3
x + 2y = 4
Answer:

Question 4.
y = -10x + 2
10x + y = 10
Answer:

Question 5.
WHAT IF?
What happens to the solution in Example 3 when the perimeter of the trapezoidal piece of land is 96 kilometers? Explain.
Answer:

Solving Special Systems of Linear Equations 5.4 Exercises

Vocabulary and Core Concept Check
Question 1.
REASONING
Is it possible for a system of linear equations to have exactly two solutions? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 1

Question 2.
WRITING
Compare the graph of a system of linear equations that has infinitely many solutions and the graph of a system of linear equations that has no solution.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3−8, match the system of linear equations with its graph. Then determine whether the system has one solution, no solution, or infinitely many solutions.
Question 3.
-x + y = 1
x – y = 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 2.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 2.2

Question 4.
2x – 2y = 4
-x + y = -2
Answer:

Question 5.
2x + y = 4
-4x – 2y = -8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 3.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 3.2

Question 6.
x – y = 0
5x – 2y = 6
Answer:

Question 7.
-2x + 4y = 1
3x – 6y = 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 7.1

Question 8.
5x + 3y = 17
x – 3y = -2
Answer:

Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 5

In Exercises 9–16, solve the system of linear equations.
Question 9.
y = -2x – 4
y = 2x – 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 9

Question 10.
y = -6x – 8
y = -6x + 8
Answer:

Question 11.
3x – y = 6
-3x + y = -6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 11

Question 12.
-x + 2y = 7
x – 2y = 7
Answer:

Question 13.
4x + 4y = -8
-2x – 2y = 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 13

Question 14.
15x – 5y = -20
-3x + y = 4
Answer:

Question 15.
9x – 15y = 24
6x – 10y = -16
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 15

Question 16.
3x – 2y = -5
4x + 5y = 47
Answer:

In Exercises 17–22, use only the slopes and y-intercepts of the graphs of the equations to determine whether the system of linear equations has one solution, no solution, or infinitely many solutions. Explain.
Question 17.
y = 7x + 13
-21x + 3y = 39
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 17.1

Question 18.
y = -6x – 2
12x + 2y = -6
Answer:

Question 19.
4x + 3y = 27
4x – 3y = -27
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 19

Question 20.
-7x + 7y = 1
2x – 2y = -18
Answer:

Question 21.
-18x + 6y = 24
3x – y = -2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 21

Question 22.
2x – 2y = 16
3x – 6y = 30
Answer:

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in solving the system of linear equations.
Question 23.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 23.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 23.2

Question 24.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 7
Answer:

Question 25.
MODELING WITH MATHEMATICS
A small bag of trail mix contains 3 cups of dried fruit and 4 cups of almonds. A large bag contains 4\(\frac{1}{2}\) cups of dried fruit and 6 cups of almonds. Write and solve a system of linear equations to find the price of 1 cup of dried fruit and 1 cup of almonds.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 25.1

Question 26.
MODELING WITH MATHEMATICS
In a canoe race, Team A is traveling 6 miles per hour and is 2 miles ahead of Team B. Team B is also traveling 6 miles per hour. The teams continue traveling at their current rates for the remainder of the race. Write a system of linear equations that represents this situation. Will Team B catch up to Team A? Explain.
Answer:

Question 27.
PROBLEM SOLVING
A train travels from New York City to Washington, D.C., and then back to New York City. The table shows the number of tickets purchased for each leg of the trip. The cost per ticket is the same for each leg of the trip. Is there enough information to determine the cost of one coach ticket? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 27.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 27.2

Question 28.
THOUGHT PROVOKING
Write a system of three linear equations in two variables so that any two of the equations have exactly one solution, but the entire system of equations has no solution.
Answer:

Question 29.
REASONING
In a system of linear equations, one equation has a slope of 2 and the other equation has a slope of –\(\frac{1}{3}\). How many solutions does the system have? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 29

Question 30.
HOW DO YOU SEE IT?
The graph shows information about the last leg of a 4 × 200-meter relay for three relay teams. Team A’s runner ran about 7.8 meters per second, Team B’s runner ran about 7.8 meters per second, and Team C’s runner ran about 8.8 meters per second.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 10
a. Estimate the distance at which Team C’s runner passed Team B’s runner.
b. If the race was longer, could Team C’s runner have passed Team A’s runner? Explain.
c. If the race was longer, could Team B’s runner have passed Team A’s runner? Explain.
Answer:

Question 31.
ABSTRACT REASONING
Consider the system of linear equations y = ax + 4 and y = bx – 2, where a and b are real numbers. Determine whether each statement is always, sometimes, or never true. Explain your reasoning.
a. The system has infinitely many solutions.
b. The system has no solution.c. When a < b, the system has one solution.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 31

Question 32.
MAKING AN ARGUMENT
One admission to an ice skating rink costs x dollars, and renting a pair of ice skates costs y dollars. Your friend says she can determine the exact cost of one admission and one skate rental. Is your friend correct? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 11
Answer:

Maintaining Mathematical Proficiency

Solve the equation. Check your solutions.
Question 33.
|2x + 6| = |x|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 33

Question 34.
|3x – 45| = |12x|
Answer:

Question 35.
|x – 7| = |2x – 8|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 35.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 35.2

Question 36.
|2x + 1| = |3x – 11|
Answer:

Solving Systems of Linear Equations Study Skills: Analyzing Your Errors

5.1 – 5.4 What Did You Learn?

Core Vocabulary
system of linear equations, p. 236
solution of a system of linear equations, p. 236

Core Concepts
Section 5.1
Solving a System of Linear Equations by Graphing, p. 237

Section 5.2
Solving a System of Linear Equations by Substitution, p. 242

Section 5.3
Solving a System of Linear Equations by Elimination, p. 248

Section 5.4
Solutions of Systems of Linear Equations, p. 254

Mathematical Practices

Question 1.
Describe the given information in Exercise 33 on page 246 and your plan for finding the solution.
Answer:

Question 2.
Describe another real-life situation similar to Exercise 22 on page 251 and the mathematics that you can apply to solve the problem.
Answer:

Question 3.
What question(s) can you ask your friend to help her understand the error in the statement she made in Exercise 32 on page 258?
Answer:

Study Skills: Analyzing Your Errors

Study Errors
What Happens: You do not study the right material or you do not learn it well enough to remember it on a test without resources such as notes.
How to Avoid This Error: Take a practice test. Work with a study group. Discuss the topics on the test with your teacher. Do not try to learn a whole chapter’s worth of material in one night.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 12

Solving Systems of Linear Equations 5.1–5.4 Quiz

Use the graph to solve the system of linear equations. Check your solution.
Question 1.
y = – \(\frac{1}{3}\)x + 2
y = x – 2
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations q 1
Answer:

Question 2.
y = \(\frac{1}{2}\)x – 1
y = 4x + 6
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations q 2
Answer:

Question 3.
y = 1
y = 2x + 1
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations q 3
Answer:

Solve the system of linear equations by substitution. Check your solution.
Question 4.
y = x – 4
-2x + y = 18
Answer:

Question 5.
2y + x = -4
y – x = -5
Answer:

Question 6.
3x – 5y = 13
x + 4y = 10
Answer:

Solve the system of linear equations by elimination. Check your solution.
Question 7.
x + y = 4
-3x – y = -8
Answer:

Question 8.
x + 3y = 1
5x + 6y = 14
Answer:

Question 9.
2x – 3y = -5
5x + 2y = 16
Answer:

Solve the system of linear equations.
Question 10.
x – y = 1
x – y = 6
Answer:

Question 11.
6x + 2y = 16
2x – y = 2
Answer:

Question 12.
3x – 3y = -2
-6x + 6y = 4
Answer:

Question 13.
You plant a spruce tree that grows 4 inches per year and a hemlock tree that grows 6 inches per year. The initial heights are shown.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations q 4
a. Write a system of linear equations that represents this situation.
b. Solve the system by graphing. Interpret your solution.
Answer:

Question 14.
It takes you 3 hours to drive to a concert 135 miles away. You drive 55 miles per hour on highways and 40 miles per hour on the rest of the roads.
a. How much time do you spend driving at each speed?
b. How many miles do you drive on highways? the rest of the roads?
Answer:

Question 15.
In a football game, all of the home team’s points are from 7-point touchdowns and 3-point field goals. The team scores six times. Write and solve a system of linear equations to find the numbers of touchdowns and field goals that the home team scores.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations q 5
Answer:

Lesson 5.5 Solving Equations by Graphing

Essential Question How can you use a system of linear equations to solve an equation with variables on both sides?

Previously, you learned how to use algebra to solve equations with variables on both sides. Another way is to use a system of linear equations.

EXPLORATION 1

Solving an Equation by Graphing
Work with a partner. Solve 2x – 1 = – \(\frac{1}{2}\) x + 4 by graphing.
a. Use the left side to write a linear equation. Then use the right side to write another linear equation.
b. Graph the two linear equations from part (a). Find the x-value of the point of intersection. Check that the x-value is the solution of
2x – 1 = –\(\frac{1}{2}\)x + 4.
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 1
c. Explain why this “graphical method” works.

EXPLORATION 2

Solving Equations Algebraically and Graphically
Work with a partner. Solve each equation using two methods.
Method 1 Use an algebraic method.
Method 2 Use a graphical method.
Is the solution the same using both methods?
a. \(\frac{1}{2}\)x + 4 = –\(\frac{1}{4}\)x + 1
b. \(\frac{2}{3}\)x + 4 = \(\frac{1}{3}\)x + 3
c. –\(\frac{2}{3}\) x – 1 = \(\frac{1}{3}\)x – 4
d. \(\frac{4}{5}\) x + \(\frac{7}{5}\) = 3x – 3
e. -x + 2.5 = 2x – 0.5
f. – 3x + 1.5 = x + 1.5

Communicate Your Answer

Question 3.
How can you use a system of linear equations to solve an equation with variables on both sides?
Answer:

Question 4.
Compare the algebraic method and the graphical method for solving a linear equation with variables on both sides. Describe the advantages and disadvantages of each method.
Answer:

Monitoring Progress

Solve the equation by graphing. Check your solution.
Question 1.
\(\frac{1}{2}\)x – 3 = 2x
Answer:

Question 2.
-4 + 9x = -3x + 2
Answer:

Solve the equation by graphing. Check your solutions.
Question 3.
|2x + 2| = |x – 2|
Answer:

Question 4.
|x – 6| = |-x + 4|
Answer:

Question 5.
WHAT IF?
Company C charges $3.30 per mile plus a flat fee of $115 per week. After how many miles are the total costs the same at Company A and Company C?
Answer:

Solving Equations by Graphing 5.5 Exercises

Vocabulary and Core Concept Check
Question 1.
REASONING
The graphs of the equations y = 3x – 20 and y = -2x + 10 intersect at the point (6, −2). Without solving, find the solution of the equation 3x – 20 = -2x + 10.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 1

Question 2.
WRITING
Explain how to rewrite the absolute value equation |2x – 4| = |-5x + 1| as two systems of linear equations.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, use the graph to solve the equation. Check your solution.
Question 3.
-2x + 3 = x
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 3

Question 4.
-3 = 4x + 1
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 3
Answer:

Question 5.
-x – 1 = \(\frac{1}{3}\)x + 3
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 5

Question 6.
–\(\frac{3}{2}\)x – 2 = -4x + 3
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 5
Answer:

In Exercises 7−14, solve the equation by graphing. Check your solution.
Question 7.
x + 4 = -x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 7

Question 8.
4x = x + 3
Answer:

Question 9.
x + 5 = -2x – 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 9

Question 10.
-2x + 6 = 5x – 1
Answer:

Question 11.
\(\frac{1}{2}\)x – 2 = 9 – 5x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 11

Question 12.
-5 + \(\frac{1}{4}\)x = 3x + 6
Answer:

Question 13.
5x – 7 = 2(x + 1)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 13

Question 14.
-6(x + 4) = -3x – 6
Answer:

In Exercises 15−20, solve the equation by graphing. Determine whether the equation has one solution, no solution, or infinitely many solutions.
Question 15.
3x – 1 = -x + 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 15

Question 16.
5x – 4 = 5x + 1
Answer:

Question 17.
-4(2 – x) = 4x – 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 17

Question 18.
-2x – 3 = 2(x – 2)
Answer:

Question 19.
-x – 5 = –\(\frac{1}{3}\) (3x + 5)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 19

Question 20.
\(\frac{1}{2}\)(8x + 3) = 4x + \(\frac{3}{2}\)
Answer:

In Exercises 21 and 22, use the graphs to solve the equation. Check your solutions.
Question 21.
|x – 4| = |3x|
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 21

Question 22.
|2x + 4| = |x – 1|
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 7
Answer:

In Exercises 23−30, solve the equation by graphing. Check your solutions.
Question 23.
|2x| = |x + 3|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 23.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 23.2

Question 24.
|2x – 6| = |x|
Answer:

Question 25.
|-x + 4| = |2x – 2|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 25.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 25.2

Question 26.
|x + 2| = |-3x + 6|
Answer:

Question 27.
|x + 1| = |x – 5|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 27.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 27.2

Question 28.
|2x + 5| = |-2x + 1|
Answer:

Question 29.
|x – 3| = 2|x|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 29.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 29.2

Question 30.
4|x + 2| = |2x + 7|
Answer:

USING TOOLS In Exercises 31 and 32, use a graphing calculator to solve the equation.
Question 31.
0.7x + 0.5 = -0.2x – 1.3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 31

Question 32.
2.1x + 0.6 = -1.4x + 6.9
Answer:

Question 33.
MODELING WITH MATHEMATICS
You need to hire a catering company to serve meals to guests at a wedding reception. Company A charges $500 plus $20 per guest. Company B charges $800 plus $16 per guest. For how many guests are the total costs the same at both companies?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 33

Question 34.
MODELING WITH MATHEMATICS
Your dog is 16 years old in dog years. Your cat is 28 years old in cat years. For every human year, your dog ages by 7 dog years and your cat ages by 4 cat years. In how many human years will both pets be the same age in their respective types of years?
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 8
Answer:

Question 35.
MODELING WITH MATHEMATICS
You and a friend race across a field to a fence and back. Your friend has a 50-meter head start. The equations shown represent you and your friend’s distances d (in meters) from the fence t seconds after the race begins. Find the time at which you catch up to your friend.
You: d = |-5t + 100|
Your friend: d = |-3\(\frac{1}{3}\)t + 50|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 35

Question 36.
MAKING AN ARGUMENT
The graphs of y = -x + 4 and y = 2x – 8 intersect at the point (4, 0). So, your friend says the solution of the equation -x + 4 = 2x – 8 is (4, 0). Is your friend correct? Explain.
Answer:

Question 37.
OPEN-ENDED
Find values for m and b so that the solution of the equation mx + b = – 2x – 1 is x = -3.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 37

Question 38.
HOW DO YOU SEE IT?
The graph shows the total revenue and expenses of a company x years after it opens for business.
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 9
a. Estimate the point of intersection of the graphs.
b. Interpret your answer in part (a).
Answer:

Question 39.
MATHEMATICAL CONNECTIONS
The value of the perimeter of the triangle (in feet) is equal to the value of the area of the triangle (in square feet). Use a graph to find x.
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 39

Question 40.
THOUGHT PROVOKING
A car has an initial value of $20,000 and decreases in value at a rate of $1500 per year. Describe a different car that will be worth the same amount as this car in exactly 5 years. Specify the initial value and the rate at which the value decreases.
Answer:

Question 41.
ABSTRACT REASONING
Use a graph to determine the sign of the solution of the equation ax + b = cx + d in each situation.
a. 0 < b < d and a < c
b. d < b < 0 and a < c Answer: Maintaining Mathematical Proficiency Graph the inequality.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 41.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 41.2

Question 42.
y > 5
Answer:

Question 43.
x ≤ -2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 43

Question 44.
n ≥ 9
Answer:

Question 45.
c < -6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 45

Use the graphs of f and g to describe the transformation from the graph of f to the graph of g.
Question 46.
f(x) = x – 5; g(x) = f(x + 2)
Answer:

Question 47.
f(x) = 6x; g(x) = -f(x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 47

Question 48.
f(x) = -2x + 1; g(x) = f(4x)
Answer:

Question 49.
f(x) = \(\frac{1}{2}\)x – 2; g(x) = f(x – 1)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 49

Lesson 5.6 Graphing Linear Inequalities in Two Variables

Essential Question How can you graph a linear inequality in two variables?
A solution of a linear inequality in two variables is an ordered pair (x, y) that makes the inequality true. The graph of a linear inequality in two variables shows all the solutions of the inequality in a coordinate plane.

EXPLORATION 1

Writing a Linear Inequality in Two Variables
Work with a partner.
a. Write an equation represented by the dashed line.
b. The solutions of an inequality are represented by the shaded region. In words, describe the solutions of the inequality.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 1
c. Write an inequality represented by the graph. Which inequality symbol did you use? Explain your reasoning.

EXPLORATION 2

Using a Graphing Calculator
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 2
Work with a partner. Use a graphing calculator to graph y ≥ \(\frac{1}{4}\)x – 3.
a. Enter the equation y = \(\frac{1}{4}\)x – 3 into your calculator.
b. The inequality has the symbol ≥. So, the region to be shaded is above the graph of y = \(\frac{1}{4}\)x – 3, as shown. Verify this by testing a point in this region, such as (0, 0), to make sure it is a solution of the inequality.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 3
Because the inequality symbol is greater than or equal to, the line is solid and not dashed. Some graphing calculators always use a solid line when graphing inequalities. In this case, you have to determine whether the line should be solid or dashed, based on the inequality symbol used in the original inequality

EXPLORATION 3

Graphing Linear Inequalities in Two Variables
Work with a partner. Graph each linear inequality in two variables. Explain your steps. Use a graphing calculator to check your graphs.
a. y > x + 5
b. y ≤ –\(\frac{1}{2}\)x + 1
c. y ≥ -x – 5

Communicate Your Answer

Question 4.
How can you graph a linear inequality in two variables?
Answer:

Question 5.
Give an example of a real-life situation that can be modeled using a linear inequality in two variables.
Answer:

Monitoring Progress

Tell whether the ordered pair is a solution of the inequality.
Question 1.
x + y > 0; (-2, 2)
Answer:

Question 2.
4x – y ≥ 5; (0, 0)
Answer:

Question 3.
5x – 2y ≤ -1; (-4, -1)
Answer:

Question 4.
-2x – 3y < 15; (5, -7)
Answer:

Graph the inequality in a coordinate plane.
Question 5.
y > -1
Answer:

Question 6.
x ≤ -4
Answer:

Question 7.
x + y ≤ -4
Answer:

Question 8.
x – 2y < 0
Answer:

Question 9.
You can spend at most $12 on red peppers and tomatoes for salsa. Red peppers cost $4 per pound, and tomatoes cost $3 per pound. Write and graph an inequality that represents the amounts of red peppers and tomatoes you can buy. Identify and interpret two solutions of the inequality.
Answer:

Graphing Linear Inequalities in Two Variables 5.6 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
How can you tell whether an ordered pair is a solution of a linear inequality?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 1

Question 2.
WRITING
Compare the graph of a linear inequality in two variables with the graph of a linear equation in two variables.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, tell whether the ordered pair is a solution of the inequality.
Question 3.
x + y < 7; (2, 3)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 3

Question 4.
x – y ≤ 0; (5, 2)
Answer:

Question 5.
x + 3y ≥ -2; (-9, 2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 5

Question 6.
8x + y > -6; (-1, 2)
Answer:

Question 7.
-6x + 4y ≤ 6; (-3, -3)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 7

Question 8.
3x – 5y ≥ 2; (-1, -1)
Answer:

Question 9.
-x – 6y > 12; (-8, 2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 9

Question 10.
-4x – 8y < 15; (-6, 3)
Answer:

In Exercises 11−16, tell whether the ordered pair is a solution of the inequality whose graph is shown.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 4
Question 11.
(0, -1)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 11

Question 12.
(-1, 3)
Answer:

Question 13.
(1, 4)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 13

Question 14.
(0, 0)
Answer:

Question 15.
(3, 3)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 15

Question 16.
(2, 1)
Answer:

Question 17.
MODELING WITH MATHEMATICS
A carpenter has at most $250 to spend on lumber. The inequality 8x + 12y ≤ 250 represents the numbers x of 2-by-8boards and the numbers y of 4-by-4 boards the carpenter can buy. Can the carpenter buy twelve 2-by-8 boards and fourteen 4-by-4 boards? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 17

Question 18.
MODELING WITH MATHEMATICS
The inequality 3x + 2y ≥ 93 represents the numbers x of multiple- choice questions and the numbers y of matching questions you can answer correctly to receive an A on a test. You answer 20 multiple-choice questions and 18 matching questions correctly. Do you receive an A on the test? Explain.
Answer:

In Exercises 19–24, graph the inequality in a coordinate plane.
Question 19.
y ≤ 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 19

Question 20.
y > 6
Answer:

Question 21.
x < 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 21

Question 22.
x ≥ -3
Answer:

Question 23.
y > -7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 23

Question 24.
x < 9
Answer:

In Exercises 25−30, graph the inequality in a coordinate plane.
Question 25.
y > -2x – 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 25

Question 26.
y ≤ 3x – 1
Answer:

Question 27.
-4x + y < -7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 27

Question 28.
3x – y ≥ 5
Answer:

Question 29.
5x – 2y ≤ 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 29

Question 30.
-x + 4y > -12
Answer:

ERROR ANALYSIS In Exercises 31 and 32, describe and correct the error in graphing the inequality.
Question 31.
y < -x + 1
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 31

Question 32.
y ≤ 3x – 2
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 7
Answer:

Question 33.
MODELING WITH MATHEMATICS
You have at most $20 to spend at an arcade. Arcade games cost $0.75 each, and snacks cost $2.25 each. Write and graph an inequality that represents the numbers of games you can play and snacks you can buy. Identify and interpret two solutions of the inequality.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 33.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 33.2
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 33.3

Question 34.
MODELING WITH MATHEMATICS
A drama club must sell at least $1500 worth of tickets to cover the expenses of producing a play. Write and graph an inequality that represents how many adult and student tickets the club must sell. Identify and interpret two solutions of the inequality.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 8
Answer:

In Exercises 35–38, write an inequality that represents the graph.
Question 35.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 35

Question 36.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 10
Answer:

Question 37.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 11
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 37

Question 38.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 12
Answer:

Question 39.
PROBLEM SOLVING
Large boxes weigh 75 pounds, and small boxes weigh 40 pounds.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 13
a. Write and graph an inequality that represents the numbers of large and small boxes a 200-pound delivery person can take on the elevator.
b. Explain why some solutions of the inequality might not be practical in real life.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 39.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 39.2
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 39.3

Question 40.
HOW DO YOU SEE IT?
Match each inequality with its graph.
a. 3x – 2y ≤ 6
b. 3x – 2y < 6 c. 3x – 2y > 6
d. 3x – 2y ≥ 6
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 14
Answer:

Question 41.
REASONING
When graphing a linear inequality in two variables, why must you choose a test point that is not on the boundary line?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 41

Question 42.
THOUGHT PROVOKING
Write a linear inequality in two variables that has the following two properties.
• (0, 0), (0, -1), and (0, 1) are not solutions.
• (1, 1), (3, -1), and (-1, 3) are solutions.
Answer:

Question 43.
WRITING
Can you always use (0, 0) as a test point when graphing an inequality? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 43

CRITICAL THINKING
In Exercises 44 and 45, write and graph an inequality whose graph is described by the given information.
Question 44.
The points (2, 5) and (−3, −5) lie on the boundary line. The points (6, 5) and (−2, −3) are solutions of the inequality.
Answer:

Question 45.
The points (−7, −16) and (1, 8) lie on the boundary line. The points (−7, 0) and (3, 14) are not solutions of the inequality.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 45.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 45.2

Maintaining Mathematical Proficiency

Write the next three terms of the arithmetic sequence.
Question 46.
0, 8, 16, 24, 32, . . .
Answer:

Question 47.
-5, -8, -11, -14, -17, . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 47.1

Question 48.
\(-\frac{3}{2},-\frac{1}{2}, \frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \ldots\)

Lesson 5.7 Systems of Linear Inequalities

Essential Question How can you graph a system of linear inequalities?

EXPLORATION 1

Graphing Linear Inequalities
Work with a partner. Match each linear inequality with its graph. Explain your reasoning.
2x + y ≤ 4 Inequality 1
2x – y ≤ 0 Inequality 2
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 1

EXPLORATION 2

Graphing a System of Linear Inequalities
Work with a partner. Consider the linear inequalities given in Exploration 1.
2x + y ≤ 4 Inequality 1
2x – y ≤ 0 Inequality 2
a. Use two different colors to graph the inequalities in the same coordinate plane. What is the result?
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 2
b. Describe each of the shaded regions of the graph. What does the unshaded region represent?

Communicate Your Answer

Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 3
Question 3.
How can you graph a system of linear inequalities?
Answer:

Question 4.
When graphing a system of linear inequalities, which region represents the solution of the system?
Answer:

Question 5.
Do you think all systems of linear inequalities have a solution? Explain your reasoning.
Answer:

Question 6.
Write a system of linear inequalities represented by the graph.
Answer:

Monitoring Progress

Tell whether the ordered pair is a solution of the system of linear inequalities.
Question 1.
(-1, 5); y < 5 y > x – 4
Answer:

Question 2.
(1, 4); y ≥ 3x + 1
y > x – 1
Answer:

Graph the system of linear inequalities.
Question 3.
y ≥ -x + 4
x + y ≤ 0
Answer:

Question 4.
y > 2x – 3
y ≥ \(\frac{1}{2}\)x + 1
Answer:

Question 5.
-2x + y < 4 2x + y > 4
Answer:

Write a system of linear inequalities represented by the graph.
Question 6.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 4
Answer:

Question 7.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 5
Answer:

Question 8.
Name another solution of Example 6.
Answer:

Question 9.
WHAT IF?
You want to spend at least 3 hours at the mall. How does this change the system? Is (2.5, 5) still a solution? Explain.
Answer:

Systems of Linear Inequalities 5.7 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
How can you verify that an ordered pair is a solution of a system of linear inequalities?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 1

Question 2.
WHICH ONE DOESN’T BELONG?
Use the graph shown. Which of the ordered pairs does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 6
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3−6, tell whether the ordered pair is a solution of the system of linear inequalities.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 7
Question 3.
(-4, 3)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 3

Question 4.
(-3, -1)
Answer:

Question 5.
(-2, 5)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 5

Question 6.
(1, 1)
Answer:

In Exercises 7−10, tell whether the ordered pair is a solution of the system of linear inequalities.
Question 7.
(-5, 2); y < 4
y > x + 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 7

Question 8.
(1, -1); y > -2
y > x – 5
Answer:

Question 9.
(0, 0); y ≤ x + 7
y ≥ 2x + 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 9

Question 10.
(4, -3); y ≤ -x + 1
y ≤ 5x – 2
Answer:

In Exercises 11−20, graph the system of linear inequalities.
Question 11.
y > -3
y ≥ 5x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 11

Question 12.
y < -1
x > 4
Answer:

Question 13.
y < -2 y > 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 13

Question 14.
y < x – 1
y ≥ x + 1
Answer:

Question 15.
y ≥ -5
y – 1 < 3x Answer: Question 16. x + y > 4
y ≥ \(\frac{3}{2}\)x – 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 15

Question 17.
x + y > 1
-x – y < -3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 17

Question 18.
2x + y ≤ 5
y + 2 ≥ -2x
Answer:

Question 19.
x < 4
y > 1
y ≥ -x + 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 19

Question 20.
x + y ≤ 10
x – y ≥ 2
y > 2
Answer:

In Exercises 21−26, write a system of linear inequalities represented by the graph.
Question 21.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 21

Question 22.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 9
Answer:

Question 23.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 23

Question 24.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 11
Answer:

Question 25.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 25

Question 26.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 13
Answer:

ERROR ANALYSIS In Exercises 27 and 28, describe and correct the error in graphing the system of linear inequalities.
Question 27.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 14
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 27

Question 28.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 15
Answer:

Question 29.
MODELING WITH MATHEMATICS
You can spend at most $21 on fruit. Blueberries cost $4 per pound, and strawberries cost $3 per pound. You need at least 3 pounds of fruit to make muffins.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 16
a. Write and graph a system of linear inequalities that represents the situation.
b. Identify and interpret a solution of the system.
c. Use the graph to determine whether you can buy 4 pounds of blueberries and 1 pound of strawberries.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 29.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 29.2
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 29.3

Question 30.
MODELING WITH MATHEMATICS
You earn $10 per hour working as a manager at a grocery store. You are required to work at the grocery store at least 8 hours per week. You also teach music lessons for $15 per hour. You need to earn at least $120 per week, but you do not want to work more than 20 hours per week.
a. Write and graph a system of linear inequalities that represents the situation.
b. Identify and interpret a solution of the system.
c. Use the graph to determine whether you can work 8 hours at the grocery store and teach 1 hour of music lessons.
Answer:

Question 31.
MODELING WITH MATHEMATICS
You are fishing for surfperch and rockfish, which are species of bottomfish. Gaming laws allow you to catch no more than 15 surfperch per day, no more than 10 rockfish per day, and no more than 20 total bottomfish per day.
a. Write and graph a system of linear inequalities that represents the situation.
b. Use the graph to determine whether you can catch 11 surfperch and 9 rockfish in 1 day.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 17
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 31.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 31.2

Question 32.
REASONING
Describe the intersection of the half-planes of the system shown.
x – y ≤ 4
x – y ≥ 4
Answer:

Question 33.
MATHEMATICAL CONNECTIONS
The following points are the vertices of a shaded rectangle.
(-1, 1), (6, 1), (6, -3), (-1, -3)
a. Write a system of linear inequalities represented by the shaded rectangle.
b. Find the area of the rectangle.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 33

Question 34.
MATHEMATICAL CONNECTIONS
The following points are the vertices of a shaded triangle.
(2, 5), (6, -3), (-2, -3)
a. Write a system of linear inequalities represented by the shaded triangle.
b. Find the area of the triangle.
Answer:

Question 35.
PROBLEM SOLVING
You plan to spend less than half of your monthly $2000 paycheck on housing and savings. You want to spend at least 10% of your paycheck on savings and at most 30% of it on housing. How much money can you spend on savings and housing?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 35.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 35.2

Question 36.
PROBLEM SOLVING
On a road trip with a friend, you drive about 70 miles per hour, and your friend drives about 60 miles per hour. The plan is to drive less than 15 hours and at least 600 miles each day. Your friend will drive more hours than you. How many hours can you and your friend each drive in 1 day?
Answer:

Question 37.
WRITING
How are solving systems of linear inequalities and solving systems of linear equations similar? How are they different?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 37

Question 38.
HOW DO YOU SEE IT?
The graphs of two linear equations are shown.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 18
Replace the equal signs with inequality symbols to create a system of linear inequalities that has point C as a solution, but not points A, B, and D. Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 19
Answer:

Question 39.
USING STRUCTURE
Write a system of linear inequalities that is equivalent to |y| < x, where x > 0. Graph the system.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 39

Question 40.
MAKING AN ARGUMENT
Your friend says that a system of linear inequalities in which the boundary lines are parallel must have no solution. Is your friend correct? Explain.
Answer:

Question 41.
CRITICAL THINKING
Is it possible for the solution set of a system of linear inequalities to be all real numbers? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 41

OPEN-ENDED In Exercises 42−44, write a system of linear inequalities with the given characteristic.
Question 42.
All solutions are in Quadrant I.
Answer:

Question 43.
All solutions have one positive coordinate and one negative coordinate.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 43

Question 44.
There are no solutions.
Answer:

Question 45.
OPEN-ENDED
One inequality in a system is -4x + 2y > 6. Write another inequality so the system has (a) no solution and (b) infinitely many solutions.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 45

Question 46.
THOUGHT PROVOKING
You receive a gift certificate for a clothing store and plan to use it to buy T-shirts and sweatshirts. Describe a situation in which you can buy 9 T-shirts and 1 sweatshirt, but you cannot buy 3 T-shirts and 8 sweatshirts. Write and graph a system of linear inequalities that represents the situation.
Answer:

Question 47.
CRITICAL THINKING
Write a system of linear inequalities that has exactly one solution.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 47

Question 48.
MODELING WITH MATHEMATICS
You make necklaces and key chains to sell at a craft fair. The table shows the amounts of time and money it takes to make a necklace and a key chain, and the amounts of time and money you have available for making them.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 20
a. Write and graph a system of four linear inequalities that represents the number x of necklaces and the number y of key chains that you can make.
b. Find the vertices (corner points) of the graph of the system.
c. You sell each necklace for $10 and each key chain for $8. The revenue R is given by the equation R = 10x + 8y. Find the revenue corresponding to each ordered pair in part (b). Which vertex results in the maximum revenue?
Answer:

Maintaining Mathematical Proficiency

Write the product using exponents.
Question 49.
4 • 4 • 4 • 4 • 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 49

Question 50.
(-13) • (-13) • (-13)
Answer:

Question 51.
x • x • x • x • x • x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 51

Write an equation of the line with the given slope and y-intercept.
Question 52.
slope: 1
y-intercept: -6
Answer:

Question 53.
slope: -3
y-intercept: 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 53

Question 54.
slope: –\(\frac{1}{4}\)
y-intercept: -1
Answer:

Question 55.
slope: \(\frac{4}{3}\)
y-intercept: 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 55

Solving Systems of Linear Equations Performance Task: Prize Patrol

5.5–5.7 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 21

Core Concepts
Section 5.5
Solving Linear Equations by Graphing, p. 262
Solving Absolute Value Equations by Graphing, p. 263

Section 5.6
Graphing a Linear Inequality in Two Variables, p. 269

Section 5.7
Graphing a System of Linear Inequalities, p. 275
Writing a System of Linear Inequalities, p. 276

Mathematical Practices

Question 1.
Why do the equations in Exercise 35 on page 266 contain absolute value expressions?
Answer:

Question 2.
Why is it important to be precise when answering part (a) of Exercise 39 on page 272?
Answer:

Question 3.
Describe the overall step-by-step process you used to solve Exercise 35 on page 279.
Answer:

Performance Task Prize Patrol

You have been selected to drive a prize patrol cart and place prizes on the competing teams’ predetermined paths. You know the teams’ routes and you can only make one pass. Where will you place the prizes so that each team will have a chance to find a prize on their route?
To explore the answers to these questions and more, go to
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 22

Solving Systems of Linear Equations Chapter Review

5.1 Solving Systems of Linear Equations by Graphing (pp. 235–240)

Solve the system of linear equations by graphing.
Question 1.
y = -3x + 1
y = x – 7
Answer:

Question 2.
y = -4x + 3
4x – 2y = 6
Answer:

Question 3.
5x + 5y = 15
2x – 2y = 10
Answer:

5.2 Solving Systems of Linear Equations by Substitution (pp. 241–246)

Solve the system of linear equations by substitution. Check your solution.
Question 4.
3x + y = -9
y = 5x + 7
Answer:

Question 5.
x + 4y = 6
x – y = 1
Answer:

Question 6.
2x + 3y = 4
y + 3x = 6
Answer:

Question 7.
You spend $20 total on tubes of paint and disposable brushes for an art project. Tubes of paint cost $4.00 each and paintbrushes cost $0.50 each. You purchase twice as many brushes as tubes of paint. How many brushes and tubes of paint do you purchase?
Answer:

5.3 Solving Systems of Linear Equations by Elimination (pp. 247 – 252)

Solve the system of linear equations by elimination. Check your solution.
Question 8.
9x – 2y = 34
5x + 2y = -6
Answer:

Question 9.
x + 6y = 28
2x – 3y = -19
Answer:

Question 10.
8x – 7y = -3
6x – 5y = -1
Answer:

5.4 Solving Special Systems of Linear Equations (pp. 253–258)

Solve the system of linear equations.
Question 11.
x = y + 2
-3x + 3y = 6
Answer:

Question 12.
3x – 6y = -9
-5x + 10y = 10
Answer:

Question 13.
-4x + 4y = 32
3x + 24 = 3y
Answer:

5.5 Solving Equations by Graphing (pp. 261–26

Solve the equation by graphing. Check your solution(s).
Question 14.
\(\frac{1}{3}\)x + 5 = -2x – 2
Answer:

Question 15.
|x + 1| = |-x – 9|
Answer:

Question 16.
|2x – 8| = |x + 5|
Answer:

5.6 Graphing Linear Inequalities in Two Variables (pp. 267–272)

Graph the inequality in a coordinate plane.
Question 17.
y > -4
Answer:

Question 18.
-9x + 3y ≥ 3
Answer:

Question 19.
5x + 10y < 40 Answer: 5.7 Systems of Linear Inequalities (pp. 273–280) Graph the system of linear inequalities. Question 20. y ≤ x – 3 y ≥ x + 1 Answer: Question 21. y > -2x + 3
y ≥ \(\frac{1}{4}\)x – 1
Answer:

Question 22.
x + 3y > 6
2x + y < 7
Answer:

Solving Systems of Linear Equations Chapter Test

Solve the system of linear equations using any method. Explain why you chose the method.

Question 1.
8x + 3y = -9
-8x + y = 29
Answer:

Question 2.
\(\frac{1}{2}\)x + y = -6
y = \(\frac{3}{5}\)x + 5
Answer:

Question 3.
y = 4x + 4
-8x + 2y = 8
Answer:

Question 4.
x = y – 11
x – 3y = 1
Answer:

Question 5.
6x – 4y = 9
9x – 6y = 15
Answer:

Question 6.
y = 5x – 7
-4x + y = -1
Answer:

Question 7.
Write a system of linear inequalities so the points (1, 2) and (4, -3) are solutions of the system, but the point (-2, 8) is not a solution of the system.
Answer:

Question 8.
How is solving the equation |2x + 1| = |x – 7| by graphing similar to solving the equation 4x + 3 = -2x + 9 by graphing? How is it different?
Answer:

Graph the system of linear inequalities.
Question 9.
y > \(\frac{1}{2}\)x + 4
2y ≤ x + 4
Answer:

Question 10.
x + y < 1 5x + y > 4
Answer:

Question 11.
y ≥ – \(\frac{2}{3}\)x + 1
-3x + y > -2
Answer:

Question 12.
You pay $45.50 for 10 gallons of gasoline and 2 quarts of oil at a gas station. Your friend pays $22.75 for 5 gallons of the same gasoline and 1 quart of the same oil.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ct 1
a. Is there enough information to determine the cost of 1 gallon of gasoline and 1 quart of oil? Explain.
b. The receipt shown is for buying the same gasoline and same oil. Is there now enough information to determine the cost of 1 gallon of gasoline and 1 quart of oil? Explain.
c. Determine the cost of 1 gallon of gasoline and 1 quart of oil.
Answer:

Question 13.
Describe the advantages and disadvantages of solving a system of linear equations by graphing.
Answer:

Question 14.
You have at most $60 to spend on trophies and medals to give as prizes for a contest.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ct 2
a. Write and graph an inequality that represents the numbers of trophies and medals you can buy. Identify and interpret a solution of the inequality.
b. You want to purchase at least 6 items. Write and graph a system that represents the situation. How many of each item can you buy?
Answer:

Question 15.
Compare the slopes and y-intercepts of the graphs of the equations in the linear system 8x + 4y = 12 and 3y = -6x – 15 to determine whether the system has one solution, no solution, or infinitely many solutions. Explain.
Answer:

Solving Systems of Linear Equations Cumulative Assessment

Question 1.
The graph of which equation is shown?
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 1
Answer:

Question 2.
A van rental company rents out 6-, 8-, 12-, and 16-passenger vans. The function C(x) = 100 + 5x represents the cost C (in dollars) of renting an x-passenger van for a day. Choose the numbers that are in the range of the function.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 2
Answer:

Question 3.
Fill in the system of linear inequalities with <, ≤, >, or ≥ so that the graph represents the system.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 3
Answer:

Question 4.
Your friend claims to be able to fill in each box with a constant so that when you set each side of the equation equal to y and graph the resulting equations, the lines will intersect exactly once. Do you support your friend’s claim? Explain.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 4
Answer:

Question 5.
Select the phrases you should use when describing the transformations from the graph of f to the graph of g.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 5
Answer:

Question 6.
Which two equations form a system of linear equations that has no solution?
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 6
Answer:

Question 7.
Fill in a value for a so that each statement is true for the equation ax – 8 = 4 – x.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 7
Answer:

Question 8.
Which ordered pair is a solution of the linear inequality whose graph is shown?
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 8
Answer:

Question 9.
Which of the systems of linear equations are equivalent?
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 9
Answer:

Question 10.
The value of x is more than 9. Which of the inequalities correctly describe the triangle? The perimeter (in feet) is represented by P, and the area (in square feet) is represented by A.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 10
Answer:

Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions

Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions

Are you in the search of Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions? If yes, you are on the correct page. Check out the detailed explanation for all the problems available on Big Ideas Grade 5 Chapter 9 Multiply Fractions Math Answers. You can easily become a math expert by referring to the BIM Grade 5 Chapter 9 Math Answers. Download Big Ideas Math Book 5th Grade Answer Key Chapter 9 Multiply Fractions now and start your preparation for the exam. You must prepare all the concepts and verify answers to test your preparation level.

Big Ideas 5th Grade Chapter 9 Multiply Fractions Math Book Answer Key

Every student must prepare each and every topic of ch 9 to score good marks in the exam. Also, take the math assignments and homework to get the perfect preparation. Prepare well by referring to our Big Ideas Math Book 5th Grade Answer Key Chapter 9 Multiply Fractions. Download Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions for free. The below links have every topic problem along with explanations. So, check out every link and practice more where you lag.

Lesson: 1 Multiply Whole Numbers by Fractions

Lesson: 2 Use Models to Multiply Fractions by Whole Numbers

Lesson: 3 Multiply Fractions and Whole Numbers

Lesson: 4 Use Models to Multiply Fractions

Lesson: 5 Multiply Fractions

Lesson: 6 Find Areas of Rectangles

Lesson: 7 Multiply Mixed Numbers

Lesson: 8 Compare Factors and Products

Chapter: 9 – Multiply Fractions

Lesson 9.1 Multiply Whole Numbers by Fractions

Explore and Grow

Write any proper fraction that is not a unit fraction. Draw a model to represent your fraction. Draw a model to find a multiple of your fraction?
Answer:
The proper fraction that is not a unit fraction is:  \(\frac{5}{8}\)
Let the proper fraction be multiplied by 5
So,
We have to find the value of 5 × \(\frac{5}{8}\)
Now,
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
5 × \(\frac{5}{8}\)
= \(\frac{5}{1}\) × \(\frac{5}{8}\)
= \(\frac{5 × 5}{1 × 8}\)
= \(\frac{25}{8}\)
Hence, from the above,
We can conclude that the multiple of your proper fraction is: \(\frac{25}{8}\)

Reasoning
How can you use a model to multiply a whole number by a fraction? Explain.
Answer:
We can multiply a whole number by a fraction using the properties of multiplication. They are:
A) a = \(\frac{a}{1}\)
B) a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
Hence, by using the above properties, we can multiply the whole number by a fraction.

Think and Grow: Multiply Whole Numbers by Fractions

Example
Find 3 × \(\frac{2}{5}\)

Show and Grow

Multiply.
Question 1.
2 × \(\frac{3}{4}\) = ______
Answer:
2 × \(\frac{3}{4}\) = \(\frac{6}{4}\)

Explanation:
The given numbers are: 2 and \(\frac{3}{4}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
2 × \(\frac{3}{4}\)
= \(\frac{2}{1}\) × \(\frac{3}{4}\)
= \(\frac{2 × 3}{1 × 4}\)
= \(\frac{6}{4}\)
Hence,
2 × \(\frac{3}{4}\) = \(\frac{6}{4}\)

Question 2.
4 × \(\frac{5}{8}\) = ____
Answer:
4 × \(\frac{5}{8}\) = \(\frac{20}{8}\)

Explanation:
The given numbers are: 4 and \(\frac{5}{8}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
4 × \(\frac{5}{8}\)
= \(\frac{4}{1}\) × \(\frac{5}{8}\)
= \(\frac{4 × 5}{1 × 8}\)
= \(\frac{20}{8}\)
Hence,
4 × \(\frac{5}{8}\) = \(\frac{20}{8}\)

Apply and Grow: Practice

Multiply.
Question 3.
5 × \(\frac{7}{10}\) = ______
Answer:
5 × \(\frac{7}{10}\) = \(\frac{35}{10}\)

Explanation:
The given numbers are: 5 and \(\frac{7}{10}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
5 × \(\frac{7}{10}\)
= \(\frac{5}{1}\) × \(\frac{7}{10}\)
= \(\frac{5 × 7}{1 × 10}\)
= \(\frac{35}{10}\)
Hence,
5 × \(\frac{7}{10}\) = \(\frac{35}{10}\)

Question 4.
8 × \(\frac{2}{3}\) = ______
Answer:
8 × \(\frac{2}{3}\) = \(\frac{16}{3}\)

Explanation:
The given numbers are: 8 and \(\frac{2}{3}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
8 × \(\frac{2}{3}\)
= \(\frac{8}{1}\) × \(\frac{2}{3}\)
= \(\frac{8 × 2}{1 × 3}\)
= \(\frac{16}{3}\)
Hence,
8 × \(\frac{2}{3}\) = \(\frac{16}{3}\)

Question 5.
7 × \(\frac{5}{6}\) = ______
Answer:
7 × \(\frac{5}{6}\) = \(\frac{35}{6}\)

Explanation:
The given numbers are: 7 and \(\frac{5}{6}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
7 × \(\frac{5}{6}\)
= \(\frac{7}{1}\) × \(\frac{5}{6}\)
= \(\frac{7 × 5}{1 × 6}\)
= \(\frac{35}{6}\)
Hence,
7 × \(\frac{5}{6}\) = \(\frac{35}{6}\)

Question 6.
9 × \(\frac{1}{2}\) = ______
Answer:
9 × \(\frac{1}{2}\) = \(\frac{9}{2}\)

Explanation:
The given numbers are: 9 and \(\frac{1}{2}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
9 × \(\frac{1}{2}\)
= \(\frac{9}{1}\) × \(\frac{1}{2}\)
= \(\frac{9 × 1}{1 × 2}\)
= \(\frac{9}{2}\)
Hence,
9 × \(\frac{1}{2}\) = \(\frac{9}{2}\)

Question 7.
6 × \(\frac{3}{100}\) = ______
Answer:
6 × \(\frac{3}{100}\) = \(\frac{18}{100}\)

Explanation:
The given numbers are: 6 and \(\frac{3}{100}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
6 × \(\frac{3}{100}\)
= \(\frac{6}{1}\) × \(\frac{3}{100}\)
= \(\frac{6 × 3}{1 × 100}\)
= \(\frac{18}{100}\)
Hence,
6 × \(\frac{3}{100}\) = \(\frac{18}{100}\)

Question 8.
15 × \(\frac{4}{7}\) = ______
Answer:
15 × \(\frac{4}{7}\) = \(\frac{60}{7}\)

Explanation:
The given numbers are: 15 and \(\frac{4}{7}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
15 × \(\frac{4}{7}\)
= \(\frac{15}{1}\) × \(\frac{4}{7}\)
= \(\frac{15 × 4}{1 × 7}\)
= \(\frac{60}{7}\)
Hence,
15 × \(\frac{4}{7}\) = \(\frac{60}{7}\)

Question 9.
10 × \(\frac{5}{3}\) = ______
Answer:
10 × \(\frac{5}{3}\) = \(\frac{50}{3}\)

Explanation:
The given numbers are: 10 and \(\frac{5}{3}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
10 × \(\frac{5}{3}\)
= \(\frac{10}{1}\) × \(\frac{5}{3}\)
= \(\frac{10 × 5}{1 × 3}\)
= \(\frac{50}{3}\)
Hence,
10 × \(\frac{5}{3}\) = \(\frac{50}{3}\)

Question 10.
4 × \(\frac{5}{2}\) = ______
Answer:
4 × \(\frac{5}{2}\) = 10

Explanation:
The given numbers are: 4 and \(\frac{5}{2}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
4 × \(\frac{5}{2}\)
= \(\frac{4}{1}\) × \(\frac{5}{2}\)
= \(\frac{4 × 5}{1 × 2}\)
= \(\frac{20}{2}\)
= 10
Hence,
4 × \(\frac{5}{2}\) = 10

Question 11.
3 × \(\frac{11}{8}\) = ______
Answer:
3 × \(\frac{11}{8}\) = \(\frac{33}{8}\)

Explanation:
The given numbers are: 3 and \(\frac{11}{8}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
3 × \(\frac{11}{8}\)
= \(\frac{3}{1}\) × \(\frac{11}{8}\)
= \(\frac{11 × 3}{1 × 8}\)
= \(\frac{33}{8}\)
Hence,
3 × \(\frac{11}{8}\) = \(\frac{33}{8}\)

Find the unknown number.
Question 12.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 2
Answer:
The missing number is: 3

Explanation:
The given fractions are: \(\frac{3}{20}\) and \(\frac{9}{20}\)
Let the missing number be X
So,
X × \(\frac{3}{20}\) = \(\frac{9}{20}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{9}{20}\) ÷ \(\frac{3}{20}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{9}{20}\) ÷ \(\frac{3}{20}\)
= \(\frac{9}{20}\) × \(\frac{20}{3}\)
= \(\frac{9 × 20}{20 × 3}\)
= 3
Hence,
The missing number is: 3

Question 13.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 3
Answer:
The missing number is: 6

Explanation:
The given fractions are: \(\frac{4}{9}\) and \(\frac{24}{9}\)
Let the missing number be X
So,
X × \(\frac{4}{9}\) = \(\frac{24}{9}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{24}{9}\) ÷ \(\frac{4}{9}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{24}{9}\) ÷ \(\frac{4}{9}\)
= \(\frac{24}{9}\) × \(\frac{9}{4}\)
= \(\frac{9 × 24}{9 × 4}\)
= 6
Hence,
The missing number is: 6

Question 14.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 4
Answer:
The missing number is: 10

Explanation:
The given fractions are: \(\frac{5}{12}\) and \(\frac{50}{12}\)
Let the missing number be X
So,
X × \(\frac{5}{12}\) = \(\frac{50}{12}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{50}{12}\) ÷ \(\frac{5}{12}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{50}{12}\) ÷ \(\frac{5}{12}\)
= \(\frac{50}{12}\) × \(\frac{12}{5}\)
= \(\frac{12 × 50}{12 × 5}\)
= 10
Hence,
The missing number is: 10

Question 15.
A recipe calls for \(\frac{3}{4}\) cup of dried rice noodles. You make 44 batches of the recipe. How many cups of dried rice noodles do you use?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 5
Answer:
The number of cups of dried rice noodles you used is: 33

Explanation:
It is given that a recipe calls for \(\frac{3}{4}\) cup of dried rice noodles.
It is also given that you make 44 batches of the recipe
So,
The number of cups of dried rice noodles you used = ( The number of batches of the recipe ) × ( The number of dried rice noodles for each batch )
= 44 × \(\frac{3}{4}\)
Now,
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
44 × \(\frac{3}{4}\)
= \(\frac{44}{1}\) × \(\frac{3}{4}\)
= \(\frac{44 × 3}{1 × 4}\)
= \(\frac{33}{1}\)
= 33
Hence, from the above,
We can conclude that the number of cups of dried noodles you used is: 33 cups

Question 16.
YOU BE THE TEACHER
Your 5 × \(\frac{3}{5}\) friend says that 5 is equal to \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\). Is your friend correct? Explain.
Answer:
Yes, your friend is correct.

Explanation:
It is given that your 5 × \(\frac{3}{5}\) friend says that 5 is equal to \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\).
Now,
We know that,
n × a = a + a + a + a + …………….. + n times
Where,
‘n’ is the multiple of a
Here,
‘a’ will be the whole number or the fraction
So,
According to the above property,
5 × \(\frac{3}{5}\) = \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) +\(\frac{3}{5}\) + \(\frac{3}{5}\)
Hence, from the above,
We can conclude that your friend is correct.

Question 17.
Patterns
Describe and complete the pattern.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 6
Answer:
The completed pattern is:

From the above pattern,
We can observe that
The whole number is constant i.e., 9
The fraction part is changing i.e., the numerator part is increasing by 1 till 4 and the denominator is constant
So,
Now,
9 × \(\frac{1}{4}\)
= \(\frac{9}{1}\) × \(\frac{1}{4}\)
= \(\frac{9 × 1}{1 × 4}\)
= \(\frac{9}{4}\)
9 × \(\frac{2}{4}\)
= \(\frac{9}{1}\) × \(\frac{2}{4}\)
= \(\frac{9 × 2}{1 × 4}\)
= \(\frac{18}{4}\)
So,
The remaining two multiplication equations will also be solved in the same way.

Think and Grow: Modeling Real Life

Example
Your goal is to make a waterslide that is at least 10 meters long. You make the waterslide using 10 plastic mats that are each \(\frac{3}{2}\) meters long. Do you reach your goal?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 7
Find the length of the waterslide by multiplying the number of mats by the length of each mat.

So,
You reached your goal.

Show and Grow

Question 18.
An excavator is moving 4 piles of dirt that are the same size. Each pile requires \(\frac{3}{4}\) hour to move. Can the excavator move all of the piles in 2 hours?
Answer:
No, the excavator can’t move all of the piles in 2 hours

Explanation:
It is given that an excavator is moving 4 piles of dirt that are of the same size and each pile requires \(\frac{3}{4}\) hour to move.
So,
We know that,
1 hour = 60 minutes
So,
\(\frac{3}{4}\) of 1 hour = \(\frac{3}{4}\) × 60
= \(\frac{3}{4}\) × \(\frac{60}{1}\)
= \(\frac{3 × 60}{4 × 1}\)
= 45 minutes
So,
The time is taken to move 4 piles of dirt = ( The time is taken to move each pile )  × ( The total number of piles )
= 45 × 4
=180 minutes
It is given that you have to move the piles of dirt in 2 hours but the time taken is 4 hours.
Hence, from the above,
We can conclude that the excavator can’t move all the piles in 2 hours.

Question 19.
You walk dogs \(\frac{5}{4}\) miles two times each day. How far do you walk the dogs in 1 week?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 9
Answer:
The number of miles you walk the dogs in 1 week is: \(\frac{35}{2}\)

Explanation:
It is given that you walk dogs \(\frac{5}{4}\) miles two times each day
So,
The total number of miles you walk dogs in 1 day = 2 × \(\frac{5}{4}\)
= \(\frac{5}{4}\) × \(\frac{2}{1}\)
= \(\frac{5 × 2}{4 × 1}\)
= \(\frac{5}{2}\) miles
We know that,
1 week = 7 days
So,
The total number of miles you walk dogs in 1 week = ( The number of miles you walk dogs in 1 day ) × 7
= \(\frac{5}{2}\) × 7
= \(\frac{5}{2}\) × \(\frac{7}{1}\)
= \(\frac{5 × 7}{2 × 1}\)
= \(\frac{35}{2}\)
Hence, from the above,
We can conclude that the number of miles you walk dogs in 1 week is: \(\frac{35}{2}\) miles

Question 20.
DIG DEEPER
You have10 feet of string. You need \(\frac{5}{3}\) feet of string to make 1 necklace. You make 5 necklaces. Do you have enough string to make another necklace? Explain.
Answer:
Yes, we have enough string to make another necklace

Explanation:
It is given that you have 10 feet of string and you need \(\frac{5}{3}\) feet of string to make 1 necklace. and you make 5 necklaces.
So,
The length of the string required for 5 necklaces = ( The length of string required for 1 necklace ) × ( The total number of necklaces )
= \(\frac{5}{3}\) × 5
= \(\frac{5}{3}\) × \(\frac{5}{1}\)
= \(\frac{5 × 5}{3 × 1}\)
= \(\frac{25}{3}\) feet
We can write 10 feet as \(\frac{30}{3}\) feet
So,
When we compare the total length of the string and the length of the string to make 5 necklaces,
We can observe that the length of the string to make 5 necklaces is less than the total length of the string.
Hence, from the above,
We can conclude that we have enough string to make another necklace.

Multiply Whole Numbers by Fractions Homework & Practice 9.1

Multiply
Question 1.
5 × \(\frac{2}{3}\) = ______
Answer:
5 × \(\frac{2}{3}\) = \(\frac{10}{3}\)

Explanation:
The given numbers are: 5 and \(\frac{2}{3}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
5 × \(\frac{2}{3}\)
= \(\frac{5}{1}\) × \(\frac{2}{3}\)
= \(\frac{2 × 5}{1 × 3}\)
= \(\frac{10}{3}\)
Hence,
5 × \(\frac{2}{3}\) = \(\frac{10}{3}\)

Question 2.
9 × \(\frac{7}{8}\) = ______
Answer:
9 × \(\frac{7}{8}\) = \(\frac{63}{8}\)

Explanation:
The given numbers are: 9 and \(\frac{7}{8}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
9 × \(\frac{7}{8}\)
= \(\frac{9}{1}\) × \(\frac{7}{8}\)
= \(\frac{9 × 7}{1 × 8}\)
= \(\frac{63}{8}\)
Hence,
9 × \(\frac{7}{8}\) = \(\frac{63}{8}\)

Question 3.
4 × \(\frac{11}{12}\) = ______
Answer:
4 × \(\frac{11}{12}\) = \(\frac{44}{12}\)

Explanation:
The given numbers are: 4 and \(\frac{11}{12}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
4 × \(\frac{11}{12}\)
= \(\frac{4}{1}\) × \(\frac{11}{12}\)
= \(\frac{4 × 11}{1 × 12}\)
= \(\frac{44}{12}\)
Hence,
4 × \(\frac{11}{12}\) = \(\frac{44}{12}\)

Question 4.
8 × \(\frac{35}{100}\) = ______
Answer:
8 × \(\frac{35}{100}\) = \(\frac{280}{100}\)

Explanation:
The given numbers are: 8 and \(\frac{35}{100}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
8 × \(\frac{35}{100}\)
= \(\frac{8}{1}\) × \(\frac{35}{100}\)
= \(\frac{8 × 35}{1 × 100}\)
= \(\frac{280}{100}\)
Hence,
8 × \(\frac{35}{100}\) = \(\frac{280}{100}\)

Question 5.
3 × \(\frac{1}{2}\) = ______
Answer:
3 × \(\frac{1}{2}\) = \(\frac{3}{2}\)

Explanation:
The given numbers are: 3 and \(\frac{1}{2}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
3 × \(\frac{1}{2}\)
= \(\frac{3}{1}\) × \(\frac{1}{2}\)
= \(\frac{1 × 3}{1 × 2}\)
= \(\frac{3}{2}\)
Hence,
3 × \(\frac{1}{2}\) = \(\frac{3}{2}\)

Question 6.
7 × \(\frac{2}{5}\) = ______
Answer:
7 × \(\frac{2}{5}\) = \(\frac{14}{5}\)

Explanation:
The given numbers are: 7 and \(\frac{2}{5}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
7 × \(\frac{2}{5}\)
= \(\frac{7}{1}\) × \(\frac{2}{5}\)
= \(\frac{2 × 7}{1 × 5}\)
= \(\frac{14}{5}\)
Hence,
7 × \(\frac{2}{5}\) = \(\frac{14}{5}\)

Question 7.
6 × \(\frac{7}{4}\) = ______
Answer:
6 × \(\frac{7}{4}\) = \(\frac{42}{4}\)

Explanation:
The given numbers are: 6 and \(\frac{7}{4}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
6 × \(\frac{7}{4}\)
= \(\frac{6}{1}\) × \(\frac{7}{4}\)
= \(\frac{6 × 7}{1 × 4}\)
= \(\frac{42}{4}\)
Hence,
6 × \(\frac{7}{4}\) = \(\frac{42}{4}\)

Question 8.
12 × \(\frac{8}{7}\) = ______
Answer:
12 × \(\frac{8}{7}\) = \(\frac{96}{7}\)

Explanation:
The given numbers are: 12 and \(\frac{8}{7}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
12 × \(\frac{8}{7}\)
= \(\frac{12}{1}\) × \(\frac{8}{7}\)
= \(\frac{12 × 8}{1 × 7}\)
= \(\frac{96}{7}\)
Hence,
12 × \(\frac{8}{7}\) = \(\frac{96}{7}\)

Question 9.
25 × \(\frac{10}{9}\) = ______
Answer:
25 × \(\frac{10}{9}\) = \(\frac{250}{9}\)

Explanation:
The given numbers are: 25 and \(\frac{10}{9}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
25 × \(\frac{10}{9}\)
= \(\frac{25}{1}\) × \(\frac{10}{9}\)
= \(\frac{25 × 10}{1 × 9}\)
= \(\frac{250}{9}\)
Hence,
25 × \(\frac{10}{9}\) = \(\frac{250}{9}\)

Find the unknown number.
Question 10.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 10
Answer:
The missing number is: 5

Explanation:
The given fractions are: \(\frac{5}{7}\) and \(\frac{25}{7}\)
Let the missing number be X
So,
X × \(\frac{5}{7}\) = \(\frac{25}{7}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{25}{7}\) ÷ \(\frac{5}{7}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{25}{7}\) ÷ \(\frac{5}{7}\)
= \(\frac{25}{7}\) × \(\frac{7}{5}\)
= \(\frac{7 × 25}{7 × 5}\)
= 5
Hence,
The missing number is: 5

Question 11.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 11
Answer:
The missing number is: 7

Explanation:
The given fractions are: \(\frac{9}{10}\) and \(\frac{63}{10}\)
Let the missing number be X
So,
X × \(\frac{9}{10}\) = \(\frac{63}{10}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{63}{10}\) ÷ \(\frac{9}{10}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{63}{10}\) ÷ \(\frac{9}{10}\)
= \(\frac{63}{10}\) × \(\frac{10}{9}\)
= \(\frac{63 × 10}{10 × 9}\)
= 7
Hence,
The missing number is: 7

Question 12.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 12
Answer:
The missing number is: 9

Explanation:
The given fractions are: \(\frac{3}{5}\) and \(\frac{27}{5}\)
Let the missing number be X
So,
X × \(\frac{3}{5}\) = \(\frac{27}{5}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{27}{5}\) ÷ \(\frac{3}{5}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{27}{5}\) ÷ \(\frac{3}{5}\)
= \(\frac{27}{5}\) × \(\frac{5}{3}\)
= \(\frac{5 × 27}{5 × 3}\)
= 9
Hence,
The missing number is: 9

Question 13.
You make 5 servings of pancakes. You top each serving with \(\frac{1}{4}\) cup of strawberries. How many cups of strawberries do you use?
Answer:
The number of cups of strawberries you used is: 20 cups

Explanation:
It is given that you make 5 servings of pancakes and you top each serving with \(\frac{1}{4}\) cup of strawberries.
So,
The number of cups of strawberries = \(\frac{The number of servings of pancakes}{Each serving in pancake}\)
= 5 ÷ \(\frac{1}{4}\)
= \(\frac{5}{1}\) ÷ \(\frac{1}{4}\)
= \(\frac{5}{1}\) × \(\frac{4}{1}\)
= \(\frac{5 × 4}{1 × 1}\)
= 20 cups
hence, from the above,
We can conclude that the number of cups of strawberries is: 20 cups

Question 14.
Which One Doesn’tBelong?
Which one does not belong with the other three?
4 × \(\frac{3}{8}\)
3 × \(\frac{1}{8}\)
\(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\)
1 + \(\frac{1}{2}\)
Answer:
Let the expresseons be named A., B., C., D.
So,
A) 4 × \(\frac{3}{8}\)
B) 3 × \(\frac{1}{8}\)
C) \(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\)
D) 1 + \(\frac{1}{2}\)
So, from the above four expressions,
Expressions A), B), C) are in the normal form and C) is the expanded form of expression A)
Hence, from the above,
We can conclude that expression D) does not belong to the other three.

Question 15.
Modeling Real Life
You complete \(\frac{5}{2}\) inches of weaving each day for 5 days. The weaving needs to be at least 11 inches long. Is your weaving complete?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 13
Answer:
Yes, your weaving is complete

Explanation:
It is given that you complete \(\frac{5}{2}\) inches of weaving each day for 5 days.
So,
The total number of inches you completed in 5 days = ( The number of inches you weaved each day ) × 5
= \(\frac{5}{2}\) × 5
= \(\frac{5}{2}\) × \(\frac{5}{1}\)
= \(\frac{5 × 5}{2 × 1}\)
= \(\frac{25}{2}\)
It is also given that the weaving needs to be at least 11 inches long.
So,
11 inches can also be written as \(\frac{22}{2}\)
So,
When we compare the total number of inches needed for weaving and the total number of inches you weaved in 5 days, we can observe that
You weaved more than the amount of the weaving.
Hence, from the above,
We can conclude that your weaving is completed.

Question 16.
DIG DEEPER!
You spend \(\frac{7}{2}\) hours playing drums each day for 2 days. Your friend spends \(\frac{5}{4}\) hours playing drums each day for 6 days. Who spends more time playing drums? How much more?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 14
Answer:
Your friend spends more time playing drums than you
The amount you played drums more than your friend is: \(\frac{1}{2}\) hours

Explanation:
It is given that you spend \(\frac{7}{2}\) hours each day playing drums for 2 days
So,
The time you played drums for 2 days = The time you played drums for each day × 2
= \(\frac{7}{2}\) × 2
= \(\frac{7}{2}\) × \(\frac{2}{1}\)
= \(\frac{7 × 2}{2 × 1}\)
= \(\frac{14}{2}\)
= \(\frac{28}{4}\)
= 7 hours
It is also given that Your friend spends \(\frac{5}{4}\) hours playing drums each day for 6 days.
So,
The time your friend played drums for 6 days = The time your friend played drums for each day × 6
= \(\frac{5}{4}\) × 6
= \(\frac{5}{4}\) × \(\frac{6}{1}\)
= \(\frac{5 × 6}{4 × 1}\)
= \(\frac{30}{4}\) hours
Now,
When we compare the time played drums by you and your friends, your friend played more time than you
So,
The amount of time more your friend played than you = \(\frac{30}{4}\) – \(\frac{28}{4}\)
= \(\frac{30 – 28}{4}\)
= \(\frac{2}{4}\)
= \(\frac{1}{2}\) hour
Hence, from the above,
We can conclude that
Your friend spends more time playing drums than you
The amount you played drums more than your friend is: \(\frac{1}{2}\) hours

Review & Refresh

Find the product.
Question 17.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 15
Answer:
0.6 × 0.4 = 0.24

Explanation:
The given decimal numbers are: 0.6 and 0.4
the representation of the decimal numbers in the fraction form is: \(\frac{6}{10}\) and \(\frac{4}{10}\)
Now,
\(\frac{6}{10}\) × \(\frac{4}{10}\)
= \(\frac{6 × 4}{10 × 10}\)
= \(\frac{24}{100}\)
So,
The representation of \(\frac{24}{100}\) in the decimal form is: 0.24
Hence, 0.6 × 0.4 = 0.24

Question 18.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 16
Answer:
2.37 × 1.9 = 4.503

Explanation:
The given decimal numbers are: 2.37 and 1.9
the representation of the decimal numbers in the fraction form is: \(\frac{237}{100}\) and \(\frac{19}{10}\)
Now,
\(\frac{237}{100}\) × \(\frac{19}{10}\)
= \(\frac{237 × 19}{100 × 10}\)
= \(\frac{4,503}{1000}\)
So,
The representation of \(\frac{4503}{1000}\) in the decimal form is: 4.503
Hence, 2.37 × 1.9 = 4.503

Question 19.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 17
Answer:
52.8 × 0.75 = 39.6

Explanation:
The given decimal numbers are: 52.8 and 0.75
the representation of the decimal numbers in the fraction form is: \(\frac{6}{10}\) and \(\frac{4}{10}\)
Now,
\(\frac{528}{10}\) × \(\frac{75}{100}\)
= \(\frac{528 × 75}{10 × 100}\)
= \(\frac{3960}{1000}\)
So,
The representation of \(\frac{3960}{1000}\) in the decimal form is: 39.6
Hence, 52.8 × 0.75 = 39.6

Lesson 9.2 Use Models to Multiply Fractions by Whole Numbers

You need to give water to \(\frac{2}{3}\) of the dogs at a shelter. There are 12 dogs at the shelter. How many dogs need water? Draw a model to support your answer.
Answer:
The number of dogs that need water is: 8 dogs

Explanation:
It is given that you need to give water to \(\frac{2}{3}\) of the dogs at a shelter and there are 12 dogs at the shelter
So,
The number of dogs that need water = ( The fraction of dogs that need water ) × ( The total number of dogs )
= \(\frac{2}{3}\) × 12
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{3}\) × 12
= \(\frac{2}{3}\) × \(\frac{12}{1}\)
= \(\frac{2 × 12}{3 × 1}\)
= \(\frac{8}{1}\)
= 8
Hence, from the above,
We can conclude that there are 8 dogs that need water

Reasoning
How can you use a model to multiply a fraction by a whole number? Explain.
Answer:
We can multiply a fraction by a whole number by using the following multiplication properties. They are:
A) \(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
B) x = \(\frac{x}{1}\)

Think and Grow: Multiply Fractions by Whole Numbers

You can use models to multiply a fraction by a whole number.
Example
Find \(\frac{3}{4}\) of 8.

Example
Find \(\frac{5}{8}\) × 4.

Show and Grow

Question 1.
Find \(\frac{2}{5}\) of 10.
Answer:
\(\frac{2}{5}\) × 10 = 4

Explanation:
The given numbers are: \(\frac{2}{5}\) and 10
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{5}\) × 10 = \(\frac{2}{5}\) × \(\frac{10}{1}\)
= \(\frac{2 × 10}{5 × 1}\)
= \(\frac{4}{1}\)
= 4
Hence,
\(\frac{2}{5}\) × 10 = 4

Question 2.
Find \(\frac{7}{12}\) × 6.
Answer:
\(\frac{7}{12}\) × 6 = \(\frac{7}{2}\)

Explanation:
The given numbers are: \(\frac{7}{12}\) and 6
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{7}{12}\) × 6 = \(\frac{7}{12}\) × \(\frac{6}{1}\)
= \(\frac{7 × 6}{12 × 1}\)
= \(\frac{7}{2}\)
Hence,
\(\frac{7}{12}\) × 6 = \(\frac{7}{2}\)

Apply and Grow: Practice

Multiply. Use a model to help.
Question 3.
\(\frac{5}{6}\) of 12
Answer:
\(\frac{5}{6}\) × 12 = 10

Explanation:
The given numbers are: \(\frac{5}{6}\) and 12
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{5}{6}\) × 12 = \(\frac{5}{6}\) × \(\frac{12}{1}\)
= \(\frac{5 × 12}{6 × 1}\)
= \(\frac{10}{1}\)
= 10
Hence,
\(\frac{5}{6}\) × 12 = 10

Question 4.
\(\frac{2}{3}\) × 9
Answer:
\(\frac{2}{3}\) × 9 = 6

Explanation:
The given numbers are: \(\frac{2}{3}\) and 9
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{3}\) × 9 = \(\frac{2}{3}\) × \(\frac{9}{1}\)
= \(\frac{2 × 9}{3 × 1}\)
= \(\frac{6}{1}\)
= 6
Hence,
\(\frac{2}{3}\) × 9 = 6

Question 5.
\(\frac{1}{5}\) × 10
Answer:
\(\frac{1}{5}\) × 10 = 2

Explanation:
The given numbers are: \(\frac{1}{5}\) and 10
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{5}\) × 10 = \(\frac{1}{5}\) × \(\frac{10}{1}\)
= \(\frac{1 × 10}{5 × 1}\)
= \(\frac{2}{1}\)
= 2
Hence,
\(\frac{1}{5}\) × 10 = 2

Question 6.
\(\frac{3}{5}\) of 5
Answer:
\(\frac{3}{5}\) × 5 = 3

Explanation:
The given numbers are: \(\frac{3}{5}\) and 5
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{5}\) × 5 = \(\frac{3}{5}\) × \(\frac{5}{1}\)
= \(\frac{3 × 5}{5 × 1}\)
= \(\frac{3}{1}\)
= 3
Hence,
\(\frac{3}{5}\) × 5 = 3

Question 7.
\(\frac{1}{6}\) of 3
Answer:
\(\frac{1}{6}\) × 3 = \(\frac{1}{2}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and 3
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{6}\) × 3 = \(\frac{1}{6}\) × \(\frac{3}{1}\)
= \(\frac{1 × 3}{6 × 1}\)
= \(\frac{1}{2}\)
Hence,
\(\frac{1}{6}\) × 3 = \(\frac{1}{2}\)

Question 8.
\(\frac{3}{8}\) × 4
Answer:
\(\frac{3}{8}\) × 4 = \(\frac{3}{2}\)

Explanation:
The given numbers are: \(\frac{3}{8}\) and 4
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{8}\) × 4 = \(\frac{3}{8}\) × \(\frac{4}{1}\)
= \(\frac{3 × 4}{8 × 1}\)
= \(\frac{3}{2}\)
Hence,
\(\frac{3}{8}\) × 4 = \(\frac{3}{2}\)

Question 9.
You have 25 beads. You use \(\frac{2}{5}\) of the beads to make a bracelet. How many beads do you use?
Answer:
The number of beads you used is: 10 beads

Explanation:
It is given that you have 25 beads and you use \(\frac{2}{5}\) of the beads to make a bracelet.
So,
The number of beads you used = ( The fraction of beads used to make 1 bracelet ) × ( The total number of beads )
= \(\frac{2}{5}\) × 25
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{5}\) × 25 = \(\frac{2}{5}\) × \(\frac{25}{1}\)
= \(\frac{2 × 25}{5 × 1}\)
= \(\frac{10}{1}\)
= 10 beads
Hence, from the above,
We can conclude that the number of beads you used to make bracelets are: 10 beads

Question 10.
Writing
Write and solve a real-life problem for the expression.
\(\frac{3}{4}\) × 20
Answer:
Suppose there are 20 passengers in a bus and they occupied the place \(\frac{3}{4}\) of the bus.
Hence,
The total number of seats in the bus = ( The fraction of the place occupied by the passengers ) × ( The total number of passengers )
= \(\frac{3}{4}\) × 20
Now,
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{4}\) × 20 = \(\frac{3}{4}\) × \(\frac{20}{1}\)
= \(\frac{3 × 20}{4 × 1}\)
= \(\frac{15}{1}\)
= 15 seats
Hence, from the above,
We can conclude that there are 15 seats in the bus.

Question 11.
YOU BE THE TEACHER
Descartes finds \(\frac{2}{3}\) × 6. Is he correct? Explain.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 3
Answer:
No, Descartes is not correct

Explanation:
According to Descartes,
The given numbers are: \(\frac{2}{3}\) and 6
So,
According to Descartes, 6 is divided into 3 equal parts
So,
6 ÷ 3 = 2 equal parts
So,
According to Descartes, these 2 equal parts also divided into 2 parts.
So,
The total number of parts = 2 equal parts + 2 equal parts = 4 parts
So,
The product of \(\frac{2}{3}\) and 6 is 4
Now,
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{3}\) × 6 = \(\frac{2}{3}\) × \(\frac{6}{1}\)
= \(\frac{6 × 2}{3 × 1}\)
= \(\frac{4}{1}\)
= 4
Hence, from the above,
We can conclude that Descartes is not correct.

Think and Grow: Modeling Real Life

Example
A recipe calls for 2 cups of rice. You only have \(\frac{3}{4}\) of that amount. How much more rice do you need?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 4
Find the number of cups of rice that you have by finding \(\frac{3}{4}\) of 2.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 5
So,
You have \(\frac{3}{2}\) cups of rice.
Subtract the amount of rice you have from the amount of rice you need.
2 – \(\frac{3}{2}\) = \(\frac{2}{1}\) – \(\frac{3}{2}\)
= \(\frac{(2 × 2) – 3}{2}\)
= \(\frac{1}{2}\)
So,
you need \(\frac{1}{2}\) cup more of rice.

Show and Grow

Question 12.
You have 12 tokens. You use \(\frac{3}{4}\) of them to play a pinball game. How many tokens do you have left?
Answer:
The number of tokens you have left is: 3 tokens

Explanation:
It is given that you have 12 tokens and you use \(\frac{3}{4}\) of them to play a pinball game.
So,
The number of tokens used to play a pinball game = ( The fraction of tokens used to play a pinball game ) × ( The total number of tokens )
= \(\frac{3}{4}\) × 12
= \(\frac{3}{4}\) × \(\frac{12}{1}\)
= \(\frac{3 × 12}{4 × 1}\)
= \(\frac{9}{1}\)
= 9 tokens
So,
The number of tokens left = ( The total number of tokens ) – ( The number of tokens used to play the pinball )
= 12 – 9
= 3 tokens
Hence, from the above,
We can conclude that the number of tokens left are: 3 tokens

Question 13.
A male lion sleeps \(\frac{5}{6}\) × 6 hours of each day. How many hours does the lion sleep in 1 week?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 6
Answer:
The number of hours the lion sleeps in 1 week is: 35 hours

Explanation:
It is given that a male lion sleeps \(\frac{5}{6}\) × 6 hours of each day
So,
The number of hours the lion sleeps in 1 day = \(\frac{5}{6}\) × 6 hours
= \(\frac{5}{6}\) × \(\frac{6}{1}\)
= \(\frac{5 × 6}{6 × 1}\)
= \(\frac{5}{1}\)
= 5 hours
We know that,
1 week = 7 days
So,
The total number of hours the lion sleeps in 1 week = ( The number of hours the lion sleeps in 1 day ) × ( The number od fays in 1 week )
= 5 × 7
= 35 hours
Hence, from the above,
We can conclude that the lion sleeps for 35 hours in 1 week

Question 14.
DIG DEEPER!
In a class of 20 students, \(\frac{1}{10}\) of the students are 10 years old, \(\frac{4}{5}\)of the students are 11 years old, and the rest are 12 years old. How many more 11-year-olds than 12-year-olds are in the class?
Answer:
The number of 11-year-olds more than 12-year-olds is: 14

Explanation:
It is given that in a class of 20 students, \(\frac{1}{10}\) of the students are 10 years old, \(\frac{4}{5}\)of the students are 11 years old, and the rest are 12 years old.
So,
The number of 10-year-olds = ( The fraction of 10-year-olds ) × ( The total number of students )
= \(\frac{1}{10}\) × 20
= \(\frac{1 × 20}{10 × 1}\)
= 2 10-year-olds
The number of 11-year-olds = ( The fraction of 11-year-olds ) × ( The total number of students )
= \(\frac{4}{5}\) × 20
= \(\frac{4}{5}\) × \(\frac{20}{1}\)
= \(\frac{4 × 20}{5 × 1}\)
= \(\frac{16}{1}\)
= 16 11-year-olds
So,
The number of 12-year-olds = ( The total number of students ) – ( The number of 10-year-olds + The number of 11-year-olds )
= 20 – ( 16 + 2 )
= 2 12-year-olds
So,
The number of 11-year-olds more than 12-year-olds = ( The number of 11-year-olds ) – ( The numebr of 12-year-olds )
= 16 – 2
= 14 students
Hence, from the above,
We can conclude that there are 14 students who are 11-years-old more than 12-years-old.

Use Models to Multiply Fractions by Whole Numbers Homework & Practice 9.2

Multiply. Use a model to help.
Question 1.
\(\frac{2}{3}\) × 6
Answer:
\(\frac{2}{3}\) × 6 = 4

Explanation:
The given numbers are: \(\frac{2}{3}\) and 6
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{3}\) × 6 = \(\frac{2}{3}\) × \(\frac{6}{1}\)
= \(\frac{2 × 6}{3 × 1}\)
= \(\frac{4}{1}\)
= 4
Hence,
\(\frac{2}{3}\) × 6 = 4

Question 2.
\(\frac{3}{5}\) of 10
Answer:
\(\frac{3}{5}\) × 10 = 6

Explanation:
The given numbers are: \(\frac{3}{5}\) and 10
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{5}\) × 10 = \(\frac{3}{5}\) × \(\frac{10}{1}\)
= \(\frac{3 × 10}{5 × 1}\)
= \(\frac{6}{1}\)
= 6
Hence,
\(\frac{3}{5}\) × 10 = 6

Question 3.
\(\frac{1}{2}\) of 4
Answer:
\(\frac{1}{2}\) × 4 = 2

Explanation:
The given numbers are: \(\frac{1}{2}\) and 4
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{2}\) × 4 = \(\frac{1}{2}\) × \(\frac{4}{1}\)
= \(\frac{1 × 4}{2 × 1}\)
= \(\frac{2}{1}\)
= 2
Hence,
\(\frac{1}{2}\) × 4 = 2

Question 4.
\(\frac{1}{4}\) × 12
Answer:
\(\frac{1}{4}\) × 12 = 3

Explanation:
The given numbers are: \(\frac{1}{4}\) and 12
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{4}\) × 12 = \(\frac{1}{4}\) × \(\frac{12}{1}\)
= \(\frac{1 × 12}{4 × 1}\)
= \(\frac{3}{1}\)
= 3
Hence,
\(\frac{1}{12}\) × 4 = 3

Question 5.
\(\frac{5}{6}\) × 3
Answer:
\(\frac{5}{6}\) × 3 = \(\frac{5}{2}\)

Explanation:
The given numbers are: \(\frac{5}{6}\) and 3
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{5}{6}\) × 3 = \(\frac{5}{6}\) × \(\frac{3}{1}\)
= \(\frac{5 × 3}{6 × 1}\)
= \(\frac{5}{2}\)
Hence,
\(\frac{5}{6}\) × 3 = \(\frac{5}{2}\)

Question 6.
\(\frac{3}{4}\) of 2
Answer:
\(\frac{3}{4}\) × 2 = \(\frac{3}{2}\)

Explanation:
The given numbers are: \(\frac{3}{4}\) and 2
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{4}\) × 2 = \(\frac{3}{4}\) × \(\frac{2}{1}\)
= \(\frac{3 × 2}{4 × 1}\)
= \(\frac{3}{2}\)
Hence,
\(\frac{3}{4}\) × 2 = \(\frac{3}{2}\)

Question 7.
You have 27 foam balls. You use \(\frac{1}{3}\) of the balls for a model. How many balls do you use?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 7
Answer:
The number of balls you used is: 9 balls

Explanation:
It is given that there are 27 foam balls and you use \(\frac{1}{3}\) of the balls for a model.
So,
The number of balls you used = ( The fraction of balls you used ) × ( The total number of balls )
= \(\frac{1}{3}\) × 27
Now,
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{3}\) × 27 = \(\frac{1}{3}\) × \(\frac{27}{1}\)
= \(\frac{1 × 27}{3 × 1}\)
= \(\frac{9}{1}\)
= 9 balls
Hence, from the above,
We can conclude that the number of foam balls you used are: 27

Question 8.
An object that weighs 1 pound on Earth weighs about \(\frac{1}{15}\) pound on Pluto. A man weighs 240 pounds on Earth. How many pounds does he weigh on Pluto?
Answer:
The number of pounds the man weigh in Pluto is: 16 pounds

Explanation:
It is given that an object that weighs 1 pound on Earth weighs about \(\frac{1}{15}\) pound on Pluto.
It is also given that a man weighs 240 pounds on Earth.
So,
The weight of a man on pluto = ( The fraction of weight of a man on pluto when compared to earth ) × ( The weight of the man on earth )
= \(\frac{1}{15}\) × 240
Now,
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{15}\) × 240 = \(\frac{1}{15}\) × \(\frac{240}{1}\)
= \(\frac{1 × 240}{15 × 1}\)
= \(\frac{16}{1}\)
= 16 pounds
Hence, from the above,
We can conclude that the weight of the man on pluto is: 16 pounds

Question 9.
Structure
Write a multiplication equation represented by the model.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 8
Answer:

Question 10.
DIG DEEPER!
Find each missing number.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 9
Answer:
Let the expression be named A), B), C), and D)
So,
The missing number of A is: 12
The missing number of B is: 6
The missing number of C is: 2
The missing number of D is: 5

Explanation:
Let the expressions be named as A), B), C), and D)
So,
A) The given fractions are: \(\frac{1}{3}\) and 4
Let the missing number be X
So,
X × \(\frac{1}{3}\) = 4
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = 4 ÷ \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{4}{1}\) ÷ \(\frac{1}{3}\)
= \(\frac{4}{1}\) × \(\frac{3}{1}\)
= \(\frac{4 × 3}{1 × 1}\)
= 12
Hence, the missing number is 12
B) The given numbers are: \(\frac{2}{5}\) and 15
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{5}\) × 15 = \(\frac{2}{5}\) × \(\frac{15}{1}\)
= \(\frac{2 × 15}{5 × 1}\)
= \(\frac{6}{1}\)
= 6
Hence,
\(\frac{2}{5}\) × 15 = 6
Like the above two expressions, the remaining two expressions can also be solved.
hence, from the above,
We can conclude that
The missing number of A is: 12
The missing number of B is: 6
The missing number of C is: 2
The missing number of D is: 5

Question 11.
Modeling Real Life
You have 28 craft sticks. You use \(\frac{4}{7}\) of them for a project. How many craft sticks do you have left?
Answer:
The number of craft sticks you have left is: 12 craft sticks

Explanation:
It is given that you have 28 craft sticks and you used \(\frac{4}{7}\) of them for a project.
So,
The number of craft sticks you used = ( The fraction of craft sticks you used ) × ( The total number of craft sticks )
= \(\frac{4}{7}\) × 28
= \(\frac{4}{7}\) × \(\frac{28}{1}\)
= \(\frac{4 × 28}{7 × 1}\)
= \(\frac{16}{1}\)
= 16
So,
The number of craft sticks you left = ( The total number of craft sticks ) – ( The number of craft sticks you used )
= 28 – 16
= 12 craft sticks
Hence, from the above,
We can conclude that the number of craft sticks you left is: 12 craft sticks

Question 12.
Modeling Real Life
A mother otter spends \(\frac{1}{3}\) of each day feeding her baby. How many hours does the mother otter spend feeding her baby in 1 week?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 10
Answer:
The number of hours the mother otter spend feeding her baby in 1 week is: 56 hours

Explanation:
It is given that a mother otter spends \(\frac{1}{3}\) of each day feeding her baby.
We know that,
1 day = 24 hours
So,
The number of hours the mother otter spends feeding her baby in 1 day = ( The fraction of time the mother otter spends in feeding milk ) × 24
= \(\frac{1}{3}\) × 24
= \(\frac{1}{3}\) × \(\frac{24}{1}\)
= \(\frac{1 × 24}{3 × 1}\)
= \(\frac{8}{1}\)
= 8 hours
Now,
We know that,
1 week = 7 days
So,
The number of hours the mother otter spends feeding her milk in 1 week = ( The number of hours the mother otter feeding her milk in 1 day ) × 7
= 8 × 7
= 56 hours
Hence, from the above,
We can conclude that the mother otter spends feeding her milk for 56 hours in 1 week

Review & Refresh

Estimate the quotient.
Question 13.
5,692 ÷ 5
Answer:
5,692 ÷ 5 =1,138 R 2

Explanation:
By using the partial quotients method,
5,692 ÷ 5 = ( 5,000 + 600 + 90 ) ÷ 5
= ( 5,000 ÷ 5 ) + ( 600 ÷ 5 ) + ( 90 ÷ 5 )
= 1,000 + 120 + 18
= 1,138 R 2
Hence, 5,692 ÷ 5 = 1,138 R 2

Question 14.
309 ÷ 12
Answer:
309 ÷ 12 = 25 R 9

Explanation:
By using the partial quotients method,
309 ÷ 12 = ( 240 + 60 ) ÷ 12
= ( 240 ÷ 12 ) + ( 60 ÷ 12 )
= 20 + 5
= 25 R 9
Hence, 309 ÷ 12 = 25 R 2

Question 15.
2,987 ÷ 53
Answer:
2,987 ÷ 53 = 2,987 ÷ ( 50 + 3 )
= ( 2,985 ÷ 50 ) + ( 2,985 ÷ 3 )
= 995 R 2 + 59.7
= 1,054.7 R 2

Lesson 9.3 Multiply Fractions and Whole Numbers

Explore and Grow

Use models to help you complete the table. What do you notice about each expression and its product?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 1
Answer:
The completed table is:

From the completed table,
We can observe the multiplication of the whole number and the fraction.
The product of a whole number and a fraction may be a whole number or a fraction.

Construct Arguments
Explain how to multiply fractions and whole numbers without using models.
Answer:
We can multiply the whole numbers and fractions by using the properties of the multiplication. They are:
A) \(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
B) p = \(\frac{p}{1}\)

Think and Grow: Multiply Fractions and Whole Numbers

Key Idea
You can find the product of a fraction and a whole number by multiplying the numerator and the whole number. Then write the result over the denominator.
Example
Find 2 × \(\frac{5}{6}\).
Multiply the numerator and the whole number.

Example
Find \(\frac{5}{6}\) × 2
Multiply the numerator and the whole number.

Show and Grow

Multiply.
Question 1.
3 × \(\frac{5}{8}\) = _______
Answer:
3 × \(\frac{5}{8}\) = \(\frac{15}{8}\)

Explanation:
The given numbers are: 3 and \(\frac{5}{8}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
3 × \(\frac{5}{8}\) = \(\frac{3}{1}\) × \(\frac{5}{8}\)
= \(\frac{5 × 3}{8 × 1}\)
= \(\frac{15}{8}\)
Hence,
3 × \(\frac{5}{8}\) = \(\frac{15}{8}\)

Question 2.
6 × \(\frac{4}{9}\) = _______
Answer:
6 × \(\frac{4}{9}\) = \(\frac{8}{3}\)

Explanation:
The given numbers are: 6 and \(\frac{4}{9}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
6 × \(\frac{4}{9}\) = \(\frac{6}{1}\) × \(\frac{4}{9}\)
= \(\frac{6 × 4}{9 × 1}\)
= \(\frac{24}{9}\)
= \(\frac{8}{3}\)
Hence,
6 × \(\frac{4}{9}\) = \(\frac{8}{3}\)

Question 3.
\(\frac{2}{5}\) × 15 = _______
Answer:
15 × \(\frac{2}{5}\) = 6

Explanation:
The given numbers are: 15 and \(\frac{2}{5}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
15× \(\frac{2}{5}\) = \(\frac{15}{1}\) × \(\frac{2}{5}\)
= \(\frac{15 × 2}{5 × 1}\)
= \(\frac{6}{1}\)
= 6
Hence,
15 × \(\frac{2}{5}\) = 6

Apply and Grow: Practice

Multiply.
Question 4.
\(\frac{3}{5}\) × 2 = _______
Answer:
2 × \(\frac{3}{5}\) = \(\frac{6}{5}\)

Explanation:
The given numbers are: 2 and \(\frac{3}{5}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
2 × \(\frac{3}{5}\) = \(\frac{2}{1}\) × \(\frac{3}{5}\)
= \(\frac{2 × 3}{5 × 1}\)
= \(\frac{6}{5}\)
Hence,
2 × \(\frac{3}{5}\) = \(\frac{6}{5}\)

Question 5.
5 × \(\frac{2}{9}\) = _______
Answer:
5 × \(\frac{2}{9}\) = \(\frac{10}{9}\)

Explanation:
The given numbers are: 5 and \(\frac{2}{9}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
5 × \(\frac{2}{9}\) = \(\frac{5}{1}\) × \(\frac{2}{9}\)
= \(\frac{5 × 2}{9 × 1}\)
= \(\frac{10}{9}\)
Hence,
5 × \(\frac{2}{9}\) = \(\frac{10}{9}\)

Question 6.
\(\frac{5}{6}\) × 4 = _______
Answer:
4 × \(\frac{5}{6}\) = \(\frac{10}{3}\)

Explanation:
The given numbers are: 4 and \(\frac{5}{6}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
4 × \(\frac{5}{6}\) = \(\frac{4}{1}\) × \(\frac{5}{6}\)
= \(\frac{5 × 4}{6 × 1}\)
= \(\frac{20}{6}\)
= \(\frac{10}{3}\)
Hence,
4 × \(\frac{5}{6}\) = \(\frac{10}{3}\)

Question 7.
8 × \(\frac{3}{10}\) = _______
Answer:
8 × \(\frac{3}{10}\) = \(\frac{12}{5}\)

Explanation:
The given numbers are: 8 and \(\frac{3}{10}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
8 × \(\frac{3}{10}\) = \(\frac{8}{1}\) × \(\frac{3}{10}\)
= \(\frac{8 × 3}{10 × 1}\)
= \(\frac{24}{10}\)
= \(\frac{12}{5}\)
Hence,
8 × \(\frac{3}{10}\) = \(\frac{12}{5}\)

Question 8.
\(\frac{1}{5}\) × 7 = _______
Answer:
7 × \(\frac{1}{5}\) = \(\frac{7}{5}\)

Explanation:
The given numbers are: 7 and \(\frac{1}{5}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
7 × \(\frac{1}{5}\) = \(\frac{7}{1}\) × \(\frac{1}{5}\)
= \(\frac{7 × 1}{5 × 1}\)
= \(\frac{7}{5}\)
Hence,
7 × \(\frac{1}{5}\) = \(\frac{7}{5}\)

Question 9.
9 × \(\frac{5}{12}\) = _______
Answer:
9 × \(\frac{5}{12}\) = \(\frac{15}{4}\)

Explanation:
The given numbers are: 9 and \(\frac{5}{12}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
9 × \(\frac{5}{12}\) = \(\frac{9}{1}\) × \(\frac{5}{12}\)
= \(\frac{9 × 5}{12 × 1}\)
= \(\frac{45}{12}\)
= \(\frac{15}{4}\)
Hence,
9 × \(\frac{5}{12}\) = \(\frac{15}{4}\)

Question 10.
15 × \(\frac{5}{8}\) = _______
Answer:
15 × \(\frac{5}{8}\) = \(\frac{75}{8}\)

Explanation:
The given numbers are: 15 and \(\frac{5}{8}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
15 × \(\frac{5}{8}\) = \(\frac{15}{1}\) × \(\frac{5}{8}\)
= \(\frac{15 × 5}{8 × 1}\)
= \(\frac{75}{8}\)
Hence,
15 × \(\frac{5}{8}\) = \(\frac{75}{8}\)

Question 11.
\(\frac{3}{4}\) × 20 = _______
Answer:
20 × \(\frac{3}{4}\) = 15

Explanation:
The given numbers are: 20 and \(\frac{3}{4}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
20 × \(\frac{3}{4}\) = \(\frac{20}{1}\) × \(\frac{3}{4}\)
= \(\frac{20 × 3}{4 × 1}\)
= \(\frac{60}{4}\)
= \(\frac{15}{1}\)
= 15
Hence,
20 × \(\frac{3}{4}\) = 15

Question 12.
\(\frac{7}{9}\) × 5 = _______
Answer:
5 × \(\frac{7}{9}\) = \(\frac{35}{9}\)

Explanation:
The given numbers are: 5 and \(\frac{7}{9}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
5 × \(\frac{7}{9}\) = \(\frac{5}{1}\) × \(\frac{7}{9}\)
= \(\frac{5 × 7}{9 × 1}\)
= \(\frac{35}{9}\)
Hence,
5 × \(\frac{7}{9}\) = \(\frac{35}{9}\)

Question 13.
One-tenth of the 50 states in the United States of America have a mockingbird as their state bird. How many states have a mockingbird as their state bird?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 4
Answer:
The number of states that have mockingbird as their state bird is: 5 states

Explanation:
It is given that one-tenth of the 50 states in the United States of America have a mockingbird as their state bird
So,
The number of states that have a mockingbird as their state bird = ( The fraction of the states that have a mockingbird as their state bird ) × ( The total number of states in the United States of America )
= \(\frac{1}{10}\) × 50
Now,
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
50 × \(\frac{1}{10}\) = \(\frac{50}{1}\) × \(\frac{1}{10}\)
= \(\frac{50 × 1}{10 × 1}\)
= \(\frac{50}{10}\)
= \(\frac{5}{1}\)
= 5 states
Hence, from the above,
We can conclude that there are 5 states that have mockingbird as their state bird

Question 14.
Writing
Explain why 9 × \(\frac{2}{3}\) is equivalent to \(\frac{2}{3}\) × 9.
Answer:
By using the Commutative property of multiplication,
a × b = b × a
So,
By using the above property,
In 9 × \(\frac{2}{3}\),
‘a’ is: 9
‘b’ is: \(\frac{2}{3}\)
Hence, from the above,
We can conclude that  9 × “\(\frac{2}{3}\)” is equivalent to “\(\frac{2}{3}\) × 9″ by using the Commutative property of multiplication.

Question 15.
Reasoning
Without calculating, determine which product is greater. Explain.
\(\frac{1}{8}\) × 24
\(\frac{7}{8}\) × 24
Answer:
The product of “\(\frac{7}{8}\) × 24″ is greater than “\(\frac{1}{8}\) × 24″

Explanation:
The given products are: \(\frac{1}{8}\) × 24 and \(\frac{7}{8}\) × 24
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So, from the above fractions,
We can observe that except numerator, the denominator and the whole number is all the same.
So,
To find which product is greater, we just have to compare the numerators of the two fractions without calculating the value of the product.
In the comparison of the numerators,
We can observe that,
1 < 7
So,
First fraction numerator < Second fraction numerator
Hence, from the above,
We can conclude that the product of “\(\frac{7}{8}\) × 24″ is greater than “\(\frac{1}{8}\) × 24″

Think and Grow: Modeling Real Life

Example
Newton buys 27 songs. Two-thirds of them are classical songs. Descartes buys 16 songs. Seven-eighths of them are classical songs. Who buys more classical songs? How many more?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 5
Multiply \(\frac{2}{3}\) by 27 to find the number of classical songs Newton buys. Multiply \(\frac{7}{8}\) by 16 to find the number Descartes buys.

So, Newton buys more classical songs.
Subtract the products to find how many more.
18 – 14 = 4
Hence,
Newton buys 4 more classical songs than Descartes.

Show and Grow

Question 16.
You take 48 pictures on a walking tour. Five-twelfths of them is of buildings. Your friend takes 45 pictures. Six-fifteenths of them are of buildings. Who takes more pictures of buildings? How many more?
Answer:
You take more pictures
You take 2 pictures more than your friend

Explanation:
It is given that you take 48 pictures on a walking tour and five-twelfths if them is of buildings.
So,
The number of buildings taken by you = ( The total number of pictures taken by you ) × ( The fraction of pictures that is of buildings )
= 48 × \(\frac{5}{12}\)
= \(\frac{48}{1}\) × \(\frac{5}{12}\)
= \(\frac{48 × 5}{1 × 12}\)
= 20 pictures
It is also given that your friend taken 45 pictures and six-fifteenth of them is buildings
So,
The number of buildings taken by your friend = ( The total number of pictures taken by your friend ) × ( The fraction of pictures that is of buildings )
= 45 × \(\frac{6}{15}\)
= \(\frac{45}{1}\) × \(\frac{6}{15}\)
= \(\frac{45 × 6}{1 × 15}\)
= 18 pictures
So,
In the comparison of the pictures of buildings,
You take more pictures than your friend.
Now,
The number of pictures more taken by you than your friend = ( The number of pictures taken by you ) – ( The number of pictures taken by your friend )
= 20 – 18
= 2 pictures
Hence, from the above,
We can conclude that
You take more pictures
You take 2 pictures more than your friend

Question 17.
You have 72 rocks in your rock collection. Five-eighths of them are sedimentary, one-sixth of them are igneous, and the rest are metamorphic. How many of your rocks are metamorphic?
Answer:
The number of rocks that are metamorphic is: 15 metamorphous rocks

Explanation:
It is given that you have 72 rocks in your rock collection and five-eighths of them are sedimentary, one-sixth of them are igneous, and the rest are metamorphic.
So,
The number of rocks that are sedimentary = ( The total number of rocks ) × ( The fraction of rocks that are sedimentary )
= 72 × \(\frac{5}{8}\)
= \(\frac{72}{1}\) × \(\frac{5}{8}\)
= \(\frac{72 × 5}{1 × 8}\)
= 45 sedimentary rocks
The number of rocks that are igneous = ( The total number of rocks ) × ( The fraction of rocks that are igneous )
= 72 × \(\frac{1}{6}\)
= \(\frac{72}{1}\) × \(\frac{1}{6}\)
= \(\frac{72 × 1}{1 × 6}\)
= 12 igneous rocks
So,
The number of metamorphous rocks = ( The total number of rocks ) – ( The number of sedimentary rocks + The number of igneous rocks)
= 72 – ( 45 + 12 )
= 15 metamorphous rocks
Hence, from the above,
We can conclude that there are 15 metamorphous rocks

Question 18.
DIG DEEPER!
Each day, you spend \(\frac{3}{4}\) hour reading and \(\frac{1}{2}\) hour writing in a journal. How many total hours do you spend reading and writing in 1 week? Describe two ways to solve the problem.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 7
Answer:
The number of hours you spend on reading and writing in 1 week is:

Explanation:
It is given that, each day you spent \(\frac{3}{4}\) hour reading and \(\frac{1}{2}\) hour writing in a journal.
So,
The number of hours you spent on reading and writing a journal in 1 day = ( The number of hours spent on reading ) + ( The number of hours spent on writing )
= \(\frac{3}{4}\) + \(\frac{1}{2}\)
Multiply \(\frac{1}{2}\) with \(\frac{2}{2}\)
So,
\(\frac{3}{4}\) + \(\frac{2}{4}\)
= \(\frac{3 + 2}{4}\)
= \(\frac{5}{4}\) hours
We know that,
1 day = 24 hours
1 week = 7 days
So,
1 week = 7 × 24 hours
So,
The number of hours you spent on reading and writing in 1 week = ( The number of hours you spent on reading and writing in 1 day ) × ( The number of hours in 1 week )
= \(\frac{5}{4}\) × 7 × 24
= \(\frac{5}{4}\) × \(\frac{7}{1}\) × \(\frac{24}{1}\)
= \(\frac{5 × 7 × 24}{4 × 1}\)
= 5 × 7 × 6
= 210 hours
Hence, from the above,
We can conclude that the number of hours you spent on reading and writing a journal in 1 week is: 210 hours

Multiply Fractions and Whole Numbers Homework & Practice 9.3

Multiply.
Question 1.
\(\frac{5}{6}\) × 3 = _______
Answer:
3 × \(\frac{5}{6}\) = \(\frac{5}{2}\)

Explanation:
The given numbers are: 3 and \(\frac{5}{6}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
3 × \(\frac{5}{6}\) = \(\frac{3}{1}\) × \(\frac{5}{6}\)
= \(\frac{3 × 5}{6 × 1}\)
= \(\frac{5}{6}\)
= \(\frac{5}{2}\)
Hence,
3 × \(\frac{5}{6}\) = \(\frac{5}{2}\)

Question 2.
\(\frac{2}{3}\) × 6 = _______
Answer:
6 × \(\frac{2}{3}\) = 4

Explanation:
The given numbers are: 6 and \(\frac{2}{3}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
6 × \(\frac{2}{3}\) = \(\frac{6}{1}\) × \(\frac{2}{3}\)
= \(\frac{6 × 2}{3 × 1}\)
= \(\frac{12}{3}\)
= \(\frac{4}{1}\)
= 4
Hence,
6 × \(\frac{2}{3}\) = 4

Question 3.
7 × \(\frac{1}{8}\) = _______
Answer:
7 × \(\frac{1}{8}\) = \(\frac{7}{8}\)

Explanation:
The given numbers are: 7 and \(\frac{1}{8}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
7 × \(\frac{1}{8}\) = \(\frac{7}{1}\) × \(\frac{1}{8}\)
= \(\frac{7 × 1}{8 × 1}\)
= \(\frac{7}{8}\)
Hence,
7 × \(\frac{1}{8}\) = \(\frac{7}{8}\)

Question 4.
2 × \(\frac{1}{2}\) = _______
Answer:
2 × \(\frac{1}{2}\) = 1

Explanation:
The given numbers are: 2 and \(\frac{1}{2}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
2 × \(\frac{1}{2}\) = \(\frac{2}{1}\) × \(\frac{1}{2}\)
= \(\frac{2 × 1}{2 × 1}\)
= \(\frac{2}{2}\)
= 1
Hence,
2 × \(\frac{1}{2}\) = 1

Question 5.
\(\frac{4}{5}\) × 9 = _______
Answer:
9 × \(\frac{4}{5}\) = \(\frac{36}{5}\)

Explanation:
The given numbers are: 9 and \(\frac{4}{5}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
9 × \(\frac{4}{5}\) = \(\frac{9}{1}\) × \(\frac{4}{5}\)
= \(\frac{9 × 4}{5 × 1}\)
= \(\frac{36}{5}\)
Hence,
9 × \(\frac{4}{5}\) = \(\frac{36}{5}\)

Question 6.
4 × \(\frac{5}{12}\) = _______
Answer:
4 × \(\frac{5}{12}\) = \(\frac{5}{3}\)

Explanation:
The given numbers are: 4 and \(\frac{5}{12}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
4 × \(\frac{5}{12}\) = \(\frac{4}{1}\) × \(\frac{5}{12}\)
= \(\frac{4 × 5}{12 × 1}\)
= \(\frac{20}{12}\)
= \(\frac{5}{3}\)
Hence,
4 × \(\frac{5}{12}\) = \(\frac{5}{3}\)

Question 7.
\(\frac{1}{4}\) × 24 = _______
Answer:
24 × \(\frac{1}{24}\) = 6

Explanation:
The given numbers are: 24 and \(\frac{1}{4}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
24 × \(\frac{1}{4}\) = \(\frac{24}{1}\) × \(\frac{1}{4}\)
= \(\frac{24 × 1}{4 × 1}\)
= \(\frac{24}{4}\)
= 6
Hence,
24 × \(\frac{1}{4}\) = 6

Question 8.
16 × \(\frac{3}{8}\) = _______
Answer:
16 × \(\frac{3}{8}\) = 6

Explanation:
The given numbers are: 16 and \(\frac{3}{8}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
16 × \(\frac{3}{8}\) = \(\frac{16}{1}\) × \(\frac{3}{8}\)
= \(\frac{16 × 3}{8 × 1}\)
= \(\frac{48}{8}\)
= 6
Hence,
16 × \(\frac{3}{8}\) = 6

Question 9.
\(\frac{7}{10}\) × 25 = _______
Answer:
25 × \(\frac{7}{10}\) = \(\frac{35}{2}\)

Explanation:
The given numbers are: 25 and \(\frac{7}{10}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
25 × \(\frac{7}{10}\) = \(\frac{25}{1}\) × \(\frac{7}{10}\)
= \(\frac{7 × 25}{10 × 1}\)
= \(\frac{175}{10}\)
= \(\frac{35}{2}\)
Hence,
25 × \(\frac{7}{10}\) = \(\frac{35}{2}\)

Question 10.
You spend \(\frac{3}{4}\) hour jumping rope every week for 8 weeks. How many hours do you jump rope altogether?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 8
Answer:
The number of hours you jump rope altogether in 8 weeks is: 6 hours

Explanation:
It is given that you spent \(\frac{3}{4}\) hour jumping rope every week for 8 weeks.
So,
The number of hours you spent jumping rope in 8 weeks = ( The number of hours you jump rope in 1 week ) × ( The total number of weeks )
= \(\frac{3}{4}\) × 8
= \(\frac{3}{4}\) × \(\frac{8}{1}\)
= \(\frac{3 × 8}{4 × 1}\)
= 6 hours
hence, from the above,
We can conclude that you spent 6 hours on jumping rope in 8 weeks.

Question 11.
Logic
Your friend finds 25 items that are either insects or flowers. She says that \(\frac{1}{6}\) of the items are insects. Can this be true? Explain.
Answer:
No, this is not true

Explanation:
It is given that your friend finds 25 items that are either insects or flowers and she says that \(\frac{1}{6}\) of the items are insects.
But,
The number of either insects or flowers will always be a whole number.
But, from the given information,
25 will not be divided by \(\frac{1}{6}\)
So,
The statement “\(\frac{1}{6}\) of the items are insects” is false.

Question 12.
Open-Ended
Write two different pairs of fractions that could represent the insects and flowers your friend finds in Exercise 11.
Answer:
The different pairs of fractions that could represent the insects and flowers are:
\(\frac{1}{5}\) and \(\frac{1}{25}\)

Explanation:
These different pairs of fractions have to divide 25
So,
We have to take the fractions the multiples of 5 i.e., 5 and 25
Hence, from the above,
We can conclude that
The different pairs of fractions that could represent the insects and flowers are:
\(\frac{1}{5}\) and \(\frac{1}{25}\)

Question 13.
Modeling Real Life
Newton bakes 56 treats. Five-eighths of them contains peanut butter. Descartes bakes 120 treats. Five-sixths of them contain peanut butter. Who bakes more peanut butter treats? How many more?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 9
Answer:
Descartes bakes more peanut butter treats
The number of peanut butter treats Descartes made more than Descartes is: 65

Explanation:
It is given that Newton bakes 56 treats and five-eights of them contains peanut butter
So,
The number of peanut butter treats made by Newton = \(\frac{5}{8}\) × 56
= \(\frac{5}{8}\) × \(\frac{56}{1}\)
= \(\frac{5 × 56}{8 × 1}\)
= 35 peanut butter treats
It is also given that Descartes bakes 120 treats and five-sixths of them contain peanut butter
So,
the number of peanut butter treats made by Descartes = \(\frac{5}{6}\) × 120
= \(\frac{5}{6}\) × \(\frac{120}{1}\)
= \(\frac{5 × 120}{6 × 1}\)
= 100 peanut butter treats
So,
In the comparison of peanut butter treats,
Descartes bakes more
Now,
The number of peanut butter treats baked by Descartes more than Newton = 100 – 35
= 65 peanut butter treats
Hence, from the above,
We can conclude that
Descartes bakes more peanut butter treats
The number of peanut butter treats Descartes made more than Descartes is: 65

Question 14.
Modeling Real Life
Your class conducts an egg-dropping experiment with 60 eggs. Three-fifths of the eggs break open, one-sixth of the eggs crack, and the rest do not break at all. How many of the eggs do not crack or break open?
Answer:
The number of eggs that do not break is:

Explanation:
It is given that your class conducts an egg-dropping experiment with 60 eggs and three-fifths of the eggs break open, one-sixth of the eggs crack and the rest do not break.
So,
The number of eggs that break open = 60 × \(\frac{3}{5}\)
= \(\frac{60}{1}\) × \(\frac{3}{5}\)
= 36
The number of eggs that cracked = 60 × \(\frac{1}{6}\)
= \(\frac{60}{1}\) × \(\frac{1}{6}\)
= 10
So,
The number of eggs that do not breaked = ( The total number of eggs ) – ( The number of eggs that break open + The number of eggs that cracked )
= 60 – ( 36 + 10 )
= 16
Hence, from the above,
We can conclude that the number of eggs that do not break is: 16 eggs

Review & Refresh

Add.
Question 15.
5\(\frac{5}{8}\) + 6\(\frac{3}{4}\) = _______
Answer:
5\(\frac{5}{8}\) + 6\(\frac{3}{4}\) = \(\frac{99}{8}\)

Explanation:
The give mixed fractions are: 6\(\frac{3}{4}\) and 5\(\frac{5}{8}\)
The representation of mixed fractions in the improper form is: \(\frac{45}{8}\) and \(\frac{27}{4}\)
In addition,
We have to make the denominators equal.
So,
Multiply \(\frac{27}{4}\) with \(\frac{2}{2}\)
So,
\(\frac{45}{8}\) + \(\frac{54}{8}\) = \(\frac{45 + 54}{8}\)
= \(\frac{99}{8}\)
Hence,
5\(\frac{5}{8}\) + 6\(\frac{3}{4}\) = \(\frac{99}{8}\)

Question 16.
1\(\frac{5}{6}\) + 8\(\frac{1}{12}\) = ________
Answer:
1\(\frac{5}{6}\) + 8\(\frac{1}{12}\) = \(\frac{119}{12}\)

Explanation:
The give mixed fractions are: 1\(\frac{5}{6}\) and 8\(\frac{1}{12}\)
The representation of mixed fractions in the improper form is: \(\frac{11}{6}\) and \(\frac{97}{12}\)
In addition,
We have to make the denominators equal.
So,
Multiply \(\frac{11}{6}\) with \(\frac{2}{2}\)
So,
\(\frac{97}{12}\) + \(\frac{22}{12}\) = \(\frac{97 + 22}{12}\)
= \(\frac{119}{12}\)
Hence,
1\(\frac{5}{6}\) + 8\(\frac{1}{12}\) = \(\frac{119}{12}\)

Question 17.
3\(\frac{1}{2}\) + \(\frac{3}{5}\) + 2\(\frac{7}{10}\) = _______
Answer:
3\(\frac{1}{2}\) + 2\(\frac{7}{10}\) + \(\frac{3}{5}\)  = \(\frac{68}{10}\)

Explanation:
The give mixed fractions are: 3\(\frac{1}{2}\), \(\frac{3}{5}\) and 2\(\frac{7}{10}\)
The representation of mixed fractions in the improper form is: \(\frac{7}{2}\) and \(\frac{27}{10}\)
In addition,
We have to make the denominators equal.
So,
Multiply \(\frac{7}{2}\) with \(\frac{5}{5}\)
Multiply \(\frac{3}{5}\) with \(\frac{2}{2}\)
So,
\(\frac{27}{10}\) + \(\frac{35}{10}\) + \(\frac{6}{10}\)  = \(\frac{27 + 35 + 6}{10}\)
= \(\frac{68}{10}\)
Hence,
3\(\frac{1}{2}\) + 2\(\frac{7}{10}\) + \(\frac{3}{5}\)  = \(\frac{68}{10}\)

Lesson 9.4 Use Models to Multiply Fractions

Explore and Grow

Fold a sheet of paper in half. Shade \(\frac{1}{4}\) of either half. What fraction of the entire sheet of paper did you shade? Draw a model to support your answer.
Answer:
The fraction of the entire sheet of paper you shaded is: \(\frac{1}{8}\)

Explanation:
Take a full sheet of paper and fold in half
So,
The number of parts of full sheet of paper = \(\frac{1}{2}\)
Now,
Shade \(\frac{1}{4}\) of the half of the paper
So,,
The fraction of the paper you shaded = \(\frac{1}{4}\) × \(\frac{1}{2}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{2}\)
= \(\frac{1 × 1}{4 × 2}\)
= \(\frac{1}{8}\)
Hence, from the above,
We can conclude that the \(\frac{1}{8}\) part of the full sheet is shaded

Reasoning
What multiplication expression does your model represent? Explain your reasoning.
Answer:
The multiplication expression your model represents is: \(\frac{1}{8}\)
We can obtain the multiplication expression by using the following multiplication properties. They are:
A) \(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
B) a = \(\frac{a}{1}\)

Think and Grow: Use Models to Multiply Fractions

You can use models to multiply a fraction by a fraction.
Example
Find \(\frac{1}{2}\) × \(\frac{1}{3}\).

Show and Grow

Multiply. Use a model to help.
Question 1.
\(\frac{1}{3}\) × \(\frac{1}{4}\) = _______
Answer:
\(\frac{1}{3}\) × \(\frac{1}{4}\) = \(\frac{1}{12}\)

Explanation:
The given fractions are: \(\frac{1}{4}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{3}\)
= \(\frac{1 × 1}{4 × 3}\)
= \(\frac{1}{12}\)
Hence,
\(\frac{1}{4}\) × \(\frac{1}{3}\) = \(\frac{1}{12}\)

Question 2.
\(\frac{2}{3}\) × \(\frac{1}{2}\) = _______
Answer:
\(\frac{2}{3}\) × \(\frac{1}{2}\) = \(\frac{1}{3}\)

Explanation:
The given fractions are: \(\frac{2}{3}\) and \(\frac{1}{2}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{2}{3}\)
= \(\frac{1 × 2}{2 × 3}\)
= \(\frac{1}{3}\)
Hence,
\(\frac{1}{2}\) × \(\frac{2}{3}\) = \(\frac{1}{3}\)

Apply and Grow: Practice

Multiply. Use a model to help.
Question 3.
\(\frac{1}{2}\) × \(\frac{1}{6}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{1}{6}\) = \(\frac{1}{12}\)

Explanation:
The given fractions are: \(\frac{1}{2}\) and \(\frac{1}{6}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{1}{6}\)
= \(\frac{1 × 1}{2 × 6}\)
= \(\frac{1}{12}\)
Hence,
\(\frac{1}{2}\) × \(\frac{1}{6}\) = \(\frac{1}{12}\)

Question 4.
\(\frac{1}{5}\) × \(\frac{1}{8}\) = _______
Answer:
\(\frac{1}{5}\) × \(\frac{1}{8}\) = \(\frac{1}{40}\)

Explanation:
The given fractions are: \(\frac{1}{5}\) and \(\frac{1}{8}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) × \(\frac{1}{8}\)
= \(\frac{1 × 1}{5 × 8}\)
= \(\frac{1}{40}\)
Hence,
\(\frac{1}{5}\) × \(\frac{1}{8}\) = \(\frac{1}{40}\)

Question 5.
\(\frac{1}{4}\) × \(\frac{1}{6}\) = _______
Answer:
\(\frac{1}{4}\) × \(\frac{1}{6}\) = \(\frac{1}{24}\)

Explanation:
The given fractions are: \(\frac{1}{4}\) and \(\frac{1}{6}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{6}\)
= \(\frac{1 × 1}{4 × 6}\)
= \(\frac{1}{24}\)
Hence,
\(\frac{1}{4}\) × \(\frac{1}{6}\) = \(\frac{1}{24}\)

Question 6.
\(\frac{2}{3}\) × \(\frac{1}{3}\) = _______
Answer:
\(\frac{2}{3}\) × \(\frac{1}{3}\) = \(\frac{2}{9}\)

Explanation:
The given fractions are: \(\frac{2}{3}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{2}{3}\) × \(\frac{1}{3}\)
= \(\frac{2 × 1}{3 × 3}\)
= \(\frac{2}{9}\)
Hence,
\(\frac{2}{3}\) × \(\frac{1}{3}\) = \(\frac{2}{9}\)

Write a multiplication equation represented by the model.
Question 7.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 2
Answer:
The multiplication equation represented by the model is:
\(\frac{1}{2}\) × \(\frac{1}{8}\)

Explanation:
The given model is:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 2
From the above model,
The number of rows is: 2
The number of columns is: 8
So,
The value of 1 row = \(\frac{1}{2}\)
The value of 1 column = \(\frac{1}{8}\)
So,
Rows × Columns = \(\frac{1}{8}\) × \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
\(\frac{1}{8}\) × \(\frac{1}{2}\)

Question 8.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 3
Answer:
The multiplication equation represented by the model is:
\(\frac{1}{4}\) × \(\frac{1}{5}\)

Explanation:
The given model is:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 3
From the above model,
The number of rows is: 4
The number of columns is: 5
So,
The value of 1 row = \(\frac{1}{4}\)
The value of 1 column = \(\frac{1}{5}\)
So,
Rows × Columns = \(\frac{1}{4}\) × \(\frac{1}{5}\)
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
\(\frac{1}{4}\) × \(\frac{1}{5}\)

Question 9.
One-fifth of the students in your school have tried skating. Of those students, \(\frac{1}{7}\) have tried ice skating. What fraction of students in your school have tried ice skating?
Answer:
The fraction of students that have tried ice skating is: \(\frac{1}{35}\)

Explanation:
It is given that one-fifth of the students in your school have tried skating and of these students, \(\frac{1}{7}\) have tried ice skating
So,
The fraction of students that have tried ice skating = ( The fraction of students that have tried skating ) × ( The fraction of students that tried ice skating out of the total students )
= \(\frac{1}{5}\) × \(\frac{1}{7}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) × \(\frac{1}{7}\)
= \(\frac{1 × 1}{5 × 7}\)
= \(\frac{1}{35}\)
Hence, from the above,
We can conclude that \(\frac{1}{35}\) of the total students tried ice skating.

Question 10.
DIG DEEPER!
Are both Newton and Descartes correct? Explain.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 4
Answer: Yes, both Newton and Descartes are correct.

Explanation:
The given models of Newton and Descartes are:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 4
From the model of Newton,
The number of rows is: 3
The number of columns is: 5
From the model of Descartes,
The number of rows is: 5
The number of columns is: 3
From the above models,
We can observe that the Descartes model is obtained by reversing the rows and columns of Newton i.e., the rows of Newton’s model become the columns of Descartes’s model and the columns of Newton’s model becomes the rows of Descartes’ model.
So,
The multiplication equation represented by Newton is:
\(\frac{1}{3}\) × \(\frac{1}{5}\)
The multiplication equation represented by Descartes is:
\(\frac{1}{5}\) × \(\frac{1}{3}\)
Hence, from the above,
We can conclude that Newton and Descartes are correct

Think and Grow: Modeling Real Life

Example
A recipe calls for \(\frac{3}{4}\) teaspoon of cinnamon. You4want to halve the recipe. What fraction of a teaspoon of cinnamon do you need?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 5
Because you want to halve the recipe, multiply \(\frac{1}{2}\) by \(\frac{3}{4}\) to find how many teaspoons of cinnamon you need.

So,
You need \(\frac{3}{8}\) teaspoon of cinnamon.

Show and Grow

Question 11.
The mass of mango is \(\frac{2}{5}\) kilogram. The mass of guava is \(\frac{1}{4}\) as much as the mango. What is the mass of the guava?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 7
Answer:
The mass of guava is: \(\frac{1}{10}\) kilograms

Explanation:
It is given that the mass of mango is \(\frac{2}{5}\) kilogram and the mass of guava is \(\frac{1}{4}\) as much as the mango.
So,
The mass of guava = ( The mass of mango ) × ( The fraction of mass of guava in the mass of mango )
= \(\frac{2}{5}\) × \(\frac{1}{4}\)
= \(\frac{1 × 2}{4 × 5}\)
= \(\frac{2}{20}\)
= \(\frac{1}{10}\) kilograms
Hnce, from the above,
We can conclude that the mass of guava is: \(\frac{1}{10}\) kilogram

Question 12.
A giant panda spends \(\frac{2}{3}\) of 1-day eating and foraging. It spends \(\frac{3}{4}\) of that time for eating bamboo. What fraction of 1 day does the panda spend eating bamboo?
Answer:
The fraction of 1 day the panda spent on eating bamboo is: \(\frac{1}{2}\)

Explanation:
It is given that a giant panda spends \(\frac{2}{3}\) of 1-day eating and foraging and it spends \(\frac{3}{4}\) of that time for eating bamboo.
So,
The fraction of 1 day the panda spent on eating bamboo = ( The fraction of time the panda spent on eating ) × ( The fraction of time the panda spent time on eating bamboo out of the total time )
= \(\frac{2}{3}\) × \(\frac{3}{4}\)
= \(\frac{3 × 2}{4 × 3}\)
= \(\frac{6}{12}\)
= \(\frac{1}{2}\) hour
Hence, from the above,
We can conclude that the time spent by the panda on eating bamboo is: \(\frac{1}{2}\) hour

Question 13.
DIG DEEPER!
You have a half-gallon carton of milk that you use only for cereal. You use the same amount each day for 5 days. There is \(\frac{3}{8}\) of the carton left. How many cups of milk do you use each day? Explain.
Answer:

Use Models to Multiply Fractions Homework & Practice 9.4

Multiply. Use a model to help.
Question 1.
\(\frac{1}{3}\) × \(\frac{1}{7}\) = _______
Answer:
\(\frac{1}{3}\) × \(\frac{1}{7}\) = \(\frac{1}{21}\)

Explanation:
The given fractions are: \(\frac{1}{7}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{7}\) × \(\frac{1}{3}\)
= \(\frac{1 × 1}{7 × 3}\)
= \(\frac{1}{21}\)
Hence,
\(\frac{1}{7}\) × \(\frac{1}{3}\) = \(\frac{1}{21}\)

Question 2.
\(\frac{1}{2}\) × \(\frac{1}{9}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{1}{9}\) = \(\frac{1}{18}\)

Explanation:
The given fractions are: \(\frac{1}{2}\) and \(\frac{1}{9}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{1}{9}\)
= \(\frac{1 × 1}{2 × 9}\)
= \(\frac{1}{18}\)
Hence,
\(\frac{1}{2}\) × \(\frac{1}{9}\) = \(\frac{1}{18}\)

Question 3.
\(\frac{2}{5}\) × \(\frac{1}{6}\) = _______
Answer:
\(\frac{2}{5}\) × \(\frac{1}{6}\) = \(\frac{1}{15}\)

Explanation:
The given fractions are: \(\frac{2}{5}\) and \(\frac{1}{6}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{2}{5}\) × \(\frac{1}{6}\)
= \(\frac{2 × 1}{5 × 6}\)
= \(\frac{1}{15}\)
Hence,
\(\frac{2}{5}\) × \(\frac{1}{6}\) = \(\frac{1}{15}\)

Question 4.
\(\frac{3}{4}\) × \(\frac{2}{7}\) = _______
Answer:
\(\frac{3}{4}\) × \(\frac{2}{7}\) = \(\frac{3}{14}\)

Explanation:
The given fractions are: \(\frac{3}{4}\) and \(\frac{2}{7}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{3}{4}\) × \(\frac{2}{7}\)
= \(\frac{3 × 2}{4 × 7}\)
= \(\frac{3}{14}\)
Hence,
\(\frac{3}{4}\) × \(\frac{2}{7}\) = \(\frac{3}{14}\)

Write a multiplication equation represented by the model.
Question 5.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 8
Answer:
The multiplication equation represented by the model is:
\(\frac{1}{2}\) × \(\frac{1}{5}\)

Explanation:
The given model is:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 8
From the above model,
The number of rows is: 2
The number of columns is: 5
So,
The value of 1 row = \(\frac{1}{2}\)
The value of 1 column = \(\frac{1}{5}\)
So,
Rows × Columns = \(\frac{1}{2}\) × \(\frac{1}{5}\)
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
\(\frac{1}{2}\) × \(\frac{1}{5}\)

Question 6.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 9
Answer:
The multiplication equation represented by the model is:
\(\frac{1}{3}\) × \(\frac{1}{6}\)

Explanation:
The given model is:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 9
From the above model,
The number of rows is: 3
The number of columns is: 6
So,
The value of 1 row = \(\frac{1}{3}\)
The value of 1 column = \(\frac{1}{6}\)
So,
Rows × Columns = \(\frac{1}{3}\) × \(\frac{1}{6}\)
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
\(\frac{1}{3}\) × \(\frac{1}{6}\)

Question 7.
One-sixth of the animals at a zoo are birds. Of the birds, \(\frac{1}{3}\) are female. What fraction of the animals at the zoo are female birds?
Answer:
The fraction of the animals at the zoo that is female birds is: \(\frac{1}{18}\)

Explanation:
It is given that one-sixth of the animals at the zoo are birds and of the birds, \(\frac{1}{3}\) is female.
So,
The fraction of the animals at the zoo are female birds = ( The fraction of animals in the zoo that is birds ) × ( The fraction of the animals in the zoo that is female birds )
= \(\frac{1}{3}\) × \(\frac{1}{6}\)
\(\frac{1 × 1}{3 × 6}\)
= \(\frac{1}{18}\)
Hence, from the above,
We can conclude that the number of animals that are female birds at the zoo is: \(\frac{1}{18}\)

Question 8.
Writing
Write and solve a real-life problem for the expression.
\(\frac{2}{3}\) × \(\frac{3}{7}\)
Answer:
Suppose in a school, there are boys ad girls.
In school, \(\frac{2}{3}\) of the boys and \(\frac{3}{7}\) of the girls passed the PET test
So,
The fraction of the students that passed the PET test out of the total number of students = ( The fraction of the boys that passed the PET test ) × ( The fraction of the girls that passed the PET test )
= \(\frac{2}{3}\) × \(\frac{3}{7}\)
= \(\frac{3 × 2}{7 × 3}\)
= \(\frac{6}{21}\)
= \(\frac{2}{7}\)
Hence, from the above,
We can conclude that the number of students who passed the PET test is: \(\frac{2}{7}\)

Question 9.
Modeling Real Life
A Gouldian finch is \(\frac{1}{2}\) the length of the sun conure. How long is the Gouldian finch?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 10
Answer:
The length of Gouldian finch is: \(\frac{11}{24}\) feet

Explanation:
It is given that a Gouldian finch is \(\frac{1}{2}\) the length of the sun conure.
From the given figure,
The length of the sun conure = \(\frac{11}{12}\) feet
So,
The length of Gouldian finch = ( The length of Sun Conure ) ÷ 2
= \(\frac{11}{12}\) ÷ 2
= \(\frac{11}{12}\) ÷ \(\frac{2}{1}\)
= \(\frac{11}{12}\) × \(\frac{1}{2}\)
= \(\frac{11 × 1}{12 × 2}\)
= \(\frac{11}{24}\) feet
Hence, from the above,
We can conclude that the length of the Gouldian finch is: \(\frac{11}{24}\) feet

Question 10.
DIG DEEPER!
A recipe calls for \(\frac{2}{3}\) cup of chopped walnuts. You chop 4 walnuts and get \(\frac{1}{4}\) of the amount you need. How much more of a cup of chopped walnuts do you need? All of your walnuts are the same size. How many more walnuts should you chop? Explain.
Answer:
The cup of chopped walnuts more you need is: \(\frac{1}{3}\)
The number of walnuts you should chop is: 6

Explanation:
It is given that a recipe calls for \(\frac{2}{3}\) cup of chopped walnuts.
It is also given that you chop 4 walnuts and get \(\frac{1}{4}\) of the amount you need.
So,
The cup more of chopped walnuts you need = 1- ( The cup of chopped walnuts you need for recipe )
= 1 – \(\frac{2}{3}\)
= \(\frac{3}{3}\) – \(\frac{2}{3}\)
= \(\frac{3 – 2}{3}\)
= \(\frac{1}{3}\) more cup
Now,
The number of walnuts you need more = ( The number of walnuts you needed to get \(\frac{1}{4}\) of cup of chopped walnuts ) ÷ ( The amount of chopped walnuts you need for recipe )
= 4 ÷ \(\frac{2}{3}\)
= \(\frac{4}{1}\) ÷ \(\frac{2}{3}\)
= \(\frac{4}{1}\) × \(\frac{3}{2}\)
= \(\frac{4 × 3}{1 × 2}\)
= \(\frac{12}{2}\)
= 6
Hence, from the above,
We can conclude that
The cup of chopped walnuts more you need is: \(\frac{1}{3}\)
The number of walnuts you should chop is: 6

Review & Refresh

Subtract.
Question 11.
5 – 1\(\frac{3}{4}\) = ______
Answer:
5 – 1\(\frac{3}{4}\) = \(\frac{13}{4}\)

Explanation:
The given numbers are: 5 and 1\(\frac{3}{4}\)
The representation of 1\(\frac{3}{4}\) in the improper fraction form is: \(\frac{7}{4}\)
So,
5 – \(\frac{7}{4}\) = \(\frac{20}{4}\) – \(\frac{7}{4}\)
= \(\frac{20 – 7}{4}\)
= \(\frac{13}{4}\)
Hence,
5 – 1\(\frac{3}{4}\) = \(\frac{13}{4}\)

Question 12.
13\(\frac{1}{4}\) – 7\(\frac{5}{8}\) = ________
Answer:
13\(\frac{1}{4}\) – 7\(\frac{5}{8}\) = \(\frac{45}{8}\)

Explanation:
The given mixed fractions are: 13\(\frac{1}{4}\) and 7\(\frac{5}{8}\)
The representation of 13\(\frac{1}{4}\) in the improper fraction form is: \(\frac{53}{4}\)
The representation of 13\(\frac{1}{4}\) in the improper fraction form is: \(\frac{61}{8}\)
To subtract the fractions, we have to make the denominators equal.
So,
Multiply \(\frac{53}{4}\) by \(\frac{2}{2}\)
So,
\(\frac{53}{4}\) = \(\frac{106}{8}\)
So,
\(\frac{106}{8}\) – \(\frac{61}{8}\)
= \(\frac{106 – 61}{8}\)
= \(\frac{45}{8}\)
Hence,
13\(\frac{1}{4}\) – 13\(\frac{1}{4}\) = \(\frac{45}{8}\)

Question 13.
12\(\frac{7}{10}\) – 5\(\frac{3}{10}\) – 1\(\frac{1}{5}\) = ________
Answer:
12\(\frac{7}{10}\) – 5\(\frac{3}{10}\) – 1\(\frac{1}{5}\) = \(\frac{62}{10}\)

Explanation:
The given mixed fractions are: 12\(\frac{7}{10}\), 5\(\frac{3}{10}\) and 1\(\frac{1}{5}\)
The representation of 12\(\frac{7}{10}\) in the improper fraction form is: \(\frac{127}{10}\)
The representation of 5\(\frac{3}{10}\) in the improper fraction form is: \(\frac{53}{10}\)
The representation of 1\(\frac{1}{5}\) in the improper fraction form is: \(\frac{6}{5}\)
To subtract the fractions, we have to make the denominators equal.
So,
Multiply \(\frac{6}{5}\) by \(\frac{2}{2}\)
So,
\(\frac{6}{5}\) = \(\frac{12}{10}\)
So,
\(\frac{127}{10}\) – \(\frac{53}{10}\) – \(\frac{12}{10}\)
= \(\frac{127 – 53 – 12}{8}\)
= \(\frac{62}{10}\)
Hence,
12\(\frac{7}{10}\) – 5\(\frac{3}{10}\) – 1\(\frac{1}{5}\) = \(\frac{62}{10}\)

Lesson 9.5 Multiply Fractions

Explore and Grow

Use models to help you complete the table. What do you notice about each expression and its product?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 1
Answer:
The completed table is:

From the given table,
We can observe the product of fractions is also a product.

Construct Arguments
Explain how to multiply two fractions without using a model.
Answer:
We can multiply the fractions by using the properties of multiplication. They are:
A) \(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
B) a = \(\frac{a}{1}\)

Think and Grow: Multiply Fractions

Key Idea
You can find the product of a fraction and a fraction by multiplying the numerators and multiplying the denominators.
Example
Find \(\frac{1}{2}\) × \(\frac{3}{2}\).
Multiply the numerators and multiply the denominators.

Example
Find \(\frac{5}{6}\) × \(\frac{3}{5}\).
Multiply the numerators and multiply the denominators.

Show and Grow

Multiply.
Question 1.
\(\frac{1}{2}\) × \(\frac{4}{3}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{4}{3}\) = \(\frac{2}{3}\)

Explanation:
The given fractions are: \(\frac{1}{2}\) and \(\frac{4}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{4}{3}\)
= \(\frac{1 × 4}{2 × 3}\)
= \(\frac{4}{6}\)
= \(\frac{2}{3}\)
Hence,
\(\frac{1}{2}\) × \(\frac{4}{3}\) = \(\frac{2}{3}\)

Question 2.
\(\frac{2}{5}\) × \(\frac{2}{3}\) = _______
Answer:
\(\frac{2}{5}\) × \(\frac{2}{3}\) = \(\frac{4}{15}\)

Explanation:
The given fractions are: \(\frac{2}{5}\) and \(\frac{2}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{2}{5}\) × \(\frac{2}{3}\)
= \(\frac{2 × 2}{5 × 3}\)
= \(\frac{4}{15}\)
Hence,
\(\frac{2}{5}\) × \(\frac{2}{3}\) = \(\frac{4}{15}\)

Question 3.
\(\frac{3}{4}\) × \(\frac{5}{8}\) = _______
Answer:
\(\frac{3}{4}\) × \(\frac{5}{8}\) = \(\frac{15}{32}\)

Explanation:
The given fractions are: \(\frac{3}{4}\) and \(\frac{5}{8}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{3}{4}\) × \(\frac{5}{8}\)
= \(\frac{3 × 5}{4 × 8}\)
= \(\frac{15}{32}\)
Hence,
\(\frac{3}{4}\) × \(\frac{5}{8}\) = \(\frac{15}{32}\)

Apply and Grow: Practice

Multiply.
Question 4.
\(\frac{1}{4}\) × \(\frac{1}{4}\) = _______
Answer:
\(\frac{1}{4}\) × \(\frac{1}{4}\) = \(\frac{1}{16}\)

Explanation:
The given fractions are: \(\frac{1}{4}\) and \(\frac{1}{4}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{4}\)
= \(\frac{1 × 1}{4 × 4}\)
= \(\frac{1}{16}\)
Hence,
\(\frac{1}{4}\) × \(\frac{1}{4}\) = \(\frac{1}{16}\)

Question 5.
\(\frac{5}{6}\) × \(\frac{7}{10}\) = _______
Answer:
\(\frac{5}{6}\) × \(\frac{7}{10}\) = \(\frac{7}{12}\)

Explanation:
The given fractions are: \(\frac{5}{6}\) and \(\frac{7}{10}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{5}{6}\) × \(\frac{7}{10}\)
= \(\frac{5 × 7}{6 × 10}\)
= \(\frac{35}{60}\)
= \(\frac{7}{12}\)
Hence,
\(\frac{5}{6}\) × \(\frac{7}{10}\) = \(\frac{7}{12}\)

Question 6.
\(\frac{6}{9}\) × \(\frac{8}{2}\) = _______
Answer:
\(\frac{6}{9}\) × \(\frac{8}{2}\) = \(\frac{8}{3}\)

Explanation:
The given fractions are: \(\frac{6}{9}\) and \(\frac{8}{2}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{6}{9}\) × \(\frac{8}{2}\)
= \(\frac{6 × 8}{2 × 9}\)
= \(\frac{48}{18}\)
= \(\frac{8}{3}\)
Hence,
\(\frac{6}{9}\) × \(\frac{8}{2}\) = \(\frac{8}{3}\)

Question 7.
\(\frac{21}{100}\) × \(\frac{3}{5}\) = _______
Answer:
\(\frac{21}{100}\) × \(\frac{3}{5}\) = \(\frac{63}{500}\)

Explanation:
The given fractions are: \(\frac{21}{100}\) and \(\frac{3}{5}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{21}{100}\) × \(\frac{3}{5}\)
= \(\frac{21 × 3}{100 × 5}\)
= \(\frac{63}{500}\)
Hence,
\(\frac{21}{100}\) × \(\frac{3}{5}\) = \(\frac{63}{500}\)

Question 8.
\(\frac{1}{12}\) × \(\frac{9}{4}\) = _______
Answer:
\(\frac{1}{12}\) × \(\frac{9}{4}\) = \(\frac{3}{16}\)

Explanation:
The given fractions are: \(\frac{1}{12}\) and \(\frac{9}{4}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{12}\) × \(\frac{9}{4}\)
= \(\frac{1 × 9}{12 × 4}\)
= \(\frac{9}{48}\)
= \(\frac{3}{16}\)
Hence,
\(\frac{1}{12}\) × \(\frac{9}{4}\) = \(\frac{3}{16}\)

Question 9.
\(\frac{4}{7}\) × \(\frac{8}{8}\) = _______
Answer:
\(\frac{4}{7}\) × \(\frac{8}{8}\) = \(\frac{4}{7}\)

Explanation:
The given fractions are: \(\frac{4}{7}\) and \(\frac{8}{8}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{4}{7}\) × \(\frac{8}{8}\)
= \(\frac{8 × 4}{7 × 8}\)
= \(\frac{4}{7}\)
Hence,
\(\frac{4}{7}\) × \(\frac{8}{8}\) = \(\frac{4}{7}\)

Evaluate
Question 10.
\(\left(\frac{1}{2} \times \frac{7}{8}\right)\) × 2 = _______
Answer:
\(\left(\frac{1}{2} \times \frac{7}{8}\right)\) × 2 = \(\frac{7}{8}\)

Explanation:
The given fractions are: \(\frac{7}{8}\) and \(\frac{1}{2}\)

We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
( \(\frac{1}{2}\) × \(\frac{7}{8}\)  ) × 2
= ( \(\frac{1 × 7}{2 × 8}\) ) × \(\frac{2}{1}\)
= \(\frac{7}{16}\) × \(\frac{2}{1}\)
= \(\frac{7 × 2}{16 × 1}\)
= \(\frac{7}{8}\)
Hence,
\(\frac{4}{7}\) × \(\frac{8}{8}\) = \(\frac{4}{7}\)

Question 11.
\(\left(\frac{7}{6}-\frac{5}{6}\right)\) × \(\frac{2}{3}\) = _______
Answer:
\(\left(\frac{7}{6}-\frac{5}{6}\right)\) × \(\frac{2}{3}\) = \(\frac{2}{9}\)

Explanation:
The given fractions are: \(\frac{7}{6}\) , \(\frac{5}{6}\), and \(\frac{2}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
( \(\frac{7}{6}\) – \(\frac{5}{6}\)  ) × \(\frac{2}{3}\)
= ( \(\frac{7 – 5}{6}\) ) × \(\frac{2}{3}\)
= \(\frac{2}{6}\) × \(\frac{2}{3}\)
= \(\frac{2 × 2}{6 × 3}\)
= \(\frac{4}{18}\)
= \(\frac{2}{9}\)
Hence,
\(\left(\frac{7}{6}-\frac{5}{6}\right)\) × \(\frac{2}{3}\) = \(\frac{2}{9}\)

Question 12.
\(\frac{9}{10}\) × \(\left(\frac{4}{9}+\frac{1}{3}\right)\) = _________
Answer:
\(\frac{9}{10}\) × \(\left(\frac{4}{9}+\frac{1}{3}\right)\) = \(\frac{7}{10}\)

Explanation:
The given fractions are: \(\frac{9}{10}\), \(\frac{4}{9}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{9}{10}\) × ( \(\frac{4}{9}\) + \(\frac{1}{3}\)  )
= ( \(\frac{4 + 3}{9}\) ) × \(\frac{9}{10}\)
= \(\frac{7}{9}\) × \(\frac{9}{10}\)
= \(\frac{7 × 9}{9 × 10}\)
= \(\frac{7}{10}\)
Hence,
\(\frac{9}{10}\) × \(\left(\frac{4}{9}+\frac{1}{3}\right)\) = \(\frac{7}{10}\)

Question 13.
At a school, \(\frac{3}{4}\) of the students play a sport. Of the students that play a sport, \(\frac{1}{5}\) play baseball. What fraction of the students at the school play baseball?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 4
Answer:
The fraction of the students at the school that play baseball is: \(\frac{3}{20}\)

Explanation:
It is given that at a school, \(\frac{3}{4}\) of the students play a sport and of the students that play a sport, \(\frac{1}{5}\) play baseball.
So,
The fraction of the students at the school that play baseall = ( The fraction of the students that play a sport ) × ( The fraction of the students that play baseball out of thetotal number of students )
= \(\frac{3}{4}\) × \(\frac{1}{5}\)
= \(\frac{3 × 1}{4 × 5}\)
= \(\frac{3}{20}\)
Hence, from the above,
We can conclude that the fraction of the students that play baseball is: \(\frac{3}{20}\)

Question 14.
Reasoning
Descartes says he can find the product of a whole number and a fraction the same way he finds the product of two fractions. Explain why his reasoning makes sense.
Answer:
The product of a whole number and a fraction follows the same procedure as that of the product of the two fractions because of the following properties of multiplication:
A) \(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
B) a= \(\frac{a}{1}\)
Hence, from the above two properties of multiplication,
We can conclude that the reasoning of Descartes makes sense

Question 15.
Writing
Explain how multiplying fractions is different than adding and subtracting fractions.
Answer:
In the multiplication of the fractions, we multiply numerators and denominators.
Example:
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
In addition of the fractions, we add only numerators making the denominators equal
Example:
\(\frac{a}{b}\) + \(\frac{p}{b}\)
= \(\frac{a + p}{b}\)
In subtraction of the fractions, we subtract only numerators making the denominators equal
Example:
\(\frac{a}{b}\) – \(\frac{p}{b}\)
= \(\frac{a – p}{b}\)

Think and Grow: Modeling Real Life

Example
A tourist is walking from the Empire State Building to Times Square. She is \(\frac{2}{3}\) of the way there. What fraction of a mile does she have left to walk?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 5
Find the distance she has walked. Because she has walked \(\frac{2}{3}\) of \(\frac{3}{4}\) mile, multiply \(\frac{2}{3}\) by \(\frac{3}{4}\).

Show and Grow

Question 16.
At a zoo, \(\frac{3}{5}\) of the animals are mammals. Of the mammals, \(\frac{5}{12}\) are primates.What fraction of the animals at the zoo are not primates?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 7
Answer:
The fractions of the animals at the zoo that are not primates is: \(\frac{3}{4}\)

Explanation:
It is given that at the zoo, \(\frac{3}{5}\) of the animals are mammals and of the animals, \(\frac{5}{12}\) are primates
So,
The fraction of the animals that are primates =( The fraction of the animals that are primates ) × ( The fraction of the animals that are primates )
= \(\frac{3}{5}\) × \(\frac{5}{12}\)
= \(\frac{3 × 5}{5 × 12}\)
= \(\frac{15}{60}\)
= \(\frac{1}{4}\)
Now,
Let the total number of animals that are mammals =1
So,
The fraction of the animals that are not primates = ( The total number of mammals ) – ( The fraction of the animals that are primates )
= 1 – \(\frac{1}{4}\)
= \(\frac{4}{4}\) – \(\frac{1}{4}\)
= \(\frac{4 – 1}{4}\)
= \(\frac{3}{4}\)
Hence, from the above,
We can conclude that the fraction of the animals that are not primates is: \(\frac{3}{4}\)

Question 17.
You have an album of 216 trading cards. One page contains \(\frac{1}{24}\) of the cards. On that page, \(\frac{2}{3}\) of the cards are epic. You only have one page with any epic cards. How many epic cards do you have?
Answer:
The number of epic cards is: 6 epic cards

Explanation:
It is given that you have an album of 216 trading cards
It is also given that one page contains \(\frac{1}{24}\) of the cards. On that page, \(\frac{2}{3}\) of the cards are epic.
So,
The fraction of the epic cards = ( The total number of cards on that page ) × ( The fraction of the epic cards that is on that page )
= \(\frac{1}{24}\) × \(\frac{2}{3}\)
= \(\frac{2 × 1}{3 × 24}\)
= \(\frac{1}{36}\)
So,
The number of epic cards = ( The total number of cards ) × ( The fraction of the epic cards )
= 216 × \(\frac{1}{36}\)
= \(\frac{216}{1}\) × \(\frac{}{36}\)
= \(\frac{216}{36}\)
= 6 epic cards
Hence, from the above,
We can conclude that there are 6 epic cards

Question 18.
DIG DEEPER!
In a class, \(\frac{2}{5}\) of the students play basketball and \(\frac{7}{10}\) play soccer.Of the students who play basketball, \(\frac{2}{3}\) also play soccer. There are 30 students in the class. How many students play soccer but do not play basketball?
Answer:
The number of students that play soccer but do not play basketball is: 4 students

Explanation:
It is given that in a class, \(\frac{2}{5}\) of the students play basketball and \(\frac{7}{10}\) play soccer.Of the students who play basketball, \(\frac{2}{3}\) also play soccer.
So,
The number of students who play basketball and soccer = ( The number of students who play basketball ) × ( The number of students playing soccer who are also playing basketball )
= \(\frac{2}{5}\) × \(\frac{2}{3}\)
= \(\frac{4}{15}\)
Now,
The number of students who only play soccer but not basketball = ( The number of students who play soccer only ) – ( The number of students who play both basketball and soccer )
= \(\frac{2}{5}\) – \(\frac{4}{15}\)
In subtraction, the denominators must be equal.
So,
Multiply \(\frac{2}{5}\) by \(\frac{3}{3}\)
So,
\(\frac{6}{15}\) – \(\frac{4}{15}\)
= \(\frac{4}{15}\)
It is also given that the total number of students are: 30
So,
The number of students who play soccer only = ( The fraction of the students who play soccer only ) × ( The total number of students )
= \(\frac{2}{15}\) × 30
= \(\frac{2}{15}\) × \(\frac{30}{1}\)
= 4 students
Hence, from the above,
We can conclude that there are 4 students who play soccer only.

Multiply Fractions Homework & Practice 9.5

Multiply.
\(\frac{1}{4}\) × \(\frac{1}{5}\) = _______
Answer:
\(\frac{1}{4}\) × \(\frac{1}{5}\) = \(\frac{1}{20}\)

Explanation:
The given fractions are: \(\frac{1}{4}\) and \(\frac{1}{5}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{5}\)
= \(\frac{1 × 1}{4 × 5}\)
= \(\frac{1}{20}\)
Hence,
\(\frac{1}{4}\) × \(\frac{1}{5}\) = \(\frac{1}{20}\)

Question 2.
\(\frac{2}{7}\) × \(\frac{1}{2}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{2}{7}\) = \(\frac{1}{7}\)

Explanation:
The given fractions are: \(\frac{1}{2}\) and \(\frac{2}{7}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{2}{7}\)
= \(\frac{1 × 2}{2 × 7}\)
= \(\frac{2}{14}\)
= \(\frac{1}{7}\)
Hence,
\(\frac{1}{2}\) × \(\frac{2}{7}\) = \(\frac{1}{7}\)

Question 3.
\(\frac{9}{10}\) × \(\frac{2}{3}\) = _______
Answer:
\(\frac{9}{10}\) × \(\frac{2}{3}\) = \(\frac{3}{5}\)

Explanation:
The given fractions are: \(\frac{9}{10}\) and \(\frac{2}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{9}{10}\) × \(\frac{2}{3}\)
= \(\frac{9 × 2}{10 × 3}\)
= \(\frac{18}{30}\)
= \(\frac{3}{5}\)
Hence,
\(\frac{9}{10}\) × \(\frac{2}{3}\) = \(\frac{3}{5}\)

Question 4.
\(\frac{5}{8}\) × \(\frac{5}{6}\) = _______
Answer:
\(\frac{5}{8}\) × \(\frac{5}{6}\) = \(\frac{25}{48}\)

Explanation:
The given fractions are: \(\frac{5}{8}\) and \(\frac{5}{6}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{5}{8}\) × \(\frac{5}{6}\)
= \(\frac{5 × 5}{8 × 6}\)
= \(\frac{25}{48}\)
Hence,
\(\frac{5}{8}\) × \(\frac{5}{6}\) = \(\frac{25}{48}\)

Question 5.
\(\frac{9}{7}\) × \(\frac{3}{4}\) = _______
Answer:
\(\frac{9}{7}\) × \(\frac{3}{4}\) = \(\frac{27}{28}\)

Explanation:
The given fractions are: \(\frac{9}{7}\) and \(\frac{3}{4}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{9}{7}\) × \(\frac{3}{4}\)
= \(\frac{9 × 3}{7 × 4}\)
= \(\frac{27}{28}\)
Hence,
\(\frac{9}{7}\) × \(\frac{3}{4}\) = \(\frac{27}{28}\)

Question 6.
\(\frac{11}{100}\) × \(\frac{2}{5}\) = _______
Answer:
\(\frac{11}{100}\) × \(\frac{2}{5}\) = \(\frac{11}{250}\)

Explanation:
The given fractions are: \(\frac{11}{100}\) and \(\frac{2}{5}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{11}{100}\) × \(\frac{2}{5}\)
= \(\frac{11 × 2}{100 × 5}\)
= \(\frac{22}{500}\)
= \(\frac{11}{250}\)
Hence,
\(\frac{11}{100}\) × \(\frac{2}{5}\) = \(\frac{11}{250}\)

Question 7.
\(\frac{7}{20}\) × \(\frac{6}{2}\) = _______
Answer:
\(\frac{7}{20}\) × \(\frac{6}{2}\) = \(\frac{21}{20}\)

Explanation:
The given fractions are: \(\frac{7}{20}\) and \(\frac{6}{2}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{7}{20}\) × \(\frac{6}{2}\)
= \(\frac{7 × 6}{20 × 2}\)
= \(\frac{42}{40}\)
= \(\frac{21}{20}\)
Hence,
\(\frac{7}{20}\) × \(\frac{6}{2}\) = \(\frac{21}{20}\)

Question 8.
\(\frac{15}{16}\) × \(\frac{1}{3}\) = _______
Answer:
\(\frac{15}{16}\) × \(\frac{1}{3}\) = \(\frac{5}{16}\)

Explanation:
The given fractions are: \(\frac{15}{16}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{15}{16}\) × \(\frac{1}{3}\)
= \(\frac{15 × 1}{16 × 3}\)
= \(\frac{15}{48}\)
= \(\frac{5}{16}\)
Hence,
\(\frac{15}{16}\) × \(\frac{1}{3}\) = \(\frac{5}{16}\)

Question 9.
\(\frac{5}{12}\) × \(\frac{3}{10}\) = _______
Answer:
\(\frac{5}{12}\) × \(\frac{3}{10}\) = \(\frac{1}{8}\)

Explanation:
The given fractions are: \(\frac{5}{12}\) and \(\frac{3}{10}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{5}{12}\) × \(\frac{3}{10}\)
= \(\frac{5 × 3}{12 × 10}\)
= \(\frac{15}{120}\)
= \(\frac{1}{8}\)
Hence,
\(\frac{5}{12}\) × \(\frac{3}{10}\) = \(\frac{1}{8}\)

Evaluate.
Question 10.
3 × \(\frac{3}{10}\) = _______
Answer:
3 ×  \(\frac{3}{10}\) = \(\frac{9}{10}\)

Explanation:
The given numbers are: 3 and \(\frac{3}{10}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
3 × \(\frac{3}{10}\)
= \(\frac{3}{1}\) × \(\frac{3}{10}\)
= \(\frac{3 × 3}{10 × 1}\)
= \(\frac{9}{10}\)
Hence,
3 × \(\frac{3}{10}\) = \(\frac{9}{10}\)

Question 11.
\(\left(\frac{1}{3}+\frac{1}{3}\right)\) × \(\frac{4}{5}\) = _______
Answer:
\(\left(\frac{1}{3}+\frac{1}{3}\right)\) × \(\frac{4}{5}\) = \(\frac{8}{15}\)

Explanation:
The given fractions are: \(\frac{1}{3}\), \(\frac{1}{3}\) and \(\frac{4}{5}\)

We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
( \(\frac{1}{3}\) + \(\frac{1}{3}\)  ) × \(\frac{4}{5}\)
= ( \(\frac{1 + 1}{3}\) ) × \(\frac{4}{5}\)
= \(\frac{2}{3}\) × \(\frac{4}{5}\)
= \(\frac{4 × 2}{3 × 5}\)
= \(\frac{8}{15}\)
Hence,
\(\left(\frac{1}{3}+\frac{1}{3}\right)\) × \(\frac{4}{5}\) = \(\frac{8}{15}\)

Question 12.
\(\frac{6}{7}\) × \(\left(\frac{3}{4}-\frac{5}{12}\right)\) = _______
Answer:
\(\frac{6}{7}\) × \(\left(\frac{3}{4}-\frac{5}{12}\right)\) = \(\frac{2}{7}\)

Explanation:
The given fractions are: \(\frac{6}{7}\), \(\frac{3}{4}\) and \(\frac{5}{12}\)

We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
( \(\frac{3}{4}\) – \(\frac{5}{12}\)  ) × \(\frac{6}{7}\)
= ( \(\frac{9}{12}\) – \(\frac{5}{12}\)  ) × \(\frac{6}{7}\)
= \(\frac{4}{12}\) × \(\frac{6}{7}\)
= \(\frac{4 × 6}{12 × 7}\)
= \(\frac{2}{7}\)
Hence,
\(\frac{6}{7}\) × \(\left(\frac{3}{4}-\frac{5}{12}\right)\) = \(\frac{2}{7}\)

Question 13.
A pancake recipe calls for \(\frac{1}{3}\) cup of water. You want to halve the recipe. What fraction of a cup of water do you need?
Answer:
The fraction of a cup of water you need is: \(\frac{1}{6}\)

Explanation:
It is given that a pancake recipe calls for \(\frac{1}{3}\) cup of water and you want to halve the recipe.
So,
The fraction of a cup of water you need for half of the recipe = ( The cup of water you need for a recipe ) × \(\frac{1}{2}\)
= \(\frac{1}{3}\) × \(\frac{1}{2}\)
= \(\frac{1 × 1}{3 × 2}\)
= \(\frac{1}{6}\)
Hence, from the above,
We can conclude that the fraction of water you need for half of the recipe is: \(\frac{1}{6}\)

Question 14.
Number Sense
Which is greater, \(\frac{3}{4}\) × \(\frac{1}{5}\) or \(\frac{3}{4}\) × \(\frac{1}{8}\)? Explain.
Answer:
\(\frac{3}{4}\) × \(\frac{1}{5}\)” is greater than “\(\frac{3}{4}\) × \(\frac{1}{8}\)

Explanation:
The given fractions are: \(\frac{3}{4}\), \(\frac{1}{5}\) and \(\frac{1}{8}\)
Now,
\(\frac{3}{4}\) × \(\frac{1}{5}\)
= \(\frac{3 × 1}{4 × 5}\)
= \(\frac{3}{20}\)
Now,
\(\frac{3}{4}\) × \(\frac{1}{8}\)
= \(\frac{3 × 1}{4 × 8}\)
= \(\frac{3}{32}\)
So,
For the comparison of the products, equate the numerators.
Multiply \(\frac{3}{20}\) by \(\frac{32}{32}\)
Multiply \(\frac{3}{32}\) by \(\frac{20}{20}\)
So,
\(\frac{3}{20}\) = \(\frac{96}{640}\)
\(\frac{3}{32}\) = \(\frac{60}{640}\)
So,
By comparison of the products,
We can observe that
96 > 60
Hence, from the above,
We can conclude that “\(\frac{3}{4}\) × \(\frac{1}{5}\)” is greater than “\(\frac{3}{4}\) × \(\frac{1}{8}\)”

Question 15.
Reasoning
Is \(\frac{17}{24}\) × \(\frac{7}{8}\) equal to \(\frac{17}{8}\) × \(\frac{7}{24}\)? Explain.
Answer:
\(\frac{17}{24}\) × \(\frac{7}{8}\) is equal to \(\frac{17}{8}\) × \(\frac{7}{24}\)

Explanation:
By the commutative property of multiplication,
a × b = b × a
So,
By the above property of multiplication,
We can interchange the numbers
Hence, from the above,
We can conclude that \(\frac{17}{24}\) × \(\frac{7}{8}\) is equal to \(\frac{17}{8}\) × \(\frac{7}{24}\)

Question 16.
Number Sense
In which equations does k = \(\frac{5}{6}\)?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 8
Answer:
Let the multiplication equations be represented A), B), C) and D)
The given expressions are:
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 8
So,
The product of the given expressions are:
A) \(\frac{1}{2}\) × \(\frac{5}{3}\) = \(\frac{5}{6}\)
B) \(\frac{1}{10}\) × k = \(\frac{1}{12}\)
C) \(\frac{1}{3}\) × k = \(\frac{2}{9}\)
D) \(\frac{5}{4}\) × \(\frac{2}{3}\) = k
Hence, from the above expressions,
We can conclude that equation A) fits the value of k perfectly.

Question 17.
Modeling Real Life
At a town hall meeting, \(\frac{37}{50}\) of the members are present. Of those who are present, \(\frac{1}{2}\) vote in favor of a new park. What fraction of the members do not vote in favor of the new park?
Answer:
The fraction of the members that do not vote in favor of the new park is: \(\frac{13}{50}\)

Explanation:
It is given that at a town hall meeting, \(\frac{37}{50}\) of the members are present. Of those who are present, \(\frac{1}{2}\) vote in favor of a new park.
So,
The fraction of the members that do not vote in favor of the new park = ( The total number of members at the town hall meeting ) – ( The fraction of the members that vote in the favor of a new park )
= 1 – \(\frac{37}{50}\)
= \(\frac{50}{50}\) – \(\frac{37}{50}\)
= \(\frac{50 – 37}{50}\)
= \(\frac{13}{50}\)
Hence, from the above,
We can conclude that the number of members that do not vote in the favor of a new park is: \(\frac{13}{50}\)

Question 18.
Modeling Real Life
There are 50 U.S. states. Seven twenty-fifths of the states share a land border with Canada or Mexico.Of those states, \(\frac{2}{7}\) share a border with Mexico. How many states share a border with Canada?
Answer:
The number of states that share a border with Canada is: 4 states

Explanation:
It is given that there are 50 U.S. states. Seven twenty-fifths of the states share a land border with Canada or Mexico and of those states, \(\frac{2}{7}\) share a border with Mexico.
So,
The number of states that share a border with Canada or Mexico = \(\frac{7}{25}\) × 50
= \(\frac{7}{25}\) × \(\frac{50}{1}\)
= 14 states
So,
The number of states that share a border with Canada = \(\frac{2}{7}\) × 14
= \(\frac{2}{7}\) × \(\frac{14}{1}\)
= 4 states
Hence, from the above,
We can conclude that there are 4 states that share a border with Canada

Review & Refresh

Evaluate. Check whether your answer is reasonable.
Question 19.
15.67 + 4 + 6.5 = _____
Answer:
15.67 + 4 + 6.5 = 26.17

Explanation:
The given numbers are: 15.67, 4, and 6.5
The representation of the numbers in the fraction form is: \(\frac{1567}{100}\), \(\frac{65}{10}\) and 4
As the highest number in the denominator is 100, make all the denominators hundred
So,
The representation of 4 with 100 in the denominator is: \(\frac{400}{100}\)
The representation of 6.5 with 100 in the denominator is: \(\frac{650}{100}\)
So,
\(\frac{1567}{100}\) + \(\frac{400}{100}\) + \(\frac{650}{100}\)
= \(\frac{  1567 + 400 + 650 }{100}\)
= \(\frac{2,617}{100}\)
= 26.17
Hence, 15.67 + 4 + 6.5 = 26.17

Question 20.
20.7 – 9.54 + 25.81 = _______
Answer:
20.7 – 9.54 + 25.81 = 36.97

Explanation:
The given numbers are: 20.7, 9.54, and 25.81
The representation of the numbers in the fraction form is: \(\frac{2581}{100}\), \(\frac{207}{10}\) and \(\frac{954}{100}\)
As the highest number in the denominator is 100, make all the denominators hundred
So,
The representation of 20.7 with 100 in the denominator is: \(\frac{2070}{100}\)
So,
\(\frac{2581}{100}\) + \(\frac{2070}{100}\) – \(\frac{954}{100}\)
= \(\frac{  2581 + 2070 – 954 }{100}\)
= \(\frac{3,697}{100}\)
= 36.97
Hence, 20.7 – 9.54 + 25.81 = 36.97

Lesson 9.6 Find Areas of Rectangles

Explore and Grow

Draw and cut out a rectangle that has any two of the side lengths below.
\(\frac{1}{2}\) ft
\(\frac{1}{3}\) ft
\(\frac{1}{4}\) ft
Use several copies of your rectangle to create a unit square. What is the area (in square feet) of each small rectangle? Explain your reasoning.
Answer:
Let there are two small rectangles with the following lengths:
\(\frac{1}{2}\) ft and \(\frac{1}{3}\) ft
\(\frac{1}{2}\) ft and \(\frac{1}{4}\) ft
So,
The area of the unit square with lengths \(\frac{1}{2}\) ft and \(\frac{1}{3}\) ft is:
\(\frac{1}{2}\) ft × \(\frac{1}{3}\) ft
= \(\frac{1 × 1}{2 × 3}\) ft
= \(\frac{1}{6}\) ft
The area of the unit square with lengths \(\frac{1}{2}\) ft and \(\frac{1}{4}\) ft is:
\(\frac{1}{2}\) ft × \(\frac{1}{4}\) ft
= \(\frac{1 × 1}{2 × 4}\) ft
= \(\frac{1}{8}\) ft
Hence, from the above,
We can conclude that the areas ofthe smaller rectngles with unit lengths is: \(\frac{1}{6}\) ft and \(\frac{1}{8}\) ft

Reasoning
How can you use a rectangle with unit fraction side lengths to find the area of the rectangle below? Explain your reasoning.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 1
Answer:
The given rectangle is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 1
To find the area of the rectangle by the unit fractions, multiply the values of the length and the breadth
So,
\(\frac{3}{4}\) × \(\frac{3}{2}\)
= 3  (\(\frac{1}{4}\) ) × 3 ( \(\frac{1}{2}\) )
= 9 ( \(\frac{1}{4}\) × \(\frac{1}{2}\) )
= 9 × \(\frac{1 × 1}{4 × 2}\)
= \(\frac{9}{1}\) × \(\frac{1}{8}\)
= \(\frac{9}{8}\)
The unit fractions are the fractions that contain the value 1 in the numerator.
Here,
\(\frac{1}{4}\) and \(\frac{1}{2}\) are the unit fractions

Think and Grow: Find Areas of Rectangles

One way to find the area of a rectangle with fractional side lengths is to fill it with smaller rectangles.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 2
Example
Find the area of the rectangle.

Show and Grow

Question 1.
Find the area of the shaded region.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 4
Answer:
The area of the shaded region is: \(\frac{3}{12}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 4
From the given figure,
The area of the shaded region = \(\frac{3}{4}\) × \(\frac{1}{3}\)
= \(\frac{3 × 1}{4 × 3}\)
= \(\frac{3}{12}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{3}{12}\)

Apply and Grow: Practice

Find the area of the shaded region.
Question 2.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 5
Answer:
The area of the shaded region is: \(\frac{2}{12}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 5
From the given figure,
The area of the shaded region = \(\frac{2}{4}\) × \(\frac{1}{3}\)
= \(\frac{2 × 1}{4 × 3}\)
= \(\frac{2}{12}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{2}{12}\)

Question 3.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 6
Answer:
The area of the shaded region is: \(\frac{3}{16}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 6
From the given figure,
The area of the shaded region = \(\frac{3}{4}\) × \(\frac{1}{4}\)
= \(\frac{3 × 1}{4 × 4}\)
= \(\frac{3}{16}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{3}{16}\)

Use rectangles with unit fraction side lengths to find the area of the rectangle.
Question 4.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 7
Answer:
The area of the rectangle is: \(\frac{6}{10}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 7
From the given figure,
The area of the rectangle = \(\frac{3}{2}\) × \(\frac{2}{5}\)
= \(\frac{3 × 2}{2 × 5}\)
= \(\frac{6}{10}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{6}{10}\)

Question 5.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 8
Answer:
The area of the rectangle is: \(\frac{10}{18}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 8
From the given figure,
The area of the rectangle = \(\frac{5}{6}\) × \(\frac{2}{3}\)
= \(\frac{5 × 2}{6 × 3}\)
= \(\frac{10}{18}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{10}{18}\)

Question 6.
Find the area of a rectangle with side lengths of \(\frac{5}{8}\) and \(\frac{4}{3}\)?
Answer:
The area of a rectangle is: \(\frac{20}{24}\)

Explanation:
The given side lengths of a rectangle are: \(\frac{5}{8}\) and \(\frac{4}{3}\)
So,
The area of the rectangle = \(\frac{5}{8}\) × \(\frac{4}{3}\)
= \(\frac{5 × 4}{8 × 3}\)
= \(\frac{20}{24}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{20}{24}\)

Question 7.
Find the area of a rectangle with side lengths of \(\frac{7}{9}\) and \(\frac{1}{2}\)?
Answer:
The area of a rectangle is: \(\frac{7}{18}\)

Explanation:
The given side lengths of a rectangle are: \(\frac{7}{9}\) and \(\frac{1}{2}\)
So,
The area of the rectangle = \(\frac{7}{9}\) × \(\frac{1}{2}\)
= \(\frac{7 × 1}{9 × 2}\)
= \(\frac{7}{18}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{7}{18}\)

Question 8.
Reasoning
Can you find the area of a rectangle with fractional side lengths the same way you find the area of a rectangle with whole-number side lengths? Explain.
Answer:
Yes, we can find the area of a rectangle with fractional side lengths the same way you find the area of a rectangle with whole-number side lengths.
Example:
Let the fractional side lengths be: \(\frac{1}{5}\) and \(\frac{1}{9}\)
Let the whole number side lengths be: 3 and 4
So,
The area of the rectangle with the fractional side lengths = \(\frac{1}{5}\) × \(\frac{1}{9}\)
= \(\frac{1 × 1}{9 × 5}\)
= \(\frac{1}{45}\)
The area of the rectangle with the whole number side lengths = 3 × 4
= \(\frac{3}{1}\) × \(\frac{4}{1}\)
= \(\frac{3 × 4}{1 × 1}\)
= \(\frac{12}{1}\)
= 12
Hence, from the above,
We can conclude that we can find the area of a rectangle with fractional side lengths the same way you find the area of a rectangle with whole-number side lengths.

Question 9.
YOU BE THE TEACHER
Our friend says she can find the area of a square given only one fractional side length. Is your friend correct? Explain.
Answer:
Yes, she can find the area of a square by only one fractional side length

Explanation:
Let the unit fractional side length of the square be: \(\frac{1}{2}\)
We know that,
The length of all the sides in a square are equal.
So,
By using this property, we can find the area of the square by taking the fractional unit side length of 1 side as the same for all the sides
So,
The area of the square = \(\frac{1}{2}\) × \(\frac{1}{2}\)
= \(\frac{1 × 1}{2 × 2}\)
= \(\frac{1}{4}\)
Hence, from the above,
We can conclude that she can find the area of a square by only one fractional side length

Think and Grow: Modeling Real Life

Example
A zoo needs an outdoor enclosure with an area of at least \(\frac{3}{10}\) square kilometer10for a camel. Is the rectangular enclosure shown large enough for a camel?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 9
Find the area of the enclosure by multiplying the length and the width.

So, the enclosure is large enough for a camel.

Show and Grow

Question 10.
The area of a square dog kennel is 4 square yards. Will the square mat fit in the kennel?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 11
Answer:
Yes, the square mat will fit in the kennel

Explanation:
It is given that the area of a square dog kennel is 4 square yards and the side of the square mat is \(\frac{5}{3}\) yd
We know that,
The length of all the sides in a square are equal.
So,
By using this property,
The area of the square mat = \(\frac{5}{3}\) × \(\frac{5}{3}\)
= \(\frac{5 × 5}{3 × 3}\)
= \(\frac{25}{9}\) yd
Now,
The area of a square dog kennel can be written as: \(\frac{36}{9}\) yd
So,
When we compare the 2 areas,
We can get 25 < 36
Hence, from the above,
We can conclude that the square mat will fit in the kennel

Question 11.
DIG DEEPER!
The side lengths of each chalk art square are \(\frac{11}{4}\) meters. The side lengths of the square zone around each chalk square are an additional \(\frac{3}{4}\) meter. How many square meters of concrete is used to create the chalk walk shown?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 12
Answer:
The amount of the concrete used to create the chalk walk is: \(\frac{196}{16}\) square meters

Explanation:
It is given that the side lengths of each chalk art square are \(\frac{11}{4}\) meters. The side lengths of the square zone around each chalk square are an additional \(\frac{3}{4}\) meter.
So,
The total side length of the chalk walk = \(\frac{11}{4}\) + \(\frac{3}{4}\)
= \(\frac{11 + 3}{4}\)
= \(\frac{14}{4}\) meters
We know that,
The length of all sides in the square are equal.
So,
By using this property,
The amount of concrete used for the chalk walk = \(\frac{14}{4}\) × \(\frac{14}{4}\)
= \(\frac{14 × 14}{4 × 4}\)
= \(\frac{196}{16}\) square meters
Hence, from the above,
We can conclude that the amount of the concrete used to create the chalk walk is: \(\frac{196}{16}\) square meters

Find Areas of Rectangles Homework & Practice 9.6

Find the area of the shaded region.
Question 1.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 13
Answer:
The area of the shaded region is: \(\frac{1}{12}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 13
From the given figure,
The area of the shaded region = \(\frac{1}{2}\) × \(\frac{1}{6}\)
= \(\frac{1 × 1}{2 × 6}\)
= \(\frac{1}{12}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{1}{12}\)

Question 2.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 14
Answer:
The area of the shaded region is: \(\frac{8}{25}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 14
From the given figure,
The area of the shaded region = \(\frac{2}{5}\) × \(\frac{4}{5}\)
= \(\frac{2 × 4}{5 × 5}\)
= \(\frac{8}{25}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{8}{25}\)

Use rectangles with unit fraction side lengths to find the area of the rectangle.
Question 3.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 15
Answer:
The area of the rectangle is: \(\frac{2}{24}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 15
From the given figure,
The area of the rectangle = \(\frac{2}{3}\) × \(\frac{1}{8}\)
= \(\frac{1 × 2}{3 × 8}\)
= \(\frac{2}{24}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{2}{24}\)

Question 4.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 16
Answer:
The area of the rectangle is: \(\frac{35}{90}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 16
From the given figure,
The area of the rectangle = \(\frac{7}{10}\) × \(\frac{5}{9}\)
= \(\frac{7 × 5}{10 × 9}\)
= \(\frac{35}{90}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{35}{90}\)

Question 5.
Find the area of a rectangle with a side length of \(\frac{3}{4}\) and \(\frac{5}{12}\).
Answer:
The area of a rectangle is: \(\frac{15}{48}\)

Explanation:
The given side lengths of a rectangle are: \(\frac{5}{12}\) and \(\frac{3}{4}\)
So,
The area of the rectangle = \(\frac{5}{12}\) × \(\frac{3}{4}\)
= \(\frac{5 × 3}{12 × 4}\)
= \(\frac{15}{48}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{15}{48}\)

Question 6.
Find the area of a square with side lengths of \(\frac{9}{16}\)
Answer:
The area of a square is: \(\frac{81}{256}\)

Explanation:
We know that,
The length of all the sides of the square is equal.
So,
The given side lengths of a square are: \(\frac{9}{16}\) and \(\frac{9}{16}\)
So,
The area of the square = \(\frac{9}{16}\) × \(\frac{9}{16}\)
= \(\frac{9 × 9}{16 × 16}\)
= \(\frac{81}{256}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{81}{256}\)

Question 7.
Open-Ended
The area of a rectangle is \(\frac{16}{24}\). What are the possible side lengths of the rectangle?
Answer:
The possible side lengths for the given area of a rectangle are:
A) \(\frac{4}{2}\) and \(\frac{4}{12}\)
B) \(\frac{1}{3}\) and \(\frac{4}{8}\)

Explanation:
The given area of a rectangle is: \(\frac{16}{24}\)
So,
To find the possible side lengths of a rectangle, find the factors for the numerator and denominator of \(\frac{16}{24}\)
So,
The factors of 16 are: 1, 2, 4, 8, 16
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
So,
The possible side lengths of a rectangle are:
A) \(\frac{4}{2}\) and \(\frac{4}{12}\)
B) \(\frac{1}{3}\) and \(\frac{4}{8}\)
There are so many possible side lengths of a rectangle like above
Hence, from the above,
We can conclude that the possible side lengths of a rectangle are:
A) \(\frac{4}{2}\) and \(\frac{4}{12}\)
B) \(\frac{1}{3}\) and \(\frac{4}{8}\)

Question 8.
Structure
Write an expression that represents the area of the shaded rectangle.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 17
Answer:
The expression that represents the area of the shaded region is: \(\frac{1}{3}\) × \(\frac{1}{3}\)

Explanation:
The given shaded rectangle is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 17
From the given shaded rectangle,
The number of shaded columns is: 3
The number of shaded rows is: 3
The total number of rows is: 1
The total number of columns are: 1
So,
The expression representing the shaded region = \(\frac{The total number of rows}{The number of shaded rows}\) × \(\frac{The total number of rows}{The number of shaded rows}\)
= \(\frac{1}{3}\) × \(\frac{1}{3}\)
Hence, from the above,
We can conclude that the expression representing the shaded region of a rectangle is: \(\frac{1}{3}\) × \(\frac{1}{3}\)

Question 9.
Modeling Real Life
The area of a square table is 9 square feet. Will the board game fit on the table?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 18
Answer:
Yes, the board game fit on the table

Explanation:
The given area of a square table is 9 square feet
So,
The representation of the area of the square table in the fraction form is: \(\frac{36}{4}\)
Now,
The given board game is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 18
The area of the board game can find out by multiplying its side lengths
So,
The area of the board game = \(\frac{3}{2}\) × \(\frac{5}{2}\)
= \(\frac{3 × 5}{2 × 2}\)
= \(\frac{15}{4}\)
So,
Compare the area of the square table and the area of the board game,
We get
Area of the square table > Area of the board game
Hence, from the above,
We can conclude that the board game fit on the table

Review & Refresh

Divide. Then check your answer.
Question 10.
365 ÷ 14 = _______
Answer:
365 ÷ 14 = 26 R 1

Explanation:
By using the partial quotients method,
365 ÷ 14 = ( 280 ÷ 84 ) ÷ 14
= ( 280 ÷ 14 ) + ( 84 ÷ 14 )
= 20 + 6
= 26 R 1
Hence, 365 ÷ 14 = 26 R 1

Question 11.
282 ÷ 27 = ______
Answer:
282 ÷ 27 = 10 R 12

Explanation:
By using the partial quotients method,
282 ÷ 27 = 270 ÷ 27
= 10 R 12
Hence, 270 ÷ 27 = 10 R 12

Question 12.
601 ÷ 72 = _____
Answer:
601 ÷ 72 = 8 R 25

Explanation:
By using the partial quotients method,
601 ÷ 72 = 576 ÷ 72
= 8 R 25
Hence, 601 ÷ 72 = 8 R 25

Lesson 9.7 Multiply Mixed Numbers

Explore and Grow

Find the area of the rectangle. Explain how you found your answer.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 1
Answer:
The area of the rectangle is: \(\frac{35}{4}\)

Explanation:
The given rectangle is:
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 1
From the above rectangle,
The side lengths are: 2\(\frac{1}{2}\) and 3\(\frac{1}{2}\)
The representation of the side lengths in the improper form is: \(\frac{5}{2}\) and \(\frac{7}{2}\)
So,
The area of rectangle = \(\frac{5}{2}\) × \(\frac{7}{2}\)
= \(\frac{5 × 7}{2 × 2}\)
= \(\frac{35}{4}\)
Hence,
2\(\frac{1}{2}\) × 3\(\frac{1}{2}\) = \(\frac{35}{4}\)

Structure
Find the area using a different method. Explain how you found your answer.
Answer:
The given mixed fractions of a rectangle are: 2\(\frac{1}{2}\) and 3\(\frac{1}{2}\)
To convert a mixed fraction into an improper fraction, we have to add the whole number part and the fractional part in the mixed fraction.
So,
2\(\frac{1}{2}\) = 2 + \(\frac{1}{2}\)
= \(\frac{4}{2}\) + \(\frac{1}{2}\)
= \(\frac{5}{2}\)
3\(\frac{1}{2}\) = 3 + \(\frac{1}{2}\)
= \(\frac{6}{2}\) + \(\frac{1}{2}\)
= \(\frac{7}{2}\)
So,
\(\frac{5}{2}\) × \(\frac{7}{2}\) = \(\frac{35}{4}\)

Think and Grow: Multiply Mixed Numbers

You can use a model to find the product of two mixed numbers. You can also write the mixed numbers as improper fractions and then multiply.
Example
Find 1\(\frac{1}{2}\) × 2\(\frac{3}{4}\).
One Way: Use an area model.
Step 1: Write each mixed number as a sum.
1\(\frac{1}{2}\) = 1 + \(\frac{1}{2}\) 2\(\frac{3}{4}\) = 2 + \(\frac{3}{4}\)
Step 2: Draw an area model that represents the product of the sums.
Step 3: Find the sum of the areas of the sections.

Another Way: Write each mixed number as an improper fraction, then multiply.

Show and Grow

Multiply.
Question 1.
2\(\frac{1}{2}\) × 1\(\frac{1}{2}\) = ________
Answer:
2\(\frac{1}{2}\) × 1\(\frac{1}{2}\) = \(\frac{15}{4}\)

Explanation:
The given fractions are: 2\(\frac{1}{2}\) and 1\(\frac{1}{2}\)
The representation of the fractions in the improper form is: \(\frac{5}{2}\) and \(\frac{3}{2}\)
So,
\(\frac{5}{2}\) × \(\frac{3}{2}\)
= \(\frac{5 × 3}{2 × 2}\)
= \(\frac{15}{4}\)
Hence,
2\(\frac{1}{2}\) × 1\(\frac{1}{2}\) = \(\frac{15}{4}\)

Question 2.
3\(\frac{1}{4}\) × 2\(\frac{2}{3}\) = ______
Answer:
3\(\frac{1}{4}\) × 2\(\frac{2}{3}\) = \(\frac{104}{12}\)

Explanation:
The given fractions are: 3\(\frac{1}{4}\) and 2\(\frac{2}{3}\)
The representation of the fractions in the improper form is: \(\frac{13}{4}\) and \(\frac{8}{3}\)
So,
\(\frac{13}{4}\) × \(\frac{8}{3}\)
= \(\frac{13 × 8}{4 × 3}\)
= \(\frac{104}{12}\)
Hence,
3\(\frac{1}{4}\) × 2\(\frac{2}{3}\) = \(\frac{104}{12}\)

Apply and Grow: Practice

Multiply.
Question 3.
1\(\frac{3}{4}\) × 2\(\frac{1}{6}\) = ______
Answer:
1\(\frac{3}{4}\) × 2\(\frac{1}{6}\) = \(\frac{91}{24}\)

Explanation:
The given fractions are: 1\(\frac{3}{4}\) and 2\(\frac{1}{6}\)
The representation of the fractions in the improper form is: \(\frac{7}{4}\) and \(\frac{13}{6}\)
So,
\(\frac{7}{4}\) × \(\frac{13}{6}\)
= \(\frac{13 × 7}{4 × 6}\)
= \(\frac{91}{24}\)
Hence,
1\(\frac{3}{4}\) × 2\(\frac{1}{6}\) = \(\frac{91}{24}\)

Question 4.
4\(\frac{1}{3}\) × 1\(\frac{5}{6}\) = ______
Answer:
4\(\frac{1}{3}\) × 1\(\frac{5}{6}\) = \(\frac{143}{18}\)

Explanation:
The given fractions are: 4\(\frac{1}{3}\) and 1\(\frac{5}{6}\)
The representation of the fractions in the improper form is: \(\frac{13}{3}\) and \(\frac{11}{6}\)
So,
\(\frac{13}{3}\) × \(\frac{11}{6}\)
= \(\frac{13 × 11}{6 × 3}\)
= \(\frac{143}{18}\)
Hence,
4\(\frac{1}{3}\) × 1\(\frac{5}{6}\) = \(\frac{143}{18}\)

Question 5.
3\(\frac{2}{5}\) × 1\(\frac{9}{10}\) = ______
Answer:
3\(\frac{2}{5}\) × 1\(\frac{9}{10}\) = \(\frac{323}{50}\)

Explanation:
The given fractions are: 3\(\frac{2}{5}\) and 1\(\frac{9}{10}\)
The representation of the fractions in the improper form is: \(\frac{17}{5}\) and \(\frac{19}{10}\)
So,
\(\frac{17}{5}\) × \(\frac{19}{10}\)
= \(\frac{17 × 19}{5 × 10}\)
= \(\frac{323}{50}\)
Hence,
3\(\frac{2}{5}\) × 1\(\frac{9}{10}\) = \(\frac{323}{50}\)

Question 6.
2\(\frac{3}{8}\) × 3\(\frac{1}{2}\) = ______
Answer:
2\(\frac{3}{8}\) × 3\(\frac{1}{2}\) = \(\frac{133}{16}\)

Explanation:
The given fractions are: 2\(\frac{3}{8}\) and 3\(\frac{1}{2}\)
The representation of the fractions in the improper form is: \(\frac{19}{8}\) and \(\frac{7}{2}\)
So,
\(\frac{19}{8}\) × \(\frac{7}{2}\)
= \(\frac{19 × 7}{8 × 2}\)
= \(\frac{133}{16}\)
Hence,
2\(\frac{3}{8}\) × 3\(\frac{1}{2}\) = \(\frac{133}{16}\)

Evaluate.
Question 7.
5\(\frac{1}{4}\) × \(\frac{2}{5}\) × 6\(\frac{1}{12}\) = _________
Answer:
5\(\frac{1}{4}\) × \(\frac{2}{5}\) × 6\(\frac{1}{12}\) = \(\frac{3,066}{240}\)

Explanation:
The given fractions are: 5\(\frac{1}{4}\), \(\frac{2}{5}\) and 6\(\frac{1}{12}\)
The representation of the fractions in the improper form is: \(\frac{21}{4}\) and \(\frac{73}{12}\)
So,
\(\frac{21}{4}\) × \(\frac{73}{12}\) × \(\frac{2}{5}\)
= \(\frac{21 × 73 × 2}{4 × 12 × 5}\)
= \(\frac{3,066}{240}\)
Hence,
5\(\frac{1}{4}\) × \(\frac{2}{5}\) × 6\(\frac{1}{12}\) = \(\frac{3,066}{240}\)

Question 8.
3\(\frac{2}{3}\) × (10\(\frac{7}{8}\) – 2\(\frac{1}{4}\)) = _________
Answer:
3\(\frac{2}{3}\) × (10\(\frac{7}{8}\) – 2\(\frac{1}{4}\)) = \(\frac{759}{24}\)

Explanation:
The given fractions are: 3\(\frac{2}{3}\), 10\(\frac{7}{8}\) and 2\(\frac{1}{4}\)
The representation of the fractions in the improper form is: \(\frac{11}{3}\), \(\frac{87}{8}\)  and \(\frac{9}{4}\)
So,
3\(\frac{2}{3}\) × (10\(\frac{7}{8}\) – 2\(\frac{1}{4}\))
= \(\frac{11}{3}\) × ( \(\frac{87}{8}\) – \(\frac{9}{4}\) )
= \(\frac{11}{3}\) × ( \(\frac{87 – 18}{8}\) )
= \(\frac{11}{3}\) × \(\frac{69}{8}\)
= \(\frac{11 × 69}{8 × 3}\)
= \(\frac{759}{24}\)
Hence,
3\(\frac{2}{3}\) × (10\(\frac{7}{8}\) – 2\(\frac{1}{4}\)) = \(\frac{759}{24}\)

Question 9.
YOU BE THE TEACHER
Your friend uses the model to find 4\(\frac{1}{2}\) × 3\(\frac{2}{3}\). Is your friend correct? Explain.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 4
Answer:
No, your friend is not correct

Explanation:
The given fractions are: 4\(\frac{1}{2}\) and 3\(\frac{2}{3}\)
The representation of the side lengths in the improper form is: \(\frac{9}{2}\) and \(\frac{11}{3}\)
So,
The area of rectangle = \(\frac{9}{2}\) × \(\frac{11}{3}\)
= \(\frac{9 × 11}{3 × 2}\)
= \(\frac{99}{6}\)
= \(\frac{33}{2}\)
= 15\(\frac{3}{2}\)
But according to your friend,
4\(\frac{1}{2}\) × 3\(\frac{2}{3}\) = 12\(\frac{1}{2}\)
Hence, from the above,
We can conclude that your friend is not correct.

Question 10.
Logic
Find the missing numbers.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 5
Answer:
The missing numbers are: 1 and \(\frac{8}{5}\)

Explanation:
According to the separation method,
To convert the mixed number into the improper fraction, we can add whole numbers and the fractions separately
Now,
Let the missing numbers be p and q
So,
p + 5 = 6
So,
p = 6 – 5 = 1
Now,
\(\frac{1}{4}\) + \(\frac{1}{q}\) = \(\frac{7}{8}\)
So,
\(\frac{1}{q}\) = \(\frac{7}{8}\) – \(\frac{1}{4}\)
= \(\frac{7}{8}\) – \(\frac{2}{8}\)
= \(\frac{7 – 2}{8}\)
= \(\frac{5}{8}\)
So,
q= \(\frac{8}{5}\)
Hence, from the above,
We can conclude that the missing numbers are: 1 and \(\frac{8}{5}\)

Think and Grow: Modeling Real Life

Example
A construction crew is paving 15 miles of a highway. The crew paves 4\(\frac{2}{10}\) miles each month. Does the crew finish paving the highway in 3\(\frac{1}{2}\) months?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 6
Find the length of the highway the crew paves by multiplying the number of months by the number of miles they pave each month. Write each mixed number as an improper fraction, then multiply.

Compare the length the crew paves to the amount that needs to be paved.
The Crew can finish paving the highway in 3\(\frac{1}{2}\) months.

Show and Grow

Question 11.
You have 3 cups of strawberries. You want to make 1\(\frac{1}{2}\) batches of the recipe. Do you have enough strawberries?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 8
Answer:
Yes, you have enough strawberries

Explanation:
The given recipe for a strawberry smoothie is:
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 8
It is given that you want to make 1\(\frac{1}{2}\) batches of the recipe
From the table,
The number of cups of strawberry is: 1\(\frac{3}{4}\)
Now,
The representation of 1\(\frac{1}{2}\) in the iproper fraction is: \(\frac{3}{2}\)
The representation of 1\(\frac{3}{4}\) in the mixed form is: \(\frac{7}{4}\)
So,
The number of strawberries for the recipe = \(\frac{3}{2}\) × \(\frac{7}{4}\)
= \(\frac{3 × 7}{4 × 2}\)
= \(\frac{21}{8}\)
It is also given that you have 3 cups of strawberries
So,
We can write 3 in the fraction form and as the multiples of 8 is: \(\frac{24}{8}\)
So,
When we compare the number of cups we obtained and given,
We can say that we have enough strawberries to make the recipe
Hence, from the above,
We can conclude that we have enough strawberries

Question 12.
On Monday,you roller-skate 6\(\frac{1}{4}\) miles. On Tuesday, you skate 1\(\frac{2}{5}\) times as far as you did on Monday. How many total miles do you roller-skate on Monday and Tuesday combined?
Answer:
The number of miles you do roller-skate on Monday and Tuesday combined is: 15 miles

Explanation:
It is given that on Monday,you roller-skate 6\(\frac{1}{4}\) miles. On Tuesday, you skate 1\(\frac{2}{5}\) times as far as you did on Monday.
So,
The number of miles you do roller-skate on Tuesday = ( The number of miles you do roller-skate on Monday ) × 1\(\frac{2}{5}\)
= 6\(\frac{1}{4}\) × 1\(\frac{2}{5}\)
= \(\frac{25}{4}\) × \(\frac{7}{5}\)
= \(\frac{25 × 7}{5 × 4}\)
= \(\frac{175}{20}\)
Now,
The number of miles you do roller-skate on Monday and Tuesday combined = 6\(\frac{1}{4}\) + \(\frac{175}{20}\)
= \(\frac{25}{4}\) + \(\frac{175}{20}\)
= \(\frac{125}{20}\) + \(\frac{175}{20}\)
= \(\frac{175 + 125}{20}\)
= \(\frac{300}{20}\)
= 15 miles
Hence, from the above,
We can conclude that the number of miles you do roller- skate on Monday and Tuesday combined is: 15 miles

Question 13.
DIG DEEPER!
An artist paints a rectangular mural. The mural is 4\(\frac{1}{3}\) feet wide. The length is 2\(\frac{1}{4}\) times the width. What is the area of the mural?
Answer:
The area of the mural is: \(\frac{117}{12}\) square feet

Explanation:
It is given that an artist paints a rectangular mural. The mural is 4\(\frac{1}{3}\) feet wide. The length is 2\(\frac{1}{4}\) times the width.
So,
The area of the mural = ( The width of the mural ) × ( The length of the mural )
= 4\(\frac{1}{3}\) × 2\(\frac{1}{4}\)
= \(\frac{13}{3}\) × \(\frac{9}{4}\)
= \(\frac{13 × 9}{4 × 3}\)
= \(\frac{117}{12}\)
Hence, from the above,
We can conclude that the area of the mural is: \(\frac{117}{12}\) square feet

Multiply Mixed Numbers Homework & Practice 9.7

Multiply.
Question 1.
1\(\frac{1}{2}\) × 1\(\frac{1}{8}\) = ______
Answer:
1\(\frac{1}{2}\) × 1\(\frac{1}{8}\) = \(\frac{18}{16}\)

Explanation:
The given fractions are: 1\(\frac{1}{2}\) and 1\(\frac{1}{8}\)
The representation of the fractions in the improper form is: \(\frac{3}{2}\) and \(\frac{9}{8}\)
So,
\(\frac{9}{8}\) × \(\frac{3}{2}\)
= \(\frac{9 × 3}{8 × 2}\)
= \(\frac{18}{16}\)
Hence,
1\(\frac{1}{2}\) × 1\(\frac{1}{8}\) = \(\frac{18}{16}\)

Question 2.
1\(\frac{5}{6}\) × 2\(\frac{1}{4}\) = ______
Answer:
1\(\frac{5}{6}\) × 2\(\frac{1}{4}\) = \(\frac{99}{24}\)

Explanation:
The given fractions are: 1\(\frac{5}{6}\) and 2\(\frac{1}{4}\)
The representation of the fractions in the improper form is: \(\frac{11}{6}\) and \(\frac{9}{4}\)
So,
\(\frac{9}{4}\) × \(\frac{11}{6}\)
= \(\frac{9 × 11}{4 × 6}\)
= \(\frac{99}{24}\)
Hence,
1\(\frac{5}{6}\) × 2\(\frac{1}{4}\) = \(\frac{99}{24}\)

Evaluate.

Question 3.
2\(\frac{3}{8}\) × 2\(\frac{3}{4}\) = ______
Answer:
2\(\frac{3}{8}\) × 2\(\frac{3}{4}\) = \(\frac{209}{32}\)

Explanation:
The given fractions are: 2\(\frac{3}{8}\) and 2\(\frac{3}{4}\)
The representation of the fractions in the improper form is: \(\frac{19}{8}\) and \(\frac{11}{4}\)
So,
\(\frac{19}{8}\) × \(\frac{11}{4}\)
= \(\frac{19 × 11}{4 × 8}\)
= \(\frac{209}{32}\)
Hence,
2\(\frac{3}{8}\) × 2\(\frac{3}{4}\) = \(\frac{209}{32}\)

Question 4.
4\(\frac{1}{6}\) × 3\(\frac{2}{7}\) = ______
Answer:
4\(\frac{1}{6}\) × 3\(\frac{2}{7}\) = \(\frac{575}{42}\)

Explanation:
The given fractions are: 4\(\frac{1}{6}\) and 3\(\frac{2}{7}\)
The representation of the fractions in the improper form is: \(\frac{25}{6}\) and \(\frac{23}{7}\)
So,
\(\frac{23}{7}\) × \(\frac{25}{6}\)
= \(\frac{23 × 25}{7 × 6}\)
= \(\frac{575}{42}\)
Hence,
4\(\frac{1}{6}\) × 3\(\frac{2}{7}\) = \(\frac{575}{42}\)

Question 5.
2\(\frac{1}{3}\) × 3\(\frac{9}{10}\) × 5\(\frac{1}{5}\) = ______
Answer:
2\(\frac{1}{3}\)× 3\(\frac{9}{10}\) ×5\(\frac{1}{5}\) = \(\frac{7,098}{150}\)

Explanation:
The given fractions are: 2\(\frac{1}{3}\), 3\(\frac{9}{10}\) and 5\(\frac{1}{5}\)
The representation of the fractions in the improper form is: \(\frac{7}{3}\), \(\frac{39}{10}\)  and \(\frac{26}{5}\)
So,
\(\frac{7}{3}\) × \(\frac{39}{10}\) ×\(\frac{26}{5}\)
= \(\frac{7 × 39 × 26}{3 × 10 × 5}\)
= \(\frac{7,098}{150}\)
Hence,
2\(\frac{1}{3}\)× 3\(\frac{9}{10}\) ×5\(\frac{1}{5}\) = \(\frac{7,098}{150}\)

Question 6.
(1\(\frac{7}{8}\) + 4\(\frac{4}{5}\)) × 2\(\frac{1}{12}\) = _______
Answer:
(1\(\frac{7}{8}\) + 4\(\frac{4}{5}\)) × 2\(\frac{1}{12}\) = \(\frac{6,675}{480}\)

Explanation:
The given fractions are: 1\(\frac{7}{8}\), 4\(\frac{4}{5}\) and 2\(\frac{1}{12}\)
The representation of the fractions in the improper form is: \(\frac{15}{8}\), \(\frac{24}{5}\) and \(\frac{25}{12}\)
So,
(1\(\frac{7}{8}\) + 4\(\frac{4}{5}\)) × \(\frac{25}{12}\)
= ( \(\frac{15}{8}\) + \(\frac{24}{5}\) ) × \(\frac{25}{12}\)
= ( \(\frac{75}{40}\) + \(\frac{192}{40}\) ) × \(\frac{25}{12}\)
= \(\frac{75 + 192}{40}\) × \(\frac{25}{12}\)
= \(\frac{267}{40}\) × \(\frac{25}{12}\)
= \(\frac{267 × 25}{40× 12 }\)
= \(\frac{6,675}{480}\)
Hence,
(1\(\frac{7}{8}\) + 4\(\frac{4}{5}\)) × 2\(\frac{1}{12}\) = \(\frac{6,675}{480}\)

Question 7.
Structure
Find the missing numbers.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 9
Answer:
The missing numbers are: 5 and \(\frac{3}{10}\)

Explanation:
Let the missing numbers be p and q
By using the partial products method,
( p × 4 ) + ( p × \(\frac{1}{7}\) ) = 20 + \(\frac{5}{7}\)
p × ( 4 + \(\frac{1}{7}\) ) = 20 + \(\frac{5}{7}\)
p × \(\frac{29}{7}\) = \(\frac{145}{7}\)
So,
p = \(\frac{145}{7}\) ÷ \(\frac{29}{7}\)
= \(\frac{145}{7}\) × \(\frac{7}{29}\)
= \(\frac{145 × 7}{7 × 29}\)
= 5
Now,
(q × 4 ) + ( q × \(\frac{1}{7}\) ) = \(\frac{12}{10}\) + \(\frac{3}{70}\)
q × ( 4 + \(\frac{1}{7}\) ) = \(\frac{12}{10}\) + \(\frac{3}{70}\)
q × \(\frac{29}{7}\) = \(\frac{84 + 3}{70}\)
q × \(\frac{29}{7}\) = \(\frac{87}{70}\)
So,
q = \(\frac{87}{70}\) ÷ \(\frac{29}{7}\)
= \(\frac{87}{70}\) × \(\frac{7}{29}\)
= \(\frac{87 × 7}{70 × 29}\)
= \(\frac{3}{10}\)
Hence, from the above,
We can conclude that the missing numbers are: 5 and \(\frac{3}{10}\)

Question 8.
YOU BE THE TEACHER
Your friend finds 1\(\frac{11}{12}\) × 2\(\frac{3}{8}\) . Is your friend correct? Explain.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 10
Answer:
Yes your friend is correct

Explanation:
The given mixed numbers are: 1\(\frac{11}{12}\) and 2\(\frac{3}{8}\)
The representation of the mixed numbers in the fraction form is: \(\frac{23}{12}\) and \(\frac{19}{8}\)
So,
\(\frac{23}{12}\) × \(\frac{19}{8}\)
= \(\frac{23 × 19}{8 × 12}\)
= \(\frac{437}{96}\)
= 4\(\frac{53}{96}\)
Hence, from the above,
We can conclude that your friend is correct

Question 9.
Modeling Real Life
Your friend earns 7\(\frac{1}{2}\) dollars each hour. Will she earn enough money to buy a $35 toy after working 4\(\frac{3}{4}\) hours?
Answer:
she can’t earn enough money to buy $35 toy after working 4\(\frac{3}{4}\) hours

Explanation:
It is given that your friend earns 7\(\frac{1}{2}\) dollars each hour.
It is given that will she earn enough money to buy a $35 toy after working 4\(\frac{3}{4}\) hours
So,
The number of dollars she can earn = 7\(\frac{1}{2}\) × 4\(\frac{3}{4}\)
= \(\frac{15}{2}\) × \(\frac{19}{4}\)
= \(\frac{15 × 19}{4 × 2}\)
= \(\frac{285}{8}\)
Now,
The representation of $35 as the multiple of 8 is: \(\frac{280}{8}\)
So,
By comparing the given money with he money she earned,
We can say that she can’t buy the toy.
Hence, from the above,
We can conclude that she can’t earn enough money to buy a $35 toy after working 4\(\frac{3}{4}\) hours

Question 10.
Modeling Real Life
One class collects 8\(\frac{1}{4}\) pounds of recyclable materials. Another class collects 1\(\frac{1}{2}\) times as many pounds as the first class. How many pounds of recyclable materials do the two classes collect altogether?
Answer:
The number of pounds of recyclable materials the two classes collects together is: \(\frac{231}{8}\) pounds

Explanation:
It is given that one class collects 8\(\frac{1}{4}\) pounds of recyclable materials. Another class collects 1\(\frac{1}{2}\) times as many pounds as the first class.
So,
The number of pounds of recyclable materials collected by another class = 8\(\frac{1}{4}\) × 1\(\frac{1}{2}\)
= \(\frac{33}{4}\) × \(\frac{3}{2}\)
= \(\frac{99}{8}\) pounds
So,
The number of pouds of recyclable materials collected by two classes together = 8\(\frac{1}{4}\) + \(\frac{99}{8}\)
= \(\frac{33}{4}\) + \(\frac{99}{8}\)
= \(\frac{132 + 99}{8}\)
= \(\frac{231}{8}\) pounds
Hence, from the above,
We can conclude that the number of pounds of recyclable materials the two classes collects together is: \(\frac{231}{8}\) pounds

Review & Refresh

Find the product.
Question 11.
6 × 5.7 = ______
Answer:
6 × 5.7 = 34.2

Explanation:
By using the partial products method,
6 × 5.7 = 6 × ( 5 + 0.7 )
= ( 6 × 5 ) + ( 6 × 0.7 )
= 30 + 4.2
= 34.2
Hence, 6 × 5.7 = 34.2

Question 12.
0.84 × 9 = ______
Answer:
0.84 × 9 = 73.17

Explanation:
By using the partial quotients method,
0.84 × 9 = ( 0.81 + 0.03 ) × 9
= ( 0.81 × 9 ) + ( 0.03 × 9 )
= 72.9 + 0.27
= 73.17
Hence, 0.84 × 9 = 73.17

Lesson 9.8 Compare Factors and Products

Explore and Grow

Without calculating, order the rectangles by area from least to greatest. Explain your reasoning.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 1
Answer:
Let the given rectangles be named A), B), C), and D)
So,
The side lengths of A are: 1 and 1\(\frac{1}{2}\)
The side lengths of B are: 1\(\frac{1}{3}\) and \(\frac{11}{12}\)
The side lengths of C are: \(\frac{3}{2}\) and \(\frac{11}{12}\)
The side lengths of D are: \(\frac{11}{10}\) and \(\frac{3}{2}\)
Now,
The area of A is: \(\frac{1}{2}\)
The area of B is: \(\frac{44}{36}\)
The area of C is: \(\frac{33}{24}\)
The area of D is: \(\frac{33}{20}\)
For comparison make the denominators of the areas of the four rectangles equal.
So,
The area of A is: \(\frac{18}{36}\)
The area of B is: \(\frac{44}{36}\)
So, by comparing these 2 areas,
We can say that B is greater
Now,
The area of C is: \(\frac{165}{120}\)
The area of D is: \(\frac{198}{120}\)
So, by comparing these 2 areas,
We can say that D is greater
Hence, from the above,
We can conclude that the order of rectangles by areas from the greatest to the least is: B > D> A > C

Construct Arguments
Explain your strategy to your partner. Compare your strategies.
Answer:
The strategy you followed is:
A) Write the side lengths of the rectangles
B) Find the areas of the four rectangles
C) Make the denominators of all the four areas of rectangles equal
D) Compare the numerators of all the four areas of the four rectangles

Think and Grow: Compare Factors and Products

Key Idea
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
Example
Without calculating, tell whether the product 3\(\frac{1}{8}\) × \(\frac{5}{6}\) is less than, greater than, or equal to or each of its factors.

Show and Grow

Without calculating, tell whether the product is less than, greater than, or equal to each of its factors.
Question 1.
8 × \(\frac{3}{10}\)
Answer:
The value of 8 × \(\frac{3}{10}\) is less than 8

Explanation:
The given numbers are: 8 and \(\frac{3}{10}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{3}{10}\) is less than 1
So,
8 × \(\frac{3}{10}\) is less than 8
Hence, from the above,
We can conclude that the value of 8 × \(\frac{3}{10}\) is less than 8

Question 2.
\(\frac{4}{4}\) × 5\(\frac{2}{3}\)
Answer:
The value of \(\frac{4}{4}\) × 5\(\frac{2}{3}\) is greater than \(\frac{4}{4}\)

Explanation:
The given fractions are: \(\frac{4}{4}\) and 5\(\frac{2}{3}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
We know that,
\(\frac{4}{4}\) is 1
Now,
The representation of 5\(\frac{2}{3}\) in the improper fraction form is: \(\frac{17}{3}\)
So,
\(\frac{17}{3}\) is greater than 1
So,
\(\frac{4}{4}\) × 5\(\frac{2}{3}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{4}{4}\) × 5\(\frac{2}{3}\) is greater than \(\frac{4}{4}\)

Question 3.
\(\frac{4}{3}\) × \(\frac{1}{6}\)
Answer:
The value of \(\frac{4}{3}\) × \(\frac{1}{6}\) is greater than \(\frac{1}{6}\) or less than \(\frac{4}{3}\)

Explanation:
The given fractions are: \(\frac{4}{3}\) and \(\frac{1}{6}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
Now,
\(\frac{4}{3}\) is greater than 1
\(\frac{1}{6}\) is less than 1
So,
\(\frac{4}{3}\) × \(\frac{1}{6}\) is less than \(\frac{4}{3}\)
Hence, from the above,
We can conclude that the value of \(\frac{4}{3}\) × \(\frac{1}{6}\) is greater than \(\frac{1}{6}\) or less than \(\frac{4}{3}\)

Apply and Grow

Without calculating, tell whether the product is less than, greater than, or equal to each of its factors.
Question 4.
\(\frac{1}{4}\) × \(\frac{1}{12}\)
Answer:
The value of \(\frac{1}{4}\) × \(\frac{1}{12}\) is less than 1

Explanation:
The given fractions are: latex]\frac{1}{4}[/latex] and \(\frac{1}{12}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{1}{4}\) is less than 1
\(\frac{1}{12}\) is less than 1
So,
\(\frac{1}{4}\) × \(\frac{1}{12}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{1}{4}\) × \(\frac{1}{12}\) is less than 1

Question 5.
3\(\frac{4}{5}\) × 6\(\frac{7}{8}\)
Answer:
The value of 3\(\frac{4}{5}\) × 6\(\frac{7}{8}\) is greater than 1

Explanation:
The given fractions are: 3\(\frac{4}{5}\) and 6\(\frac{7}{8}\)
The representation of 3\(\frac{4}{5}\) and 6\(\frac{7}{8}\) in the improper fractions form is: \(\frac{19}{5}\) and \(\frac{55}{8}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{19}{5}\) is greater than 1
\(\frac{55}{8}\) is greater than 1
So,
3\(\frac{4}{5}\) × 6\(\frac{7}{8}\) is greater than 1
Hence, from the above,
We can conclude that the value of 3\(\frac{4}{5}\) × 6\(\frac{7}{8}\) is greater than 1

Question 6.
\(\frac{1}{6}\) × \(\frac{10}{10}\)
Answer:
The value of \(\frac{1}{6}\) × \(\frac{10}{10}\) is less than 1

Explanation:
The given fractions are: \(\frac{1}{6}\) and \(\frac{10}{10}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
We know that,
\(\frac{10}{10}\) is 1
So,
\(\frac{1}{6}\) is less than 1
So,
\(\frac{1}{6}\) × \(\frac{10}{10}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{1}{6}\) × \(\frac{10}{10}\) is less than 1

Question 7.
\(\frac{2}{3}\) × 5
Answer:
The value of \(\frac{2}{3}\) × 5 is less than 5

Explanation:
The given fractions are: \(\frac{2}{3}\) and 5
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{2}{3}\) is less than 1
So,
\(\frac{2}{3}\) × 5 is less than 5
Hence, from the above,
We can conclude that the value of \(\frac{2}{3}\) × 5 is less than 5

Question 8.
\(\frac{7}{10}\) × 4\(\frac{8}{9}\)
Answer:
The value of \(\frac{7}{10}\) × 4\(\frac{8}{9}\) is greater than 1

Explanation:
The given fractions are: \(\frac{7}{10}\) and 4\(\frac{8}{9}\)
The representation of 4\(\frac{8}{9}\) in the improper fraction form is: \(\frac{44}{9}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{7}{10}\) is less than 1
\(\frac{44}{9}\) is greater than 1
So,
\(\frac{7}{10}\) × 4\(\frac{8}{9}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{7}{10}\) × 4\(\frac{8}{9}\) is greater than 1

Question 9.
\(\frac{9}{2}\) × 1\(\frac{3}{4}\)
Answer:
The value of \(\frac{9}{2}\) × 1\(\frac{3}{4}\) is greater than 1

Explanation:
The given fractions are: \(\frac{9}{2}\) × 1\(\frac{3}{4}\)
The representation of 1\(\frac{3}{4}\) in the improper fraction form is: \(\frac{7}{4}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{9}{2}\) is greater than 1
\(\frac{7}{4}\) is greater than 1
So,
\(\frac{9}{2}\) × 1\(\frac{3}{4}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{9}{2}\) × 1\(\frac{3}{4}\) is greater than 1

Without calculating, order the products from least to greatest.
Question 10.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 3
Answer:
The order of products from the least to the greatest is:
\(\frac{5}{6}\) × \(\frac{1}{3}\) < \(\frac{5}{6}\) × \(\frac{7}{7}\) < \(\frac{5}{6}\) × 1\(\frac{8}{9}\)

Explanation:
The given products are:
A) \(\frac{5}{6}\) × \(\frac{1}{3}\)
B) \(\frac{5}{6}\) × \(\frac{7}{7}\)
C) \(\frac{5}{6}\) × 1\(\frac{8}{9}\)
So,
In A),
\(\frac{5}{6}\) and \(\frac{1}{3}\) are less than 1
So,
\(\frac{5}{6}\) × \(\frac{1}{3}\) is less than 1
In B),
\(\frac{5}{6}\) is less than 1
\(\frac{7}{7}\) is 1
So,
\(\frac{5}{6}\) × \(\frac{7}{7}\) is less than 1
In C),
\(\frac{5}{6}\) is less than1
The representation of  1\(\frac{8}{9}\) in the improper fraction form is: \(\frac{17}{9}\)
So,
\(\frac{17}{9}\) is greater than 1
So,
\(\frac{5}{6}\) × 1\(\frac{8}{9}\) is greater than 1
Hence, from the above,
We can conclude that the order of the products from the least to the greatest is:
\(\frac{5}{6}\) × \(\frac{1}{3}\) < \(\frac{5}{6}\) × \(\frac{7}{7}\) < \(\frac{5}{6}\) × 1\(\frac{8}{9}\)

Question 11.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 4
Answer:
The order of products from the least to the greatest is:
\(\frac{1}{6}\) × \(\frac{1}{4}\) < \(\frac{1}{10}\) × \(\frac{1}{4}\) < 5\(\frac{7}{10}\) × \(\frac{1}{4}\)

Explanation:
The given products are:
A) \(\frac{1}{6}\) × \(\frac{1}{4}\)
B) \(\frac{1}{10}\) × \(\frac{1}{4}\)
C) \(\frac{1}{4}\) × 5\(\frac{7}{10}\)
So,
In A),
\(\frac{1}{6}\) and \(\frac{1}{4}\) are less than 1
So,
\(\frac{1}{6}\) × \(\frac{1}{4}\) is less than 1
In B),
\(\frac{1}{10}\) is less than 1
\(\frac{1}{4}\) is less than 1
So,
\(\frac{1}{10}\) × \(\frac{1}{4}\) is less than 1
In C),
\(\frac{1}{4}\) is less than1
The representation of  5\(\frac{7}{10}\) in the improper fraction form is: \(\frac{57}{10}\)
So,
\(\frac{57}{10}\) is greater than 1
So,
\(\frac{1}{4}\) × 5\(\frac{7}{10}\) is greater than 1
Hence, from the above,
We can conclude that the order of the products from the least to the greatest is:
\(\frac{1}{6}\) × \(\frac{1}{4}\) < \(\frac{1}{10}\) × \(\frac{1}{4}\) < 5\(\frac{7}{10}\) × \(\frac{1}{4}\)

Question 12.
YOU BE THE TEACHER
Your friend says that \(\frac{1}{2}\) × 8 is half as much as 8. Is your friend correct? Explain.
Answer:
Yes, your friend is correct

Explanation:
The given numbers are: \(\frac{1}{2}\) and 8
So,
\(\frac{1}{2}\) × 8 = \(\frac{1}{2}\) × \(\frac{8}{1}\)
= \(\frac{1 × 8}{2 × 1}\)
= 4
It is also given that according to your friend,
\(\frac{1}{2}\) × 8 is half as much as 8.
So,
8 ÷ 2 = 4
Hence, from the above,
We can conclude that your friend is correct.

Question 13.
DIG DEEPER!
Without calculating, tell whether the product is less than, greater than, or equal to 3\(\frac{3}{4}\). Explain.
\(\left(\frac{1}{2} \times 3 \frac{3}{4}\right)\) × \(\frac{2}{7}\)
Answer:
\(\left(\frac{1}{2} \times 3 \frac{3}{4}\right)\) × \(\frac{2}{7}\) is less than  3\(\frac{3}{4}\)

Explanation:
The representation of 3\(\frac{3}{4}\) in the improper fraction form is: \(\frac{15}{4}\)
Now,
\(\left(\frac{1}{2} \times 3 \frac{3}{4}\right)\) × \(\frac{2}{7}\)
= ( \(\frac{1}{2}\) × \(\frac{15}{4}\) ) × \(\frac{2}{7}\)
= \(\frac{15}{8}\) × \(\frac{2}{7}\)
= \(\frac{2 × 15}{7 × 8}\)
= \(\frac{30}{56}\)
= \(\frac{15}{28}\)
Now,
\(\frac{15}{4}\) is multiplied by \(\frac{7}{7}\)
So,
\(\frac{15}{4}\) = \(\frac{105}{28}\)
So,
When we compare \(\frac{105}{28}\) and \(\frac{15}{28}\)
We will get \(\frac{105}{28}\) is greater
Hence, from the above,
We can conclude that \(\left(\frac{1}{2} \times 3 \frac{3}{4}\right)\) × \(\frac{2}{7}\) is less than  3\(\frac{3}{4}\)

Think and Grow: Modeling Real Life

Example
Men’s shot put competitions use a shot with a mass of 7\(\frac{1}{4}\) kilograms. The mass of a bowling ball is \(\frac{7}{8}\) as much as the mass of the shot. Is the mass of the bowling ball less than, greater than, or equal to the mass of the shot?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 5
Write an expression to represent the mass of the bowling ball.

So, the mass of the bowling ball is greater than the mass of the shot.

Show and Grow

Question 14.
You practice playing the keyboard for 3\(\frac{1}{2}\) hours. Your friend practices playing the keyboard for \(\frac{5}{4}\) as many hours as you. Does your friend practice for fewer hours, more hours, or the same number of hours as you?
Answer:
You practice more hours than your friend

Explanation:
It is given that you practice playing the keyboard for 3\(\frac{1}{2}\) hours. Your friend practices playing the keyboard for \(\frac{5}{4}\) as many hours as you.
So,
Now,
The representation of 3\(\frac{1}{2}\) in the improper fraction form is: \(\frac{7}{2}\)
So,
The number of hours you practice playing the keyboard is: \(\frac{7}{2}\) hours
The number of hours your friend practice playing the keyboard is: \(\frac{5}{4}\) hours
So,
For comparison, make the denominators equal.
So,
Multiply \(\frac{7}{2}\) with \(\frac{2}{2}\)
So,
\(\frac{7}{2}\) = \(\frac{14}{4}\)
By comparing the timings ,
We can say that
\(\frac{14}{4}\) > \(\frac{5}{4}\)
Hence, from the above,
We can conclude that you practice more hours than your friend

Question 15.
The original price of a telescope is $99. The sale price is \(\frac{4}{5}\) of the original price. An astrologist buys the telescope at its sale price and uses a half-off-coupon. What fraction of the original price does the astrologist pay for the telescope?
Answer:
The fraction of the original price the astrologist pays for the telescope is: \(\frac{4}{10}\)

Explanation:
It is given that the original price of a telescope is $99. The sale price is \(\frac{4}{5}\) of the original price. and an astrologist buys the telescope at its sale price and uses a half-off-coupon.
So,
The cost of the telescope that an astrologist bought = ( The \(\frac{4}{5}\)th of the original price ) × ( Half-off -coupon on the \(\frac{4}{5}\)th of the original price )
= \(\frac{4}{5}\) × \(\frac{1}{2}\)
= \(\frac{4 × 1}{5 × 2}\)
= \(\frac{4}{10}\)
Hence, from the above,
We can conclude that the fraction of the original price the astrologist pays for the telescope is: \(\frac{4}{10}\)

Question 16.
The Abraj Al-Bait Clock Tower is \(\frac{6}{10}\) kilometer tall. Zifeng Tower is \(\frac{3}{4}\) as tall as the clock tower. Is Zifeng Tower shorter than, taller than, or the same height as the Abraj Al-Bait Clock Tower? What is the height of each tower in meters?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 7
Answer:
The Abraj al-Bait Clock tower is taller than the Zifeng tower
The height of Abraj al-Bait Clock tower is: \(\frac{24}{40}\)
The height of Zifeng tower is: \(\frac{18}{40}\)

Explanation:
It is given that the Abraj Al-Bait Clock Tower is \(\frac{6}{10}\) kilometer tall and Zifeng Tower is \(\frac{3}{4}\) as tall as the clock tower.
So,
The height of Abraj al-Bait Clock tower is: \(\frac{6}{10}\) kilometer
Now,
The height of Zifeng tower is = \(\frac{6}{10}\) × \(\frac{3}{4}\)
= \(\frac{6 × 3}{10 × 4}\)
= \(\frac{18}{40}\)
Now, for comparison, make the denomonators of both the towers equal.
So,
\(\frac{6}{10}\) is multiplied by \(\frac{4}{4}\)
So,
\(\frac{6}{10}\) = \(\frac{24}{40}\)
By comparing the heights of the two towers,
We can say that
\(\frac{24}{40}\) > \(\frac{18}{40}\)
Hence, from the above,
We can conclude that
The Abraj al-Bait Clock tower is taller than the Zifeng tower
The height of Abraj al-Bait Clock tower is: \(\frac{24}{40}\)
The height of Zifeng tower is: \(\frac{18}{40}\)

Compare Factors and Products Homework & Practice 9.8

Without calculating, tell whether the product is less than, greater than, or equal to each of its factors.
Question 1.
1\(\frac{3}{4}\) × 6
Answer:
The value of 1\(\frac{3}{4}\) × 6 is greater than 6

Explanation:
The given numbers are: 1\(\frac{3}{4}\) and 6
The representation of 1\(\frac{3}{4}\) in the improper fraction form is: \(\frac{7}{4}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{7}{4}\) is greater than 1
So,
1\(\frac{3}{4}\) × 6 is greater than 6
Hence, from the above,
We can conclude that the value of 1\(\frac{3}{4}\) × 6 is greater than 6

Question 2.
\(\frac{5}{12}\) × \(\frac{1}{6}\)
Answer:
The value of \(\frac{5}{12}\) × \(\frac{1}{6}\) is less than 1

Explanation;
The given fractions are: \(\frac{5}{12}\) and \(\frac{1}{6}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{5}{12}\) is less than 1
\(\frac{1}{6}\) is less than 1
So,
\(\frac{5}{12}\) × \(\frac{1}{6}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{5}{12}\) × \(\frac{1}{6}\) is less than 1

Question 3.
\(\frac{2}{7}\) × \(\frac{5}{5}\)
Answer:
The value of \(\frac{2}{7}\) × \(\frac{5}{5}\) is less than 1

Explanation:
The given fractions are: \(\frac{2}{7}\) and \(\frac{5}{5}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{2}{7}\) is less than 1
We know that
\(\frac{5}{5}\) is 1
So,
\(\frac{2}{7}\) × \(\frac{5}{5}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{2}{7}\) × \(\frac{5}{5}\) is less than 1

Question 4.
3\(\frac{4}{5}\) × 2\(\frac{9}{10}\)
Answer:
The value of 3\(\frac{4}{5}\) × 2\(\frac{9}{10}\) is greater than 1

Explanation:
the given mixed fractions are: 3\(\frac{4}{5}\) and 2\(\frac{9}{10}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
The representation of the mixed numbers in the fraction form is: \(\frac{19}{5}\) and \(\frac{29}{10}\)
So,
\(\frac{19}{5}\) is greater than 1
\(\frac{29}{10}\) is greater than
So,
3\(\frac{4}{5}\) × 2\(\frac{9}{10}\) is greater than 1
Hence, from the above,
We can conclude that the value of 3\(\frac{4}{5}\) × 2\(\frac{9}{10}\) is greater than 1

Question 5.
8 × \(\frac{2}{3}\)
Answer:
The value of  8 is greater than \(\frac{2}{3}\)

Explanation:
The given numbers are: 8 and \(\frac{2}{3}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{2}{3}\) is less than 1
So,
8 is greater than \(\frac{2}{3}\)
Hence, from the above,
We can conclude that the value of 8 is greater than \(\frac{2}{3}\)

Question 6.
1\(\frac{7}{8}\) × \(\frac{1}{4}\)
Answer:
The value of 1\(\frac{7}{8}\) × \(\frac{1}{4}\) is less than 1

Explanation:
The given fractions are: 1\(\frac{7}{8}\) and \(\frac{1}{4}\)
The representation of 1\(\frac{7}{8}\) in the fraction form is: \(\frac{15}{8}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{15}{8}\) is greater than 1
\(\frac{1}{4}\) is less than 1
So,
1\(\frac{7}{8}\) × \(\frac{1}{4}\) is less than 1
Hence, from the above,
We can conclude that the value of 1\(\frac{7}{8}\) × \(\frac{1}{4}\) is less than 1

Without calculating, order the products from least to greatest.
Question 7.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 8
Answer:
The order of products from the least to the greatest is:
\(\frac{1}{3}\) × \(\frac{4}{5}\) < \(\frac{1}{3}\) × \(\frac{6}{6}\) < 8\(\frac{2}{9}\) × \(\frac{1}{3}\)

Explanation:
The given products are:
A) \(\frac{1}{3}\) × \(\frac{4}{5}\)
B) \(\frac{1}{3}\) × \(\frac{6}{6}\)
C) 8\(\frac{2}{9}\) × \(\frac{1}{3}\)
So,
In A),
\(\frac{1}{3}\) and \(\frac{4}{5}\) are less than 1
So,
\(\frac{1}{3}\) × \(\frac{4}{5}\) is less than 1
In B),
\(\frac{1}{3}\) is less than 1
\(\frac{6}{6}\) is 1
So,
\(\frac{1}{3}\) × \(\frac{6}{6}\) is less than 1
In C),
\(\frac{1}{3}\) is less than1
The representation of  8\(\frac{2}{9}\) in the improper fraction form is: \(\frac{74}{9}\)
So,
\(\frac{74}{9}\) is greater than 1
So,
\(\frac{1}{3}\) × 8\(\frac{2}{9}\) is greater than 1
Hence, from the above,
We can conclude that the order of the products from the least to the greatest is:
\(\frac{1}{3}\) × \(\frac{4}{5}\) < \(\frac{1}{3}\) × \(\frac{6}{6}\) < 8\(\frac{2}{9}\) × \(\frac{1}{3}\)

Question 8.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 9
Answer:
The order of products from the least to the greatest is:
\(\frac{1}{12}\) × \(\frac{3}{5}\) < 4 × \(\frac{3}{5}\) < 2\(\frac{1}{2}\) × \(\frac{3}{5}\)

Explanation:
The given products are:
A) \(\frac{1}{12}\) × \(\frac{3}{5}\)
B) 4 × \(\frac{3}{5}\)
C) 2\(\frac{1}{2}\) × \(\frac{3}{5}\)
So,
In A),
\(\frac{1}{12}\) and \(\frac{3}{5}\) are less than 1
So,
\(\frac{1}{12}\) × \(\frac{3}{5}\) is less than 1
In B),
\(\frac{3}{5}\) is less than 1
So,
4 × \(\frac{3}{5}\) is less than 4
In C),
\(\frac{3}{5}\) is less than1
The representation of  2\(\frac{1}{2}\) in the improper fraction form is: \(\frac{5}{2}\)
So,
\(\frac{5}{2}\) is greater than 1
So,
\(\frac{3}{5}\) × 2\(\frac{1}{2}\) is greater than 1
Hence, from the above,
We can conclude that the order of the products from the least to the greatest is:
\(\frac{1}{12}\) × \(\frac{3}{5}\) < 4 × \(\frac{3}{5}\) < 2\(\frac{1}{2}\) × \(\frac{3}{5}\)

Question 9.
Logic
Without calculating, use >, <, or = to make the statement true. Explain.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 10
Answer:
The value of 3\(\frac{1}{3}\) × \(\frac{1}{6}\) is less than 3\(\frac{1}{2}\)

Explanation:
The given fractions are: 3\(\frac{1}{3}\), \(\frac{1}{6}\) and 3\(\frac{1}{2}\)
The representations of the 3\(\frac{1}{3}\) and  3\(\frac{1}{2}\)  in the fraction form is: \(\frac{7}{2}\) and \(\frac{10}{3}\)
So,
3\(\frac{1}{3}\) × \(\frac{1}{6}\)
= \(\frac{10}{3}\)  × \(\frac{1}{6}\)
= \(\frac{10}{18}\)
To compare the fractions, equate the denominators
So,
Multiply \(\frac{7}{2}\) by \(\frac{9}{9}\)
= \(\frac{63}{18}\)
So,
By comparison,
We can say that
\(\frac{10}{18}\) < \(\frac{63}{18}\)
Hence, from the above,
We can conclude that the value of 3\(\frac{1}{3}\) × \(\frac{1}{6}\) is less than 3\(\frac{1}{2}\)

Question 10.
Logic
Without calculating, determine which number makes the statement true.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 11
Answer:
Let the missing number be x.
The options for x are given as:
A) 1 B) \(\frac{1}{2}\) and C) 1\(\frac{4}{5}\)
So,
From the given three fractions,
The value of x is: 1\(\frac{4}{5}\)

Explanation:
The given multiplication equation is:
x × 1\(\frac{7}{8}\) is greater than 1\(\frac{7}{8}\)
So,
To find the value of x, there are 3 options. They are:
A) 1 B) \(\frac{1}{2}\) and C) 1\(\frac{4}{5}\)
Now,
Let x = 1\(\frac{4}{5}\)
So,
1\(\frac{4}{5}\) × 1\(\frac{7}{8}\)
= \(\frac{9}{5}\) × \(\frac{15}{8}\)
= \(\frac{27}{8}\)
So, by comparison,
We will get,
\(\frac{27}{8}\) is greater than \(\frac{15}{8}\)
Hence, from the above,
We can conclude that the missing number is: 1\(\frac{4}{5}\)

Question 11.
Reasoning
Why does multiplying by a fraction greater than one result in a product greater than the original number?
Answer:
In the given fraction, if the numerator is greater than the denominator, then the given fraction is greater than 1
So,
When the whole number is multiplied by the fraction which is greater than 1, the result in a product will be greater than the original number i.e, the whole number

Question 12.
Modeling Real Life
You snowboard 1\(\frac{7}{8}\) miles. Your friend snowboards 3\(\frac{2}{3}\) times as far as you. Does your friend snowboard fewer miles, more miles, or the same number of miles as you?
Answer:
The number of miles you snowboarded is less than your friend

Explanation:
It is given that you snowboard 1\(\frac{7}{8}\) miles. Your friend snowboards 3\(\frac{2}{3}\) times as far as you.
So,
The number of miles you snowboard is: 1\(\frac{7}{8}\) miles
So,
The number of miles your friend snowboard is = 1\(\frac{7}{8}\) × 3\(\frac{2}{3}\)
= \(\frac{15}{8}\) × \(\frac{11}{3}\)
= \(\frac{15 × 11}{3 × 8}\)
= \(\frac{165}{24}\) miles
So,
For comparison we have to make the denominators equal.
So,
Multiply \(\frac{15}{8}\) by \(\frac{3}{3}\)
So,
\(\frac{15}{8}\) = \(\frac{45}{24}\)
So,
By comparing, we will get
\(\frac{45}{24}\) is less than \(\frac{165}{24}\) miles
Hence, from the above,
We can conclude that the number of miles you snowboarded is less than your friend

Question 13.
Modeling Real Life
A pet owner has three dogs. The youngest dog weighs \(\frac{1}{4}\) as much as the second oldest dog. The oldest dog weighs 1\(\frac{1}{4}\) as much as the second oldest. The second oldest weighs 20 pounds. Which dog weighs the most? the least?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 12

Answer:
The dog that weighs the most is: The oldest dog
The dog that weighs the least is: The youngest dog

Explanation:
It is given that a pet owner has three dogs. The youngest dog weighs \(\frac{1}{4}\) as much as the second oldest dog. The oldest dog weighs 1\(\frac{1}{4}\) as much as the second oldest. The second oldest weighs 20 pounds.
So,
The weight of the second oldest dog is: 20 pounds
Now,
The weight of the oldest dog = 1\(\frac{1}{4}\) × ( The weight of the oldest second dog )
= 1\(\frac{1}{4}\) × 20
= \(\frac{5}{4}\) × 20
= \(\frac{5 × 20}{4 × 1}\)
= 25 pounds
Now,
The weight of the youngest dog = \(\frac{1}{4}\) × ( The weight of the second oldest dog )
= \(\frac{1}{4}\) × 20
= \(\frac{1 × 20}{4}\)
= 5 pounds
Hence, from the above,
We can conclude that
The dog that weighs the most is: The oldest dog
The dog that weighs the least is: The youngest dog

Review & Refresh

Compare.
Question 14.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 13
Answer:
40.5 is greater than 40.13

Explanation:
The given decimal numbers are: 40.5 and 40.13
For comparison, compare the place value of the given digits
If there are more than 2 digits after the decimal, first compare the tenths position and after that only compare the hundredths position
If we can get the result in the tenths position only, then there is no need for further comparison
Hence, from the above,
We can conclude that 40.5 is greater than 40.13

Question 15.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 14
Answer:
13.90 is equal to 13.9

Explanation:
The given decimal numbers are: 13.90 and 13.9
For comparison, compare the place value of the given digits
If there are more than 2 digits after the decimal, first compare the tenths position and after that only compare the hundredths position
If we can get the result in the tenths position only, then there is no need for further comparison
Hence, from the above,
We can conclude that 13.90 is equal to 13.9

Question 16.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 15
Answer:
32.006 is less than 32.06

Explanation:
The given decimal numbers are: 32.006 and 32.06
For comparison, compare the place value of the given digits
If there are more than 2 digits after the decimal, first compare the tenths position and after that only compare the hundredths position
If we can get the result in the tenths position only, then there is no need for further comparison
Hence, from the above,
We can conclude that 32.006 is less than 32.06

Multiply Fractions Performance Task

Question 1.
You see a rock formation at a national park. The formation has layers that formed millions of years ago when particles settled in the water and became rock. You make a model of the rock formation using \(\frac{3}{16}\) -inch foam sheets.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 1
a. The three types of sedimentary rocks are limestone, sandstone, and shale. Use the number of foam sheets to find the height of each sedimentary rock layer.
Answer:
The height of limestone is: \(\frac{9}{4}\) inches
The height of sandstone is: \(\frac{3}{2}\) inches
The height of shale is: \(\frac{15}{4}\) inches

Explanation:
It is given that the height of each foam sheet is: \(\frac{3}{16}\) -inches
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 1
From the table,
The number of foam sheets for Limestone is: 12
The number of foam sheets for sandstone is: 8
The number of foam sheets for shale is: 20
So,
The height of limestone is: ( The height of each foam sheet ) × ( The total number of foam sheets )
= \(\frac{3}{16}\) × 12
= \(\frac{3}{16}\) × \(\frac{12}{1}\)
= \(\frac{9}{4}\) inches
The height of sandstone is: ( The height of each foam sheet ) × ( The total number of foam sheets )
= \(\frac{3}{16}\) × 8
= \(\frac{3}{16}\) × \(\frac{8}{1}\)
= \(\frac{3}{2}\) inches
The height of shale is: ( The height of each foam sheet ) × ( The total number of foam sheets )
= \(\frac{3}{16}\) × 20
= \(\frac{3}{16}\) × \(\frac{20}{1}\)
= \(\frac{15}{4}\) inches
Hence, from the above,
We can conclude that
The height of limestone is: \(\frac{9}{4}\) inches
The height of sandstone is: \(\frac{3}{2}\) inches
The height of shale is: \(\frac{15}{4}\) inches

b. What is the combined height of the sedimentary rock layers?
Answer:
The combined height of the sedimentary layers is: \(\frac{15}{2}\) inches

Explanation:
The combined height of the sedimentary rock layers = The height of limestone + The height of sandstone + The height of the shale
= \(\frac{9}{4}\) + \(\frac{3}{2}\) + \(\frac{15}{4}\)
= \(\frac{24}{4}\) + \(\frac{3}{2}\)
= \(\frac{12}{2}\) + \(\frac{3}{2}\)
= \(\frac{15}{2}\) inches
Hence, from the above,
We can conclude that the combined height of the sedimentary layers is: \(\frac{15}{2}\) inches

c. Will you use more foam sheets for the granite layers of the shale layers? Explain.
Answer:
We use  more foam sheets for the granite layers

Explanation:
From the above,
The height of the shale layers is: \(\frac{15}{4}\) inches
From the table,
The height of the granite layers is: 3 inches
When we compare the two values,
We can observe that the height of shale layers is greater than the granite layers
Hence, from the above,
We can conclude that we will use more foam sheets for granite layers

d. The height of the topsoil layer is 1\(\frac{1}{4}\) times the height of the sandstone layer. How many foam sheets do you use in the topsoil layer?
Answer:
The height of the topsoil layer is: \(\frac{15}{8}\) inches

Explanation:
From the above,
The height of the sandstone layer is: \(\frac{3}{2}\) inches
So,
The height of topsoil layer = 1\(\frac{1}{4}\) × \(\frac{3}{2}\)
= \(\frac{5}{4}\) × \(\frac{3}{2}\)
= \(\frac{15}{8}\) inches
Hence, from the above,
We can conclude that the height of the topsoil layer is: \(\frac{15}{8}\) inches

e. On your model, 1 inch represents 40 feet. What is the actual height of the rock formation?
Answer:
The actual height of the rock formation is: 495 feet

Explanation:
From the above,
The height of the sedimentary layers is: \(\frac{15}{2}\) inches
The height of the topsoil layer is: \(\frac{15}{8}\) inches
The height of the granite layer is: 3 inches
So,
The combined height of the rock formation = \(\frac{15}{2}\) + \(\frac{15}{8}\) + 3
= \(\frac{99}{8}\) inches
But, it is given that,
1 inch = 40 feet
So,
The combined height of the rock formation in feet = \(\frac{99}{8}\) inches × 40
= 495 feet
Hence, from the above,
We can conclude that the height of the rock formation in feet is: 495 feet

f. Why do you think the rock formation has layers?
Answer:
The rock formation has layers because of the tectonic plates.

Multiply Fractions Activity

Fraction Connection: Multiplication
Directions:
1. Players take turns rolling three dice.
2. On your turn, evaluate the expression indicated by your roll and cover the answer.
3. The first player to get four in a row, horizontally, vertically, or diagonally, wins?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 2
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 3

Multiply Fractions Chapter Practice

9.1 Multiply Whole Numbers by Fractions

Multiply.
Question 1.
5 × \(\frac{1}{2}\) = _______
Answer:
5 × \(\frac{1}{2}\) = \(\frac{5}{2}\)

Explanation:
The given numbers are: 5 and \(\frac{1}{2}\)
So,
5 × \(\frac{1}{2}\)
= \(\frac{5}{1}\) × \(\frac{1}{2}\)
= \(\frac{5 × 1}{1 × 2}\)
= \(\frac{5}{2}\)
Hence,
5 × \(\frac{1}{2}\) = \(\frac{5}{2}\)

Question 2.
2 × \(\frac{7}{10}\) = _______
Answer:
2 × \(\frac{7}{10}\) = \(\frac{14}{10}\)

Explanation:
The given numbers are: 2 and \(\frac{7}{10}\)
So,
2 × \(\frac{7}{10}\)
= \(\frac{2}{1}\) × \(\frac{7}{10}\)
= \(\frac{2 × 7}{1 × 10}\)
= \(\frac{14}{10}\)
Hence,
2 × \(\frac{7}{10}\) = \(\frac{14}{10}\)

Question 3.
9 × \(\frac{5}{8}\) = _______
Answer:
9 × \(\frac{5}{8}\) = \(\frac{45}{8}\)

Explanation:
The given numbers are: 9 and \(\frac{5}{8}\)
So,
9 × \(\frac{5}{8}\)
= \(\frac{9}{1}\) × \(\frac{5}{8}\)
= \(\frac{5 × 9}{1 × 8}\)
= \(\frac{45}{8}\)
Hence,
9 × \(\frac{5}{8}\) = \(\frac{45}{8}\)

Question 4.
6 × \(\frac{71}{100}\) = _______
Answer:
6 × \(\frac{71}{100}\) = \(\frac{426}{100}\)

Explanation:
The given numbers are: 6 and \(\frac{71}{100}\)
So,
6 × \(\frac{71}{100}\)
= \(\frac{6}{1}\) × \(\frac{71}{100}\)
= \(\frac{6 × 71}{1 × 100}\)
= \(\frac{426}{100}\)
Hence,
6 × \(\frac{71}{100}\) = \(\frac{426}{100}\)

Question 5.
4 × \(\frac{8}{5}\) = _______
Answer:
4 × \(\frac{8}{5}\) = \(\frac{32}{5}\)

Explanation:
The given numbers are: 4 and \(\frac{8}{5}\)
So,
4 × \(\frac{8}{5}\)
= \(\frac{4}{1}\) × \(\frac{8}{5}\)
= \(\frac{4 × 8}{1 × 5}\)
= \(\frac{32}{5}\)
Hence,
4 × \(\frac{8}{5}\) = \(\frac{32}{5}\)

Question 6.
7 × \(\frac{5}{3}\) = _______
Answer:
7 × \(\frac{5}{3}\) = \(\frac{35}{3}\)

Explanation:
The given numbers are: 7 and \(\frac{5}{3}\)
So,
7 × \(\frac{5}{3}\)
= \(\frac{7}{1}\) × \(\frac{5}{3}\)
= \(\frac{5 × 7}{1 × 3}\)
= \(\frac{35}{3}\)
Hence,
7 × \(\frac{5}{3}\) = \(\frac{35}{3}\)

9.2 Use Models to Multiply Fractions by Whole Numbers

Multiply. Use a model to help.
Question 7.
\(\frac{2}{5}\) of 20
Answer:
20 × \(\frac{2}{5}\) = 8

Explanation:
The given numbers are: 20 and \(\frac{2}{5}\)
So,
20 × \(\frac{2}{5}\)
= \(\frac{20}{1}\) × \(\frac{2}{5}\)
= \(\frac{20 × 2}{1 × 5}\)
= \(\frac{8}{1}\)
= 8
Hence,
20 × \(\frac{2}{5}\) = 8

Question 8.
\(\frac{1}{6}\) × 12
Answer:
12 × \(\frac{1}{6}\) = 2

Explanation:
The given numbers are: 12 and \(\frac{1}{6}\)
So,
12 × \(\frac{1}{6}\)
= \(\frac{12}{1}\) × \(\frac{1}{6}\)
= \(\frac{12 × 1}{1 × 6}\)
= \(\frac{2}{1}\)
= 2
Hence,
12 × \(\frac{1}{6}\) = 2

Question 9.
\(\frac{1}{3}\) × 6
Answer:
6 × \(\frac{1}{3}\) = 2

Explanation:
The given numbers are: 6 and \(\frac{1}{3}\)
So,
6 × \(\frac{1}{3}\)
= \(\frac{6}{1}\) × \(\frac{1}{3}\)
= \(\frac{6 × 1}{1 × 3}\)
= \(\frac{2}{1}\)
= 2
Hence,
6 × \(\frac{1}{3}\) = 2

Question 10.
\(\frac{5}{16}\) of 8
Answer:
8 × \(\frac{5}{16}\) = \(\frac{5}{2}\)

Explanation:
The given numbers are: 8 and \(\frac{5}{16}\)
So,
8 × \(\frac{5}{16}\)
= \(\frac{8}{1}\) × \(\frac{5}{16}\)
= \(\frac{5 × 8}{1 × 16}\)
= \(\frac{5}{2}\)
Hence,
8 × \(\frac{5}{16}\) = \(\frac{5}{2}\)

Question 11.
Modeling Real Life
You have 24 apples. You use \(\frac{1}{4}\) of them to make a single serving of applesauce. How many apples do you not use?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 11
Answer:
The number of apples you do not use is: 18 apples

Explanation:
It is given that you have 24 apples and you use \(\frac{1}{4}\) of them to make a single serving of applesauce.
So,
The number of apples you used = ( The total number of apples ) × ( The fraction of apples you used to make a single serving of applesauce )
= 24 × \(\frac{1}{4}\)
= \(\frac{24}{1}\) × \(\frac{1}{4}\)
= \(\frac{6}{1}\)
= 6 apples
Now,
The number of apples that do not use = ( The total number of apples ) – ( The number of apples you used )
= 24 – 6
= 18 apples
Hence, from the above,
We can conclude that the number of apples that do not use is: 18 apples

9.3 Multiply Fractions and Whole Numbers

Multiply.
Question 12.
\(\frac{3}{5}\) × 15 = _______
Answer:
15 × \(\frac{3}{5}\) = 9

Explanation:
The given numbers are: 15 and \(\frac{3}{5}\)
So,
15 × \(\frac{3}{5}\)
= \(\frac{15}{1}\) × \(\frac{3}{5}\)
= \(\frac{15 × 3}{1 × 5}\)
= \(\frac{9}{1}\)
= 9
Hence,
15 × \(\frac{3}{5}\) = 9

Question 13.
\(\frac{9}{10}\) × 30 = _______
Answer:
30 × \(\frac{9}{10}\) = 27

Explanation:
The given numbers are: 30 and \(\frac{9}{10}\)
So,
30 × \(\frac{9}{10}\)
= \(\frac{30}{1}\) × \(\frac{9}{10}\)
= \(\frac{30 × 9}{1 × 10}\)
= \(\frac{27}{1}\)
= 27
Hence,
30 × \(\frac{9}{10}\) = 27

Question 14.
48 × \(\frac{3}{4}\) = _______
Answer:
48 × \(\frac{3}{4}\) = 36

Explanation:
The given numbers are: 48 and \(\frac{3}{4}\)
So,
48 × \(\frac{3}{4}\)
= \(\frac{48}{1}\) × \(\frac{3}{4}\)
= \(\frac{48 × 3}{1 × 4}\)
= \(\frac{36}{1}\)
= 36
Hence,
48 × \(\frac{3}{4}\) = 36

Question 15.
11 × \(\frac{5}{9}\) = _______
Answer:
11 × \(\frac{5}{9}\) = \(\frac{55}{9}\)

Explanation:
The given numbers are: 11 and \(\frac{5}{9}\)
So,
11 × \(\frac{5}{9}\)
= \(\frac{11}{1}\) × \(\frac{5}{9}\)
= \(\frac{5 × 11}{1 × 9}\)
= \(\frac{55}{9}\)
Hence,
11 × \(\frac{5}{9}\) = \(\frac{55}{9}\)

Question 16.
\(\frac{1}{6}\) × 19 = _______
Answer:
19 × \(\frac{1}{6}\) = \(\frac{19}{6}\)

Explanation:
The given numbers are: 19 and \(\frac{1}{6}\)
So,
19 × \(\frac{1}{6}\)
= \(\frac{19}{1}\) × \(\frac{1}{6}\)
= \(\frac{19 × 1}{1 × 6}\)
= \(\frac{19}{6}\)
Hence,
19 × \(\frac{1}{6}\) = \(\frac{19}{6}\)

Question 17.
7 × \(\frac{13}{50}\) = _______
Answer:
7 × \(\frac{13}{50}\) = \(\frac{91}{50}\)

Explanation:
The given numbers are: 7 and \(\frac{13}{50}\)
So,
7 × \(\frac{13}{50}\)
= \(\frac{7}{1}\) × \(\frac{13}{50}\)
= \(\frac{7 × 13}{1 × 50}\)
= \(\frac{91}{50}\)
Hence,
7 × \(\frac{13}{50}\) = \(\frac{91}{50}\)

9.4 Use Models to Multiply Fractions

Multiply. Use a model to help.
Question 18.
\(\frac{1}{2}\) × \(\frac{1}{10}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{1}{10}\) = \(\frac{1}{20}\)

Explanation:
The given numbers are: \(\frac{1}{2}\) and \(\frac{1}{10}\)
So,
\(\frac{1}{2}\) × \(\frac{1}{10}\)
= \(\frac{1 × 1}{2 × 10}\)
= \(\frac{1}{20}\)
Hence,
\(\frac{1}{2}\) × \(\frac{1}{10}\) = \(\frac{1}{20}\)

Question 19.
\(\frac{1}{5}\) × \(\frac{1}{9}\) = _______
Answer:
\(\frac{1}{5}\) × \(\frac{1}{9}\) = \(\frac{1}{45}\)

Explanation:
The given numbers are: \(\frac{1}{5}\) and \(\frac{1}{9}\)
So,
\(\frac{1}{5}\) × \(\frac{1}{9}\)
= \(\frac{1 × 1}{5 × 9}\)
= \(\frac{1}{45}\)
Hence,
\(\frac{1}{5}\) × \(\frac{1}{9}\) = \(\frac{1}{45}\)

Question 20.
\(\frac{1}{6}\) × \(\frac{1}{7}\) = _______
Answer:
\(\frac{1}{6}\) × \(\frac{1}{7}\) = \(\frac{1}{42}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and \(\frac{1}{7}\)
So,
\(\frac{1}{6}\) × \(\frac{1}{7}\)
= \(\frac{1 × 1}{6 × 7}\)
= \(\frac{1}{42}\)
Hence,
\(\frac{1}{6}\) × \(\frac{1}{7}\) = \(\frac{1}{42}\)

Question 21.
\(\frac{1}{3}\) × \(\frac{1}{8}\) = _______
Answer:
\(\frac{1}{3}\) × \(\frac{1}{8}\) = \(\frac{1}{24}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and \(\frac{1}{8}\)
So,
\(\frac{1}{3}\) × \(\frac{1}{8}\)
= \(\frac{1 × 1}{3 × 8}\)
= \(\frac{1}{24}\)
Hence,
\(\frac{1}{3}\) × \(\frac{1}{8}\) = \(\frac{1}{24}\)

Question 22.
\(\frac{2}{5}\) × \(\frac{1}{3}\) = _______
Answer:
\(\frac{2}{5}\) × \(\frac{1}{3}\) = \(\frac{2}{15}\)

Explanation:
The given numbers are: \(\frac{2}{5}\) and \(\frac{1}{3}\)
So,
\(\frac{2}{5}\) × \(\frac{1}{3}\)
= \(\frac{2 × 1}{5 × 3}\)
= \(\frac{2}{15}\)
Hence,
\(\frac{2}{5}\) × \(\frac{1}{3}\) = \(\frac{2}{15}\)

Question 23.
\(\frac{2}{3}\) × \(\frac{3}{5}\) = _______
Answer:
\(\frac{2}{3}\) × \(\frac{3}{5}\) = \(\frac{2}{5}\)

Explanation:
The given numbers are: \(\frac{2}{3}\) and \(\frac{3}{5}\)
So,
\(\frac{2}{3}\) × \(\frac{3}{5}\)
= \(\frac{2 × 3}{3 × 5}\)
= \(\frac{2}{5}\)
Hence,
\(\frac{2}{3}\) × \(\frac{3}{5}\) = \(\frac{2}{5}\)

9.5 Multiply Fractions

Evaluate.
Question 24.
\(\frac{1}{3}\) × \(\frac{1}{8}\) = _______
Answer:
\(\frac{1}{3}\) × \(\frac{1}{8}\) = \(\frac{1}{24}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and \(\frac{1}{8}\)
So,
\(\frac{1}{3}\) × \(\frac{1}{8}\)
= \(\frac{1 × 1}{3 × 8}\)
= \(\frac{1}{24}\)
Hence,
\(\frac{1}{3}\) × \(\frac{1}{8}\) = \(\frac{1}{24}\)

Question 25.
\(\frac{5}{6}\) × \(\frac{1}{4}\) = _______
Answer:
\(\frac{5}{6}\) × \(\frac{1}{4}\) = \(\frac{5}{24}\)

Explanation:
The given numbers are: \(\frac{5}{6}\) and \(\frac{1}{4}\)
So,
\(\frac{5}{6}\) × \(\frac{1}{4}\)
= \(\frac{5 × 1}{6 × 4}\)
= \(\frac{5}{24}\)
Hence,
\(\frac{5}{6}\) × \(\frac{1}{4}\) = \(\frac{5}{24}\)

Question 26.
\(\frac{7}{2}\) × \(\frac{2}{5}\) = _______
Answer:
\(\frac{7}{2}\) × \(\frac{2}{5}\) = \(\frac{7}{5}\)

Explanation:
The given numbers are: \(\frac{7}{2}\) and \(\frac{2}{5}\)
So,
\(\frac{7}{2}\) × \(\frac{2}{5}\)
= \(\frac{7 × 2}{2 × 5}\)
= \(\frac{7}{5}\)
Hence,
\(\frac{7}{2}\) × \(\frac{2}{5}\) = \(\frac{7}{5}\)

Question 27.
\(\frac{9}{10}\) × \(\frac{3}{7}\) = _______
Answer:
\(\frac{9}{10}\) × \(\frac{3}{7}\) = \(\frac{27}{70}\)

Explanation:
The given numbers are: \(\frac{9}{10}\) and \(\frac{3}{7}\)
So,
\(\frac{9}{10}\) × \(\frac{3}{7}\)
= \(\frac{9 × 3}{7 × 10}\)
= \(\frac{27}{70}\)
Hence,
\(\frac{9}{10}\) × \(\frac{3}{7}\) = \(\frac{27}{70}\)

Question 28.
\(\frac{4}{5}\) × \(\frac{13}{100}\) = _______
Answer:
\(\frac{4}{5}\) × \(\frac{13}{100}\) = \(\frac{52}{500}\)

Explanation:
The given numbers are: \(\frac{4}{5}\) and \(\frac{13}{100}\)
So,
\(\frac{4}{5}\) × \(\frac{13}{100}\)
= \(\frac{4 × 13}{5 × 100}\)
= \(\frac{52}{500}\)
Hence,
\(\frac{4}{5}\) × \(\frac{13}{100}\) = \(\frac{52}{500}\)

Question 29.
\(\frac{11}{25}\) × \(\frac{3}{4}\) = _______
Answer:
\(\frac{11}{25}\) × \(\frac{3}{4}\) = \(\frac{33}{100}\)

Explanation:
The given numbers are: \(\frac{11}{25}\) and \(\frac{3}{4}\)
So,
\(\frac{11}{25}\) × \(\frac{3}{4}\)
= \(\frac{11 × 3}{25 × 4}\)
= \(\frac{33}{100}\)
Hence,
\(\frac{11}{25}\) × \(\frac{3}{4}\) = \(\frac{33}{100}\)

Question 30.
9 × \(\left(\frac{2}{9} \times \frac{1}{2}\right)\) = _______
Answer:
9 × ( \(\frac{1}{2}\) × \(\frac{2}{9}\) ) = 1

Explanation:
The given numbers are: 9, \(\frac{1}{2}\) and \(\frac{2}{9}\)
So,
9 × (\(\frac{1}{2}\) × \(\frac{2}{9}\) )
= 9 × (\(\frac{1 × 2}{2 × 9}\) )
=  9 × \(\frac{1}{9}\)
= 1
Hence,
9 × ( \(\frac{1}{2}\) × \(\frac{2}{9}\) ) = 1

Question 31.
\(\left(\frac{1}{10}+\frac{7}{10}\right)\) × \(\frac{3}{4}\) = _______
Answer:
\(\left(\frac{1}{10}+\frac{7}{10}\right)\) × \(\frac{3}{4}\) = \(\frac{3}{5}\)

Explanation:
The given fractions are: \(\frac{1}{10}\), \(\frac{7}{10}\) and \(\frac{3}{4}\)
So,
\(\frac{3}{4}\) × (\(\frac{1}{10}\) + \(\frac{7}{10}\) )
= \(\frac{3}{4}\) × (\(\frac{7 × 1}{10}\) )
=  \(\frac{3}{4}\) × \(\frac{8}{10}\)
= \(\frac{3 × 8}{4 × 10}\)
= \(\frac{24}{40}\)
= \(\frac{3}{5}\)
Hence,
\(\left(\frac{1}{10}+\frac{7}{10}\right)\) × \(\frac{3}{4}\) = \(\frac{3}{5}\)

Question 32.
\(\frac{4}{7}\) × \(\left(\frac{5}{8}-\frac{1}{4}\right)\) = _______
Answer:
\(\frac{4}{7}\) × \(\left(\frac{5}{8}-\frac{1}{4}\right)\) = \(\frac{3}{14}\)

Explanation:
The given fractions are: \(\frac{4}{7}\), \(\frac{5}{8}\) and \(\frac{1}{4}\)
So,
\(\frac{4}{7}\) × (\(\frac{5}{8}\) – \(\frac{1}{4}\) )
= \(\frac{4}{7}\) × (\(\frac{5 – 2}{8}\) )
=  \(\frac{4}{7}\) × \(\frac{3}{8}\)
= \(\frac{3 × 4}{7 × 8}\)
= \(\frac{12}{56}\)
= \(\frac{3}{14}\)
Hence,
\(\frac{4}{7}\) × \(\left(\frac{5}{8}-\frac{1}{4}\right)\) = \(\frac{3}{14}\)

9.6 Find Areas of Rectangles

Use rectangles with unit fraction side lengths to find the area of the rectangle.
Question 33.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 33
Answer:
The area of the rectangle is: \(\frac{1}{4}\)

Explanation:
The given figure is:
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 33
From the given figure,
The side lengths of the rectangle are: \(\frac{3}{8}\) and \(\frac{2}{3}\)
So,
The area of the rectangle = \(\frac{3}{8}\) × \(\frac{2}{3}\)
= \(\frac{2 × 3}{3 × 8}\)
= \(\frac{1}{4}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{1}{4}\)

Question 34.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 34
Answer:
The area of the rectangle is: \(\frac{35}{48}\)

Explanation:
The given figure is:
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 34
From the given figure,
The side lengths of the rectangle are: \(\frac{7}{12}\) and \(\frac{5}{4}\)
So,
The area of the rectangle = \(\frac{7}{12}\) × \(\frac{5}{4}\)
= \(\frac{7 × 5}{12 × 4}\)
= \(\frac{35}{48}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{35}{48}\)

Question 35.
Find the area of a rectangle with side lengths of \(\frac{2}{9}\) and \(\frac{1}{10}\).
Answer:
The area of the rectangle is: \(\frac{1}{45}\)

Explanation:
The given side lengths of a rectangle are: \(\frac{2}{9}\) and \(\frac{1}{10}\)
So,
The area of the rectangle = \(\frac{2}{9}\) × \(\frac{1}{10}\)
= \(\frac{2 × 1}{10 × 9}\)
= \(\frac{1}{45}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{1}{45}\)

Question 36.
Find the area of a with a side length of \(\frac{3}{4}\).
Answer:
The area of the square is: \(\frac{9}{16}\)

Explanation:
The given side length of a square is: \(\frac{3}{4}\)
We know that,
The length of all sides of a square are equal
So,
the area of the square = \(\frac{3}{4}\) × \(\frac{3}{4}\)
= \(\frac{3 × 3}{4 × 4}\)
= \(\frac{9}{16}\)
Hence, from the above,
We can conclude that the area of the square is: \(\frac{9}{16}\)

9.7 Multiply Mixed Numbers

Multiply.
Question 37.
1\(\frac{1}{4}\) × 2\(\frac{1}{4}\) = _______
Answer:
1\(\frac{1}{4}\) × 2\(\frac{1}{4}\) =\(\frac{45}{16}\)

Explanation:
The given mixed fractions are: 1\(\frac{1}{4}\) and 2\(\frac{1}{4}\)
The representation of 1\(\frac{1}{4}\) and 2\(\frac{1}{4}\) in the improper fraction form is: \(\frac{5}{4}\) and \(\frac{9}{4}\)
So,
\(\frac{5}{4}\) × \(\frac{9}{4}\)
= \(\frac{5 × 9}{4 × 4}\)
= \(\frac{45}{16}\)
Hence,
1\(\frac{1}{4}\) × 2\(\frac{1}{4}\) =\(\frac{45}{16}\)

Question 38.
3\(\frac{4}{5}\) × 1\(\frac{1}{2}\) = _______
Answer:
3\(\frac{4}{5}\) × 1\(\frac{1}{2}\) = \(\frac{57}{10}\)

Explanation:
The given mixed fractions are: 3\(\frac{4}{5}\) and 1\(\frac{1}{2}\)
The representation of 3\(\frac{4}{5}\) and 1\(\frac{1}{2}\) in the improper fraction form is: \(\frac{19}{5}\)  and \(\frac{3}{2}\)
So,
\(\frac{19}{5}\) × \(\frac{3}{2}\)
= \(\frac{19 × 3}{5 × 2}\)
= \(\frac{57}{10}\)
Hence,
3\(\frac{4}{5}\) × 1\(\frac{1}{2}\) = \(\frac{57}{10}\)

Question 39.
5\(\frac{1}{3}\) × 4\(\frac{7}{8}\) = _______
Answer:
5\(\frac{1}{3}\) × 4\(\frac{7}{8}\) = \(\frac{624}{24}\)

Explanation:
The given mixed fractions are: 5\(\frac{1}{3}\) and 4\(\frac{7}{8}\)
The representation of 5\(\frac{1}{3}\) and 4\(\frac{7}{8}\) in the improper fraction form is: \(\frac{16}{3}\)  and \(\frac{39}{8}\)
So,
\(\frac{16}{3}\) × \(\frac{39}{8}\)
= \(\frac{39 × 16}{3 × 8}\)
= \(\frac{624}{24}\)
Hence,
5\(\frac{1}{3}\) × 4\(\frac{7}{8}\) = \(\frac{624}{24}\)

Question 40.
4\(\frac{5}{6}\) × 2\(\frac{9}{10}\) × \(\frac{1}{8}\) = _______
Answer:
4\(\frac{5}{6}\) × 2\(\frac{9}{10}\) × \(\frac{1}{8}\) = \(\frac{841}{480}\)

Explanation:
The given mixed fractions are: 4\(\frac{5}{6}\), 2\(\frac{9}{10}\) and \(\frac{1}{8}\)
The representation of 4\(\frac{5}{6}\) and 2\(\frac{9}{10}\) in the improper fraction form is: \(\frac{29}{6}\) and \(\frac{29}{10}\)
So,
\(\frac{29}{6}\) × \(\frac{29}{10}\)  × \(\frac{1}{8}\)
= \(\frac{29 × 29 × 1}{8 × 10 × 6}\)
= \(\frac{841}{480}\)
Hence,
4\(\frac{5}{6}\) × 2\(\frac{9}{10}\) × \(\frac{1}{8}\) = \(\frac{841}{480}\)

Question 41.
Logic
Find the missing numbers.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 41
Answer:
The missing numbers are: 2 and 11

Explanation:
Let the missing numbers be: p and q
So,
4\(\frac{1}{p}\) × 2\(\frac{1}{5}\) = 9\(\frac{q}{10}\)
So,
To solve the mixed numbers, we can separate the whole numbers and the fraction numbers
So,
\(\frac{1}{p}\) × 2\(\frac{1}{5}\) = \(\frac{q}{10}\)
\(\frac{1}{p}\) × \(\frac{11}{5}\) = \(\frac{q}{10}\)
\(\frac{1 × 11}{p × 5}\) = \(\frac{q}{10}\)
\(\frac{11}{5p}\) = \(\frac{q}{10}\)
By comparing LHS and RHS
q = 11 and p = 2
Hence, from the above,
We can conclude that the missing numbers are: 2 and 11

9.8 Compare Factors and Products

Without calculating, tell whether the product is less than, greater than, or equal to each of its factors.
Question 42.
1\(\frac{1}{2}\) × 3
Answer:
The value of 1\(\frac{1}{2}\) × 3 is greater than 3

Explanation:
The given numbers are: 3 and 1\(\frac{1}{2}\)
The representation of 1\(\frac{1}{2}\) in the mixed form is: \(\frac{3}{2}\)
So,
\(\frac{3}{2}\) is greater than 1
So,
\(\frac{3}{2}\) × 3 is greater than 3
Hence, from the above,
We can conclude that the value of 1\(\frac{1}{2}\) × 3 is greater than 3

Question 43.
\(\frac{1}{8}\) × \(\frac{1}{6}\)
Answer:
The value of \(\frac{1}{8}\) × \(\frac{1}{6}\) is less than 1

Explanation:
The given numbers are: \(\frac{1}{8}\) and \(\frac{1}{6}\)
So,
\(\frac{1}{8}\) is less than 1
\(\frac{1}{6}\) is less than 1
So,
\(\frac{1}{8}\) × \(\frac{1}{6}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{1}{8}\) × \(\frac{1}{6}\) is less than 1

Question 44.
\(\frac{5}{5}\) × 2\(\frac{4}{7}\)
Answer:
The value of \(\frac{5}{5}\) × 2\(\frac{4}{7}\) is greater than 1

Explanation:
The given numbers are: \(\frac{5}{5}\) and 2\(\frac{4}{7}\)
The representation of 2\(\frac{4}{7}\) in the improper fraction form is: \(\frac{18}{7}\)
So,
\(\frac{18}{7}\) is greater than 1
\(\frac{5}{5}\) is 1
So,
\(\frac{5}{5}\) × 2\(\frac{4}{7}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{5}{5}\) × 2\(\frac{4}{7}\) is greater than 1

Question 45.
\(\frac{3}{4}\) × 2\(\frac{11}{12}\)
Answer:
The value of \(\frac{3}{4}\) × 2\(\frac{11}{12}\) is greater than 1

Explanation:
The given numbers are: \(\frac{3}{4}\) and 2\(\frac{11}{12}\)
The representation of 2\(\frac{11}{12}\) in the improper fraction form is: \(\frac{35}{12}\)
So,
\(\frac{35}{12}\) is greater than 1
\(\frac{3}{4}\) is less than 1
So,
\(\frac{3}{4}\) × 2\(\frac{11}{12}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{3}{4}\) × 2\(\frac{11}{12}\) is greater than 1

Question 46.
3\(\frac{1}{9}\) × 2\(\frac{7}{8}\)
Answer:
The value of 3\(\frac{1}{9}\) × 2\(\frac{7}{8}\) is greater than 1

Explanation:
The given mixed fractions are: 3\(\frac{1}{9}\) and 2\(\frac{7}{8}\)
The representation of 3\(\frac{1}{9}\) and 2\(\frac{7}{8}\) in the improper fractions form is: \(\frac{28}{9}\) and \(\frac{23}{8}\)
So,
\(\frac{28}{9}\) is greater than 1
\(\frac{23}{8}\) is greater than 1
So,
3\(\frac{1}{9}\) × 2\(\frac{7}{8}\) is greater than 1
Hence, from the above,
We can conclude that the value of 3\(\frac{1}{9}\) × 2\(\frac{7}{8}\) is greater than 1

Question 47.
\(\frac{9}{8}\) × \(\frac{5}{2}\)
Answer:
The value of \(\frac{9}{8}\) × \(\frac{5}{2}\) is greater than 1

Explanation:
The given fractions are: \(\frac{9}{8}\) and \(\frac{5}{2}\)
So,
\(\frac{9}{8}\) is greater than 1
So,
\(\frac{9}{8}\) × \(\frac{5}{2}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{9}{8}\) × \(\frac{5}{2}\) is greater than 1

Conclusion:

Hope the information provided in the above answer key is beneficial for all the students of grade 5. Follow our Big Ideas Math Answers and make your learning fun and interesting.