Eureka Math

EngageNY Pre K Math Module 1 Answer Key | Pre K Eureka Math Module 1 Answer Key

Eureka Math Pre K Module 1 Counting to 5

Eureka Math Pre K Module 1 Topic A How Many Questions with up to 7 Objects

• Eureka Math Pre K Module 1 Lesson 1 Answer Key
• Eureka Math Pre K Module 1 Lesson 2 Answer Key
• Eureka Math Pre K Module 1 Lesson 3 Answer Key
• Eureka Math Pre K Module 1 Lesson 4 Answer Key

Engage NY Math Pre K Module 1 Topic B Sorting

• Eureka Math Pre K Module 1 Lesson 5 Answer Key
• Eureka Math Pre K Module 1 Lesson 6 Answer Key
• Eureka Math Pre K Module 1 Lesson 7 Answer Key

Pre K Eureka Math Module 1 Topic C How Many Questions with 1, 2, or 3 Objects

• Eureka Math Pre K Module 1 Lesson 8 Answer Key
• Eureka Math Pre K Module 1 Lesson 9 Answer Key
• Eureka Math Pre K Module 1 Lesson 10 Answer Key
• Eureka Math Pre K Module 1 Lesson 11 Answer Key

EngageNY Pre K Math Module 1 Topic D Matching 1 Numeral with up to 3 Objects

• Eureka Math Pre K Module 1 Lesson 12 Answer Key
• Eureka Math Pre K Module 1 Lesson 13 Answer Key
• Eureka Math Pre K Module 1 Lesson 14 Answer Key

Eureka Math Pre K Module 1 Mid Module Assessment Answer Key

Great Minds Eureka Math Pre K Module 1 Topic E How Many Questions with 4 or 5 Objects

• Eureka Math Pre K Module 1 Lesson 15 Answer Key
• Eureka Math Pre K Module 1 Lesson 16 Answer Key
• Eureka Math Pre K Module 1 Lesson 17 Answer Key
• Eureka Math Pre K Module 1 Lesson 18 Answer Key
• Eureka Math Pre K Module 1 Lesson 19 Answer Key
• Eureka Math Pre K Module 1 Lesson 20 Answer Key

Engage NY Pre K Module 1 Topic F Matching 1 Numeral with up to 5 Objects

• Eureka Math Pre K Module 1 Lesson 21 Answer Key
• Eureka Math Pre K Module 1 Lesson 22 Answer Key
• Eureka Math Pre K Module 1 Lesson 23 Answer Key
• Eureka Math Pre K Module 1 Lesson 24 Answer Key
• Eureka Math Pre K Module 1 Lesson 25 Answer Key
• Eureka Math Pre K Module 1 Lesson 26 Answer Key
• Eureka Math Pre K Module 1 Lesson 27 Answer Key

Eureka Math Pre K Module 1 Topic G One More with Numbers 1 to 5

• Eureka Math Pre K Module 1 Lesson 28 Answer Key
• Eureka Math Pre K Module 1 Lesson 29 Answer Key
• Eureka Math Pre K Module 1 Lesson 30 Answer Key
• Eureka Math Pre K Module 1 Lesson 31 Answer Key
• Eureka Math Pre K Module 1 Lesson 32 Answer Key

Engage NY Math Pre K Module 1 Topic H Counting 5, 4, 3, 2, 1

• Eureka Math Pre K Module 1 Lesson 33 Answer Key
• Eureka Math Pre K Module 1 Lesson 34 Answer Key
• Eureka Math Pre K Module 1 Lesson 35 Answer Key
• Eureka Math Pre K Module 1 Lesson 36 Answer Key
• Eureka Math Pre K Module 1 Lesson 37 Answer Key

Eureka Math Pre K Module 1 End of Module Assessment Answer Key

Engage NY Eureka Math 6th Grade Module 1 Lesson 14 Answer Key

Eureka Math Grade 6 Module 1 Lesson 14 Example Answer Key

Example 1.
Dinner service starts once the train is 250 miles away from Yonkers. What is the minimum time the players will have to wait before they can have their meal?

Answer:


The minimum time is 5 hours.

Eureka Math Grade 6 Module 1 Lesson 14 Exercise Answer Key

Exercise 1.
Create a table to show the time it will take Kelli and her team to travel from Yonkers to each town listed in the
schedule assuming that the ratio of the amount of time traveled to the distance traveled is the same for each city.
Then, extend the table to include the cumulative time it will take to reach each destination on the ride home.

Answer:

Exercise 2.
Create a double number line diagram to show the time It will take Kelli and her team to travel from Yonkers to each town listed in the schedule. Then, extend the double number line diagram to include the cumulative time it will take to reach each destination on the ride home. Represent the ratio of the distance traveled on the round trip to the amount of time taken with an equation.

Answer:

Using the information from the double number line diagram, how many miles would be traveled in one hour?
Answer:
50

How do you know?
Answer:
If the train is moving at a constant speed, half of 2 hours is 1 hour, and half of 100 miles is 50 miles.

Eureka Math Grade 6 Module 1 Lesson 14 Problem Set Answer Key

Question 1.
Complete the table of values to find the following:
Find the number of cups of sugar needed if for each pie Karrie makes, she has to use 3 cups of sugar.

Answer:

Use a graph to represent the relationship

Answer:

Create a double number line diagram to show the relationship.

Answer:

Question 2.
Write a story context that would be represented by the ratio 1: 4.
Answer:
Answers will vary. Example: Kendra’s mom pays her four dollars for every load of laundry she washes and dries.

Complete a table of values for this equation and graph.

Answer:

Answer:

Eureka Math Grade 6 Module 1 Lesson 14 Exit Ticket Answer Key

Dominic works on the weekends and on vacations from school mowing lawns in his neighborhood. For every lawn he mows, he charges $12. Complete the table. Then determine ordered pairs, and create a labeled graph.

Answer:

Question 1.
How many lawns wIll Dominic need to mow In order to make $240?
Answer:
20 lawns

Question 2.
How much money wIll Dominic make If he mows 9 lawns?
Answer:
$108

Engage NY Eureka Math 6th Grade Module 1 Lesson 13 Answer Key

Eureka Math Grade 6 Module 1 Lesson 13 Exercise Answer Key

Exercise 1
Jorge is mixing a special shade of orange paint. He mixed 1 gallon of red paint with 3 gallons of yellow paint. Based on this ratio, which of the following statements are true?

→ \(\frac{3}{4}\) of a 4 gallon mix would be yellow paint.
Answer:
True

→ Every 1 gallon of yellow paint requires \(\frac{1}{3}\) gallon of red paint. .
Answer:
True

→ Every 1 gallon of red paint requires 3 gallons of yellow paint.
Answer:
True

→ There is 1 gallon of red paint in a 4-gallon mix of orange paint,
Answer:
True

→ There are 2 gallons of yellow paint in an 8-gallon mix of orange paint.
Answer:
False

Use the space below to determine If each statement is true or false.

Answer:
Allow students to discuss each question with a partner or group. When the class comes back together as a whole group, each group is responsible for explaining to the class one of the statements and whether the group feels the statement is true or false and why. (The first four statements are true while the fifth statement is false. To be made true, the fifth statement should read “There are 6 gallons of yellow paint in an 8 galIon mix of orange paint.”)

Exercise 2.
Based on the information on red and yellow paint given in Exercise 1, complete the table below.

Answer:

Exercise 3
a. Jorge now plans to mix red paint and blue paint to create purple paint. The color of purple he has decided to make combines red paint and blue paint in the ratio 4: 1. If Jorge can only purchase paint in one gallon containers, construct a ratio table for all possible combinations for red and blue paint that will give Jorge no more than 25 gallons of purple paint.

Answer:

Write an equation that will let Jorge calculate the amount of red paint he will need for any given amount of blue paint.
Answer:
R = 4B

Write an equation that will let Jorge calculate the amount of blue paint he will need for any given amount o
red paint.
Answer:
B = \(\frac{1}{4}\)R

If Jorge has 24 gallons of red paint, how much blue paint will he have to use to create the desired color of purple?
Answer:
Jorge will have to use 6 gallons of blue paint.

If Jorge has 24 gallons of blue paint, how much red paint will he have to use to create the desired color of purple?
Answer:
Jorge will have to use 96 gallons of red paint.

b. Using the same relationship of red to blue from above, create a table that models the relationship of the three colors blue, red, and purple (total) paint. Let B represent the number of gallons of blue paint, let R represent the number of gallons of red paint, and let T represent the total number of gallons of (purple) paint. Then write an equation that models the relationship between the blue paint and the total amount of paint, and answer the questions.

Answer:

Equation:
Answer:
T = 5B

Value of the ratio of total paint to blue paint:
Answer:
\(\frac{5}{1}\)

How is the value of the ratio related to the equation?
Answer:
The value of the ratio is used to determine the total paint value by multiplying it with the blue paint value.

Exercise 4.
During a particular U.S. Air Force training exercise, the ratio of the number of men to the number of women was 6: 1. Use the ratio table provided below to create at least two equations that model the relationship between the number of men and the number of women participating in this training exercise.

Answer:

Equations:
Answer:
M = 6w
W = \(\left(\frac{1}{6}\right)\)M
\(\frac{M}{W}\) = 6
\(\frac{W}{M}\) = \(\frac{1}{6}\)

If 200 women participated in the training exercise, use one of your equations to calculate the number of men who participated.
Answer:
I can substitute 200 for the value of women and multiply by 6, the value of the ratio, to get the number of men. There would be 1,200 men participating in the training exercise.

Exercise 5.
Malia is on a road trip. During the first five minutes of Malia’s trip, she sees 18 cars and 6 trucks. Assuming this ratio of cars to trucks remains constant over the duration of the trip, complete the ratio table using this comparison. Let T represent the number of trucks she sees, and let C represent the number of cars she sees.

Answer:

What is the value of the ratio of the number of cars to the number of trucks?
Answer:
\(\frac{3}{1}\)

What equation would model the relationship between cars and trucks?
Answer:
C = 3T and T = \(\left(\frac{1}{3}\right)\)C

At the end of the trip, Malla had counted 1, 254 trucks. How many cars did she see?
Answer:
C = 1,254.3; C = 3, 762 cars

Exercise 6
Kevin is training to run a half-marathon. His training program recommends that he run for 5 minutes and walk for 1 minute. Let R represent the number of minutes running, and let W represent the number of minutes walking.

Answer:

What is the value of the ratio of the number of minutes walking to the number of minutes running?
Answer:
\(\frac{1}{5}\)

What equation could you use to calculate the minutes spent walking if you know the minutes spent running?
Answer:
W = \(\frac{1}{5}\)R; Answers will vary.

Eureka Math Grade 6 Module 1 Lesson 13 Problem Set Answer Key

A cookie recipe calls for 1 cup of white sugar and 3 cups of brown sugar.
Make a table showing the comparison of the amount of white sugar to the amount of brown sugar.

Answer:

Question 1.
Write the value of the ratio of the amount of white sugar to the amount of brown sugar.
Answer:
\(\frac{1}{3}\)

Question 2.
Write an equation that shows the relationship of the amount of white sugar to the amount of brown sugar.
Answer:
B = 3W or W = \(\frac{1}{3}\) B

Question 3.
Explain how the value of the ratio can be seen In the table.
Answer:
The values in the first row show the values in the ratio. The ratio of the amount of brown sugar to the amount of white sugar is 3: 1. The value of the ratio is \(\frac{3}{1}\).

Question 4.
Explain how the value of the ratio can be seen in the equation.
Answer:
The amount of brown sugar is represented as B in the equation. The amount of white sugar is represented as W. The value is represented because the amount of brown sugar is three times as much as the amount of white sugar, or B = 3W.

Using the same recipe, compare the amount of white sugar to the amount of total sugars used in the recipe.
Make a table showing the comparison of the amount of white sugar to the amount of total sugar.

Answer:

Question 5.
Write the value of the ratio of the amount of total sugar to the amount of white sugar.
Answer:
\(\frac{4}{1}\)

Question 6.
Write an equation that shows the relationship of total sugar to white sugar.
Answer:
T = 4W

Eureka Math Grade 6 Module 1 Lesson 13 Exit Ticket Answer Key

Question 1.
A carpenter uses four nails to install each shelf. Complete the table to represent the relationship between the number of nails (N) and the number of shelves (S). Write the ratio that describes the number of nails per number of shelves. Write as many different equations as you can that describe the relationship between the two quantities.

Answer:

\(\left(\frac{N}{S}\right)=\left(\frac{4}{1}\right)\)
Equations:
N = 4s
S = \(\left(\frac{1}{4}\right)\)N

Engage NY Eureka Math 6th Grade Module 1 Lesson 12 Answer Key

Eureka Math Grade 6 Module 1 Lesson 12 Exercise Answer Key

Exercise 2.
The amount of sugary beverages Americans consume is a leading health concern. For a given brand of cola, a 12 oz. serving of cola contains about 40 g of sugar. Complete the ratio table, using the given ratio to find equivalent ratios.

Answer:
Answers may vary but are found by either multiplying or dividing both 12 and 40 by the same number.

Exercise 3.
A 1 L bottle of cola contains approximately 34 fluid ounces. How many grams of sugar would be in a 1 L bottle of the cola? Explain and show how to arrive at the solution.
Answer:

Exercise 4.
A school cafeteria has a restriction on the amount of sugary drinks available to students. Drinks may not have more than 25 g of sugar. Based on this restriction, what is the largest size cola (in ounces) the cafeteria can offer to students?
Answer:

My estimate is between 6 and 12 oz. but closer to 6 ounces. I need to find \(\frac{1}{4}\) of 6 and add it to 6.
\(\frac{1}{4} \times \frac{6}{1}=\frac{6}{4}=1 \frac{1}{2}\)
6 + 1\(\frac{1}{2}\) = 7\(\frac{1}{2}\)
As 7\(\frac{1}{2}\) oz. cola is the largest size that the school cafeteria can offer to students.

Exercise 5.
Shontelle solves three math problems in four minutes.
a. Use this information to complete the table below.

Answer:

b. Shontelle has soccer practice on Thursday evening. She has a half hour before practice to work on her math homework and to talk to her friends. She has 20 math skill-work questions for homework, and she wants to complete them before talking with her friends. How many minutes will Shontelle have left after completing her math homework to talk to her friends?
Use a double number line diagram to support your answer, and show all work.
Answer:

Step 1: \(\frac{2}{3} \times 4=\frac{8}{3}=2 \frac{2}{3}\)
Step 2: \(24+2 \frac{2}{3}=26 \frac{2}{3}\)
Step 3: \(30-26 \frac{2}{3}=3 \frac{1}{3}\)
Shontelle can talk to her friends for 3\(\frac{1}{3}\) minutes.

Eureka Math Grade 6 Module 1 Lesson 12 Problem Set Answer Key

Question 1.
While shopping, Kyla found a dress that she would like to purchase, but it costs $52.25 more than she has. Kyla charges $5.50 an hour for babysitting. She wants to figure out how many hours she must babysit to earn $52.25 to buy the dress. Use a double number line to support your answer.
Answer:
9.5 hours

Question 2.
Frank has been driving at a constant speed for 3 hours, during which time he traveled 195 miles. Frank would like to know how long it will take him to complete the remaining 455 miles, assuming he maintains the same constant speed. Help Frank determine how long the remainder of the trip will take. Include a table or diagram to support your answer.
Answer:
7 hours

Eureka Math Grade 6 Module 1 Lesson 12 Exit Ticket Answer Key

Question 1.
Kyra is participating in a fundraiser walk-a-thon. She walks 2 miles in 30 minutes. If she continues to walk at the same rate, determine how many minutes it will take her to walk 7 miles. Use a double number line diagram to support your answer.
Answer:

It will take kyra 105 minutes to walk 7 miles.

Engage NY Eureka Math 6th Grade Module 1 Lesson 11 Answer Key

Eureka Math Grade 6 Module 1 Lesson 11 Example Answer Key

Example 1.
Create four equivalent ratios (2 by scaling up and 2 by scaling down) using the ratio 30 to 80.
Answer:
There are various possible answers.
Some examples of scaling down are 3: 8, 6: 16, 9: 24, 12: 32, 15: 40, 18: 48, 21: 56, 24: 64, and 27: 72.
Some examples of scaling up are 60: 160, 90: 240, 120: 320, etc.

Write a ratio to describe the relationship shown in the table.

Answer:
The ratio used to create the table is 1:8, which means that there are 8 pizzas being sold every hour.

Eureka Math Grade 6 Module 1 Lesson 11 Exercise Answer Key

Exercise 1
The following tables show how many words each person can text in a given amount of time. Compare the rates of texting for each person using the ratio table.

Michaela

Jenna

Maria

Answer:
Michaela texts the fastest because she texts 50 words per minute, next is Jenna who texts 45 words per minute, and last is Maria who texts 40 words per minute.

Complete the table so that it shows Max has a texting rate of 55 words per minute.

Max

Answer:

Exercise 2:
Making Juice (Comparing Juice to Water)

a. The tables below show the comparison of the amount of water to the amount of juice concentrate (JC) in grape juice made by three different people. Whose juice has the greatest water-to-juice concentrate ratio, and whose juice would taste strongest? Be sure to justify your answer.
Answer:
Franca’s juice has the greatest amount of water in comparison to juice concentrate, followed by Milton, and then Laredo, Because Laredo’s juice has the least amount of water in comparison to juice concentrate, his juice would taste the strongest.

Answer:

Put the juices in order from the juice containing the most water to the juice containing the least water.
Answer:
Franca, Milton, Laredo

Explain how you used the values in the table to determine the order.
Answer:
Laredo makes his juice by combining three cups of water for every one cup of juice concentrate.
Franca makes her juice by combining five cups of water for every one cup of juice concentrate.
Milton makes his juice by combining four cups of water for every one cup of juice concentrate.

What ratio was used to create each table?
Answer:
Laredo 3: 1, Franca 5: 1, Milton 4: 1

Explain how the ratio could help you compare the juices.
Answer:
Answers will vary.

b. The next day, each of the three people made juice again, but this time they were making apple juice. Whose juice has the greatest water-to-juice concentrate ratio, and whose juice would taste the strongest? Be sure to justify your answer.

Answer:

Put the Juices in order from the strongest apple taste to the weakest apple taste.
Answer:
Franca, Milton, Laredo

Explain how you used the values in the table to determine the order.
Answer:
Answers will vary.

What ratio was used to create each table?
Answer:
Laredo: 6: 1, Franca: 5: 2, Milton: 8: 3

Explain how the ratio could help you compare the juices.
Answer:
Answers will vary.

How was this problem different than the grape juice questions in part (a)?
Answer:
Answers will vary.

c. Max and Sheila are making orange juice. Max has mixed 15 cups of water with 4 cups of juice concentrate. Sheila has made her juice by mixing 8 cups water with 3 cups of juice concentrate. Compare the ratios of juice concentrate to water using ratio tables. State which beverage has a higher juice concentrate-to-water ratio.
Answer:

Sheila has a higher juice concentrate-to-water ratio because she mixed 12 cups of juice concentrate to only 32 cups of water. Max’s juice would be more watery because he would have 45 cups of water with the 12 cups of juice concentrate.

d. Victor is making recipes for smoothies. His first recipe calls for 2 cups of strawberries and 7 cups of other ingredients. His second recipe says that 3 cups of strawberries are combined with 9 cups of other Ingredients. Which smoothie recipe has more strawberries compared to other ingredients? Use ratio tables to justify your answer.
Answer:
Recipe 2 has more strawberries compared to other ingredients.

Recipe 2 has more strawberries compared to the other ingredients. When comparing 6 cups of strawberries, there were fewer other ingredients added in Recipe 2 than in Recipe 1.

Eureka Math Grade 6 Module 1 Lesson 11 Problem Set Answer Key

Question 1.
Sarah and Eva were swimming.
a. Use the ratio tables below to determine who the faster swimmer is.

Sarah

Eva

Answer:
Eva is the faster swimmer because she swims 26 meters in 1 minute, which is faster than Sarah who swims 25 meters in 1 minute.

b. Explain the method that you used to determine your answer.
Answer:
Answers will vary.

Question 2.
A 120 lb. person would weigh about 20 lb. on the earth’s moon. A 150 lb. person would weigh about 28 lb. on lo, a moon of Jupiter. Use ratio tables to determine which moon would make a person weigh the most.
Answer:
Answers will vary. A person on lo will weigh more than a person on our moon.

Eureka Math Grade 6 Module 1 Lesson 11 Exit Ticket Answer Key

Question 1.
Beekeepers sometimes supplement the diet of honey bees with sugar water to help promote colony growth in the spring and help the bees survive through fall and winter months. The tables below show the amount of water and the amount of sugar used in the Spring and in the Fall.

Write a sentence that compares the ratios of the number of cups of sugar to the number of cups of water in each table.
Answer:
The value of the ratio for the Spring sugar water is \(\frac{1.5}{1}\), while the value of the ratio of the Fall sugar water is \(\frac{2}{1}\). Therefore, the Fall sugar water mixture has more sugar mixed in for every cup of water added to the mixture than the Spring sugar water mixture.

Explain how you determined your answer.
Answer:
Spring: \(\frac{6}{4}\) = \(\frac{3}{2}\) = \(\frac{1.5}{1}\)
Fall: \(\frac{4}{2}\) = \(\frac{2}{1}\)

Engage NY Eureka Math 6th Grade Module 1 Lesson 10 Answer Key

Eureka Math Grade 6 Module 1 Lesson 10 Exploratory Challenge Answer Key

Imagine that you are making a fruit salad. For every quart of blueberries you add, you would like to put in 10 quarts of strawberries. Create three ratio tables that show the amounts of blueberries and strawberries you would use If you needed to make fruit salad for greater numbers of people.
Table 1 should contain amounts where you have added fewer than 10 quarts of blueberries to the salad.
Table 2 should contain amounts of blueberries between and including 10 and 50 quarts.
Table 10 should contain amounts of blueberries greater than or equal to 100 quarts.
Answer:
Student answers may vary. Here are possible solutions:

Answer:

Answer:

Answer:

The answers to the questions will depend on the variation of the table that students have created.

a. Describe any patterns you see in the tables. Be specific in your descriptions.
Answer:
The value in the second column is always three times as much as the corresponding value in the first column. In the first table, the entries in the first column increase by 1, and the entries in the second column increase by 3. In the second table, the entries in the first column increase by 10, and the entries in the second column increase by 30. In the third table, the entries in the first column increase by 100, and the entries in the second column increase by 300.

b. How are the amounts of blueberries and strawberries related to each other?
Answer:
The amount of strawberries is always three times the amount of blueberries. Students could also respond that the ratio of the number of quarts of blueberries to the number of quarts of strawberries is always equivalent to 1:3.

c. How are the values in the Blueberries column related to each other?
Answer:
Answers will vary. However, students could use the chart paper and write on the table to see the patterns. Most tables should have addition repeating throughout.

d. How are the values in the Strawberries column related to each other?
Answer:
Answers will vary. However, students could use the chart paper and write on the table to see the patterns. Most tables should have addition repeating throughout.

e. If we know we want to add 7 quarts of blueberries to the fruit salad in Table 1, how can we use the table to help us determine how many strawberries to add?
Answer:
We could extend our table until we get to 7 in the blueberry column.

f. If we know we used 70 quarts of blueberries to make our salad, how can we use a ratio table to find out how many quarts of strawberries were used?
Answer:
We could start with the ratio 1: 3 that was given in the description and then multiply by ten to get 10 and 30. These would be the first values in our table. Then, we would count up by tens in the Blueberries column and count up by 30’s in the Strawberries column.

Eureka Math Grade 6 Module 1 Lesson 10 Exercise Answer Key

Exercise 1.
The following tables were made incorrectly. Find the mistakes that were made, create the correct ratio table, and state the ratio that was used to make the correct ratio table.
a.

Answer:

b.

Answer:

Eureka Math Grade 6 Module 1 Lesson 10 Problem Set Answer Key

Question 1.
a. Create a ratio table for making lemonade with a lemon juice-to-water ratio of 1: 3. Show how much lemon juice would be needed if you use 36 cups of water to make lemonade.
Answer:

12 cups of lemon Juice would be needed if 36 cups of water is used.

b. How is the value of the ratio used to create the table?
Answer:
The value of the ratio is \(\frac{1}{3}\). If we know the amount of lemon juice, we can multiply that amount by 3 to get the amount of water. If we know the amount of water, we can multiply that amount by \(\frac{1}{3}\) (or divide by 3) to get the amount of lemon juice.

Question 2.
Ryan made a table to show how much blue and red paint he mixed to get the shade of purple he will use to paint the room. He wants to use the table to make larger and smaller batches of purple paint.

a. What ratio was used to create this table? Support your answer.
Answer:
The ratio of the amount of blue paint to the amount of red point is 4: 1. I know this because 12: 3, 20: 5, 28: 7, and 36:9 are all equivalent to 4: 1.

b. How are the values in each row related to each other?
Answer:
In each row, the amount of red paint is \(\frac{1}{4}\) times the amount of blue paint, or the amount of blue paint is 4 times the amount of red paint.

c. How are the values in each column related to each other?
Answer:
The values in the columns are increasing using the ratio. Since the ratio of the amount of blue paint to the amount of red paint is 4: 1, we have used 4 × 2: 1 × 2, or 8: 2, and repeatedly added to form the table. 8 was added to the entries in the blue column while 2 was added to the entries in the red column.

Eureka Math Grade 6 Module 1 Lesson 10 Exit Ticket Answer Key

Question 1.
Show more than one way you could use the structure of the table to find the unknown value. Fill in the unknown values.

Answer:

I can add two to the weeks each time to get the next number. I can add $350 to the money to get the next values.

In the rows, we have 2:350, which is equal to 1: 175. So the money is 175 times larger than the week. I can just multiply the week by 175 to get the amount of money in the account.

The ratio used to create the table was 1: 175.

Engage NY Eureka Math 6th Grade Module 1 Lesson 9 Answer Key

Eureka Math Grade 6 Module 1 Lesson 9 Example Answer Key

Example 1.
To make paper mache, the art teacher mixes water and flour. For every two cups of water, she needs to mix in three cups of flour to make the paste.
Find equivalent ratios for the ratio relationship 2 cups of water to 3 cups of flour. Represent the equivalent ratios in the table below:

Answer:

Example 2.
Javier has a new job designing websites. He is paid at a rate of $700 for every 3 pages of web content that he builds. Create a ratio table to show the total amount of money Javier has earned in ratio to the number of pages he has built.

Answer:

Javier is saving up to purchase a used car that costs $4,200. How many web pages will Javier need to build before he can pay for the car?
Answer:
Javier will need to build 18 web pages in order to pay for the car.

Eureka Math Grade 6 Module 1 Lesson 9 Exercise Answer Key

Exercise 1.
Spraying plants with cornmeal juice is a natural way to prevent fungal growth on the plants. It is made by soaking cornmeal in water, using two cups of cornmeal for every nine gallons of water. Complete the ratio table to answer the questions that follow.

Answer:

a. How many cups of cornmeal should be added to 45 gallons of water?
Answer:
10 cups of cornmeal should be added t0 45 gallons of water.

b. Paul has only 8 cups of cornmeal. How many gallons of water should he add if he wants to make as much cornmeal juice as he can?
Answer:
Paul should add 36 gallons of water.

c. What can you say about the values of the ratios in the table?
Answer:
The values of the ratios are equivalent.

Exercise 2.
James is setting up a fish tank. He is buying a breed of goldfish that typically grows to be 12 inches long, It is recommended that there be 1 gallon of water for every inch of fish length in the tank. What is the recommended ratio of gallons of water per fully-grown goldfish in the tank?
Complete the ratio table to help answer the following questions:

Answer:

a. What size tank (in gallons) Is needed for James to have 5 full-grown goldfish?
Answer:
James needs a tank that holds 60 gallons of water in order to have 5 full-grown goldfish.

b. How many full-grown goldfish can go in a 40-gallon tank?
Answer:
3 full-grown goldfish can go in a 40-gallon tank.

c. What can you say about the values of the ratios in the table?
Answer:
The values of the ratios are equivalent.

Eureka Math Grade 6 Module 1 Lesson 9 Problem Set Answer Key

Assume each of the following represents a table of equivalent ratios. Fill in the missing values. Then choose one of the tables, and create a real-world context for the ratios shown in the table.

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Eureka Math Grade 6 Module 1 Lesson 9 Exit Ticket Answer Key

Question 1.
A father and his young toddler are walking along the sidewalk. For every 3 steps the father takes, the son takes 5 steps just to keep up. What is the ratio of the number of steps the father takes to the number of steps the son takes? Add labels to the columns of the table, and place the ratio into the first row of data. Add equivalent ratios to build a ratio table.

Answer:

What can you say about the values of the ratios in the table?
Answer:
The values of the ratios in the table should all be equal since the ratios in the table are equivalent.

Engage NY Eureka Math 6th Grade Module 1 Lesson 8 Answer Key

Eureka Math Grade 6 Module 1 Lesson 8 Exercise Answer Key

Exercise 1
Circle any equivalent ratios from the list below.
Ratio: 1: 2
Ratio: 5: 10
Ratio: 6: 16
Ratio: 12: 32
Answer:

Revisit this when discussing the value of the equivalent ratios.

Find the value of the following ratios, leaving your answer as a fraction, but rewrite the fraction using the largest possible unit.
Ratio: 1: 2             Value of the Ratio:
Ratio: 5: 10            Value of the Ratio:
Ratio: 6: 16             Value of the Ratio:
Ratio: 12: 32           Value of the Ratio:
Answer:
Ratio: 1: 2           Value of the Ratio: \(\frac{1}{2}\)
Ratio: 5: 10          Value of the Ratio: \(\frac{1}{2}\)
Ratio: 6: 16           Value of the Ratio: \(\frac{3}{8}\)
Ratio: 12: 32          Value of the Ratio: \(\frac{3}{8}\)

What do you notice about the value of the equivalent ratios?
Answer:
The value of the ratio is the same for equivalent ratios.

Exercise 2
Here is a theorem: If A: B with B ≠ 0 and C: D with D ≠ 0 are equivalent, then they have the same value: \(\frac{A}{B}=\frac{C}{D}\). This is essentially stating that if two ratios are equivalent, then their values are the same (when they have values). Can you provide any counterexamples to the theorem above?
Answer:
Allow students to try this in pairs. Observe the progress of students and question students’ counterexamples. Ask for further clarification or proof that the two ratios are equivalent but do not have the same value. If students still think they have discovered a counterexample, share the example with the class and discuss why it is not a counterexample.

Ask entire class if anyone thought of a counterexample. If students share examples, have others explain why they are not counterexamples. Then discuss why there are no possible counterexamples to the given theorem. It is important for students to understand that the theorem is always true, so it is not possible to come up with a counterexample.

Exercise 3
Taivon is training for a duathlon, which is a race that consists of running and cycling. The cycling leg is longer than the running leg of the race, so while Taivon trains, he rides his bike more than he runs. During training, Taivon runs 4 miles for every 14 miles he rides his bike.

a. Identify the ratio associated with this problem and find its value.
Answer:
The ratio of the number of miles he ran to the number of miles he cycled is 4: 14, and the value of the ratio is \(\frac{2}{7}\) The ratio of the number of miles he cycled to the number of miles he ranis 14: 4, and the value of the ratio is \(\frac{7}{2}\).

Use the value of each ratio to solve the following.

b. When Taivon completed all of his training for the duathlon, the ratio of total number of miles he ran to total number of miles he cycled was 80: 280. Is this consistent with Taivon’s training schedule? Explain why or why not.
Answer:
This is consistent because the ratio of the number of miles he ran to the number of miles he cycled, 80: 280, has the value of \(\frac{2}{7}\), which is the same value as the ratio 4: 14.

c. In one training session, Taivon ran 4 miles and cycled 7 miles. Did this training session represent an equivalent ratio of the distance he ran to the distance he cycled? Explain why or why not.
Answer:
This training session does not represent an equivalent ratio of the distance he ran to the distance he cycled because the value of the ratio in this instance is \(\frac{4}{7}\), which is not equal to \(\frac{2}{7}\).

Eureka Math Grade 6 Module 1 Lesson 8 Problem Set Answer Key

Question 1.
The ratio of the number of shaded sections to the number of unshaded sections is 4 to 2. What is the value of the ratio of the number of shaded pieces to the number of unshaded pieces?

Answer:
\(\frac{4}{2}\) = \(\frac{2}{1}\) or 2

Question 2.
Use the value of the ratio to determine which ratios are equivalent to 7: 15.
a. 21:45
b. 14:45
c. 3:5
d. 63: 135
Answer:
Both (a) and (d) are equivalent to 7: 15.

Question 3.
Sean was at batting practice. He swung 25 times but only hit the ball 15 times.
a. Describe and write more than one ratio related to this situation.
Answer:
Ratio of the number of hits to the total number of swings is 15: 25.
Ratio of the number hits to the number of misses is 15: 10.
Ratio of the number of misses to the number of hits is 10: 15.
Ratio of the number of misses to the total number of swings is 10: 25.

b. For each ratio you oeated, use the value of the ratio to express one quantity as a fraction of the other quantity.
Answer:
The number of hits is \(\frac{15}{25}\) or \(\frac{3}{5}\) of the total number of swh,gs.
The number of hits is \(\frac{15}{10}\) or \(\frac{3}{2}\) the number of misses
The number of misses is \(\frac{10}{15}\) or \(\frac{2}{3}\) the number of hits.
The number of misses is \(\frac{10}{25}\) or \(\frac{2}{5}\) of the total number of swings.

c. Make up a word problem that a student can solve using one of the ratios and its value.
Answer:
If Sean estimates he will take 10 swings in his next game how many hits would he expect to get assuming his ratio of hits-to-swings does not change.

Question 4.
Your middle school has 900 students. \(\frac{1}{3}\) of students bring their lunch instead of buying lunch at school. What is the value of the ratio of the number of students who do bring their lunch to the number of students who do not?

Answer:
First, I created a tape diagram. In the tape diagram, \(\frac{1}{3}\) of students bring their lunch instead of buying lunch at school. I determined that 300 students bring their lunch, leaving 600 students who buy their lunch. One unit of the tape diagram represents 300, and 2 units of the tape diagram represent 600. This creates a ratio of 1: 2. As such, the value of the ratio of the number of students who bring their lunch to the number of students who buy their lunch is \(\frac{1}{2}\)

Eureka Math Grade 6 Module 1 Lesson 8 Exit Ticket Answer Key

Question 1.
You created a new playlist, and 100 of your friends listened to it and shared if they liked the new playlist or not. Nadhii said the ratio of the number of people who liked the playlist to the number of people who did not like the playlist is 75: 25. Dylan said that for every three people who liked the playlist, one person did not.

Do Nadhii and Dy’an agree? Prove your answer using the values of the ratios.
Answer:
Dylan and Nadhii agree. The value of both of their ratios is equivalent, so their ratios are also equivalent.

Engage NY Eureka Math 6th Grade Module 1 Lesson 7 Answer Key

Eureka Math Grade 6 Module 1 Lesson 7 Example Answer Key

Example 1
Which of the following correctly models that the number of red gumballs is \(\frac{5}{3}\) the number of white gumballs?
a.

b.

c.

d.

Answer:
b.

Example 2.
The duration of two films are modeled below.

a. The ratio of the length of Film A to the length of Film B is _______ : _______.
Answer:
The ratio of the length of Film A to the length of Film B is 5: 7.

b. The length of Film A is of the length of Film B.
Answer:
The length of Film A is of the length of Film B.

c. The length of Film B is of the length of Film A.
Answer:
The length of Film B is of the length of Film A.

Eureka Math Grade 6 Module 1 Lesson 7 Exercise Answer Key

Exercise 1
Sammy and Kaden went fishing using live shrimp as bait. Sammy brought 8 more shrimp than Kaden brought. When they combined their shrimp they had 32 shrimp altogether.
a. How many shrimp did each boy bring?
Answer:
Kaden brought 12 shrimp. Sammy brought 20 shrimp.

b. What is the ratio of the number of shrimp Sammy brought to the number of shrimp Kaden brought?
Answer:
20: 12

c. Express the number of shrimp Sammy brought as a fraction of the number of shrimp Kaden brought.
Answer:
\(\frac{20}{12}\)

d. What is the ratio of the number of shrimp Sammy brought to the total number of shrimp?
Answer:
20: 32

e. What fraction of the total shrimp did Sammy bring?
Answer:
\(\frac{20}{32}\)

Exercise 2.
A food company that produces peanut butter decides to try out a new version of its peanut butter that is extra crunchy, using twice the number of peanut chunks as normal. The company hosts a sampling of its new product at grocery stores and finds that 5 out of every 9 customers prefer the new extra crunchy version.

a. Let’s make a list of ratios that might be relevant for this situation.
i. The ratio of number preferring new extra crunchy to total number surveyed is __________.
Answer:
The ratio of number preferring new extra crunchy to total number surveyed is  5 to 9  .

ii. The ratio of number preferring regular crunchy to the total number surveyed is __________.
Answer:
The ratio of number preferring regular crunchy to the total number surveyed is   4 to 9  .

iii. The ratio of number preferring regular crunchy to number preferring new extra crunchy is __________.
Answer:
The ratio of number preferring regular crunchy to number preferring new extra crunchy is   4 to 5   .

iv. The ratio of number preferring new extra crunchy to number preferring regular crunchy is __________.
Answer:
The ratio of number preferring new extra crunchy to number preferring regular crunchy is   5 to 4    .

b. Let’s use the value of each ratio to make multiplicative comparisons for each of the ratios we described here.
i. The number preferring new extra crunchy is _________ of the total number surveyed.
Answer:
The number preferring new extra crunchy is \(\frac{5}{9}\)  of the total number surveyed.

ii. The number preferring regular crunchy is _________ of the total number surveyed.
Answer:
The number preferring regular crunchy is \(\frac{4}{9}\) of the total number surveyed.

iii. The number preferring regular crunchy is __________ of those preferring new extra crunchy.
Answer:
The number preferring regular crunchy is \(\frac{4}{5}\) of those preferring new extra crunchy.

iv. The number preferring new extra crunchy is _________ of those preferring regular crunchy.
Answer:
The number preferring new extra crunchy is \(\frac{5}{4}\) of those preferring regular crunchy.

c. If the company is planning to produce 90,000 containers of crunchy peanut butter, how many of these containers should be the new extra crunchy variety, and how many of these containers should be the regular crunchy peanut butter? What would be helpful in solving this problem? Does one of our comparison statements above help us?
Answer:
The company should produce 50,000 containers of new crunchy peanut butter and 40,000 containers of regular crunchy peanut butter.

d. If the company decides to produce 2000 containers of regular crunchy peanut butter, how many containers of new extra crunchy peanut butter would it produce?
Answer:
2,500 new extra crunchy peanut butter containers

e. If the company decides to produce 10,000 containers of new extra crunchy peanut butter, how containers of regular crunchy peanut butter would it produce?
Answer:
8,000 regular crunchy peanut butter containers

f. If the company decides to only produce 3,000 containers of new extra crunchy peanut butter, how many containers of regular crunchy peanut butter would it produce?
Answer:
2,400 regular crunchy peanut butter containers

Eureka Math Grade 6 Module 1 Lesson 7 Problem Set Answer Key

Question 1.
Maritza is baking cookies to bring to school and share with her friends on her birthday. The recipe requires 3 eggs for every 2 cups of sugar. To have enough cookies for all of her friends, Maritza determined she would need 12 eggs. If her mom bought 6 cups of sugar, does Maritza have enough sugar to make the cookies? Why or why not?
Answer:
Maritza will NOT have enough sugar to make all the cookies because she needs 8 cups of sugar and only has 6 cups of sugar.

Question 2.
Hamza bought 8 gallons of brown paint to paint his kitchen and dining room. Unfortunately, when Hamza started painting, he thought the paint was too dark for his house, so he wanted to make It lighter. The store manager would not let Hamza return the paint but did inform him that if he used \(\frac{1}{4}\) of a gallon of white paint mixed with 2 gallons of brown paint, he would get the shade of brown he desired. If Hamza decided to take this approach, how many gallons of white paint would Hamza have to buy to lighten the 8 gallons of brown paint?
Answer:
Hamza would need 1 gallon of white paint to make the shade of brown he desires.

Eureka Math Grade 6 Module 1 Lesson 7 Exit Ticket Answer Key

Alyssa’s extended family Is staying at the lake house this weekend for a family reunion. She is in charge of making homemade pancakes for the entire group. The pancake mix requires 2 cups of flour for every 10 pancakes.

Question 1.
Write a ratio to show the relationship between the number of cups of flour and the number of pancakes made.
Answer:
2: 10

Question 2.
Determine the value of the ratio.
Answer:
\(\frac{2}{10}=\frac{1}{5}\)

Question 3.
Use the value of the ratio to make a multiplicative comparison statement.
a. The number of pancakes made is _______ times the number of cups of flour needed.
Answer:
The number of pancakes made is 5 times the number of cups of flour needed.

b. The number of cups of flour needed is _______ of the number of pancakes made.
Answer:
The number of cups of flour needed is \(\frac{1}{5}\) of the number of pancakes made.

Question 4.
If Alyssa has to make 70 pancakes, how many cups of flour will she have to use?
Answer:
Alyssa will have to use 14 cups of flour.

Engage NY Eureka Math 7th Grade Module 3 Lesson 2 Answer Key

Eureka Math Grade 7 Module 3 Lesson 2 Example Answer Key

Example 1.
Subtracting Expressions
a. Subtract: (40 + 9) – (30 + 2).
Answer:
The opposite of a sum is the sum of its opposites. Order of operations

40 + 9 + (- (30 + 2))
40 + 9 + (- 30) + (- 2)
49 + (- 30) + (- 2)
19 + (- 2)
17

(40 + 9)- (30 + 2)
(49)- (32)
17

b. Subtract: (3x + 5y – 4) – (4x + 11).
Answer:
3x + 5y + ( – 4) + ( – (4x + 11)) Subtraction as adding the opposite
3x + 5y + ( – 4) + ( – 4x) + ( – 11) The opposite of a sum is the sum of its opposites.
3x + ( – 4x) + 5y + ( – 4) + ( – 11) Any order, any grouping
– x + 5y + ( – 15) Combining like terms
– x + 5y – 15 Subtraction replaces adding the opposite.
Have students check the equivalency of the expressions by substituting 2 for x and 6 for y.
(3x + 5y – 4) – (4x + 11)
(3(2) + 5(6) – 4) – (4(2) + 11)
(6 + 30 – 4) – (8 + 11)
(36 – 4) – (19)
32 – 19
13
– x + 5y – 15
– (2) + 5(6) – 15
– 2 + 30 + ( – 15)
28 + ( – 15)
13
→ When writing the difference as adding the expression’s opposite in Example 1(b), what happens to the grouped terms that are being subtracted?
→ When the subtraction is changed to addition, every term in the parentheses that follows must be converted to its opposite.

Example 2.
Combining Expressions Vertically
a. Find the sum by aligning the expressions vertically.
(5a + 3b – 6c) + (2a – 4b + 13c)
Answer:
(5a + 3b + ( – 6c)) + (2a + ( – 4b) + 13c) Subtraction as adding the opposite

7a – b + 7c Adding the opposite is equivalent to subtraction.

b. Find the difference by aligning the expressions vertically.
(2x + 3y – 4) – (5x + 2)
Answer:
(2x + 3y + ( – 4)) + ( – 5x + ( – 2)) Subtraction as adding the opposite

– 3x + 3y – 6 Adding the opposite is equivalent to subtraction.
Students should recognize that the subtracted expression in Example 1(b) did not include a term containing the variable y, so the 3y from the first grouped expression remains unchanged in the answer.

Example 3.
Using Expressions to Solve Problems
A stick is x meters long. A string is 4 times as long as the stick.
a. Express the length of the string in terms of x.
Answer:
The length of the stick in meters is x meters, so the string is 4∙x, or 4x, meters long.

b. If the total length of the string and the stick is 15 meters long, how long is the string?
Answer:
The length of the stick and the string together in meters can be represented by x + 4x, or 5x. If the length of the stick and string together is 15 meters, the length of the stick is 3 meters, and the length of the string is 12 meters.

Example 4.
Expressions from Word Problems
It costs Margo a processing fee of $3 to rent a storage unit, plus $17 per month to keep her belongings in the unit. Her friend Carissa wants to store a box of her belongings in Margo’s storage unit and tells her that she will pay her $1 toward the processing fee and $3 for every month that she keeps the box in storage. Write an expression in standard form that represents how much Margo will have to pay for the storage unit if Carissa contributes. Then, determine how much Margo will pay if she uses the storage unit for 6 months.
Answer:
Let m represent the number of months that the storage unit is rented.
(17m + 3) – (3m + 1) Original expression
17m + 3 + ( – (3m + 1)) Subtraction as adding the opposite
17m + 3 + ( – 3m) + ( – 1) The opposite of the sum is the sum of its opposites.
17m + ( – 3m) + 3 + ( – 1) Any order, any grouping
14m + 2 Combined like terms
This means that Margo will have to pay only $2 of the processing fee and $14 per month that the storage unit is used.
14(6) + 2
84 + 2
86
Margo will pay $86 toward the storage unit rental for 6 months of use.

Example 5.
Extending Use of the Inverse to Division

Answer:

Eureka Math Grade 7 Module 3 Lesson 2 Opening Exercise Answer Key

Opening Exercise
Additive inverses have a sum of zero. Fill in the center column of the table with the opposite of the given number or expression, then show the proof that they are opposites. The first row is completed for you.

Answer:

Eureka Math Grade 7 Module 3 Lesson 2 Problem Set Answer Key

Question 1.
Write each expression in standard form. Verify that your expression is equivalent to the one given by evaluating each expression using x=5.

a. 3x + (2 – 4x)
Answer:
3x + (2 – 4x)
– x + 2
– 5 + 2
– 3

3(5) + (2 – 4(5))
15 + (2 + ( – 20))
15 + ( – 18)
– 3

b. 3x + ( – 2 + 4x)
Answer:
3x + ( – 2 + 4x)
7x – 2
7(5) – 2
35 – 2
33

3(5) + ( – 2 + 4(5))
15 + ( – 2 + 20)
15 + 18
33

c. – 3x + (2 + 4x)
Answer:
– 3x + (2 + 4x)
x + 2
5 + 2
7

– 3(5) + (2 + 4(5))
– 15 + (2 + 20)
– 15 + 22
7

d. 3x + ( – 2 – 4x)
Answer:
3x + ( – 2 – 4x)
– x – 2
– 5 – 2
– 7

3(5) + ( – 2 – 4(5))
15 + ( – 2 + ( – 4(5)))
15 + ( – 2 + ( – 20))
15 + ( – 22)
– 7

e. 3x – (2 + 4x)
Answer:
3x – (2 + 4x)
– x – 2
– 5 – 2
– 7

3(5) – (2 + 4(5))
15 – (2 + 20)
15 – 22
15 + ( – 22)
– 7

f. 3x – ( – 2 + 4x)
Answer:
3x – ( – 2 + 4x)
– x + 2
– 5 + 2
– 3

3(5) – ( – 2 + 4(5))
15 – ( – 2 + 20)
15 – (18)
15 + ( – 18)
– 3

g. 3x – ( – 2 – 4x)
Answer:
3x – ( – 2 – 4x)
7x + 2
7(5) + 2
35 + 2
37

3(5) – ( – 2 – 4(5))
15 – ( – 2 + ( – 4(5)))
15 – ( – 2 + ( – 20))
15 – ( – 22)
15 + 22
37

h. 3x – (2 – 4x)
Answer:
3x – (2 – 4x)
7x – 2
7(5) – 2
35 – 2
33

3(5) – (2 – 4(5))
15 – (2 + ( – 4(5)))
15 – (2 + ( – 20))
15 – ( – 18)
15 + 18
33

i. – 3x – ( – 2 – 4x)
Answer:
– 3x – ( – 2 – 4x)
x + 2
5 + 2
7

– 3(5) – ( – 2 – 4(5))
– 15 – ( – 2 + ( – 4(5)))
– 15 – ( – 2 + ( – 20))
– 15 – ( – 22)
– 15 + 22
7

j. In problems (a)–(d) above, what effect does addition have on the terms in parentheses when you removed the parentheses?
Answer:
By the any grouping property, the terms remained the same with or without the parentheses.

k. In problems (e)–(i), what effect does subtraction have on the terms in parentheses when you removed the parentheses?
Answer:
The opposite of a sum is the sum of the opposites; each term within the parentheses is changed to its opposite.

Question 2.
Write each expression in standard form. Verify that your expression is equivalent to the one given by evaluating each expression for the given value of the variable.
a. 4y – (3 + y); y=2
Answer:
4y – (3 + y)
3y – 3
3(2) – 3
6 – 3
3

4(2) – (3 + 2)
8 – 5
8 + ( – 5)
3

b. (2b + 1) – b; b= – 4
Answer:
(2b + 1) – b
b + 1
– 4 + 1
– 3

(2( – 4) + 1) – ( – 4)
( – 8 + 1) + 4
( – 7) + 4
– 3

c. (6c – 4) – (c – 3); c= – 7
Answer:
(6c – 4) – (c – 3)
5c – 1
5( – 7) – 1
– 35 – 1
– 36

(6( – 7) – 4) – ( – 7 – 3)
( – 42 – 4) – ( – 10)
– 42 + ( – 4) + (10)
– 46 + 10
– 36

d. (d + 3d) – ( – d + 2); d=3
Answer:
(d + 3d) – ( – d + 2)
5d – 2
5(3) – 2
15 – 2
13

(3 + 3(3)) – ( – 3 + 2)
(3 + 9) – ( – 1)
12 + 1
13

e. ( – 5x – 4) – ( – 2 – 5x); x=3
Answer:
– 2
( – 5(3) – 4) – ( – 2 – 5(3))
( – 15 – 4) – ( – 2 – 15)
( – 19) – ( – 17)
( – 19) + 17
– 2

f. 11f – ( – 2f + 2); f=\(\frac{1}{2}\)
13f – 2
13(\(\frac{1}{2}\) ) – 2
\(\frac{13}{2}\) – 2
6 \(\frac{1}{2}\) – 2
4 \(\frac{1}{2}\)

11(\(\frac{1}{2}\) ) – ( – 2(\(\frac{1}{2}\) ) + 2)
1\(\frac{1}{2}\) – ( – 1 + 2)
1\(\frac{1}{2}\) – 1
1\(\frac{1}{2}\) + ( – \(\frac{2}{2}\) )
\(\frac{9}{2}\)
4 \(\frac{1}{2}\)

g. – 5g + (6g – 4); g= – 2
Answer:
– 5g + (6g – 4)
g – 4
– 2 – 4
– 6

– 5( – 2) + (6( – 2) – 4)
10 + ( – 12 – 4)
10 + ( – 12 + ( – 4))
10 + ( – 16)
– 6

h. (8h – 1) – (h + 3); h= – 3
Answer:
(8h – 1) – (h + 3)
7h – 4
7( – 3) – 4
– 21 – 4
– 25

(8( – 3) – 1) – ( – 3 + 3)
( – 24 – 1) – (0)
( – 25) – 0
– 25

i. (7 + w) – (w + 7); w= – 4
Answer
(7 + w) – (w + 7)
0
(7 + ( – 4)) – ( – 4 + 7)
3 – 3
3 + ( – 3)
0

j. (2g + 9h – 5) – (6g – 4h + 2); g= – 2 and h=5
Answer:
(2g + 9h – 5) – (6g – 4h + 2)
– 4g + 13h – 7
– 4( – 2) + 13(5) – 7
8 + 65 + ( – 7)
73 + ( – 7)
66

(2( – 2) + 9(5) – 5) – (6( – 2) – 4(5) + 2)
( – 4 + 45 – 5) – ( – 12 + ( – 4(5)) + 2)
(41 – 5) – ( – 12 + ( – 20) + 2)
(41 + ( – 5)) – ( – 32 + 2)
36 – ( – 30)
36 + 30
66

Question 3.
Write each expression in standard form. Verify that your expression is equivalent to the one given by evaluating both expressions for the given value of the variable.

a. – 3(8x); x=\(\frac{1}{4}\)
Answer:
– 3(8x)
– 24x
– 24(\(\frac{1}{4}\) )
– 24/4
– 6

– 3(8(\(\frac{1}{4}\) ))
– 3(2)
– 6

b. 5∙k∙( – 7); k=\(\frac{3}{5}\)
Answer:
5∙k∙( – 7)
– 35k
– 35(\(\frac{3}{5}\) )
– 105/5
– 21

5(\(\frac{3}{5}\) )( – 7)
3( – 7)
– 21

c. 2( – 6x)∙2; x=\(\frac{3}{4}\)
Answer:
2( – 6x)∙2
– 24x
– 24(\(\frac{3}{4}\) )
– \(\frac{72}{4}\)
– 18

2( – 6(\(\frac{3}{4}\) ))∙2
2( – 3(\(\frac{3}{2}\) ))∙2
2( – 3)(\(\frac{3}{2}\) )(2)
– 6(3)
– 18

d. – 3(8x) + 6(4x); x=2
Answer:
– 3(8x) + 6(4x)
0

– 3(8(2)) + 6(4(2))
– 3(16) + 6(8)
– 48 + 48
0

e. 8(5m) + 2(3m); m= – 2
Answer:
8(5m) + 2(3m)
46m
46( – 2)
– 92

8(5( – 2)) + 2(3( – 2))
8( – 10) + 2( – 6)
– 80 + ( – 12)
– 92

f. – 6(2v) + 3a(3); v=\(\frac{1}{3}\) ; a=\(\frac{2}{3}\)
Answer:
– 6(2v) + 3a(3); v=\(\frac{1}{3}\)
– 6(2v) + 3a(3)
– 12v + 9a
– 12(\(\frac{1}{3}\) ) + 9(\(\frac{2}{3}\) )
– 1\(\frac{2}{3}\) + \(\frac{18}{3}\)
– 4 + 6
2

– 6(2(\(\frac{1}{3}\) )) + 3(\(\frac{2}{3}\) )(3)
– 6(\(\frac{2}{3}\) ) + 2(3)
– 4 + 6
2

Question 4.
Write each expression in standard form. Verify that your expression is equivalent to the one given by evaluating both expressions for the given value of the variable

a. 8x ÷ 2; x= – \(\frac{1}{4}\)
Answer:
8x ÷ 2
4x
4( – \(\frac{1}{4}\) )
– 1

8( – \(\frac{1}{4}\) ) ÷ 2
– 2 ÷ 2
– 1

b. 18w ÷ 6; w=6
Answer:
18w ÷ 6
3w
3(6)
18

18(6) ÷ 6
108 ÷ 6
18

c. 25r ÷ 5r; r= – 2
Answer:
25r ÷ 5r
5

25( – 2) ÷ (5( – 2))
– 50 ÷ ( – 10)
5

d. 33y ÷ 11y; y= – 2
Answer:
33y ÷ 11y
3

33( – 2) ÷ (11( – 2))
( – 66) ÷ ( – 22)
3

e. 56k ÷ 2k; k=3
Answer:
56k ÷ 2k
28

56(3) ÷ (2(3))
168 ÷ 6
28

f. 24xy ÷ 6y; x= – 2;y=3
Answer:
24xy ÷ 6y
4x
4( – 2)
– 8

24( – 2)(3) ÷ (6(3))
– 48(3) ÷ 18
– 144 ÷ 18
– 8

Question 5.
For each problem (a)–(g), write an expression in standard form.
a. Find the sum of – 3x and 8x.
Answer:
– 3x + 8x
5x

b. Find the sum of – 7g and 4g + 2.
Answer:
– 7g + (4g + 2)
– 3g + 2

c. Find the difference when 6h is subtracted from 2h – 4.
Answer:
(2h – 4) – 6h
– 4h – 4

d. Find the difference when – 3n – 7 is subtracted from n + 4.
Answer:
(n + 4) – ( – 3n – 7)
4n + 11

e. Find the result when 13v + 2 is subtracted from 11 + 5v.
Answer:
(11 + 5v) – (13v + 2)
– 8v + 9

f. Find the result when – 18m – 4 is added to 4m – 14.
Answer:
(4m – 14) + ( – 18m – 4)
– 14m – 18

g. What is the result when – 2x + 9 is taken away from – 7x + 2?
Answer:
( – 7x + 2) – ( – 2x + 9)
– 5x – 7

Question 6.
Marty and Stewart are stuffing envelopes with index cards. They are putting x index cards in each envelope. When they are finished, Marty has 15 stuffed envelopes and 4 extra index cards, and Stewart has 12 stuffed envelopes and 6 extra index cards. Write an expression in standard form that represents the number of index cards the boys started with. Explain what your expression means.
Answer:
They inserted the same number of index cards in each envelope, but that number is unknown, x. An expression that represents Marty’s index cards is 15x + 4 because he had 15 envelopes and 4 cards left over. An expression that represents Stewart’s index cards is 12x + 6 because he had 12 envelopes and 6 left over cards. Their total number of cards together would be:
15x + 4 + 12x + 6
15x + 12x + 4 + 6
27x + 10
This means that altogether, they have 27 envelopes with x index cards in each, plus another 10 leftover index cards.

Question 7.
The area of the pictured rectangle below is 24b ft². Its width is 2b ft. Find the height of the rectangle and name any properties used with the appropriate step.
Answer:
24b ÷ 2b

24b∙\(\frac{1}{2b}\) Multiplying the reciprocal
\(\frac{27b}{2b}\) Multiplication
\(\frac{24}{2}\) ∙\(\frac{b}{b}\) Any order, any grouping in multiplication
12∙1
12
The height of the rectangle is 12 ft.

Eureka Math Grade 7 Module 3 Lesson 2 Generating Equivalent Expressions—Round 1 Answer Key

Directions: Write each as an equivalent expression in standard form as quickly and as accurately as possible within the allotted time.

Answer:

Eureka Math Grade 7 Module 3 Lesson 2 Generating Equivalent Expressions—Round 2 Answer Key

Directions: Write each as an equivalent expression in standard form as quickly and as accurately as possible within the allotted time.

Answer:

Eureka Math Grade 7 Module 3 Lesson 2 Exit Ticket Answer Key

Question 1.
Write the expression in standard form.
Answer:
(4f – 3 + 2g) – ( – 4g + 2)
4f + ( – 3) + 2g + ( – ( – 4g + 2)) Subtraction as adding the opposite
4f + ( – 3) + 2g + 4g + ( – 2) The opposite of a sum is the sum of its opposites.
4f + 2g + 4g + ( – 3) + ( – 2) Any order, any grouping
4f + 6g + ( – 5) Combined like terms
4f + 6g – 5 Subtraction as adding the opposite

Question 2.
Find the result when 5m + 2 is subtracted from 9m.
Answer:
9m – (5m + 2) Original expression
9m + ( – (5m + 2)) Subtraction as adding the opposite
9m + ( – 5m) + ( – 2) The opposite of a sum is the sum of its opposites.
4m + ( – 2) Combined like terms
4m – 2 Subtraction as adding the opposite

Question 3.
Write the expression in standard form.
27h ÷ 3h
Answer:
27h∙\(\frac{1}{3 h}\) Multiplying by the reciprocal
\(\frac{27h}{3 h}\) Multiplication
\(\frac{27}{3}\)∙\(\frac{h}{h}\) Any order, any grouping
9∙1
9

Engage NY Eureka Math 7th Grade Module 3 Lesson 1 Answer Key

Eureka Math Grade 7 Module 3 Lesson 1 Example Answer Key

Example 1.
Any Order, Any Grouping Property with Addition
a. Rewrite 5x+3x and 5x-3x by combining like terms.
Write the original expressions and expand each term using addition. What are the new expressions equivalent to?
Answer:

Because both terms have the common factor of x, we can use the distributive property to create an equivalent expression.
5x+3x=(5+3)x=8x
5x-3x=(5-3)x=2x

Ask students to try to find an example (a value for x) where 5x+3x≠8x or where
5x-3x≠2x. Encourage them to use a variety of positive and negative rational numbers. Their failure to find a counterexample helps students realize what equivalence means.

In Example 1, part (b), students see that the commutative and associative properties of addition are regularly used in consecutive steps to reorder and regroup like terms so that they can be combined. Because the use of these properties does not change the value of an expression or any of the terms within the expression, the commutative and associative properties of addition can be used simultaneously. The simultaneous use of these properties is referred to as the any order, any grouping property.

b. Find the sum of 2x+1 and 5x.
Answer:

7x+1 Equivalent expression to the given problem
→ Why did we use the associative and commutative properties of addition?
→ We reordered the terms in the expression to group together like terms so that they could be combined.
→ Did the use of these properties change the value of the expression? How do you know?
→ The properties did not change the value of the expression because each equivalent expression includes the same terms as the original expression, just in a different order and grouping.
→ If a sequence of terms is being added, the any order, any grouping property allows us to add those terms in any order by grouping them together in any way.
→ How can we confirm that the expressions (2x+1)+5x and 7x+1 are equivalent expressions?
→ When a number is substituted for the x in both expressions, they both should yield equal results.
The teacher and student should choose a number, such as 3, to substitute for the value of x and together check to see if both expressions evaluate to the same result.

Given Expression
((2x+1)+5x
(2∙3+1)+5∙3
(6+1)+15
(7)+15
22

Equivalent Expression?
7x+1
7∙3+1
21+1
22

→ The expressions both evaluate to 22; however, this is only one possible value of x. Challenge students to find a value for x for which the expressions do not yield the same number. Students find that the expressions evaluate to equal results no matter what value is chosen for x.
→ What prevents us from using any order, any grouping in part (c), and what can we do about it?
→ The second expression, (5a-3), involves subtraction, which is not commutative or associative; however, subtracting a number x can be written as adding the opposite of that number. So, by changing subtraction to addition, we can use any order and any grouping.

c. Find the sum of -3a+2 and 5a-3.
Answer:
(-3a+2)+(5a-3) Original expression
-3a+2+5a+(-3) Add the opposite (additive inverse)
-3a+5a+2+(-3) Any order, any grouping
2a+(-1) Combined like terms (Stress to students that the expression is not yet simplified.)
2a-1 Adding the inverse is subtracting.

→ What was the only difference between this problem and those involving all addition?
→ We first had to rewrite subtraction as addition; then, this problem was just like the others.

Example 2.
Any Order, Any Grouping with Multiplication
Find the product of 2x and 3.
Answer:

→ Why did we use the associative and commutative properties of multiplication?

→ We reordered the factors to group together the numbers so that they could be multiplied.

→ Did the use of these properties change the value of the expression? How do you know?

→ The properties did not change the value of the expression because each equivalent expression includes the same factors as the original expression, just in a different order or grouping.

→ If a product of factors is being multiplied, the any order, any grouping property allows us to multiply those factors in any order by grouping them together in any way.

Example 3.
Any Order, Any Grouping in Expressions with Addition and Multiplication
Use any order, any grouping to write equivalent expressions.
a. 3(2x)
Answer:
(3∙2)x
6x

b. 4y(5)
Answer:
(4∙5)y
20y

c. 4∙2∙z
Answer:
(4∙2)z
8z

d. 3(2x)+4y(5)
Answer:

(3∙2)x+(4∙5)y
6x+20y

e. 3(2x)+4y(5)+4∙2∙z
Answer:

(3∙2)x+(4∙5)y+(4∙2)z
6x+20y+8z

f. Alexander says that 3x+4y is equivalent to (3)(4)+xy because of any order, any grouping. Is he correct? Why or why not?

Encourage students to substitute a variety of positive and negative rational numbers for x and y because in order for the expressions to be equivalent, the expressions must evaluate to equal numbers for every substitution of numbers into all the letters in both expressions.

Alexander is incorrect; the expressions are not equivalent because if we, for example, let x=-2 and let
y=-3, then we get the following:

-18≠18, so the expressions cannot be equivalent.

→ What can be concluded as a result of part (f)?
→ Any order, any grouping cannot be used to mix multiplication with addition. Numbers and letters that are factors within a given term must remain factors within that term.

Eureka Math Grade 7 Module 3 Lesson 1 Problem Set Answer Key

For Problems 1–9, write equivalent expressions by combining like terms. Verify the equivalence of your expression and the given expression by evaluating each for the given values: a=2, b=5, and c=-3.

Question 1.
3a+5a
Answer:
8a
8(2)
16

3(2)+5(2)
6+10
16

Question 2.
8b – 4b
Answer:
4b
4(5)
20

8(5)-4(5)
40-20
20

Question 3.
5c+4c+c
10c
10(-3)
-30

5(-3)+4(-3)+(-3)
-15+(-12)+(-3)
-27+(-3)
-30

Question 4.
3a+6+5a
8a+6
8(2)+6
16+6
22

3(2)+6+5(2)
6+6+10
12+10
22

Question 5.
8b + 8 – 4b
4b+8
4(5)+8
20+8
28

8(5)+8-4(5)
40+8-20
48-20
28

Question 6.
2c
2(-3)
-6

5(-3)-4(-3)+(-3)
-15+(-4(-3))+(-3)
-15+(12)+(-3)
-3+(-3)
-6

Eureka Math Grade 7 Module 3 Lesson 1 Exit Ticket Answer Key

Question 1.
Write an equivalent expression to 2x+3+5x+6 by combining like terms.
Answer:
2x+3+5x+6
2x+5x+3+6
7x+9

Question 2.
Find the sum of (8a+2b-4) and (3b-5).
Answer:
(8a+2b-4)+(3b-5)
8a+2b+(-4)+3b+(-5)
8a+2b+3b+(-4)+(-5)
8a+(5b)+(-9)
8a+5b-9

Question 3.
Write the expression in standard form: 4(2a)+7(-4b)+(3∙c∙5).
Answer:
(4∙2)a+(7∙(-4))b+(3∙5)c
8a+(-28)b+15c
8a-28b+15c

Eureka Math Grade 7 Module 3 Lesson 1 Opening Exercise Answer Key

Each envelope contains a number of triangles and a number of quadrilaterals. For this exercise, let t represent the number of triangles, and let q represent the number of quadrilaterals.
a. Write an expression using t and q that represents the total number of sides in your envelope. Explain what the terms in your expression represent.
Answer:
3t+4q. Triangles have 3 sides, so there will be 3 sides for each triangle in the envelope. This is represented by 3t. Quadrilaterals have 4 sides, so there will be 4 sides for each quadrilateral in the envelope. This is represented by 4q. The total number of sides will be the number of triangle sides and the number of quadrilateral sides together.

b. You and your partner have the same number of triangles and quadrilaterals in your envelopes. Write an expression that represents the total number of sides that you and your partner have. If possible, write more than one expression to represent this total.
Answer:
3t+4q+3t+4q; 2(3t+4q); 6t+8q

c. Each envelope in the class contains the same number of triangles and quadrilaterals. Write an expression that represents the total number of sides in the room.
Answer:
Answer depends on the number of students in the classroom. For example, if there are 12 students in the classroom, the expression would be 12(3t+4q), or an equivalent expression

d. Use the given values of t and q and your expression from part (a) to determine the number of sides that should be found in your envelope.
Answer:
3t+4q
3(4)+4(2)
12+8
20
There should be 20 sides contained in my envelope.

e. Use the same values for t and q and your expression from part (b) to determine the number of sides that should be contained in your envelope and your partner’s envelope combined.

My partner and I have a combined total of 40 sides.

f. Use the same values for t and q and your expression from part (c) to determine the number of sides that should be contained in all of the envelopes combined.
Answer:
Answer will depend on the seat size of your classroom. Sample responses for a class size of 12:

For a class size of 12 students, there should be 240 sides in all of the envelopes combined.

g. What do you notice about the various expressions in parts (e) and (f)?
Answer:
The expressions in part (e) are all equivalent because they evaluate to the same number: 40. The expressions in part (f) are all equivalent because they evaluate to the same number: 240. The expressions themselves all involve the expression 3t+4q in different ways. In part (e), 3t+3t is equivalent to 6t, and 4q+4q is equivalent to 8q. There appear to be several relationships among the representations involving the commutative, associative, and distributive properties.

Engage NY Eureka Math 7th Grade Module 2 Lesson 23 Answer Key

Eureka Math Grade 7 Module 2 Lesson 23 Exercise Answer Key

Exercises
Youth Group Trip

Exercise 1.
The youth group is going on a trip to an amusement park in another part of the state. The trip costs each group member $150, which includes $85 for the hotel and two one – day combination entrance and meal plan passes.
a. Write an equation representing the cost of the trip. Let P be the cost of the park pass.
Answer:
85 + 2P = 150

b. Solve the equation algebraically to find the cost of the park pass. Then write the reason that justifies each step using if – then statements.
Answer:
If: 85 + 2P = 150,
Then: 85 – 85 + 2P = 150 – 85 Subtraction property of equality for the additive inverse of 85
If: 0 + 2P = 65
Then: 2P = 65 Additive identity
If: 2P = 65
Then: (\(\frac{1}{2}\))2P = (\(\frac{1}{2}\))65 Multiplication property of equality using the multiplicative inverse of 2
If: 1P = 32.5
Then: P = 32.5 Multiplicative identity
The park pass costs $32.50.

c. Model the problem using a tape diagram to check your work.
Answer:
150 – 85 = 65
65÷2 = 32.50

Suppose you want to buy your favorite ice cream bar while at the amusement park, and it costs $2.89. If you purchase the ice cream bar and 3 bottles of water, pay with a $10 bill, and receive no change, then how much did each bottle of water cost?

d. Write an equation to model this situation.
Answer:
W: the cost of one bottle of water
2.89 + 3W = 10

e. Solve the equation to determine the cost of one water bottle. Then write the reason that justifies each step using if – then statements.
Answer:
If: 2.89 + 3W = 10
Then: 2.89 – 2 .89 + 3W = 10 – 2.89 Subtraction property of equality for the additive inverse of 2.89
If: 0 + 3W = 7.11
Then: 3W = 7.11 Additive identity
If: 3W = 7.11
Then: \(\frac{1}{3}\) (3W) = \(\frac{1}{3}\) (7.11) Multiplication property of equality using the multiplicative inverse of 3
If: 1W = 2.37
Then: W = 2.37 Multiplicative identity

A bottle of water costs $2.37.

f. Model the problem using a tape diagram to check your work.
Answer:

10 – 2.89 = 7.11
\(\frac{7.11}{3}\) = 2.37

Question 2.
Weekly Allowance
Charlotte receives a weekly allowance from her parents. She spent half of this week’s allowance at the movies but earned an additional $4 for performing extra chores. If she did not spend any additional money and finished the week with $12, what is Charlotte’s weekly allowance?
a. Write an equation that can be used to find the original amount of Charlotte’s weekly allowance. Let A be the value of Charlotte’s original weekly allowance.
Answer:
\(\frac{1}{2}\) A + 4 = 12

b. Solve the equation to find the original amount of allowance. Then write the reason that justifies each step using if – then statements.
Answer:
If: \(\frac{1}{2}\) A + 4 = 12
Then: \(\frac{1}{2}\) A + 4 – 4 = 12 – 4 Subtraction property of equality for the additive inverse of 4
If: \(\frac{1}{2}\) A + 0 = 8
Then: \(\frac{1}{2}\) A = 8 Additive identity
If : \(\frac{1}{2}\) A = 8
Then: (2) \(\frac{1}{2}\) A = (2)8 Multiplication property of equality using the multiplicative inverse of \(\frac{1}{2}\)
If: 1A = 16
Then: A = 16 Multiplicative identity
The original allowance was $16.

c. Explain your answer in the context of this problem.
Answer:
Charlotte’s weekly allowance is $16.

d. Charlotte’s goal is to save $100 for her beach trip at the end of the summer. Use the amount of weekly allowance you found in part (c) to write an equation to determine the number of weeks that Charlotte must work to meet her goal. Let w represent the number of weeks.
Answer:
16 w = 100
(\(\frac{1}{16}\))16w = (\(\frac{1}{16}\))100
1w = 6.25
w = 6.25

e. In looking at your answer to part (d) and based on the story above, do you think it will take Charlotte that many weeks to meet her goal? Why or why not?
Answer:
Charlotte needs more than 6 weeks’ allowance, so she will need to save 7 weeks’ allowance (and not spend any of it). There are 10–12 weeks in the summer; so, yes, she can do it.

Exercise 3.
Travel Baseball Team
Allen is very excited about joining a travel baseball team for the fall season. He wants to determine how much money he should save to pay for the expenses related to this new team. Players are required to pay for uniforms, travel expenses, and meals.
a. If Allen buys 4 uniform shirts at one time, he gets a $10.00 discount so that the total cost of 4 shirts would be $44. Write an algebraic equation that represents the regular price of one shirt. Solve the equation. Write the reason that justifies each step using if – then statements.
Answer:
s: the cost of one shirt
If: 4s – 10 = 44
Then: 4s – 10 + 10 = 44 + 10 Addition property of equality using the additive inverse of – 10
If: 4s + 0 = 54
Then: 4s = 54 Additive identity
If: 4s = 54
Then: (\(\frac{1}{4}\))4s = (\(\frac{1}{4}\))54 Multiplication property of equality using multiplicative inverse of 4
If: 1s = 13.50
Then: s = 13.50 Multiplicative identity

b. What is the cost of one shirt without the discount?
Answer:
The cost of one shirt is $13.50.

c. What is the cost of one shirt with the discount?
Answer:
4s = 44
(\(\frac{1}{4}\))4s = (\(\frac{1}{4}\))44
1s = 11
s = 11
The cost of one shirt with the discount is $11.00.

d. How much more do you pay per shirt if you buy them one at a time (rather than in bulk)?
Answer:
13.50 – 11.00 = 2.50
One shirt costs $11 if you buy them in bulk. So, Allen would pay $2.50 more per shirt if he bought them one at a time.

Allen’s team was also required to buy two pairs of uniform pants and two baseball caps, which total $68. A pair of pants costs $12 more than a baseball cap.

e. Write an equation that models this situation. Let c represent the cost of a baseball cap.
Answer:
2(cap + 1 pair of pants) = 68
2(c + c + 12) = 68 or 2 ( 2c + 12) = 68 or 4c + 24 = 68

f. Solve the equation algebraically to find the cost of a baseball cap. Write the reason that justifies each step using if – then statements.
Answer:
If: 2 (2 c + 12) = 68
Then: (\(\frac{1}{2}\))(2 )(2 c + 12) = (\(\frac{1}{2}\))68 Multiplication property of equality using the multiplicative inverse of 2
If: 1(2c + 12) = 34
Then: 2c + 12 = 34 Multiplicative identity
If: 2c + 12 = 34
Then: 2c + 12 – 12 = 34 – 12 Subtraction property of equality for the additive inverse of 12
If: 2c + 0 = 22
Then: 2c = 22 Additive identity
If: 2c = 22
Then: (\(\frac{1}{2}\))2c = (\(\frac{1}{2}\))22 Multiplication property of equality using the multiplicative inverse of 2
If: 1c = 11
Then: c = 11 Multiplicative identity

g. Model the problem using a tape diagram in order to check your work from part (f).
Answer:

h. What is the cost of one cap?
Answer:
The cost of one cap is $11.

i. What is the cost of one pair of pants?
Answer:
11 + 12 = 23 The cost of one pair of pants is $23.

Eureka Math Grade 7 Module 2 Lesson 23 Problem Set Answer Key

For Exercises 1–4, solve each equation algebraically using if – then statements to justify your steps.

Question 1.
\(\frac{2}{3}\) x – 4 = 20
Answer:
If: \(\frac{2}{3}\) x – 4 = 20
Then: \(\frac{2}{3}\) x – 4 + 4 = 20 + 4 Addition property of equality using the additive inverse of – 4
If: \(\frac{2}{3}\) x + 0 = 24
Then: \(\frac{2}{3}\) x = 24 Additive identity
If: \(\frac{2}{3}\) x = 24
Then: (\(\frac{3}{2}\)) \(\frac{2}{3}\) x = (\(\frac{3}{2}\))24 Multiplication property of equality using the multiplicative inverse of \(\frac{2}{3}\)
If: 1x = 36
Then: x = 36 Multiplicative identity

Question 2.
4 = \(\frac{ – 1 + x}{2}\)
Answer:
If: 4 = \(\frac{ – 1 + x}{2}\)
Then: 2 (4) = 2 (\(\frac{ – 1 + x}{2}\)) Multiplication property of equality using the multiplicative inverse of \(\frac{1}{2}\)
If: 8 = 1 ( – 1 + x)
Then: 8 = – 1 + x Multiplicative identity
If: 8 = – 1 + x
Then: 8 – ( – 1) = – 1 – ( – 1) + x Subtraction property of equality for the additive inverse of – 1
If: 9 = 0 + x
Then: 9 = x Additive identity

Question 3.
12(x + 9) = – 108
Answer:
If: 12(x + 9) = – 108
Then: (\(\frac{1}{12}\))12(x + 9) = (\(\frac{1}{12}\))( – 108) Multiplication property of equality using the multiplicative inverse of 12
If: 1 (x + 9) = – 9
Then: x + 9 = – 9 Multiplicative identity
If: x + 9 = – 9
Then: x + 9 – 9 = – 9 – 9 Subtraction property of equality for the additive inverse of 9
If: x + 0 = – 18
Then: x = – 18 Additive identity

Question 4.
5x + 14 = – 7
Answer:
If: 5x + 14 = – 7
Then: 5x + 14 – 14 = – 7 – 14 Subtraction property of equality for the additive inverse of 14
If: 5x + 0 = – 21
Then: 5x = – 21 Additive identity
If: 5x = – 21
Then: (\(\frac{1}{5}\))5x = (\(\frac{1}{5}\))( – 21) Multiplication property of equality using the multiplicative inverse of 5
If: 1x = – 4.2
Then: x = – 4.2 Multiplicative identity

For Exercises 5–7, write an equation to represent each word problem. Solve the equation showing the steps, and then state the value of the variable in the context of the situation.

Question 5.
A plumber has a very long piece of pipe that is used to run city water parallel to a major roadway. The pipe is cut into two sections. One section of pipe is 12 ft. shorter than the other. If \(\frac{3}{4}\) of the length of the shorter pipe is
120 ft., how long is the longer piece of the pipe?
Answer:
Let x represent the longer piece of pipe.
If: \(\frac{3}{4}\)(x – 12) = 120
Then: \(\frac{4}{3}\) (\(\frac{3}{4}\))(x – 12) = (\(\frac{4}{3}\))120 Multiplication property of equality using the multiplicative inverse of \(\frac{3}{4}\)
If: 1(x – 12) = 160
Then: x – 12 = 160 Multiplicative identity
If: x – 12 = 160
Then: x – 12 + 12 = 160 + 12 Addition property of equality for the additive inverse of – 12
If: x + 0 = 172
Then: x = 172 Additive identity
The longer piece of pipe is 172 ft.

Question 6.
Bob’s monthly phone bill is made up of a $10 fee plus $0.05 per minute. Bob’s phone bill for July was $22. Write an equation to model the situation using m to represent the number of minutes. Solve the equation to determine the number of phone minutes Bob used in July.
Answer:
Let m represent the number of phone minutes Bob used.
If: 10 + 0.05m = 22
Then: 10 – 10 + 0.05m = 22 – 10 Subtraction property of equality for the additive inverse of 10
If: 0 + 0.05m = 12
Then: 0.05m = 12 Additive identity
If: 0.05m = 12
Then: (\(\frac{1}{0.05}\))0.05m = (\(\frac{1}{0.05}\))12 Multiplication property of equality using the multiplicative inverse of 0.05
If: 1m = 240
Then: m = 240 Multiplicative identity
Bob used 240 phone minutes in July.

Question 7.
Kym switched cell phone plans. She signed up for a new plan that will save her $3.50 per month compared to her old cell phone plan. The cost of the new phone plan for an entire year is $294. How much did Kym pay per month under her old phone plan?
Answer:
Let n represent the amount Kym paid per month for her old cell phone plan.
If: 294 = 12(n – 3.50)
Then: (\(\frac{1}{12}\))(294) = (\(\frac{1}{12}\))12(n – 3.50) Multiplication property of equality using the multiplicative inverse of 12
If: 24.5 = 1 (n – 3.50)
Then: 24.5 = n – 3.50 Multiplicative identity
If: 24.5 = n – 3.50
Then: 24.5 + 3.50 = n – 3.50 + 3.50 Addition property of equality for the additive inverse of – 3.50
If: 28 = n + 0
Then: 28 = n Additive identity
Kym paid $28 per month for her old cell phone plan.

Eureka Math Grade 7 Module 2 Lesson 23 Exit Ticket Answer Key

Andrew’s math teacher entered the seventh – grade students in a math competition. There was an enrollment fee of $30 and also an $11 charge for each packet of 10 tests. The total cost was $151. How many tests were purchased?
Set up an equation to model this situation, solve it using if – then statements, and justify the reasons for each step in your solution.
Answer:
Let p represent the number of test packets.
Enrollment fee + cost of test = 151
If: 30 + 11p = 151
Then: 30 – 30 + 11p = 151 – 30 Subtraction property of equality for the additive inverse of 30
If: 0 + 11p = 121
Then: 11p = 121 Additive identity
If: 11p = 121
Then: \(\frac{1}{11}\) (11p) = \(\frac{1}{11}\) (121) Multiplication property of equality using the multiplicative inverse of 11
If: 1p = 11
Then: p = 11 Multiplicative identity
Andrew’s math teacher bought 11 packets of tests. There were 10 tests in each packet, and 10×11 = 110.
So, there were 110 tests purchased.

Engage NY Eureka Math 7th Grade Module 2 Lesson 22 Answer Key

Eureka Math Grade 7 Module 2 Lesson 22 Example Answer Key

Example 1.
Yoshiro’s New Puppy
Yoshiro has a new puppy. She decides to create an enclosure for her puppy in her backyard. The enclosure is in the shape of a hexagon (six-sided polygon) with one pair of opposite sides running the same distance along the length of two parallel flower beds. There are two boundaries at one end of the flower beds that are 10 ft. and 12 ft., respectively, and at the other end, the two boundaries are 15 ft. and 20 ft., respectively. If the perimeter of the enclosure is 137 ft., what is the length of each side that runs along the flower bed?

→ What is the general shape of the puppy yard? Draw a sketch of the puppy yard.

→ Write an equation that would model finding the perimeter of the puppy yard.
→ The sum of the lengths of the sides = Perimeter
n + n + 10 + 12 + 20 + 15 = 137

→ Model and solve this equation with a tape diagram.
→ Sample response:

→ Now review making zero in an equation and making one in an equation. Explicitly connect making zero and making one in the next question to the bar model diagram. Subtracting 57 from 137 in the bar diagram is the same as using the subtraction property of equality (i.e., subtracting 57 from both sides of the equation in order to make zero). Dividing 80 by 2 to find the size of two equal groups that total 80 is the same as using the multiplicative property of equality (i.e., multiplying each side of the equation by \(\frac{1}{2}\) to make one group of n).
→ Use algebra to solve this equation.
→ First, use the additive inverse to find out what the lengths of the two missing sides are together. Then, use the multiplicative inverse to find the length of one of the two equal sides. Sum of missing sides + Sum of known sides = Perimeter
If: 2n + 57 = 137
Then: 2n + 57-57 = 137-57 Subtraction property of equality
If: 2n + 0 = 80
Then: 2n = 80 Additive identity
If: 2n = 80
Then: \(\frac{1}{2}\) (2n) = \(\frac{1}{2}\)(80) Multiplication property of equality
If: 1n = 40
Then: n = 40 Multiplicative identity

→ Does your solution make sense in this context? Why?
→ Yes, 40 ft. makes sense because when you replace the two missing sides of the hexagon with 40 in the number sentence ( 40 + 40 + 10 + 12 + 20 + 15 = 137), the lengths of the sides reach a total of 137.

Example 2.
Swim Practice
Jenny is on the local swim team for the summer and has swim practice four days per week. The schedule is the same each day. The team swims in the morning and then again for 2 hours in the evening. If she swims 12 hours per week, how long does she swim each morning?
Answer:
→ Write an algebraic equation to model this problem. Draw a tape diagram to model this problem.
→ Let x = number of hours of swimming each morning
Model days per week (number of hours swimming a.m. and p.m.) = hours of swimming total

Recall in the last problem that students used making zero first and then making one to solve the equation. Explicitly connect making zero and making one in the previous statement to the tape diagram.

→ Solve the equations algebraically and graphically with the help of the tape diagram.
→ Sample response:

→ Does your solution make sense in this context? Why?
→ Yes, if Jenny swims 1 hour in the morning and 2 hours in the evening for a total of 3 hours per day and swims 4 days per week, then 3(4) = 12 hours for the entire week.

Eureka Math Grade 7 Module 2 Lesson 22 Exercise Answer Key

Solve each equation algebraically using if–then statements to justify each step.

Question 1.
5x + 4 = 19
Answer:
If: 5x + 4 = 19
Then: 5x + 4-4 = 19-4 Subtraction property of equality for the additive inverse of 4
If: 5x + 0 = 15
Then: 5x = 15 Additive identity
If: 5x = 15
Then: \(\frac{1}{5}\) (5x) = (\(\frac{1}{5}\))15 Multiplication property of equality for the multiplicative inverse of 5
If: 1x = 3
Then: x = 3 Multiplicative identity

Question 2.
15x + 14 = 19
Answer:
If: 15x + 14 = 19
Then: 15x + 14-14 = 19-14 Subtraction property of equality for the additive inverse of 14
If: 15x + 0 = 5
Then: 15x = 5 Additive Identity
If: 15x = 5
Then: \(\frac{1}{15}\) (15x) = (\(\frac{1}{15}\))5 Multiplication property of equality for the multiplicative inverse of 15
If: 1x = \(\frac{1}{3}\)
Then: x = \(\frac{1}{3}\) Multiplicative identity

Question 3.
Claire’s mom found a very good price on a large computer monitor. She paid $325 for a monitor that was only $65 more than half the original price. What was the original price?
Answer:
x: the original price of the monitor
If: \(\frac{1}{2}\) x + 65 = 325
Then: \(\frac{1}{2}\) x + 65-65 = 325-65 Subtraction property of equality for the additive inverse of 65
If: \(\frac{1}{2}\) x + 0 = 260
Then: \(\frac{1}{2}\) x = 260 Additive identity
If: \(\frac{1}{2}\) x = 260
Then: (2)\(\frac{1}{2}\) x = (2)260 Multiplication property of equality for the multiplicative inverse of \(\frac{1}{2}\)
If: 1x = 520
Then: x = 520 Multiplicative identity
The original price was $520.

Question 4.
2(x + 4) = 18
Answer:
If: 2(x + 4) = 18
Then: \(\frac{1}{2}\) (2 (x + 4)) = \(\frac{1}{2}\) (18) Multiplication property of equality using the multiplicative inverse of 2
If: 1(x + 4) = 9
Then: x + 4 = 9 Multiplicative identity
If: x + 4 = 9
Then: x + 4-4 = 9-4 Subtraction property of equality for the additive inverse of 4
If: x + 0 = 5
Then: x = 5 Additive identity

Question 5.
Ben’s family left for vacation after his dad came home from work on Friday. The entire trip was 600 mi. Dad was very tired after working a long day and decided to stop and spend the night in a hotel after 4 hours of driving. The next morning, Dad drove the remainder of the trip. If the average speed of the car was 60 miles per hour, what was the remaining time left to drive on the second part of the trip? Remember: Distance = rate multiplied by time.
Answer:
m: the number of miles driven on the second day
60 (m + 4) = 600
If: 60 (m + 4) = 600
Then: (\(\frac{1}{60}\))60 (m + 4) = (\(\frac{1}{60}\))600 Multiplication property of equality for the multiplicative inverse of 60
If: 1 ( m + 4) = 10
Then: m + 4 = 10 Multiplicative identity
If: m + 4 = 10
Then: m + 4-4 = 10-4 Subtraction property of equality for the additive inverse of 4
If: m + 0 = 6
Then: m = 6 Additive identity
There were 6 hours left to drive.

Eureka Math Grade 7 Module 2 Lesson 22 Problem Set Answer Key

For each problem below, explain the steps in finding the value of the variable. Then find the value of the variable, showing each step. Write if–then statements to justify each step in solving the equation.

Question 1.
7(m + 5) = 21
Answer:
Multiply both sides of the equation by \(\frac{1}{7}\) , and then subtract 5 from both sides of the equation; m = -2.
If: 7(m + 5) = 21
Then: \(\frac{1}{7}\) (7(m + 5)) = \(\frac{1}{7}\) (21) Multiplication property of equality using the multiplicative inverse of 7
If: 1 (m + 5) = 3
Then: m + 5 = 3 Multiplicative identity
If: m + 5 = 3
Then: m + 5-5 = 3-5 Subtraction property of equality for the additive inverse of 5
If: m + 0 = -2
Then: m = -2 Additive identity

Question 2.
-2v + 9 = 25
Answer:
Subtract 9 from both sides of the equation, and then multiply both sides of the equation by –\(\frac{1}{2}\); v = -8.
If: -2v + 9 = 25
Then: -2v + 9-9 = 25-9 Subtraction property of equality for the additive inverse of 9
If: -2v + 0 = 16
Then: -2v = 16 Additive identity
If: -2v = 16
Then: –\(\frac{1}{2}\) (-2v) = –\(\frac{1}{2}\) (16) Multiplication property of equality using the multiplicative inverse of -2
If: 1 v = -8
Then: v = -8 Multiplicative identity

Question 3.
\(\frac{1}{3}\) y-18 = 2
Answer:
Add 18 to both sides of the equation, and then multiply both sides of the equation by 3; y = 60.
If: \(\frac{1}{3}\) y-18 = 2
Then: \(\frac{1}{3}\) y-18 + 18 = 2 + 18 Addition property of equality for the additive inverse of –18
If: \(\frac{1}{3}\) y + 0 = 20
Then: \(\frac{1}{3}\) y = 20 Additive identity
If: \(\frac{1}{3}\) y = 20
Then: 3(\(\frac{1}{3}\) y) = 3 (20) Multiplication property of equality using the multiplicative inverse of \(\frac{1}{3}\)
If: 1 y = 60
Then: y = 60 Multiplicative identity

Question 4.
6-8p = 38
Answer:
Subtract 6 from both sides of the equation, and then multiply both sides of the equation by –\(\frac{1}{8}\); p = -4.
If: 6-8p = 38
Then: 6-6-8p = 38-6 Subtraction property of equality for the additive inverse of 6
If: 0 + (-8p) = 32
Then: -8p = 32 Additive identity
If: -8p = 32
Then: (-\(\frac{1}{8}\))(-8p) = (-\(\frac{1}{8}\))32 Multiplication property of equality using the multiplicative inverse of -8
If: 1 p = -4
Then: p = -4 Multiplicative identity

Question 5.
15 = 5k-13
Answer:
Add 13 to both sides of the equation, and then multiply both sides of the equation by \(\frac{1}{5}\); k = 5.6.
If: 15 = 5k-13
Then: 15 + 13 = 5k-13 + 13 Addition property of equality for the additive inverse of -13
If: 28 = 5k + 0
Then: 28 = 5k Additive identity
If: 28 = 5k
Then: (\(\frac{1}{5}\))28 = (\(\frac{1}{5}\))5k Multiplication property of equality using the multiplicative inverse of 5
If: 5.6 = 1k
Then: 5.6 = k Multiplicative identity

Eureka Math Grade 7 Module 2 Lesson 22 Exit Ticket Answer Key

Susan and Bonnie are shopping for school clothes. Susan has $50 and a coupon for a $10 discount at a clothing store where each shirt costs $12.
Susan thinks that she can buy three shirts, but Bonnie says that Susan can buy five shirts. The equations they used to model the problem are listed below. Solve each equation algebraically, justify your steps, and determine who is correct and why?

Answer:
Bonnie is correct. The equation that would model this situation is 12n-10 = 50. Solving this equation would involve making zero by adding 10. And by doing so, 12n-10 + 10 = 50 + 10, we arrive at 12n = 60. So, if a group of shirts that cost $12 each totals $60, then there must be five shirts since \(\frac{60}{12}\) equals 5.
Bonnie’s Equation:
12n-10 = 50
12n-10 + 10 = 50 + 10 Addition property of equality for the additive inverse of -10
12n + 0 = 60
12n = 60 Additive identity
(\(\frac{1}{12}\))12n = (\(\frac{1}{12}\))60 Multiplication property of equality using the multiplicative inverse of 12
1n = 5
n = 5 Multiplicative identity

Susan’s Equation:
12n + 10 = 50
12n + 10-10 = 50-10 Subtraction property of equality for the additive inverse of 10
12n + 0 = 40
12n = 40 Additive identity
(\(\frac{1}{12}\))12n = (\(\frac{1}{12}\))40 Multiplication property of equality using the multiplicative inverse of 12
1n = 3 \(\frac{1}{3}\)
n = 3 \(\frac{1}{3}\) Multiplicative identity

Engage NY Eureka Math 7th Grade Module 2 Lesson 21 Answer Key

Eureka Math Grade 7 Module 2 Lesson 21 Exercise Answer Key

Question 1.
The table below shows two hands from the Integer Game and a series of changes that occurred to each hand. Part of the table is completed for you. Complete the remaining part of the table; then summarize the results.

Answer:

Since the sums of each original hand are the same, the same cards can be added, subtracted, multiplied, and divided, and the sums will remain equal to each other.

Question 2.
Complete the table below using the multiplication property of equality.

Answer:

Eureka Math Grade 7 Module 2 Lesson 21 Exploratory Challenge Answer Key

Exploratory Challenge: Integer Game Revisited
Let’s investigate what happens if a card is added or removed from a hand of integers.

Event 1.

Answer:

Repeat this process with one minor change; this time both students receive one integer card containing the same negative value. Have students record their new scores and, after comparing with their partners, write a conclusion using an if–then statement.

Event 1 (both partners receive the card – 1)

Series of questions leading to the conclusion:
→ Were your scores the same when we began?
→ Yes
→ Did you add the same values to your hand each time?
→ Yes
→ Did the value of your hand change each time you added a new card?
→ Yes
→ Was the value of your hand still the same as your partner’s after each card was added?
→ Yes
→ Why did the value of your hand remain the same after you added the new cards?
→ We started with the same sum; therefore, when we added a new card, we had equivalent expressions, which resulted in the same sum.
→ Since your original cards were different, but your original sum was the same, write a conclusion that was exemplified by this event.
→ If the original sums were equal, you can add a number, either positive or negative, and the sums will remain equal.

Event 2.

Answer:

Event 3.

Answer:

If the sums are the same, then the product of the sums will remain equal when both are multiplied by the same rational number.

Event 4.

Expression:

Conclusion:
Answer:

Eureka Math Grade 7 Module 2 Lesson 21 Problem Set Answer Key

Question 1.
Evaluate the following numerical expressions.
a. 2 + ( – 3) + 7=
Answer:
6

b. – 4 – 1=
Answer:
– 5

c. – \(\frac{5}{2}\) ×2=
Answer:
– 5

d. – 10÷2 + 3=
Answer:
– 2

e. (\(\frac{1}{2}\) )(8) + 2 =
Answer:
6

f. 3 + ( – 4) – 1=
Answer:
– 2

Question 2.
Which expressions from Exercise 1 are equal?
Answer:
Expressions (a) and (e) are equivalent.
Expressions (b) and (c) are equivalent.
Expressions (d) and (f) are equivalent.

Question 3.
If two of the equivalent expressions from Exercise 1 are divided by 3, write an if–then statement using the properties of equality.
Answer:
If 2 + ( – 3) + 7=(\(\frac{1}{2}\) )(8) + 2, then (2 + ( – 3) + 7)÷3=((\(\frac{1}{2}\) )(8) + 2)÷3.

Question 4.
Write an if–then statement if – 3 is multiplied by the following equation: – 1 – 3= – 4.
Answer:
If – 1 – 3 = – 4, then – 3( – 1 – 3) = – 3( – 4)

Question 5.
Simplify the expression.
5 + 6 – 5 + 4 + 7 – 3 + 6 – 3
Answer:
=17

Using the expression, write an equation.
Answer:
5 + 6 – 5 + 4 + 7 – 3 + 6 – 3=17

Rewrite the equation if 5 is added to both expressions.
Answer:
5 + 6 – 5 + 4 + 7 – 3 + 6 – 3 + 5=17 + 5

Write an if–then statement using the properties of equality.
Answer:
If 5 + 6 – 5 + 4 + 7 – 3 + 6 – 3 = 17, then 5 + 6 – 5 + 4 + 7 – 3 + 6 – 3 + 5
= 17 + 5

Eureka Math Grade 7 Module 2 Lesson 21 Exit Ticket Answer Key

Compare the two expressions.
Expression 1: 6 + 7 + – 5
Expression 2: – 5 + 10 + 3

Question 1.
Are the two expressions equivalent? How do you know?
Answer:
Yes, the expressions are equivalent because Expression 1 is equal to 8 and Expression 2 is equal to 8, as well. When two expressions evaluate to the same number, they are equivalent.

Question 2.
Subtract – 5 from each expression. Write the new numerical expression, and write a conclusion as an if–then statement.
Answer:
Expression 1:
6 + 7±5 – ( – 5)
13
Expression 2: – 5 + 10 + 3 – ( – 5)
13
If 6 + 7 + – 5= – 5 + 10 + 3, then 6 + 7 + – 5 – ( – 5)= – 5 + 10 + 3 – ( – 5).
If Expression 1= Expression 2, then (Expression 1 – ( – 5))=(Expression 2 – ( – 5)).

Question 3.
Add 4 to each expression. Write the new numerical expression, and write a conclusion as an if–then statement.
Answer:
Expression 1: 6 + 7±5 + 4
12
Expression 2: – 5 + 10 + 3 + 4
12
If 6 + 7 + – 5= – 5 + 10 + 3, then 6 + 7 + – 5 + 4= – 5 + 10 + 3 + 4.
If Expression 1 = Expression 2, then (Expression 1 + 4)=(Expression 2 + 4).

Question 4.
Divide each expression by – 2. Write the new numerical expression, and write a conclusion as an if–then statement.
Answer:
Expression 1: (6 + 7 + – 5)÷ – 2
8÷ – 2
– 4
Expression 2: ( – 5 + 10 + 3)÷ – 2
8÷ – 2
– 4
If 6 + 7 + – 5= – 5 + 10 + 3, then (6 + 7 + – 5)÷ – 2=( – 5 + 10 + 3)÷ – 2
If Expression 1= Expression 2, then (Expression 1 ÷ – 2)=(Expression 2 ÷ – 2).