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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).

Engage NY Eureka Math 7th Grade Module 2 Lesson 20 Answer Key

Eureka Math Grade 7 Module 2 Lesson 20 Task Answer Key

TASK: Using the above information, semiannual statements, register, and beginning balance, do the following:
1. Record the beginning balance and all transactions from the account statements into the register.
2. Determine the annual gain or loss as well as the overall 5-year gain or loss.
3. Determine if there is enough money in the account after 5 years to cover $30,000 of college expenses for Justin and Adrienne’s daughter. Write a summary to defend your answer. Be sure to indicate how much money is in excess, or the shortage that exists.
4. Answer the related questions that follow.

Answer:

Question 5.
Register

Answer:

→ Describe the process of completing the register.
→ Starting with the beginning balance, fill in the description of the transaction and the amount. If the transaction is an investment loss,
withdrawal, or fee, then the amount is recorded in the payment column. If the transaction is an investment gain or deposit, then the amount is recorded in the deposit column. To obtain the new balance, subtract the payment amount, or add the deposit amount, from the balance on the preceding line. Record the new balance, and use that balance to complete the next line.
→ Describe how to find the broker’s fee.
→ The broker’s fee is 2% of the transaction amount. To find the broker’s fee, you must first find the total
of the transaction amount. Once you have that, write the percent as a fraction out of 100 and multiply
the fraction by the transaction amount. This result is the amount of the broker’s fee, which is then
subtracted from the preceding balance.
Example: 2% of $2,500
\(\frac{2}{100}\) × 2,500 = \(\frac{1}{50}\) × 2,500 = 50
→ Compare your register with the person next to you. Did each of you list the transactions in the same order?
Does it make a difference?
→ The order is probably not the same. The order of the transactions for each 6-month period does not
make a difference.
→ Continue to compare your registers. Do you both get the same balance at the end of 2012? If not, switch papers and check to see if you can find your neighbor’s mistake.

Question 6.
Annual Gain/Loss Summary

Answer:

Question 7.
Summary
Answer:
There is not enough money in the account at the end of 5 years to cover the college expenses, but it is close. They needed at least $30,000 in the account to cover the expenses, and there was $28,118.59, leaving a shortage of $1,881.41.

Question 8.
Related Questions
a. For the first half of 2009, there was a $700 gain on the initial investment of $20,000. Represent the gain as a percentage of the initial investment.
Answer:
\(\frac{x}{100}\) =700/(20,000) The gain was 3.5% of $20,000.

b. Based on the gains and losses on their investment during this 5-year period, over what period of time was their investment not doing well? How do you know? What factors might contribute to this?
Answer:
The investment was not doing well in 2009. There were losses on the investment for both halves of the year, and $500 was taken out of the account. It could be because the economy was doing badly, and a recession affected the investment’s performance.

c. In math class, Jaheim and Frank were working on finding the total amount of the investment after 5 years. As a final step, Jaheim subtracted $150 for administrative fees from the balance he arrived at after adding in all the deposits and subtracting out the one withdrawal and broker’s fee. For every semiannual statement, Frank subtracted $15 from the account balance for the administrative fee. Both boys arrived at the same ending 5-year balance. How is this possible? Explain.
Answer:
Jaheim took the $15 fee and multiplied it by 10, since there were 10 statements, and deducted the $150 total. Frank subtracted $15 from the account balance for each statement. That was 10 times. So, both ways produce the same result: reducing the account balance by $150 overall.

d. Based on the past statements for their investment account, predict what activity you might expect to see on Adrienne and Justin’s January–June 2013 account statement. Then record it in the register to arrive at the balance as of June 30, 2013.
Answer:
I predict the account will continue to produce gains. The gains have been around $900 for the past four statements, so I predict it will be about $900 again, since it decreased by a little bit the last time, and there was a $909.71 gain the last time. If I take away $15 for the administrative fee, the balance would go up by $885, and it would be $29,003.59.

e. Using the answer from part (d), if their daughter’s college bill is due in September of 2013, how much money do you estimate will be in their investment account at the end of August 2013 before the college bill is paid? Support your answer.
Answer:
Their investment could gain more money for July and August. Right now, it is gaining about $900 per statement. If I divide that by 6, it equals $150 (which is the average gain per month). So, for July and August I estimate that it will earn about another $300 (including the $15 fee), so there might be $29,333.59 in the account.

Eureka Math Grade 7 Module 2 Lesson 20 Exercise Answer Key

Below is a transaction log of a business entertainment account. The transactions are completed, and the ending balance in the account is $525.55. Determine the beginning balance.

Answer:

Eureka Math Grade 7 Module 2 Lesson 20 Exit Ticket Answer Key

Question 1.
Using the incomplete register below, work forward and backward to determine the beginning and ending balances after the series of transactions listed.

Answer:

Question 2.
Write an expression to represent the balance after the paycheck was deposited on 1/31/12. Let x represent the beginning balance.
Answer:
x+350.55

Question 3.
Write a numerical expression to represent the balance after the transaction for Main Street Jeweler’s was made.
Answer:
685.26-425.30

Engage NY Eureka Math 7th Grade Module 2 Lesson 19 Answer Key

Eureka Math Grade 7 Module 2 Lesson 19 Example Answer Key

Example 1.
Tic-Tac-Toe Review
Fill in the 9 spaces with one expression from the list below. Use one expression per space. You will use 9 of the expressions:
12-4x
8x+4-12x
8(\(\frac{1}{2}\) x -2)

12-6x+2x
-4x+4
x-2+2x-4
4x-12
4(x-4)
3(x-2)
0.1(40x)-\(\frac{1}{2}\) (24)

Example 2.

Answer:

Example 3.
An item that has an original price of x dollars is discounted 33%.
a. Write an expression that represents the amount of the discount.
Answer:
0.33x

b. Write two equivalent expressions that represent the new, discounted price.
Answer:
x-0 .33x
x(1-0.33)
x(0.67)

c. Use one of your expressions to calculate the new, discounted price if the original price was $56.
Answer:
0.67x
0.67(56)
37.52
The new discounted price is $37.52.

d. How would the expressions you created in parts (a) and (b) have to change if the item’s price had increased by 33% instead of decreased by 33%?
Answer:
Instead of subtracting 0.33x, you would have to add for the increase. The expression would be
x+0.33x
1.33x.

Example 4.

Answer:

Eureka Math Grade 7 Module 2 Lesson 19 Problem Set Answer Key

Solve the following problems. If necessary, round to the nearest penny.

Question 1.
A family of 12 went to the local Italian restaurant for dinner. Every family member ordered a drink and meal, 3 ordered an appetizer, and 6 people ordered cake for dessert.
a. Write an expression that can be used to figure out the cost of the bill. Include the definitions for the variables the server used.
Answer:
d= drink
m = meal
a = appetizer
c = cake
12d+12m+3a+6c

b. The waitress wrote on her ordering pad the following expression: 3(4d+4m+a+2c). Was she correct? Explain why or why not.
Answer:
Yes, she was correct because her expression is equivalent to the expression from part (a). If the distributive property is applied, the expressions would be exact.

c. What is the cost of the bill if a drink costs $3, a meal costs $20, an appetizer costs $5.50, and a slice of cake costs $3.75?
Answer:
12d+12m+3a+6c
12(3)+12(20)+3(5.50)+6(3.75)
36+240+16.50+22.50
315
The cost of the bill is $315.

d. Suppose the family had a 10% discount coupon for the entire check and then left an 18% tip. What is the total?
Answer:
(315-315(0.10))+0.18(315-315(0.10))
1.18(315-315(0.10))
1.18(315(0.90))
334.53
After the discount and tip, the new total is $334.53.

Question 2.
Sally designs web pages for customers. She charges $135.50 per web page; however, she must pay a monthly rental fee of $650 for her office. Write an expression to determine her take-home pay after expenses. If Sally designed 5 web pages last month, what was her take-home pay after expenses?
w= number of webpages Sally designs
135.50w-650
135.50(5)-650
27.50
After expenses, Sally’s take-home pay is $27.50.

Question 3.
While shopping, Megan and her friend Rylie find a pair of boots on sale for 25% off the original price. Megan calculates the final cost of the boots by first deducting the 25% and then adding the 6% sales tax. Rylie thinks Megan will pay less if she pays the 6% sales tax first and then takes the 25% discount.
a. Write an expression to represent each girl’s scenario if the original price of the boots was x dollars.
Answer:

b. Evaluate each expression if the boots originally cost $200.
Answer:

Using both Megan’s and Rylie’s methods would show that the boots would cost $159.

c. Who was right? Explain how you know.
Answer:
Neither girl was right. They both pay the same amount.

d. Explain how both girls’ expressions are equivalent.
Answer:
Two expressions are equivalent if they yield the same number for every substitution of numbers for the variables in each expression. Since multiplication is commutative, the order of the multiplication can be reversed, and the result will remain the same.

Eureka Math Grade 7 Module 2 Lesson 19 Exit Ticket Answer Key

Question 1.
Write three equivalent expressions that can be used to find the final price of an item costing g dollars that is on sale for 15% off and charged 7% sales tax.
Answer:
(x-0.15x) + 0.07(x-0.15x) 1.07(x -0 .15x) 1.07(0.85x) or 0.85(1.07)x

Question 2.
Using all of the expressions, determine the final price for an item that costs $75. If necessary, round to the nearest penny.
Answer:

The final price of an item that costs $75 is $68.21.

Question 3.
If each expression yields the same final sale price, is there anything to be gained by using one over the other?
Answer:
Using the final two expressions makes the problem shorter and offers fewer areas to make errors. However, all three expressions are correct.

Question 4.
Describe the benefits, special characteristics, and properties of each expression.
Answer:
The second and third expressions collect like terms. The third expression can be written either way using the commutative property of multiplication. The first and second expressions find the discount price first, whereas the third expression is written in terms of percent paid.

Engage NY Eureka Math 7th Grade Module 2 Lesson 18 Answer Key

Eureka Math Grade 7 Module 2 Lesson 18 Exercise Answer Key

Exercise 1.
John’s father asked him to compare several different cell phone plans and identify which plan will be the least expensive for the family. Each phone company charges a monthly fee, but this fee does not cover any services: phone lines, texting, or internet access. Use the information contained in the table below to answer the following questions.
Cell Phone Plans

All members of the family may not want identical plans; therefore, we will let x represent the number of phone lines, y represent the number of phone lines with unlimited texting, and z represent the number of phone lines with internet access.
Expression
Company A __

Company B __

Company C ___
Answer:
Company A 70 + 20x + 15y + 15z
Company B 90 + 15x + 10y + 20z
Company C 200 + 10x

Using the expressions above, find the cost to the family of each company’s phone plan if:
a. Four people want a phone line, four people want unlimited texting, and the family needs two internet lines.

Answer:

Which cell phone company should John’s family use? Why?
Answer:
The family should choose Company B since it is cheaper than the others for the given values

b. Four people want a phone line, four people want unlimited texting, and all four people want internet lines.

Answer:

Which cell phone company should John’s family use? Why?
Answer:
The family should choose Company C since it is cheaper than the other companies for the given values.

c. Two people want a phone line, two people want unlimited texting, and the family needs two internet lines.

Answer:

Which cell phone company should John’s family use? Why?
Answer:
The family should choose Company A since it is cheaper than the other companies for the given values.

Exercise 2.
Three friends went to the movies. Each purchased a medium – sized popcorn for p dollars and a small soft drink for s dollars.
a. Write the expression that represents the total amount of money (in dollars) the three friends spent at the concession stand.
Answer:
3(p + s)

b. If the concession stand charges $6.50 for a medium – sized popcorn and $4.00 for a small soft drink, how much did the three friends spend on their refreshments altogether?
Answer:
One possible solution is shown here; more solution methods are shown in the discussion that follows.
3(p + s)
3(6.50 + 4)
3(10.50)
31.50
They spent $31.50.

Exercise 3.
Complete the table below by writing equivalent expressions to the given expression and evaluating each expression with the given values.


Answer:

Eureka Math Grade 7 Module 2 Lesson 18 Problem Set Answer Key

Question 1.
Sally is paid a fixed amount of money to walk her neighbor’s dog every day after school. When she is paid each month, she puts aside $20 to spend and saves the remaining amount. Write an expression that represents the amount Sally will save in 6 months if she earns m dollars each month. If Sally is paid $65 each month, how much will she save in 6 months?
Answer:
m= monthly pay
6(m – 20)
6m – 120
For m=65
6(m – 20)
6(65 – 20)
6(45)
270
or
6(m – 20)
6(65 – 20)
390 – 120
270
Sally will save $270 in 6 months.

Question 2.
A football team scored 3 touchdowns, 3 extra points, and 4 field goals.
a. Write an expression to represent the total points the football team scored.
Answer:
t= number of points for a touchdown.
e= number of points for the extra point.
f= number of points for a field goal.
3t + 3e + 4f

b. Write another expression that is equivalent to the one written above.
Answer:
Answers may vary. Sample response: 3t + 3e + 2f + 2f

c. If each touchdown is worth 6 points, each extra point is 1 point, and each field goal is 3 points, how many total points did the team score?
Answer:
3t + 3e + 4f
3(6) + 3(1) + 4(3)
18 + 3 + 12
33

Question 3.
Write three other expressions that are equivalent to 8x – 12.
Answer:
Answers may vary.
4(2x – 3)
6x + 2x – 12
8(x – 1) – 4
– 12 + 8x

Question 4.
Profit is defined as earnings less expenses (earnings – expenses). At the local hot – air balloon festival, the Ma & Pops Ice Cream Truck sells ice cream pops, which cost them $0.75 each but are sold for $2 each. They also paid $50 to the festival’s organizers for a vendor permit. The table below shows the earnings, expenses, and profit earned when 50, 75, and 100 ice cream pops were sold at the festival.

a. Write an expression that represents the profit (in dollars) Ma & Pops earned by selling ice cream pops at the festival.
Answer:
p represents the number of pops sold.
2p – 0.75p – 50

b. Write an equivalent expression.
Answer:
1.25p – 50

c. How much of a profit did Ma & Pops Ice Cream Truck make if it sold 20 ice cream pops? What does this mean? Explain why this might be the case.
Answer:
1.25p – 50
1.25(20) – 50
25 – 50
– 25
They did not make any money; they lost $25. A possible reason is it could have been cold or rainy and people were not buying ice cream.

d. How much of a profit did Ma & Pops Ice Cream Truck make if it sold 75 ice cream pops? What does this mean? Explain why this might be the case.
Answer:
1.25p – 50
1.25(75) – 50
93.75 – 50
43.75
They made a profit of $43.75. Possible reasons are the weather could have been warmer and people bought the ice cream, or people just like to eat ice cream no matter what the weather is.

Eureka Math Grade 7 Module 2 Lesson 18 Exit Ticket Answer Key

Bradley and Louie are roommates at college. At the beginning of the semester, they each paid a security deposit of A dollars. When they move out, their landlord will deduct from this deposit any expenses (B) for excessive wear and tear and refund the remaining amount. Bradley and Louie will share the expenses equally.
→ Write an expression that describes the amount each roommate will receive from the landlord when the lease expires.
→ Evaluate the expression using the following information: Each roommate paid a $125 deposit, and the landlord deducted $50 total for damages.
Answer:
Deposit each person paid: A
Total damages: B
Each roommate receives: A – \(\frac{B}{2}\)
A=125, B=50
A – \(\frac{B}{2}\)
125 – \(\frac{50}{2}\)
125 – 25
100

Engage NY Eureka Math 7th Grade Module 2 Lesson 17 Answer Key

Eureka Math Grade 7 Module 2 Lesson 17 Opening Exercise Answer Key

For his birthday, Zack and three of his friends went to a movie. They each got a ticket for $8.00 and the same snack from the concession stand. If Zack’s mom paid $48 for the group’s tickets and snacks, how much did each snack cost?
The equation 4(s+8)=48 represents the situation when s represents the cost, in dollars, of one snack.
Answer:
4(s + 8) = 48
\(\frac{1}{4}\) (4(s+8))=\(\frac{1}{4}\) (48)
s + 8 = 12
s + 8 – 8 = 12 – 8
s+0=4
s=4

OR
4(s+8)=48
4s+32=48
4s+32-32=48-32
4s+0=16
4s=16
\(\frac{1}{4}\) (4s)=\(\frac{1}{4}\) (16)
1s=4
s=4

Eureka Math Grade 7 Module 2 Lesson 17 Exercise Answer Key

Exercise
The cost of a babysitting service on a cruise is $10 for the first hour and $12 for each additional hour. If the total cost of babysitting baby Aaron was $58, how many hours was Aaron at the sitter?
Algebraic Solution
h = number of additional hours
12h + 10 = 58
12h + 10-10 = 58-10
12h + 0 = 48
(\(\frac{1}{12}\) )(12h)=(48)(\(\frac{1}{12}\) )
1h = 4
h=4

1+4=5
Aaron was with the babysitter for 5 hours

Eureka Math Grade 7 Module 2 Lesson 17 Exploratory Challenge Answer Key

Exploratory Challenge: Expenses on Your Family Vacation
John and Ag are summarizing some of the expenses of their family vacation for themselves and their three children, Louie, Missy, and Bonnie. Write an algebraic equation, create a model to determine how much each item will cost using all of the given information, and answer the questions that follow.
Expenses:

Your Group’s Scenario Solution:
Answer:
Scenario 1
During one rainy day on the vacation, the entire family decided to go watch a matinee movie in the morning and a drive-in movie in the evening. The price for a matinee movie in the morning is different than the cost of a drive-in movie in the evening. The tickets for the matinee morning movie cost $6 each. How much did each person spend that day on movie tickets if the ticket cost for each family member was the same? What was the cost for a ticket for the drive-in movie in the evening?
Answer:
Algebraic Equation & Solution

Morning matinee movie: $6 each
Evening drive-in movie: e each
5(e+6)=75
5e+30=75
5e+30-30=75-30
5e+0=45
(\(\frac{1}{5}\))5e=45(\(\frac{1}{5}\))
1e=9
e=9
OR
5(e+6)=75
(\(\frac{1}{5}\) )5(e+6)=75(\(\frac{1}{5}\) )
e+6=15
e+6-6=15-6
e =9

The total each person spent on movies in one day was $15. The evening drive-in movie costs $9 each

Scenario 2
For dinner one night, the family went to the local pizza parlor. The cost of a soda was $3. If each member of the family had a soda and one slice of pizza, how much did one slice of pizza cost?
Answer:
Algebraic Equation & Solution
One Soda: $3
Slice of Pizza: p dollars
5(p+3)=37.95
5p +15 = 37.95
5p+15-15 =37.95 -15
5p+0=22.95
(\(\frac{1}{5}\) )5p=22.95(\(\frac{1}{5}\) )
1p = 4.59
p= 4.59
OR
5(p+3)=37.95
(\(\frac{1}{5}\) )5(p+3)=(37.95)(\(\frac{1}{5}\) )
p + 3 = 7.59
p + 3-3 = 7.59-3
p = 4.59

one slice of pizza cost $4.59

Scenario 3
One night, John, Louie, and Bonnie went to see the local baseball team play a game. They each bought a game ticket and a hat that cost $10. How much was each ticket to enter the ballpark?
Answer:
Algebraic Equation & Solution
Ticket: t dollars
Hat: $10
3(t +10)=103.83
3t + 30 = 103.83
3t+30-30=103.83-30
3t + 0 = 73.83
(\(\frac{1}{3}\) )3t=73.83(\(\frac{1}{3}\) )
1t = 24.61
t = 24.61
OR
3(t +10)=103.83
(\(\frac{1}{3}\) )3(t +10)=(103.83)(\(\frac{1}{3}\) )
t+10=34.61
t+10-10=34.61-10
t=24.61
One ticket costs $24.61.

Scenario 4.
While John, Louie, and Bonnie went to see the baseball game, Ag and Missy went shopping. They bought a T-shirt for each member of the family and bought two pairs of sandals that cost $10 a pair. How much was each T-shirt?
Answer:
Algebraic Equation & Solution
T-Shirt: t dollars
Sandals: 2 × $10 = $20
5t + 20 = 120
5t +20-20 =120- 20
5t + 0 = 100
(\(\frac{1}{5}\) )5t=100(\(\frac{1}{5}\) )
1t=20
t=20

One T-shirt costs $20.

Scenario 5.
The family flew in an airplane to their vacation destination. Each person had to have his own ticket for the plane and also pay $25 in insurance fees per person. What was the cost of one ticket?
Answer:
Algebraic Equation & Solution
One ticket: t dollars
Insurance: $25 per person
5(t + 25) =875
5t +125-125=875 -125
5t + 0 = 750
5t=750(\(\frac{1}{5}\) )
t = 150
OR
5(t + 25)=875
(\(\frac{1}{5}\) )5(t+25)=(\(\frac{1}{5}\) )(875)
t+25=175
t+25-25=175-25
t = 150
One ticket costs $150.

Scenario 6
While on vacation, the family rented a car to get them to all the places they wanted to see for five days. The car costs a certain amount each day, plus a one-time insurance fee of $50. How much was the daily cost of the car (not including the insurance fees)?
Answer:
Algebraic Equation & Solution
Daily fee: d dollars
Insurance fee: $50
5d + 50 = 400
5d +50-50 =400 –50
5d+0=350
(\(\frac{1}{5}\) )5d=350(\(\frac{1}{5}\) )
1d = 70
d = 70

one day costs $70

Scenario 7.
The family decided to stay in a motel for four nights. The motel charges a nightly fee plus $60 in state taxes. What is the nightly charge with no taxes included?
Answer:
Algebraic Equation & Solution
Nightly charge: n dollars
Taxes: $60
4n + 60 = 400
4n +60-60 =400-60
4n + 0=340
(\(\frac{1}{4}\) )4n=340(\(\frac{1}{4}\) )
1n = 85
n = 85
One night costs $85.

After collaborating with all of the groups, summarize the findings in the table below.

Answer:

Using the results, determine the cost of the following:

Question 1.
A slice of pizza, 1 plane ticket, 2 nights in the motel, and 1 evening movie.
Answer:
4.59 + 150 + 2(85) + 9 = 333.59

Question 2.
One T-shirt, 1 ticket to the baseball game, and 1 day of the rental car.
Answer:
20 + 24.61 + 70 = 114.61

Eureka Math Grade 7 Module 2 Lesson 17 Problem Set Answer Key

Question 1.
A taxi cab in Myrtle Beach charges $2 per mile and $1 for every person. If a taxi cab ride for two people costs $12, how far did the taxi cab travel?
Answer:
Algebraic Equation & Solution
Number of miles: m
People: 2
12-2 = 10 10 ÷2 = 5
2m+2=12
2m+2-2=12-2
2m+0=10
(1/2)2m=10(1/2)
1m=5
m = 5
The taxi cab traveled 5 miles.

Question 2.
Heather works as a waitress at her family’s restaurant. She works 2 hours every morning during the breakfast shift and returns to work each evening for the dinner shift. In the last four days, she worked 28 hours. If Heather works the same number of hours every evening, how many hours did she work during each dinner shift?
Answer:
Algebraic Equation & Solution
Number of morning hours: 2
Number of evening hours: e
4(e +2)=28
4e+8-8=28-8
4e+0=20
(\(\frac{1}{4}\) )4e=20(\(\frac{1}{4}\) )
1e=5
e=5
OR
(\(\frac{1}{4}\) )4(e+2)=28(\(\frac{1}{4}\) )
e+2=7
e+2-2=7-2
e=5
Heather worked 5 hours in the evening.

Question 3.
Jillian exercises 5 times a week. She runs 3 miles each morning and bikes in the evening. If she exercises a total of 30 miles for the week, how many miles does she bike each evening?
Answer:
Algebraic Equation & Solution

Run: 3 mi.
Bikes: b mi.
OR
(\(\frac{1}{5}\) )(b+3)=30(\(\frac{1}{5}\) )
b+3=6
b+3-3=6-3
b=3
Jillian bikes 3 miles every evening.

Question 4.
Marc eats an egg sandwich for breakfast and a big burger for lunch every day. The egg sandwich has 250 calories. If Marc has 5,250 calories for breakfast and lunch for the week in total, how many calories are in one big burger?
Answer:
Algebraic Equation & Solution

Egg Sandwich: 250 cal.
Hamburger: m cal
7(m+250)=5,250
7m+1,750-1750=5250-1750
7m+0=3,500
(\(\frac{1}{7}\) )7m=3,500(\(\frac{1}{7}\) )
1m = 500
m=500
OR
(\(\frac{1}{7}\) )7(m+250)=(\(\frac{1}{7}\) )5,250
m+250=750
m+250-250=750-250
m=500
Each hamburger has 500 calories.

Question 5.
Jackie won tickets playing the bowling game at the local arcade. The first time, she won 60 tickets. The second time, she won a bonus, which was 4 times the number of tickets of the original second prize. Altogether she won 200 tickets. How many tickets was the original second prize?
Answer:
Algebraic Equation & Solution

First Prize: 60 tickets
Second Prize: p tickets
4p+60=200
4p+60-60=200-60
4p+0=140
(\(\frac{1}{4}\) )4p=140(\(\frac{1}{4}\) )
1p=35
p=35
The original second prize was 35 tickets.

Eureka Math Grade 7 Module 2 Lesson 17 Exit Ticket Answer Key

Question 1.
Eric’s father works two part-time jobs, one in the morning and one in the afternoon, and works a total of 40 hours each 5-day workweek. If his schedule is the same each day, and he works 3 hours each morning, how many hours does Eric’s father work each afternoon?
Answer:
Algebraic Equation & Solution
Number of afternoon hours: a
Number of morning hours: 3
5(a+3)=40
5a+15-15=40-15
5a+0= 25
(\(\frac{1}{5}\) )5a=25(\(\frac{1}{5}\) )
a=5

OR
5(a+3)=40
(\(\frac{1}{5}\) )5(a+3)=40(\(\frac{1}{5}\) )
a+3=8
a+3-3=8-3
a=5
Eric’s father works 5 hours in the afternoon.

Question 2.
Henry is using a total of 16 ft. of lumber to make a bookcase. The left and right sides of the bookcase are each 4 ft. high. The top, bottom, and two shelves are all the same length, labeled S. How long is each shelf?
Answer:
Algebraic Equation & Solution
Shelves: s ft.
Sides: 8 ft.
4s+8=16
4s+8-8=16-8
4s+0=8
(\(\frac{1}{4}\) )4s=8(\(\frac{1}{4}\) )
1s=2
s=2
Each shelf is 2 ft. long.

Engage NY Eureka Math 7th Grade Module 4 Lesson 11 Answer Key

Eureka Math Grade 7 Module 4 Lesson 11 Exercise Answer Key

Opening Exercise: Tax, Commission, Gratuity, and Fees
How are each of the following percent applications different, and how are they the same? Solve each problem, and then compare your solution process for each problem.
a. Silvio earns 10% for each car sale he makes while working at a used car dealership. If he sells a used car for $2,000, what is his commission?
Answer:
His commission is $200.

b. Tu’s family stayed at a hotel for 10 nights on their vacation. The hotel charged a 10% room tax, per night. How much did they pay in room taxes if the room cost $200 per night?
Answer:
They paid $200.

c. Eric bought a new computer and printer online. He had to pay 10% in shipping fees. The items totaled $2,000. How much did the shipping cost?
Answer:
The shipping cost $200.

d. Selena had her wedding rehearsal dinner at a restaurant. The restaurant’s policy is that gratuity is included in the bill for large parties. Her father said the food and service were exceptional, so he wanted to leave an extra 10% tip on the total amount of the bill. If the dinner bill totaled $2,000, how much money did her father leave as the extra tip?
Answer:
Her father left $200 as the extra tip.
For each problem, I had to find 10% of the total ($2,000). Even though each problem was different—one was a commission, one was a tax, one was a fee, and one was a gratuity—I arrived at the answer in the same manner, by taking 10% of $2,000 means \(\frac{1}{10}\) of $2,000, which is $200.

Exercises 1 – 4
Show all work; a calculator may be used for calculations.
The school board has approved the addition of a new sports team at your school.
Exercise 1.
The district ordered 30 team uniforms and received a bill for $2,992.50. The total included a 5% discount.
a. The school needs to place another order for two more uniforms. The company said the discount will not apply because the discount only applies to orders of $1,000 or more. How much will the two uniforms cost?
Answer:
Quantity = Percent∙Whole
2,992.50 = 0.95W
2,992.50(\(\frac{1}{0.95}\)) = 0.95(\(\frac{1}{0.95}\))W
3,150 = W
30 uniforms cost $3,150 before the discount. ($3,150)/30 per uniform means each uniform costs $105.
$105 × 2 = $210, so it will cost $210 for 2 uniforms without a discount.

b. The school district does not have to pay the 8% sales tax on the $2,992.50 purchase. Estimate the amount of sales tax the district saved on the $2,992.50 purchase. Explain how you arrived at your estimate.
Answer:
$2,992.50≈$3,000. To find 8% of $3,000, I know 8% of 100 is 8, since percent means per hundred. 8% of 1,000 is ten times as much since 1,000 is ten times as much as 100. 8(10) = 80. Then, I multiplied that by 3 since it is $3,000, so 3(80) = 240. The district saved about $240 in sales tax.

c. A student who loses a uniform must pay a fee equal to 75% of the school’s cost of the uniform. For a uniform that cost the school $105, will the student owe more or less than $75 for the lost uniform? Explain how to use mental math to determine the answer.
Answer:
75% means 75 per hundred. Since the uniform cost more than $100, a 75% fee will be more than $75.

d. Write an equation to represent the proportional relationship between the school’s cost of a uniform and the amount a student must pay for a lost uniform. Use u to represent the uniform cost and s to represent the amount a student must pay for a lost uniform. What is the constant of proportionality?
Answer:
s = 0.75u; the constant of proportionality is 75% = 0.75.

Exercise 2.
A taxpayer claims the new sports team caused his school taxes to increase by 2%.
a. Write an equation to show the relationship between the school taxes before and after a 2% increase. Use b to represent the dollar amount of school tax before the 2% increase and t to represent the dollar amount of school tax after the 2% increase.
Answer:
t = 1.02b

b. Use your equation to complete the table below, listing at least 5 pairs of values.

Answer:

c. On graph paper, graph the relationship modeled by the equation in part (a). Be sure to label the axes and scale.
Answer:

d. Is the relationship proportional? Explain how you know.
Answer:
Yes. The graph is a straight line that touches the point (0,0).

e. What is the constant of proportionality? What does it mean in the context of the situation?
Answer:
The constant of proportionality is 1.02. It means that after the 2% tax increase, $1.02 will be paid for every dollar of tax paid before the increase.

f. If a taxpayers’ school taxes rose from $4,000 to $4,020, was there a 2% increase? Justify your answer using your graph, table, or equation.
Answer:
No. The change represents less than a 2% increase. On my graph, the point (4000,4020) does not fall on the line; it falls below the line, which means 4,020 is too low for the second coordinate (the new tax amount). If I examined my table, when b is 4,000, t is 4,080. The equation would be
4,000(1.02) = 4,080, which is not equivalent to 4,020.

Exercise 3.
The sports booster club is selling candles as a fundraiser to support the new team. The club earns a commission on its candle sales (which means it receives a certain percentage of the total dollar amount sold). If the club gets to keep 30% of the money from the candle sales, what would the club’s total sales have to be in order to make at least $500?
Answer:
Part = Percent∙Whole
500 = 0.3W
500(\(\frac{1}{0.3}\)) = 0.3(\(\frac{1}{0.3}\))W
1,666.67 ≈ W
They will need candle sales totaling at least $1,666.67.

Exercise 4.
Christian’s mom works at the concession stand during sporting events. She told him they buy candy bars for $0.75 each and mark them up 40% to sell at the concession stand. What is the amount of the markup? How much does the concession stand charge for each candy bar?
Answer:
Let N represent the new price of a candy after the markup. Let M represent the percent or markup rate.
N = M∙Whole
N = (100% + 40%)(0.75)
N = (1 + 0.4)(0.75)
N = 1.05
The candy bars cost $1.05 at the concession stand. $1.05 – $0.75 = $0.30, so there is a markup of $0.30.

With your group, brainstorm solutions to the problems below. Prepare a poster that shows your solutions and math work. A calculator may be used for calculations.

Exercise 5.
For the next school year, the new soccer team will need to come up with $600.
a. Suppose the team earns $500 from the fundraiser at the start of the current school year, and the money is placed for one calendar year in a savings account earning 0.5% simple interest annually. How much money will the team still need to raise to meet next year’s expenses?
Answer:
Interest = Principal × Interest Rate × Time
Interest = $500 × 0.005 × 1
Interest = $2.50
Total Money Saved = Interest + Principal = $500.00 + $2.50 = $502.50
Total Money Needed For Next Year = $600.00 – $502.50 = $97.50
The team will need to raise $97.50 more toward their goal.

b. Jeff is a member of the new sports team. His dad owns a bakery. To help raise money for the team, Jeff’s dad agrees to provide the team with cookies to sell at the concession stand for next year’s opening game. The team must pay back the bakery $0.25 for each cookie it sells. The concession stand usually sells about 60 to 80 baked goods per game. Using your answer from part (a), determine a percent markup for the cookies the team plans to sell at next year’s opening game. Justify your answer.
Answer:
The team needs to raise $97.50. Based on past data for the typical number of baked goods sold, we estimate that we will sell 60 cookies, so we need to divide 97.50 by 60. 97.5÷60 is about 1.63. That means we need to make a profit of $1.63 per cookie after we pay back the bakery $0.25 per cookie. So, if we add $0.25 to $1.63, we arrive at a markup price of $1.88. We decide to round that up to $2.00 since we want to be sure we raise enough money. We may sell fewer than 60 cookies (especially if the data for the typical number of baked goods sold includes items other than cookies, such as cupcakes or muffins).
To find the percent markup, we used the following equation with $0.25 as the original price; since
$2.00 – $0.25 = $1.75, then $1.75 is the markup.
Markup = Markup Rate∙Original Price
1.75 = Markup Rate∙(0.25)
1.75(\(\frac{1}{0.25}\)) = Markup Rate∙(0.25)(\(\frac{1}{0.25}\))
7 = Markup Rate
7 = \(\frac{7}{1}\) = \(\frac{700}{100}\) = 700% markup

c. Suppose the team ends up selling 78 cookies at next year’s opening game. Find the percent error in the number of cookies that you estimated would be sold in your solution to part (b).
Percent Error = \(\frac{|a – x|}{|x|}\) ∙100%, where x is the exact value and a is the approximate value.
Answer:
We estimated 60 cookies would be sold, but if 78 are sold, then 78 is the actual value. Next, we used the percent error formula:
Percent Error = \(\frac{|a – x|}{|x|}\)∙100%
Percent Error = \(\frac{|60 – 78|}{|78|}\)∙100%
Percent Error = \(\frac{18}{78}\)∙100%
Percent Error ≈ 23%
There was about a 23% error in our estimate for the number of cookies sold.

Eureka Math Grade 7 Module 4 Lesson 11 Problem Set Answer Key

Question 1.
A school district’s property tax rate rises from 2.5% to 2.7% to cover a $300,000 budget deficit (shortage of money). What is the value of the property in the school district to the nearest dollar? (Note: Property is assessed at 100% of its value.)
Answer:
Let W represent the worth of the property in the district, in dollars.
300,000 = 0.002W
300,000(\(\frac{1}{0.002}\)) = 0.002(\(\frac{1}{0.002}\))W
150,000,000 = W
The property is worth $150,000,000.

Question 2.
Jake’s older brother, Sam, has a choice of two summer jobs. He can either work at an electronics store or at the school’s bus garage. The electronics store would pay him to work 15 hours per week. He would make $8 per hour plus a 2% commission on his electronics sales. At the school’s bus garage, Sam could earn $300 per week working 15 hours cleaning buses. Sam wants to take the job that pays him the most. How much in electronics would Sam have to sell for the job at the electronics store to be the better choice for his summer job?
Answer:
Let S represent the amount, in dollars, sold in electronics.
300 < 8(15) + 0.02(S)
300 < 120 + 0.02S
180 < 0.02S
180(\(\frac{1}{0.02}\)) < 0.02(\(\frac{1}{0.05}\))S
9,000 < S
Sam would have to sell more than $9,000 in electronics for the electronics store to be the better choice.

Question 3.
Sarah lost her science book. Her school charges a lost book fee equal to 75% of the cost of the book. Sarah received a notice stating she owed the school $60 for the lost book.
a. Write an equation to represent the proportional relationship between the school’s cost for the book and the amount a student must pay for a lost book. Let B represent the school’s cost of the book in dollars and N represent the student’s cost in dollars.
Answer:
N = 0.75B

b. What is the constant or proportionality? What does it mean in the context of this situation?
Answer:
The constant of proportionality is 75% = 0.75. It means that for every $1 the school spends to purchase a textbook, a student must pay $0.75 for a lost book.

c. How much did the school pay for the book?
Answer:
60 = 0.75B
60(\(\frac{1}{0.75}\)) = 0.75(\(\frac{1}{0.75}\))B
\(\frac{60}{0.75}\) = B
80 = B
The school paid $80 for the science book.

Question 4.
In the month of May, a certain middle school has an average daily absentee rate of 8% each school day. The absentee rate is the percent of students who are absent from school each day.
a. Write an equation that shows the proportional relationship between the number of students enrolled in the middle school and the average number of students absent each day during the month of May. Let s represent the number of students enrolled in school, and let a represent the average number of students absent each day in May.
Answer:
a = 0.08s

b. Use your equation to complete the table. List 5 possible values for s and a.

Answer:

c. Identify the constant of proportionality, and explain what it means in the context of this situation.
Answer:
The constant of proportionality is 0.08. 0.08 = 8%, so on average, for every 100 students enrolled in school, 8 are absent from school.

d. Based on the absentee rate, determine the number of students absent on average from school during the month of May if there are 350 students enrolled in the middle school.
Answer:
28 students; 350 is halfway between 300 and 400. So, I used the table of values and looked at the numbers of students absent that correspond to 300 and 400 students at the school, which are 24 and 32. Halfway between 24 and 32 is 28.

Question 5.
The equation shown in the box below could relate to many different percent problems. Put an X next to each problem that could be represented by this equation. For any problem that does not match this equation, explain why it does not. Quantity = 1.05∙Whole

_________ Find the amount of an investment after 1 year with 0.5% interest paid annually.
Answer:
The equation should be Quantity = 1.005∙Whole.

________ Write an equation to show the amount paid for an item including tax, if the tax rate is 5%.
Answer:
X

________ A proportional relationship has a constant of proportionality equal to 105%.
X

Answer:

_______ Mr. Hendrickson sells cars and earns a 5% commission on every car he sells. Write an equation to show the relationship between the price of a car Mr. Hendrickson sold and the amount of commission he earns.
Answer:
The equation should be Quantity = 0.05∙Whole.

Eureka Math Grade 7 Module 4 Lesson 11 Exit Ticket Answer Key

Lee sells electronics. He earns a 5% commission on each sale he makes.
a. Write an equation that shows the proportional relationship between the dollar amount of electronics Lee sells, d, and the amount of money he makes in commission, c.
Answer:
c = \(\frac{1}{20}\)d or c = 0.05d

b. Express the constant of proportionality as a decimal.
Answer:
0.05

c. Explain what the constant of proportionality means in the context of this situation.
Answer:
The constant of proportionality of 0.05 means that Lee would earn five cents for every dollar of electronics that he sells.

d. If Lee wants to make $100 in commission, what is the dollar amount of electronics he must sell?
Answer:
c = 0.05 d
100 = 0.05 d
\(\frac{1}{0.05}\)(100) = \(\frac{1}{0.05}\)(0.05) d
2,000 = d
Lee must sell $2,000 worth of electronics.