## Engage NY Eureka Math 7th Grade Module 4 Lesson 6 Answer Key

### Eureka Math Grade 7 Module 4 Lesson 6 Example Answer Key

Example 1: Mental Math and Percents

a. 75% of the students in Jesse’s class are 60 inches or taller. If there are 20 students in her class, how many students are 60 inches or taller?

Answer:

→ Is this question a comparison of two separate quantities, or is it part of the whole? How do you know?

The problem says that the students make up 75% of Jesse’s class, which means they are part of the whole class; this is a part of the whole problem.

→ What numbers represent the part, whole, and percent?

The part is the number of students that are 60 inches or taller, the whole is the 20 students that make up Jesse’s class, and the percent is 75%.

Instruct students to discuss the problem with a partner; challenge them to solve it using mental math only. After 1–2 minutes of discussion, ask for students to share their mental strategies with the class.

Possible strategies:

75% is the same as \(\frac{3}{4}\) of 100%; 20 → 100% and 20 = 4(5), so 3(5) = 15, which means 15 is \(\frac{3}{4}\) of 20.

100% → 20

25% → 5

75% → 15

Have students write a description of how to mentally solve the problem (including the math involved) in their student materials.

→ Was this problem easy to solve mentally? Why?

The numbers involved in the problem shared factors with 100 that were easy to work with.

b. Bobbie wants to leave a tip for her waitress equal to 15% of her bill. Bobbie’s bill for her lunch is $18. How much money represents 15% of the bill?

Answer:

→ Is this question a comparison of two separate quantities, or is it part of a whole? How do you know?

She is leaving a quantity that is equal to 15% of her bill, so this is a comparison of two separate quantities.

→ What numbers represent the part, the whole, and the percent? Is the part actually part of her lunch bill?

The part is the amount that she plans to leave for her waitress and is not part of her lunch bill but is calculated as if it is a part of her bill; the whole is the $18 lunch bill, and the percent is 15%.

Instruct students to discuss the problem with a partner; challenge them to solve it using mental math only. After 1–2 minutes of discussion, ask for students to share their mental strategies with the class.

Possible strategy includes the following:

15% = 10%+5%; 10% of $18 is $1.80; half of 10% is 5%, so 5% → 1/2( $1.80) = $0.90;

$1.80+$0.90 = $2.70.

→ Was this problem easy to solve mentally? Why?

The numbers involved in the problem shared factors with 100 that were easy to work with.

→ Could you use this strategy to find 7% of Bobbie’s bill?

Yes; 7% = 5%+2(1%); 1% of $18 is $0.18, so 2% → $0.36; $0.90+$0.36 = $1.26, so

7% → $1.26.

Have students write a description of how to mentally solve the problem in their student materials including the math involved.

### Eureka Math Grade 7 Module 4 Lesson 6 Exercise Answer Key

Exercise 1.

Express 9 hours as a percentage of 3 days.

Answer:

3 days is the equivalent of 72 hours since 3(24) = 72.

72 hours represents the whole.

Quantity = Percent × Whole. Let p represent the unknown percent.

9 = p(72)

\(\frac{1}{72}\) (9) = p(72) ∙ \(\frac{1}{72}\)

\(\frac{9}{72}\) = p(1)

\(\frac{1}{8}\) = p

\(\frac{1}{8}\) (100%) = 12.5%

Exercise 2.

Richard works from 11:00 a.m. to 3:00 a.m. His dinner break is 75% of the way through his work shift. What time is Richard’s dinner break?

Answer:

The total amount of time in Richard’s work shift is 16 hours since 1+12+3 = 16.

16 hours represents the whole.

Quantity = Percent × Whole. Let b represent the number of hours until Richard’s dinner break.

b = 0.75(16)

b = 12

Richard’s dinner break is 12 hours after his shift begins.

12 hours after 11:00 a.m. is 11:00 p.m.

Richard’s dinner break is at 11:00 p.m.

Exercise 3.

At a playoff basketball game, there were 370 fans cheering for school A and 555 fans cheering for school B.

a. Express the number of fans cheering for school A as a percent of the number of fans cheering for school B.

Answer:

The number of fans for school B is the whole.

Quantity = Percent × Whole. Let p represent the unknown percent.

370 = p(555)

\(\frac{1}{555}\) (370) = p(555)\(\frac{1}{555}\)

\(\frac{370}{555}\) = p(1)

\(\frac{2}{3}\) = p

\(\frac{2}{3}\) (100%) = 66\(\frac{2}{3}\)%

The number of fans cheering for school A is 66 2/3% of the number of fans cheering for school B.

b. Express the number of fans cheering for school B as a percent of the number of fans cheering for school A.

Answer:

The number of fans cheering for school A is the whole.

Quantity = Percent × Whole. Let p represent the unknown percent.

555 = p(370)

\(\frac{1}{370}\) (555) = p(370)\(\frac{1}{370}\)

\(\frac{555}{370}\) = p(1)

\(\frac{3}{2}\) = p

\(\frac{3}{2}\) (100%) = 150%

The number of fans cheering for school B is 150% of the number of fans cheering for school A.

c. What percent more fans were there for school B than for school A?

Answer:

There were 50% more fans cheering for school B than for school A.

Exercise 4.

Rectangle A has a width of 8 cm and a length of 16 cm. Rectangle B has the same area as the first, but its width is 62.5% of the width of the first rectangle. Express the length of Rectangle B as a percent of the length of Rectangle A. What percent more or less is the length of Rectangle B than the length of Rectangle A?

Answer:

To find the width of Rectangle B:

The width of Rectangle A is the whole.

Quantity = Percent × Whole. Let w represent the unknown width of Rectangle B.

w = 0.625(8) = 5

The width of Rectangle B is 5 cm.

To find the length of Rectangle B:

The area of Rectangle B is 100% of the area of Rectangle A because the problem says the areas are the same.

Area = Width × Length. Let A represent the unknown area of Rectangle A.

A = 8 cm(16 cm) = 128 cm^{2}

Area = Width × Length. Let l represent the unknown length of Rectangle B.

128 cm^{2} = 5 cm (l)

25.6 cm = l

The length of Rectangle B is 25.6 cm.

To express the length of Rectangle B as a percent of the length of Rectangle A:

The length of Rectangle A is the whole.

Quantity = Percent × Whole. Let p represent the unknown percent.

25.6 cm = p(16 cm)

1.6 = p

1.6(100%) = 160%; The length of Rectangle B is 160% of the length of Rectangle A.

Therefore, the length of Rectangle B is 60% more than the length of Rectangle A.

Exercise 5.

A plant in Mikayla’s garden was 40 inches tall one day and was 4 feet tall one week later. By what percent did the plant’s height increase over one week?

Answer:

4 feet is equivalent to 48 inches since 4(12) = 48.

40 inches is the whole.

Quantity = Percent × Whole. Let p represent the unknown percent.

8 = p(40)

\(\frac{1}{5}\) = p

\(\frac{1}{5}\) = 20/100 = 20%

The plant’s height increased by 20% in one week.

Exercise 6.

Loren must obtain a minimum number of signatures on a petition before it can be submitted. She was able to obtain 672 signatures, which is 40% more than she needs. How many signatures does she need?

Answer:

The number of signatures needed represents the whole.

Quantity = Percent × Whole. Let s represent the number of signatures needed.

672 = 1.4(s)

480 = s

Loren needs to obtain 480 signatures on her petition.

### Eureka Math Grade 7 Module 4 Lesson 6 Problem Set Answer Key

Question 1.

Micah has 294 songs stored in his phone, which is 70% of the songs that Jorge has stored in his phone. How many songs are stored on Jorge’s phone?

Answer:

Quantity = Percent × Whole. Let s represent the number of songs on Jorge’s phone.

294 = \(\frac{70}{100}\) ∙ s

294 = \(\frac{7}{10}\) ∙ s

294 ∙ \(\frac{10}{7}\) = \(\frac{7}{10}\) ∙ \(\frac{10}{7}\) ∙ s

42 ∙ 10 = 1 ∙ s

420 = s

There are 420 songs stored on Jorge’s phone.

Question 2.

Lisa sold 81 magazine subscriptions, which is 27% of her class’s fundraising goal. How many magazine subscriptions does her class hope to sell?

Answer:

Quantity = Percent × Whole. Let s represent the number of magazine subscriptions Lisa’s class wants to sell.

81 = \(\frac{27}{100}\) ∙ s

81 ∙ \(\frac{100}{27}\) = \(\frac{27}{100}\) ∙ \(\frac{100}{27}\) ∙ s

3 ∙ 100 = 1 ∙ s

300 = s

Lisa’s class hopes to sell 300 magazine subscriptions.

Question 3.

Theresa and Isaiah are comparing the number of pages that they read for pleasure over the summer. Theresa read 2,210 pages, which was 85% of the number of pages that Isaiah read. How many pages did Isaiah read?

Answer:

Quantity = Percent × Whole. Let p represent the number of pages that Isaiah read.

2,210 = \(\frac{85}{100}\) ∙ p

2,210 = \(\frac{17}{20}\) ∙ p

2,210 ∙ \(\frac{20}{17}\) = \(\frac{17}{20}\) ∙ \(\frac{20}{17}\) ∙ p

130 ∙ 20 = 1 ∙ p

2,600 = p

Isaiah read 2,600 pages over the summer.

Question 4.

In a parking garage, the number of SUVs is 40% greater than the number of non-SUVs. Gina counted 98 SUVs in the parking garage. How many vehicles were parked in the garage?

Answer:

40% greater means 100% of the non-SUVs plus another 40% of that number, or 140%.

Quantity = Percent × Whole. Let d represent the number of non-SUVs in the parking garage.

98 = \(\frac{140}{100}\) ∙ d

98 = \(\frac{7}{5}\) ∙ d

98 ∙ \(\frac{5}{7}\) = \(\frac{7}{5}\) ∙ \(\frac{5}{7}\) ∙ d

14 ∙ 5 = 1 ∙ d

70 = d

There are 70 non-SUVs in the parking garage.

The total number of vehicles is the sum of the number of the SUVs and non-SUVs.

70+98 = 168. There is a total of 168 vehicles in the parking garage.

Question 5.

The price of a tent was decreased by 15% and sold for $76.49. What was the original price of the tent in dollars?

Answer:

If the price was decreased by 15%, then the sale price is 15% less than 100% of the original price, or 85%.

Quantity = Percent × Whole. Let t represent the original price of the tent.

76.49 = \(\frac{85}{100}\) ∙ t

76.49 = \(\frac{17}{20}\) ∙ t

76.49 ∙ \(\frac{20}{17}\) = \(\frac{17}{20}\) ∙ \(\frac{20}{17}\) ∙ t

\(\frac{1,529.8}{17}\) = 1 ∙ t

89.988 ≈ t

Because this quantity represents money, the original price was $89.99 after rounding to the nearest hundredth.

Question 6.

40% of the students at Rockledge Middle School are musicians. 75% of those musicians have to read sheet music when they play their instruments. If 38 of the students can play their instruments without reading sheet music, how many students are there at Rockledge Middle School?

Answer:

Let m represent the number of musicians at the school, and let s represent the total number of students. There are two whole quantities in this problem. The first whole quantity is the number of musicians. The 38 students who can play an instrument without reading sheet music represent 25% of the musicians.

Quantity = Percent × Whole

38 = 25/100 ∙ m

38 = \(\frac{1}{4}\) ∙ m

38 ∙ \(\frac{4}{1}\) = \(\frac{1}{4}\) ∙ \(\frac{4}{1}\) ∙ m

\(\frac{152}{1}\) = 1 ∙ m

152 = m

There are 152 musicians in the school.

Quantity = Percent × Whole

152 = \(\frac{40}{100}\) ∙ s

152 = \(\frac{2}{5}\) ∙ s

152 ∙ \(\frac{5}{2}\) = \(\frac{2}{5}\) ∙ \(\frac{5}{2}\) ∙ s

\(\frac{760}{2}\) = 1 ∙ s

380 = s

There are 380 students at Rockledge Middle School.

Question 7.

At Longbridge Middle School, 240 students said that they are an only child, which is 48% of the school’s student enrollment. How many students attend Longbridge Middle School?

Answer:

Quantity → 100%

240 → 48%

\(\frac{240}{48}\) → 1%

\(\frac{240}{48}\) (100) → 100%

5(100) → 100%

500 → 100%

There are 500 students attending Longbridge Middle School.

Question 8.

Grace and her father spent 4 1/2 hours over the weekend restoring their fishing boat. This time makes up 6% of the time needed to fully restore the boat. How much total time is needed to fully restore the boat?

Answer:

Quantity → %

4 \(\frac{1}{2}\) → 6%

\(\frac{\frac{9}{2}}{6}\) → 6%

\(\frac{\frac{9}{2}}{6}\) → 1%

\(\frac{\frac{9}{2}}{6}\) (100) → 100%

(\(\frac{9}{2}\))(\(\frac{1}{6}\))100 → 100%

(\(\frac{9}{12}\))100 → 100%

(\(\frac{3}{4}\))100 → 100%

300/4 → 100%

75 → 100%

The total amount of time to restore the boat is 75 hours.

Question 9.

Bethany’s mother was upset with her because Bethany’s text messages from the previous month were 218% of the amount allowed at no extra cost under her phone plan. Her mother had to pay for each text message over the allowance. Bethany had 5,450 text messages last month. How many text messages is she allowed under her phone plan at no extra cost?

Answer:

Quantity → %

5,450 → 218%

\(\frac{5,450}{218}\) → 1%

\(\frac{5,450}{218}\) (100) → 100%

25(100) → 100%

2,500 → 100%

Bethany is allowed 2,500 text messages without extra cost.

Question 10.

Harry used 84% of the money in his savings account to buy a used dirt bike that cost him $1,050. How much money is left in Harry’s savings account?

Answer:

Quantity → %

1,050 → 84%

\(\frac{1,050}{84}\) → 1%

\(\frac{1,050}{84}\) (100) → 100%

12.5(100) → 100%

1,250 → 100%

Harry started with $1,250 in his account but then spent $1,050 of it on the dirt bike.

1,250-1,050 = 200

Harry has $200 left in his savings account.

Question 11.

15% of the students in Mr. Riley’s social studies classes watch the local news every night. Mr. Riley found that 136 of his students do not watch the local news. How many students are in Mr. Riley’s social studies classes?

Answer:

If 15% of his students do watch their local news, then 85% do not.

Quantity → %

136 → 85%

\(\frac{136}{85}\) → 1%

(\(\frac{136}{85}\))(100) → 100%

1.6(100) → 100%

160 → 100%

There are 160 total students in Mr. Riley’s social studies classes.

Question 12.

Grandma Bailey and her children represent about 9.1% of the Bailey family. If Grandma Bailey has 12 children, how many members are there in the Bailey family?

Answer:

Quantity → %

13 → 9.1%

(\(\frac{1}{9.1}\))(13) → 1%

100(\(\frac{13}{9.1}\)) → 100%

\(\frac{1,300}{9.1}\) → 100%

142.857″…” → 100%

The Bailey family has 143 members.

Question 13.

Shelley earned 20% more money in tips waitressing this week than last week. This week she earned $72.00 in tips waitressing. How much money did Shelley earn last week in tips?

Answer:

Quantity = Percent × Whole. Let m represent the number of dollars Shelley earned waitressing last week.

72 = \(\frac{120}{100}\) m

72(\(\frac{100}{120}\)) = \(\frac{120}{100}\) (\(\frac{100}{120}\))m

60 = m

Shelley earned $60 waitressing last week.

Question 14.

Lucy’s savings account has 35% more money than her sister Edy’s. Together, the girls have saved a total of $206.80. How much money has each girl saved?

Answer:

The money in Edy’s account corresponds to 100%. Lucy has 35% more than Edy, so the money in Lucy’s account corresponds to 135%. Together, the girls have a total of $206.80, which is 235% of Edy’s account balance.

Quantity = Pecent × Whole. Let b represent Edy’s savings account balance in dollars.

206.8 = \(\frac{235}{100}\) ∙ b

206.8 = \(\frac{47}{20}\) ∙ b

206.8 ∙ \(\frac{20}{47}\) = \(\frac{47}{20}\) ∙ \(\frac{20}{47}\) ∙ b

\(\frac{4,136}{47}\) = 1 ∙ b

88 = b

Edy has saved $88 in her account. Lucy has saved the remainder of the $206.80, so 206.8-88 = 118.8.

Therefore, Lucy has $118.80 saved in her account.

Question 15.

Bella spent 15% of her paycheck at the mall, and 40% of that was spent at the movie theater. Bella spent a total of $13.74 at the movie theater for her movie ticket, popcorn, and a soft drink. How much money was in Bella’s paycheck?

Answer:

$13.74 → 40%

$3.435 → 10%

$34.35 → 100%

Bella spent $34.35 at the mall.

$34.35 → 15%

$11.45 → 5%

$229 → 100%

Bella’s paycheck was $229.

Question 16.

On a road trip, Sara’s brother drove 47.5% of the trip, and Sara drove 80% of the remainder. If Sara drove for 4 hours and 12 minutes, how long was the road trip?

Answer:

There are two whole quantities in this problem. First, Sara drove 80% of the remainder of the trip; the remainder is the first whole quantity. 4 hr.12 min. is equivalent to 4 12/60 hr. = 4.2 hr.

Quantity → %

4.2 → 80%

\(\frac{4.2}{80}\) → 1%

\(\frac{4.2}{80}\) (100) → 100%

\(\frac{420}{80}\) → 100%

\(\frac{42}{8}\) → 100%

5.25 → 100%

The remainder of the trip that Sara’s brother did not drive was 5.25 hours. He drove 47.5% of the trip, so the remainder of the trip was 52.5% of the trip, and the whole quantity is the time for the whole road trip.

Quantity → %

5.25 → 52.5%

\(\frac{5.25}{52.5}\) → 1%

(\(\frac{5.25}{52.5}\))(100) → 100%

\(\frac{525}{52.5}\) → 100%

10 → 100%

The road trip was a total of 10 hours.

### Eureka Math Grade 7 Module 4 Lesson 6 Exit Ticket Answer Key

Question 1.

Parker was able to pay for 44% of his college tuition with his scholarship. The remaining $10,054.52 he paid for with a student loan. What was the cost of Parker’s tuition?

Answer:

Parker’s tuition is the whole; 56% represents the amount paid by a student loan.

Quantity = Percent × Whole. Let t represent the cost of Parker’s tuition.

10,054.52 = 0.56(t)

\(\frac{10,054.52}{0.56}\) = t

17,954.50 = t

Parker’s tuition was $17,954.50.

Question 2.

Two bags contain marbles. Bag A contains 112 marbles, and Bag B contains 140 marbles. What percent fewer marbles does Bag A have than Bag B?

Answer:

The number of marbles in Bag B is the whole.

There are 28 fewer marbles in Bag A.

Quantity = Percent × Whole. Let p represent the unknown percent.

28 = p(140)

\(\frac{2}{10}\) = p

\(\frac{2}{10}\) = \(\frac{20}{100}\) = 20%

Bag A contains 20% fewer marbles than Bag B.

Question 3.

There are 42 students on a large bus, and the rest are on a smaller bus. If 40% of the students are on the smaller bus, how many total students are on the two buses?

Answer:

The 42 students on the larger bus represent 60% of the students. If I divide both 60% and 42 by 6, then I get 7 → 10%. Multiplying both by 10, I get 70 → 100%. There are 70 total students on the buses.

### Eureka Math Grade 7 Module 4 Lesson 6 Percent More or Less—Round 1 Answer Key

Directions: Find each missing value.

Answer:

### Eureka Math Grade 7 Module 4 Lesson 6 Percent More or Less—Round 2 Answer Key

Directions: Find each missing value.

Answer: