Eureka Math

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.

Engage NY Eureka Math 7th Grade Module 4 Lesson 10 Answer Key

Eureka Math Grade 7 Module 4 Lesson 10 Example Answer Key

Example 1: Can Money Grow? A Look at Simple Interest
Larry invests $100 in a savings plan. The plan pays 4 \(\frac{1}{2}\)% interest each year on his $100 account balance.
a. How much money will Larry earn in interest after 3 years? After 5 years?
Answer:
3 years:
I = Prt
I = 100(0.045)(3)
I = 13.50
Larry will earn $13.50 in interest after 3 years.

5 years:
I = Prt
I = 100(0.045)(5)
I = 22.50
Larry will earn $22.50 in interest after 5 years.

b. How can you find the balance of Larry’s account at the end of 5 years?
Answer:
You would add the interest earned after 5 years to the beginning balance. $22.50 + $100 = $122.50.

Example 2: Time Other Than One Year
A $1,000 savings bond earns simple interest at the rate of 3% each year. The interest is paid at the end of every month. How much interest will the bond have earned after 3 months?
Answer:
Step 1: Convert 3 months to a year.
12 months = 1 year. So, divide both sides by 4 to get 3 months = \(\frac{1}{4}\) year.
Step 2: Use the interest formula to find the answer.
I = Prt
I = ($1000)(0.03)(0.25)
I = $7.50
The interest earned after 3 months is $7.50.

Example 3: Solving for P, r, or t
Mrs. Williams wants to know how long it will take an investment of $450 to earn $200 in interest if the yearly interest rate is 6.5%, paid at the end of each year.
Answer:
I = Prt
$200 = ($450)(0.065)t
$200 = $29.25t
$200(1/($29.25)) = (1/($29.25))$29.25t
6.8376 = t
Six years is not enough time to earn $200. At the end of seven years, the interest will be over $200. It will take seven years since the interest is paid at the end of each year.

Eureka Math Grade 7 Module 4 Lesson 10 Exercise Answer Key

Exercise 1.
Find the balance of a savings account at the end of 10 years if the interest earned each year is 7.5%. The principal is $500.
Answer:
I = Prt
I = $500(0.075)(10)
I = $375
The interest earned after 10 years is $375. So, the balance at the end of 10 years is $375 + $500 = $875.

Exercise 2.
Write an equation to find the amount of simple interest, A, earned on a $600 investment after 1 1/2 years if the semi – annual (6 – month) interest rate is 2%.
Answer:
1 \(\frac{1}{2}\)years is the same as

Interest = Principal × Rate × Time
A = 600(0.02)(3) 1.5 years is 1 year and 6 months, so t = 3.
A = 36
The amount of interest earned is $36.

Exercise 3.
A $1,500 loan has an annual interest rate of 4 \(\frac{1}{4}\)% on the amount borrowed. How much time has elapsed if the interest is now $127.50?
Answer:
Interest = Principal × Rate × Time
Let t be time in years.
127.50 = (1,500)(0.0425)t
127.50 = 63.75t
(127.50)(\(\frac{1}{63.75}\)) = (\(\frac{1}{63.75}\))(63.75)t
2 = t
Two years have elapsed.

Eureka Math Grade 7 Module 4 Lesson 10 Problem Set Answer Key

Question 1.
Enrique takes out a student loan to pay for his college tuition this year. Find the interest on the loan if he borrowed $2,500 at an annual interest rate of 6% for 15 years.
Answer:
I = 2,500(0.06)(15)
I = 2,250
Enrique would have to pay $2,250 in interest.

Question 2.
Your family plans to start a small business in your neighborhood. Your father borrows $10,000 from the bank at an annual interest rate of 8% rate for 36 months. What is the amount of interest he will pay on this loan?
Answer:
I = 10,000(0.08)(3)
I = 2,400
He will pay $2,400 in interest.

Question 3.
Mr. Rodriguez invests $2,000 in a savings plan. The savings account pays an annual interest rate of 5.75% on the amount he put in at the end of each year.
a. How much will Mr. Rodriguez earn if he leaves his money in the savings plan for 10 years?
Answer:
I = 2,000(0.0575)(10)
I = 1,150
He will earn $1,150.

b. How much money will be in his savings plan at the end of 10 years?
Answer:
At the end of 10 years, he will have $3,150 because $2,000 + $1,150 = $3,150.

c. Create (and label) a graph in the coordinate plane to show the relationship between time and the amount of interest earned for 10 years. Is the relationship proportional? Why or why not? If so, what is the constant of proportionality?
Answer:

Yes, the relationship is proportional because the graph shows a straight line touching the origin. The constant of proportionality is 115 because the amount of interest earned increases by $115 for every one year.

d. Explain what the points (0,0) and (1,115) mean on the graph.
Answer:
(0, 0) means that no time has elapsed and no interest has been earned. (1, 115) means that after 1 year, the savings plan would have earned $115. 115 is also the constant of proportionality.

e. Using the graph, find the balance of the savings plan at the end of seven years.
Answer:
From the table, the point (7,805) means that the balance would be $2,000 + $805 = $2,805.

f. After how many years will Mr. Rodriguez have increased his original investment by more than 50%? Show your work to support your answer.
Answer:
Quantity = Percent × Whole
Let Q be the account balance that is 50% more than the original investment.
Q > (1 + 0.50)(2,000)
Q > 3,000
The balance will be greater than $3,000 beginning between 8 and 9 years because the graph shows (8, 920) and (9, 1035), so $2,000 + $920 = $2,920<$3,000, and $2,000 + $1,035 = $3,035 > $3,000.

Challenge Problem:
Question 4.
George went on a game show and won $60,000. He wanted to invest it and found two funds that he liked. Fund 250 earns 15% interest annually, and Fund 100 earns 8% interest annually. George does not want to earn more than $7,500 in interest income this year. He made the table below to show how he could invest the money.

Answer:

a. Explain what value x is in this situation.
Answer:
x is the principal, in dollars, that George could invest in Fund 100.

b. Explain what the expression 60,000 – x represents in this situation.
Answer:
60,000 – x is the principal, in dollars, that George could invest in Fund 250. It is the money he would have left over once he invests in Fund 100.

c. Using the simple interest formula, complete the table for the amount of interest earned.
Answer:
See the table above.

d. Write an inequality to show the total amount of interest earned from both funds.
Answer:
0.08x + 0.15(60,000 – x)≤7,500

e. Use algebraic properties to solve for x and the principal, in dollars, George could invest in Fund 100. Show your work.
Answer:
0.08x + 9,000 – 0.15x ≤ 7,500
9,000 – 0.07x ≤ 7,500
9,000 – 9,000 – 0.07x ≤ 7,500 – 9,000
– 0.07x ≤ – 1,500
(\(\frac{1}{ – 0.07}\))( – 0.07x) ≤ (\(\frac{1}{ – 0.07}\))( – 1,500)
x ≥ 21,428.57
x approximately equals $21,428.57. George could invest $21,428.57 or more in Fund 100.

f. Use your answer from part (e) to determine how much George could invest in Fund 250.
Answer:
He could invest $38,571.43 or less in Fund 250 because 60,000 – 21,428.57 = 38,571.43.

g. Using your answers to parts (e) and (f), how much interest would George earn from each fund?
Answer:
Fund 100: 0.08 × 21,428.57 × 1 approximately equals $1,714.29.
Fund 250: 0.15 × 38,571.43 × 1 approximately equals $5,785.71 or $7,500 – $1,714.29.

Eureka Math Grade 7 Module 4 Lesson 10 Exit Ticket Answer Key

Question 1.
Erica’s parents gave her $500 for her high school graduation. She put the money into a savings account that earned 7.5% annual interest. She left the money in the account for nine months before she withdrew it. How much interest did the account earn if interest is paid monthly?
Answer:
I = Prt
I = (500)(0.075)(\(\frac{9}{12}\))
I = 28.125
The interest earned is $28.13.

Question 2.
If she would have left the money in the account for another nine months before withdrawing, how much interest would the account have earned?
Answer:
I = Prt
I = (500)(0.075)(\(\frac{18}{12}\))
I = 56.25
The account would have earned $56.25.

Question 3.
About how many years and months would she have to leave the money in the account if she wants to reach her goal of saving $750?
Answer:
750 – 500 = 250 She would need to earn $250 in interest.
I = Prt
250 = (500)(0.075)t
250 = 37.5t
250(\(\frac{1}{37.5}\)) = (\(\frac{1}{37.5}\))(37.5)t
6 \(\frac{2}{3}\) = t
It would take her 6 years and 8 months to reach her goal because \(\frac{2}{3}\) × 12 months is 8 months.

Eureka Math Grade 7 Module 4 Lesson 10 Fractional Percents—Round 1 Answer Key

Directions: Find the part that corresponds with each percent.

Answer:

Eureka Math Grade 7 Module 4 Lesson 10 Fractional Percents—Round 2 Answer Key

Directions: Find the part that corresponds with each percent.

Answer:

Engage NY Eureka Math 7th Grade Module 4 Lesson 9 Answer Key

Eureka Math Grade 7 Module 4 Lesson 9 Example Answer Key

Example 1.
The amount of money Tom has is 75% of Sally’s amount of money. After Sally spent $120 and Tom saved all his money, Tom’s amount of money is 50% more than Sally’s. How much money did each have at the beginning? Use a visual model and a percent line to solve the problem.
Answer:

Example 2.
Erin and Sasha went to a candy shop. Sasha bought 50% more candies than Erin. After Erin bought 8 more candies, Sasha had 20% more. How many candies did Erin and Sasha have at first?
a. Model the situation using a visual model.

Answer:

b. How many candies did Erin have at first? Explain.
Answer:
Each bar in the after tape diagram is 8 candies. Sasha has 48 candies. Each bar in the before tape diagram is 16 candies. Erin started with 32 candies.

Example 3.
Kimberly and Mike have an equal amount of money. After Kimberly spent $50 and Mike spent $25, Mike’s money is 50% more than Kimberly’s. How much did Kimberly and Mike have at first?
a. Use an equation to solve the problem.
Answer:
Equation Method:
Let x be the amount of Kimberly’s money, in dollars, after she spent $50. After Mike spent $25, his money is 50% more than Kimberly’s. Mike’s money is also $25 more than Kimberly’s.
0.5x = 25
x = 50
Kimberly started with $100 because 100 – 50 = 50. Mike has $75 because (1.5)50 = 75.
They each started with $100.

b. Use a visual model to solve the problem.
Answer:

Each bar is $25. They both started with $100.

c. Which method do you prefer and why?
Answer:
Answers will vary. I prefer the visual method because it is easier for me to draw the problem out instead of using the algebraic properties.

Eureka Math Grade 7 Module 4 Lesson 9 Exercise Answer Key

Exercise 1.
Todd has 250% more video games than Jaylon. Todd has 56 video games in his collection. He gives Jaylon 8 of his games. How many video games did Todd and Jaylon have in the beginning? How many do they have now?
Answer:
Answers may vary. Sample answer is provided below.
Visual Model:

Each bar in the dark box is 8 games.
Equation Method:
Let z be the number of video games that Jaylon had at the beginning. Then, Todd started with 3.5z video games.
3.5z = 56
z = 16
In the beginning, Jaylon had 16, and Todd had 56. After Todd gave Jaylon 8 of his games, Jaylon had 24, and Todd had 48.

Eureka Math Grade 7 Module 4 Lesson 9 Problem Set Answer Key

Question 1.
Solve each problem using an equation.
a. What is 150% of 625?
Answer:
n = 1.5(625)
n = 937.5

b. 90 is 40% of what number?
Answer:
90 = 0.4(n)
n = 225

c. What percent of 520 is 40? Round to the nearest hundredth of a percent.
Answer:
40 = p(520)
p ≈ 0.0769 = 7.69%

Question 2.
The actual length of a machine is 12.25 cm. The measured length is 12.2 cm. Round the answer to part (b) to the nearest hundredth of a percent.
a. Find the absolute error.
Answer:
|12.2 – 12.25| = 0.05
The absolute error is 0.05 cm.

b. Find the percent error.
Answer:
\(\frac{0.05}{|12.25|}\) × 100% = 0.4082%
percent error ≈ 0.41%

Question 3.
A rowing club has 600 members. 60% of them are women. After 200 new members joined the club, the percentage of women was reduced to 50%. How many of the new members are women?
Answer:
40 of the new members are women.

Question 4.
40% of the marbles in a bag are yellow. The rest are orange and green. The ratio of the number of orange to the number of green is 4:5. If there are 30 green marbles, how many yellow marbles are there? Use a visual model to show your answer.
Answer:
5 units = 30 marbles
1 unit = 30 marbles ÷5 = 6 marbles
4 units = 4 × 6 marbles = 24 marbles
30+24 = 54→60%
18 → 20%
36 → 40%
There are 36 yellow marbles because 40% of the marbles are yellow.

Question 5.
Susan has 50% more books than Michael. Michael has 40 books. If Michael buys 8 more books, will Susan have more or less books than Michael? What percent more or less will Susan’s books be? Use any method to solve the problem.
Answer:
Susan has 25% more.

Question 6.
Harry’s amount of money is 75% of Kayla’s amount of money. After Harry earned $30 and Kayla earned 25% more of her money, Harry’s amount of money is 80% of Kayla’s money. How much money did each have at the beginning? Use a visual model to solve the problem.
Answer:

Each bar is $30. Harry started with $90, and Kayla started with $120.

Eureka Math Grade 7 Module 4 Lesson 9 Exit Ticket Answer Key

Question 1.
Terrence and Lee were selling magazines for a charity. In the first week, Terrance sold 30% more than Lee. In the second week, Terrance sold 8 magazines, but Lee did not sell any. If Terrance sold 50% more than Lee by the end of the second week, how many magazines did Lee sell?
Choose any model to solve the problem. Show your work to justify your answer.
Answer:
Answers may vary.
Equation Model:
Let m be the number of magazines Lee sold.
150% – 130% = 20%, so 0.2m = 8 and m = 40
Visual Model:

20% → 8
100% → 40

Engage NY Eureka Math 7th Grade Module 4 Lesson 8 Answer Key

Eureka Math Grade 7 Module 4 Lesson 8 Example Answer Key

Example 1: How Far Off?
Find the absolute error for the following problems. Explain what the absolute error means in context.
a. Taylor’s Measurement 1
Answer:
|15 \(\frac{2}{8}\)in. – 15 in.| = |0.25 in| = 0.25 in.
Taylor’s Measurement 1 was 0.25 in. away from the actual value of 15 in.

b. Connor’s Measurement 1
Answer:
|15\(\frac{4}{8}\)in. – 15 in.| = |0.5 in.| = 0.5 in.
Connor’s Measurement 1 was 0.5 in. away from the actual value of 15 in.

c. Jordan’s Measurement 2
Answer:
|14\(\frac{6}{8}\) in. – 15 in.| = 0.25 in.
Jordan’s Measurement 2 was 0.25 in. away from the actual value of 15 in.

Example 2: How Right Is Wrong?
a. Find the percent error for Taylor’s Measurement 1. What does this mean?
Answer:
\(\frac{\left|15 \frac{2}{8} – 15\right|}{|15|}\) × 100%
\(\frac{|0.25|}{|15|}\) × 100%
\(\frac{1}{60}\) × 100%
1\(\frac{2}{3}\)%
This means that Taylor’s measurement of 15.25 in. has an error that is 1\(\frac{2}{3}\)% of the actual value.

b. From Example 1, part (b), find the percent error for Connor’s Measurement 1. What does this mean?
Answer:
\(\frac{0.5 \mathrm{in} .}{15 \mathrm{in} .}\) × 100% = 3\(\frac{1}{3}\)%
This means that Connor’s measurement of 15\(\frac{4}{8}\)in. has an error that is 3\(\frac{1}{3}\)% of the actual value.

c. From Example 1, part (c), find the percent error for Jordan’s Measurement 2. What does it mean?
Answer:
\(\frac{0.25 \mathrm{in} .}{15 \mathrm{in} .}\) × 100% = 1\(\frac{2}{3}\)%
This means that Jordan’s measurement of 14\(\frac{6}{8}\)in. has an error that is 1\(\frac{2}{3}\)% of the actual value.

d. What is the purpose of finding percent error?
Answer:
It tells you how big your error is compared to the true value. An error of 1 cm is very small when measuring the distance for a marathon, but an error of 1 cm is very large if you are a heart surgeon. In evaluating the seriousness of an error, we usually compare it to the exact value.

Example 3: Estimating Percent Error
The attendance at a musical event was counted several times. All counts were between 573 and 589. If the actual attendance number is between 573 and 589, inclusive, what is the most the percent error could be? Explain your answer.
Answer:
The most the absolute error could be is |589 – 573| = 16. The percent error will be largest when the exact value is smallest. Therefore, the most the percent error could be is \(\frac{16}{573}\) × 100%<2.8%. In this case, the percent error is less than 2.8%.

Eureka Math Grade 7 Module 4 Lesson 8 Exercise Answer Key

Calculate the percent error for Problems 1–3. Leave your final answer in fraction form, if necessary.
Exercise 1.
A real estate agent expected 18 people to show up for an open house, but 25 attended.
Answer:
\(\frac{|18 – 25|}{|25|}\) × 100% = 28%

Exercise 2.
In science class, Mrs. Moore’s students were directed to weigh a 300 – gram mass on the balance scale. Tina weighed the object and reported 328 grams.
Answer:
\(\frac{|328 – 300|}{|300|}\) × 100% = 9\(\frac{1}{3}\)%

Exercise 3.
Darwin’s coach recorded that he had bowled 250 points out of 300 in a bowling tournament. However, the official scoreboard showed that Darwin actually bowled 225 points out of 300.
Answer:
\(\frac{|250 – 225|}{|225|}\) × 100% = 11 \(\frac{1}{9}\)%

Eureka Math Grade 7 Module 4 Lesson 8 Problem Set Answer Key

Question 1.
The odometer in Mr. Washington’s car does not work correctly. The odometer recorded 13.2 miles for his last trip to the hardware store, but he knows the distance traveled is 15 miles. What is the percent error? Use a calculator and the percent error formula to help find the answer. Show your steps.
Answer:
15 is the exact value, and 13.2 is the approximate value. Using the percent error formula, \(\frac{|a – x|}{|x|}\) × 100%, the percent error is
\(\frac{|13.2 – 15|}{|15|}\) × 100% = 12%.
The percent error is equal to 12%.

Question 2.
The actual length of a soccer field is 500 feet. A measuring instrument shows the length to be 493 feet. The actual width of the field is 250 feet, but the recorded width is 246.5 feet. Answer the following questions based on this information. Round all decimals to the nearest tenth.

a. Find the percent error for the length of the soccer field.
Answer:
\(\frac{|493 – 500|}{|500|}\) × 100% = 1.4%

b. Find the percent error of the area of the soccer field.
Answer:
Actual area:
A = l × w
A = (500)(250) = 125,000
The actual area is 125,000 square feet.

Approximate area:
A = l × w
A = (493)(246.5)
The approximate area is 121,524.5 square feet.

Percent error of the area:
\(\frac{|121,524.5 – 125,000|}{|125,000|}\) × 100% = 2.8%

c. Explain why the values from parts (a) and (b) are different.
Answer:
In part (a), 1.4% is the percent error for the length, which is one dimension of area. Part (b) is the percent error for the area, which includes two dimensions―length and width. The percent error for the width of the soccer field should be the same as the percent error for the length if the same measuring tool is used. So, 2.8% = 1.4% × 2. However, this is not always the case. Percent error for the width is not always the same as the percent error for the length. It is possible to have an error for both the length and the width, yet the area has no error. For example: publicized length = 100 feet, publicized width = 90 feet,
actual length = 150 feet, and actual width = 60 feet.

Question 3.
Kayla’s class went on a field trip to an aquarium. One tank had 30 clown fish. She miscounted the total number of clown fish in the tank and recorded it as 24 fish. What is Kayla’s percent error?
Answer:
\(\frac{|24 – 30|}{|30|}\) × 100% = 20%

Question 4.
Sid used geometry software to draw a circle of radius 4 units on a grid. He estimated the area of the circle by counting the squares that were mostly inside the circle and got an answer of 52 square units.

a. Is his estimate too large or too small?
Answer:
A = πr²
A = 4² π = 16π
The exact area of the circle is 16π square units. 16π is approximately 50.3. His estimate is too large.

b. Find the percent error in Sid’s estimation to the nearest hundredth using the π key on your calculator.
\(\frac{|52 – 16 \pi|}{|16 \pi|}\) × 100% ≈ 3.45%

Question 5.
The exact value for the density of aluminum is 2.699 g/cm³. Working in the science lab at school, Joseph finds the density of a piece of aluminum to be 2.75 g/cm³. What is Joseph’s percent error? (Round to the nearest hundredth.)
Answer:
\(\frac{|2.75 – 2.699|}{|2.699|}\) × 100% ≈ 1.89%

Question 6.
The world’s largest marathon, The New York City Marathon, is held on the first Sunday in November each year. Between 2 million and 2.5 million spectators will line the streets to cheer on the marathon runners. At most, what is the percent error?
Answer:
\(\frac{|2.5 – 2|}{|2|}\) × 100% = 25%

Question 7.
A circle is inscribed inside a square, which has a side length of 12.6 cm. Jared estimates the area of the circle to be about 80% of the area of the square and comes up with an estimate of 127 cm².

a. Find the absolute error from Jared’s estimate to two decimal places using the π key on your calculator.
|127 – π6.3²|≈ 2.31
The absolute error is approximately 2.31 cm.

b. Find the percent error of Jared’s estimate to two decimal places using the π key on your calculator.
Answer:
\(\frac{\left|127 – \pi 6 \cdot 3^{2}\right|}{\left|\pi 6 \cdot 3^{2}\right|}\) × 100% ≈ 1.85%. The percent error is approximately 1.85%.

c. Do you think Jared’s estimate was reasonable?
Answer:
Yes. The percent error is less than 2%.

d. Would this method of computing the area of a circle always be too large?
Answer:
Yes. If the circle has radius r, then the area of the circle is πr², and the area of the square is 4r².
\(\frac{\pi r^{2}}{4 r^{2}}\) = \(\frac{\pi}{4}\). The area approximately equals 0.785 = 78.5%<80%.

Question 8.
In a school library, 52% of the books are paperback. If there are 2,658 books in the library, how many of them are not paperback to the nearest whole number?
Answer:
100% – 52% = 48%
Let n represent the number of books that are not paperback.
n = 0.48(2,658)
n = 1,275.84
About 1,276 books are not paperback.

Question 9.
Shaniqua has 25% less money than her older sister Jennifer. If Shaniqua has $180, how much money does Jennifer have?
Answer:
100% – 25% = 75%
Let j represent the amount of money that Jennifer has.
180 = \(\frac{3}{4}\)j
\(\frac{4}{3}\)(180) = (\(\frac{3}{4}\))(\(\frac{4}{3}\))j
240 = j
Jennifer has $240.

Question 10.
An item that was selling for $1,102 is reduced to $806. To the nearest whole, what is the percent decrease?
Answer:
Let p represent the percent decrease.
1,102 – 806 = 296
296 = p ∙ 1,102
\(\frac{296}{1,10.2}\) = p
0.2686 = p
The percent decrease is approximately 27%.

Question 11.
If 60 calories from fat is 75% of the total number of calories in a bag of chips, find the total number of calories in the bag of chips.
Answer:
Let t represent the total number of calories in a bag of chips.
60 = \(\frac{3}{4}\) t
\(\frac{4}{3}\) ∙ 60 = \(\frac{3}{4}\) ∙ \(\frac{4}{3}\) ∙ t
80 = t
The total number of calories in the bag of chips is 80 calories.

Eureka Math Grade 7 Module 4 Lesson 8 Exit Ticket Answer Key

Question 1.
The veterinarian weighed Oliver’s new puppy, Boaz, on a defective scale. He weighed 36 pounds. However, Boaz weighs exactly 34.5 pounds. What is the percent of error in measurement of the defective scale to the nearest tenth?
Answer:
\(\frac{|36 – 34.5|}{|34.5|}\) × 100% = 4 \(\frac{8}{23}\)%
≈ 4.3%

Question 2.
Use the π key on a scientific or graphing calculator to compute the percent of error of the approximation of pi, 3.14, to the value π. Show your steps, and round your answer to the nearest hundredth of a percent.
Answer:
\(\frac{|3 \cdot 14 – \pi|}{|\pi|}\) × 100% = 0.05%

Question 3.
Connor and Angie helped take attendance during their school’s practice fire drill. If the actual count was between 77 and 89, inclusive, what is the most the absolute error could be? What is the most the percent error could be? Round your answer to the nearest tenth of a percent.
Answer:
The most the absolute error could be is |89 – 77| = |12| = 12.
The percent error will be largest when the exact value is smallest. The most the percent error could be is
\(\frac{|12|}{|77|}\) × 100% < 15.6%. The percent error is less than 15.6%.

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

Eureka Math Grade 7 Module 4 Lesson 7 Example Answer Key

Example 1: A Video Game Markup
Games Galore Super Store buys the latest video game at a wholesale price of $30.00. The markup rate at Game’s Galore Super Store is 40%. You use your allowance to purchase the game at the store. How much will you pay, not including tax?
a. Write an equation to find the price of the game at Games Galore Super Store. Explain your equation.
Answer:
Let P represent the price of the video game.
Quantity = Percent × Whole
P = (100% + 40%)(30)
The equation shows that the price of the game at the store is equal to the wholesale cost, which is 100% and the 40% increase. This makes the new price 140% of the wholesale price.

b. Solve the equation from part (a).
Answer:
P = (100% + 40%)(30)
P = (1.40)(30)
P = 42
I would pay $42.00 if I bought it from Games Galore Super Store.

c. What was the total markup of the video game? Explain.
Answer:
The markup was $12.00 because $42 – $30 = $12.

d. You and a friend are discussing markup rate. He says that an easier way to find the total markup is by multiplying the wholesale price of $30.00 by 40%. Do you agree with him? Why or why not?
Answer:
Yes, I agree with him because (0.40)(30) = 12. The markup rate is a percent of the wholesale price. Therefore, it makes sense to multiply them together because Quantity = Percent × Whole.

→ Which quantity is the whole quantity in this problem?
The wholesale price is the whole quantity.

→ How do 140% and 1.4 correspond in this situation?
The markup price of the video game is 140% times the wholesale price. 140% and 1.4 are equivalent forms of the same number. In order to find the markup price, convert the percent to a decimal or fraction, and multiply it by the whole.

→ What does a markup mean?
A markup is the amount of increase in a price.

Example 2: Black Friday
A $300 mountain bike is discounted by 30% and then discounted an additional 10% for shoppers who arrive before
5:00 a.m.
a. Find the sales price of the bicycle.
Answer:
Find the price with the 30% discount:
Let D represent the discount price of the bicycle with the 30% discount rate.
Quantity = Percent × Whole
D = (100% – 30%)(300)
D = (0.70)(300)
D = 210
$210 is the discount price of the bicycle with the 30% discount rate.
Find the price with the additional 10% discount:
Let A represent the discount price of the bicycle with the additional 10% discount.
A = (100% – 10%)(210)
D = (1 – 0.10)(210)
D = (0.90)(210)
D = 189
$189 is the discount price of the bicycle with the additional 10% discount.

b. In all, by how much has the bicycle been discounted in dollars? Explain.
Answer:
$300 – $189 = $111. The bicycle has been discounted $111 because the original price was $300. With both discounts applied, the new price is $189.

c. After both discounts were taken, what was the total percent discount?
Answer:
A final discount of 40% means that you would add 30% to 10% and apply it to the same whole. This is not the case because the additional 10% discount is taken after the 30% discount has been applied, so you are only receiving that 10% discount on 70% of the original price. A 40% discount would make the final price $180 because 180 = (0.60)(300).
However, the actual final discount as a percent is 37%.
Let P be the percent the sales price is of the original price. Let F represent the actual final discount as a percent.
Part = Percent × Whole
189 = P × 300
(\(\frac{1}{300}\))189 = P × 300(\(\frac{1}{300}\))
0.63 = 63% = P
F = 100% – 63% = 37%

d. Instead of purchasing the bike for $300, how much would you save if you bought it before 5:00 a.m.?
Answer:
You would save $111 if you bought the bike before 5:00 a.m. because $300 – $189 is $111.

Example 3: Working Backward
A car that normally sells for $20,000 is on sale for $16,000. The sales tax is 7.5%.

→ What is the whole quantity in this problem?
a. The whole quantity is the original price of the car, $20,000.
Answer:
What percent of the original price of the car is the final price?
Quantity = Percent × Whole
16,000 = P(20,000)
16,000(\(\frac{1}{20,000}\)) = P(20,000)(\(\frac{1}{20,000}\))
0.8 = P
0.8 = \(\frac{80}{100}\) = 80%
The final price is 80% of the original price.

b. Find the discount rate.
Answer:
The discount rate is 20% because 100% – 80% = 20%.

c. By law, sales tax has to be applied to the discount price. However, would it be better for the consumer if the 7.5% sales tax was calculated before the 20% discount was applied? Why or why not?
Answer:
Apply Sales Tax First
Apply the sales tax to the whole.
(100% + 7.5%)(20,000)
(1 + 0.075)(20,000)
(1.075)(20,000)
$21,500 is the price of the car, including tax, before the discount.
Apply the discount to the new whole.
(100% – 20%)(21,500)
(1 – 0.2)(21,500)
$17,200 is the final price, including the discount and tax.

Apply the Discount First
(100% + 7.5%)(16,000)
(1 + 0.075)(16,000)
(1.075)(16,000)
$17,200 is the final price, including the discount and tax.

Because both final prices are the same, it does not matter which is applied first. This is because multiplication is commutative. The discount rate and sales tax rate are both being applied to the whole, $20,000.

d. Write an equation applying the commutative property to support your answer to part (c).
Answer:
20,000(1.075)(0.8) = 20,000(0.8)(1.075)

Eureka Math Grade 7 Module 4 Lesson 7 Exercise Answer Key

Exercises 1–3

Exercise 1.
Sasha went shopping and decided to purchase a set of bracelets for 25% off the regular price. If Sasha buys the bracelets today, she will save an additional 5%. Find the sales price of the set of bracelets with both discounts. How much money will Sasha save if she buys the bracelets today?

Answer:
Let B be the sales price with both discounts in dollars.
B = (0.95)(0.75)(44) = 31.35. The sales price of the set of bracelets with both discounts is $31.35. Sasha will save $12.65.

Exercise 2.
A golf store purchases a set of clubs at a wholesale price of $250. Mr. Edmond learned that the clubs were marked up 200%. Is it possible to have a percent increase greater than 100%? What is the retail price of the clubs?
Answer:
Yes, it is possible. Let C represent the retail price of the clubs, in dollars.
C = (100% + 200%)(250)
C = (1 + 2)(250)
C = (3)(250)
C = 750
The retail price of the clubs is $750.

Exercise 3.
Is a percent increase of a set of golf clubs from $250 to $750 the same as a markup rate of 200%? Explain.
Answer:
Yes, it is the same. In both cases, the percent increase and markup rate show by how much (in terms of percent) the new price is over the original price. The whole is $250 and corresponds to 100%. \(\frac{750}{250}\) = \(\frac{3}{1}\) × 100% = 300%. $750 is 300% of $250. 300% – 100% = 200%. From Exercise 2, the markup is 200%. So, percent increase is the same as markup.

Exercise 4.
a. Write an equation to determine the selling price in dollars, p, on an item that is originally priced s dollars after a markup of 25%.
Answer:
p = 1.25s or p = (0.25 + 1)s

b. Create and label a table showing five possible pairs of solutions to the equation.
Answer:

c. Create and label a graph of the equation.

Answer:

d. Interpret the points (0, 0) and (1, r).
Answer:
The point (0, 0) means that a $0 (free) item will cost $0 because the 25% markup is also $0. The point (1, r) is (1, 1.25). It means that a $1.00 item will cost $1.25 after it is marked up by 25%; r is the unit rate.

Exercise 5.
Use the following table to calculate the markup or markdown rate. Show your work. Is the relationship between the original price and the selling price proportional or not? Explain.

Answer:
Because the selling price is less than the original price, use the equation: Selling Price = (1 – m) × Whole.
1,400 = (1 – m)(1,750)
\(\frac{1,400}{1,750}\) = (1 – m)\(\frac{1,750}{1,750}\)
0.80 = 1 – m
0.20 = m
The markdown rate is 20%. The relationship between the original price and selling price is proportional because the table shows the ratio \(\frac{p}{m}\) = \(\frac{0.80}{1}\) for all possible pairs of solutions.

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

Question 1.
You have a coupon for an additional 25% off the price of any sale item at a store. The store has put a robotics kit on sale for 15% off the original price of $40. What is the price of the robotics kit after both discounts?
Answer:
(0.75)(0.85)(40) = 25.50. The price of the robotics kit after both discounts is $25.50.

Question 2.
A sign says that the price marked on all music equipment is 30% off the original price. You buy an electric guitar for the sale price of $315.
a. What is the original price?
Answer:
\(\frac{315}{1 – 0.30}\) = \(\frac{315}{0.70}\) = 450. The original price is $450.

b. How much money did you save off the original price of the guitar?
Answer:
450 – 315 = 135. I saved $135 off the original price of the guitar.

c. What percent of the original price is the sale price?
Answer:
\(\frac{315}{450}\) = \(\frac{70}{100}\) = 70%. The sale price is 70% of the original price.

Question 3.
The cost of a New York Yankee baseball cap is $24.00. The local sporting goods store sells it for $30.00. Find the markup rate.
Answer:
Let P represent the unknown percent.
30 = P(24)
P = \(\frac{30}{24}\) = 1.25 = (100% + 25%). The markup rate is 25%.

Question 4.
Write an equation to determine the selling price in dollars, p, on an item that is originally priced s dollars after a markdown of 15%.
Answer:
p = 0.85s or p = (1 – 0.15)s

a. Create and label a table showing five possible pairs of solutions to the equation.
Answer:

b. Create and label a graph of the equation.

Answer:

c. Interpret the points (0,0) and (1,r).
Answer:
The point (0, 0) means that a $0 (free) item will cost $0 because the 15% markdown is also $0. The point (1, r) is (1,0.85), which represents the unit rate. It means that a $1.00 item will cost $0.85 after it is marked down by 15%.

Question 5.
At the amusement park, Laura paid $6.00 for a small cotton candy. Her older brother works at the park, and he told her they mark up the cotton candy by 300%. Laura does not think that is mathematically possible. Is it possible, and if so, what is the price of the cotton candy before the markup?
Answer:
Yes, it is possible. \(\frac{6.00}{1 + 3}\) = \(\frac{6}{4}\) = 1.50. The price of the cotton candy before the markup is $1.50.

Question 6.
A store advertises that customers can take 25% off the original price and then take an extra 10% off. Is this the same as a 35% off discount? Explain.
Answer:
No, because the 25% is taken first off the original price to get a new whole. Then, the extra 10% off is multiplied to the new whole. For example, (1 – 0.25)(1 – 0.10) = 0.675 or (0.75)(0.90) = 0.675. This is multiplied to the whole, which is the original price of the item. This is not the same as adding 25% and 10% to get 35% and then multiplying by (1 – 0.35), or 0.65.

Question 7.
An item that costs $50.00 is marked 20% off. Sales tax for the item is 8%. What is the final price, including tax?
a. Solve the problem with the discount applied before the sales tax.
Answer:
(1.08)(0.80)(50) = 43.20. The final price is $43.20.

b. Solve the problem with the discount applied after the sales tax.
Answer:
(0.80)(1.08)(50) = 43.20. The final price is $43.20.

c. Compare your answers in parts (a) and (b). Explain.
Answer:
My answers are the same. The final price is $43.20. This is because multiplication is commutative.

Question 8.
The sale price for a bicycle is $315. The original price was first discounted by 50% and then discounted an additional 10%. Find the original price of the bicycle.
Answer:
(315 ÷ 0.9) ÷ 0.5 = 700. The original price was $700.

Question 9.
A ski shop has a markup rate of 50%. Find the selling price of skis that cost the storeowner $300.
Answer:
Solution 1: Use the original price of $300 as the whole. The markup rate is 50% of $300 or $150.
The selling price is $300 + $150 = $450.
Solution 2: Multiply $300 by 1 plus the markup rate (i.e., the selling price is (1.5)($300) = $450).

Question 10.
A tennis supply store pays a wholesaler $90 for a tennis racquet and sells it for $144. What is the markup rate?
Answer:
Solution 1: Let the original price of $90 be the whole. Quantity = Percent × Whole.
144 = Percent(90)
\(\frac{144}{90}\) = Percent
1.6 = 160%. This is a 60% increase. The markup rate is 60%.
Solution 2:
Selling Price = (1 + m)(Whole)
144 = (1 + m)90
1 + m = \(\frac{144}{90}\)
m = 1.6 – 1 = 0.6 = 60%
The markup rate is 60%.

Question 11.
A shoe store is selling a pair of shoes for $60 that has been discounted by 25%. What was the original selling price?
Answer:
Solution 1:
$60 → 75%
$20 → 25%
$80 → 100%
The original price was $80.

Solution 2: Let x be the original cost in dollars.
(1 – 0.25)x = 60
\(\frac{3}{4}\) x = 60
(\(\frac{4}{3}\))(\(\frac{3}{4}\) x) = \(\frac{4}{3}\) (60)
x = 80
The original price was $80.

Question 12.
A shoe store has a markup rate of 75% and is selling a pair of shoes for $133. Find the price the store paid for the shoes.
Answer:
Solution 1:
$133 → 175%
$19 → 25%
$76 → 100%
The store paid $76.

Solution 2: Divide the selling price by 1.75.
\(\frac{133}{1.75}\) = 76
The store paid $76.

Question 13.
Write 5 \(\frac{1}{4}\)% as a simple fraction.
Answer:
\(\frac{21}{400}\)

Question 14.
Write \(\frac{3}{8}\) as a percent.
Answer:
37.5%

Question 15.
If 20% of the 70 faculty members at John F. Kennedy Middle School are male, what is the number of male faculty members?
Answer:
(0.20)(70) = 14. Therefore, 14 faculty members are male.

Question 16.
If a bag contains 400 coins, and 33 1/2% are nickels, how many nickels are there? What percent of the coins are not nickels?
Answer:
(400)(0.335) = 134. Therefore, 134 of the coins are nickels. The percent of coins that are not nickels is 66 1/2%.

Question 17.
The temperature outside is 60 degrees Fahrenheit. What would be the temperature if it is increased by 20%?
Answer:
(60)(1.2) = 72. Therefore, the temperature would be 72 degrees Fahrenheit.

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

A store that sells skis buys them from a manufacturer at a wholesale price of $57. The store’s markup rate is 50%.
a. What price does the store charge its customers for the skis?
Answer:
57 × (1 + 0.50) = 85.50. The store charges $85.50 for the skis.

b. What percent of the original price is the final price? Show your work.
Answer:
Quantity = Percent × Whole. Let P represent the unknown percent.
85.50 = P(57)
85.50(\(\frac{1}{57}\)) = P(57)(\(\frac{1}{57}\))
1.50 = P
1.50 = \(\frac{150}{100}\) = 150%. The final price is 150% of the original price

c. What is the percent increase from the original price to the final price?
Answer:
The percent increase is 50% because 150% – 100% = 50%.

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 cm²
Area = Width × Length. Let l represent the unknown length of Rectangle B.
128 cm² = 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:

Engage NY Eureka Math 7th Grade Module 4 Lesson 5 Answer Key

Eureka Math Grade 7 Module 4 Lesson 5 Example Answer Key

Example 1: Using a Modified Double Number Line with Percents
The 42 students who play wind instruments represent 75% of the students who are in band. How many students are in band?
Answer:
→ Which quantity in this problem represents the whole?
The total number of students in band is the whole, or 100%.

→ Draw the visual model shown with a percent number line and a tape diagram.

→ Use the number line and tape diagram to find the total number of students in band.
100% represents the total number of students in band, and 75% is 3/4 of 100%. The greatest common factor of 75 and 100 is 25.
42→75%
\(\frac{42}{3}\)→25%
4(\(\frac{42}{3}\))→100%
4(14)→100%
56→100%

Example 2: Mental Math Using Factors of 100
Answer each part below using only mental math, and describe your method.
a. If 39 is 1% of a number, what is that number? How did you find your answer?
Answer:
39 is 1% of 3,900. I found my answer by multiplying 39∙100 because 39 corresponds with each 1% in 100%, and 1%∙100 = 100%, so 39∙100 = 3,900.

b. If 39 is 10% of a number, what is that number? How did you find your answer?
Answer:
39 is 10% of 390. 10 is a factor of 100, and there are ten 10% intervals in 100%. The quantity 39 corresponds to 10%, so there are 39∙10 in the whole quantity, and 39∙10 = 390.

c. If 39 is 5% of a number, what is that number? How did you find your answer?
Answer:
39 is 5% of 780. 5 is a factor of 100, and there are twenty 5% intervals in 100%. The quantity 39 corresponds to 5%, so there are twenty intervals of 39 in the whole quantity.
39∙20
39∙2∙10 Factored 20 for easier mental math
78∙10
780

d. If 39 is 15% of a number, what is that number? How did you find your answer?
Answer:
39 is 15% of 260. 15 is not a factor of 100, but 15 and 100 have a common factor of 5. If 15% is 39, then because 5 = 15/3 , 5% is 13 = 39/3. There are twenty 5% intervals in 100%, so there are twenty intervals of 13 in the whole.
13∙20
13∙2∙10 Factored 20 for easier mental math
26∙10
260

e. If 39 is 25% of a number, what is that number? How did you find your answer?
Answer:
39 is 25% of 156. 25 is a factor of 100, and there are four intervals of 25% in 100%. The quantity 39 corresponds with 25%, so there are 39∙4 in the whole quantity.
39∙4
39∙2∙2 Factored 4 for easier mental math
78∙2
156

Eureka Math Grade 7 Module 4 Lesson 5 Exercise Answer Key

Opening Exercise
What are the whole number factors of 100? What are the multiples of those factors? How many multiples are there of each factor (up to 100)?

Answer:

→ How do you think we can use these whole number factors in calculating percents on a double number line?
The factors represent all ways by which we could break 100% into equal – sized whole number intervals. The multiples listed would be the percents representing each cumulative interval. The number of multiples would be the number of intervals.

Exercises 1–3

Exercise 1.
Bob’s Tire Outlet sold a record number of tires last month. One salesman sold 165 tires, which was 60% of the tires sold in the month. What was the record number of tires sold?
Answer:

The salesman’s total is being compared to the total number of tires sold by the store, so the total number of tires sold is the whole quantity. The greatest common factor of 60 and 100 is 20, so I divided the percent line into five equal – sized intervals of 20%. 60% is three of the 20% intervals, so I divided the salesman’s 165 tires by 3 and found that 55 tires corresponds with each 20% interval. 100% consists of five 20% intervals, which corresponds to five groups of 55 tires. Since 5∙55 = 275, the record number of tires sold was 275 tires.

Exercise 2.
Nick currently has 7,200 points in his fantasy baseball league, which is 20% more points than Adam. How many points does Adam have?
Answer:

Nick’s points are being compared to Adam’s points, so Adam’s points are the whole quantity. Nick has 20% more points than Adam, so Nick really has 120% of Adam’s points. The greatest common factor of 120 and 100 is 20, so I divided the 120% on the percent line into six equal – sized intervals. I divided Nick’s 7,200 points by 6 and found that 1,200 points corresponds to each 20% interval. Five intervals of 20% make 100%, and five intervals of 1,200 points totals 6,000 points. Adam has 6,000 points in the fantasy baseball league.

Exercise 3.
Kurt has driven 276 miles of his road trip but has 70% of the trip left to go. How many more miles does Kurt have to drive to get to his destination?
Answer:

With 70% of his trip left to go, Kurt has only driven 30% of the way to his destination. The greatest common factor of 30 and 100 is 10, so I divided the percent line into ten equal – sized intervals. 30% is three of the 10% intervals, so I divided 276 miles by 3 and found that 92 miles corresponds to each 10% interval. Ten intervals of 10% make 100%, and ten intervals of 92 miles totals 920 miles. Kurt has already driven 276 miles, and 920 – 276 = 644, so Kurt has 644 miles left to get to his destination.

Exercises 4–5

Exercise 4.
Derrick had a 0.250 batting average at the end of his last baseball season, which means that he got a hit 25% of the times he was up to bat. If Derrick had 47 hits last season, how many times did he bat?
Answer:
The decimal 0.250 is 25%, which means that Derrick had a hit 25% of the times that he batted. His number of hits is being compared to the total number of times he was up to bat. The 47 hits corresponds with 25%, and since 25 is a factor of 100, 100 = 25∙4. I used mental math to multiply the following:
47∙4
(50 – 3)∙4 Used the distributive property for easier mental math
200 – 12
188
Derrick was up to bat 188 times last season.

Exercise 5.
Nelson used 35% of his savings account for his class trip in May. If he used $140 from his savings account while on his class trip, how much money was in his savings account before the trip?
Answer:
35% of Nelson’s account was spent on the trip, which was $140. The amount that he spent is being compared to the total amount of savings, so the total savings represents the whole. The greatest common factor of 35 and 100 is 5. 35% is seven intervals of 5%, so I divided $140 by 7 to find that $20 corresponds to 5%.
100% = 5%∙20, so the whole quantity is $20∙20 = $400. Nelson’s savings account had $400 in it before his class trip.

Eureka Math Grade 7 Module 4 Lesson 5 Problem Set Answer Key

Use a double number line to answer Problems 1–5.
Question 1.
Tanner collected 360 cans and bottles while fundraising for his baseball team. This was 40% of what Reggie collected. How many cans and bottles did Reggie collect?
Answer:

The greatest common factor of 40 and 100 is 20.
\(\frac{1}{2}\) (40%) = 20%, and \(\frac{1}{2}\) (360) = 180, so 180 corresponds with 20%. There are five intervals of 20% in 100%, and 5(180) = 900, so Reggie collected 900 cans and bottles.

Question 2.
Emilio paid $287.50 in taxes to the school district that he lives in this year. This year’s taxes were a 15% increase from last year. What did Emilio pay in school taxes last year?
Answer:

The greatest common factor of 100 and 115 is 5. There are 23 intervals of 5% in 115%, and \(\frac{287.5}{23}\) = 12.5, so 12.5 corresponds with 5%. There are 20 intervals of 5% in 100%, and 20(12.5) = 250, so Emilio paid $250 in school taxes last year.

Question 3.
A snowmobile manufacturer claims that its newest model is 15% lighter than last year’s model. If this year’s model weighs 799 lb., how much did last year’s model weigh?
Answer:

15% lighter than last year’s model means 15% less than 100% of last year’s model’s weight, which is 85%. The greatest common factor of 85 and 100 is 5. There are 17 intervals of 5% in 85%, and \(\frac{799}{17}\) = 47, so 47 corresponds with 5%. There are 20 intervals of 5% in 100%, and 20(47) = 940, so last year’s model weighed 940 pounds.

Question 4.
Student enrollment at a local school is concerning the community because the number of students has dropped to 504, which is a 20% decrease from the previous year. What was the student enrollment the previous year?
Answer:

A 20% decrease implies that this year’s enrollment is 80% of last year’s enrollment. The greatest common factor of 80 and 100 is 20. There are 4 intervals of 20% in 80%, and \(\frac{504}{4}\) = 126, so 126 corresponds to 20%. There are 5 intervals of 20% in 100%, and 5(126) = 630, so the student enrollment from the previous year was 630 students.

Question 5.
The color of paint used to paint a race car includes a mixture of yellow and green paint. Scotty wants to lighten the color by increasing the amount of yellow paint 30%. If a new mixture contains 3.9 liters of yellow paint, how many liters of yellow paint did he use in the previous mixture?
Answer:

The greatest common factor of 130 and 100 is 10. There are 13 intervals of 10% in 130%, and \(\frac{3.9}{13}\) = 0.3, so 0.3 corresponds to 10%. There are 10 intervals of 10% in 100%, and 10(0.3) = 3, so the previous mixture included 3 liters of yellow paint.

Use factors of 100 and mental math to answer Problems 6–10. Describe the method you used.
Question 6.
Alexis and Tasha challenged each other to a typing test. Alexis typed 54 words in one minute, which was 120% of what Tasha typed. How many words did Tasha type in one minute?
Answer:
The greatest common factor of 120 and 100 is 20, and there are 6 intervals of 20% in 120%, so I divided 54 into 6 equal – sized intervals to find that 9 corresponds to 20%. There are five intervals of 20% in 100%, so there are five intervals of 9 words in the whole quantity. 9∙5 = 45, so Tasha typed 45 words in one minute.

Question 7.
Yoshi is 5% taller today than she was one year ago. Her current height is 168 cm. How tall was she one year ago?
Answer:
5% taller means that Yoshi’s height is 105% of her height one year ago. The greatest common factor of 105 and 100 is 5, and there are 21 intervals of 5% in 105%, so I divided 168 into 21 equal – sized intervals to find that 8 cm corresponds to 5%. There are 20 intervals of 5% in 100%, so there are 20 intervals of 8 cm in the whole quantity. 20∙8 cm = 160 cm, so Yoshi was 160 cm tall one year ago.

Question 8.
Toya can run one lap of the track in 1 min.3 sec., which is 90% of her younger sister Niki’s time. What is Niki’s time for one lap of the track?
Answer:
1 min.3 sec = 63 sec. The greatest common factor of 90 and 100 is 10, and there are nine intervals of 10 in 90, so I divided 63 sec. by 9 to find that 7 sec. corresponds to 10%. There are 10 intervals of 10% in 100%, so 10 intervals of 7 sec. represents the whole quantity, which is 70 sec. 70 sec. = 1 min.10 sec. Niki can run one lap of the track in 1 min.10 sec.

Question 9.
An animal shelter houses only cats and dogs, and there are 25% more cats than dogs. If there are 40 cats, how many dogs are there, and how many animals are there total?
Answer:
25% more cats than dogs means that the number of cats is 125% the number of dogs. The greatest common factor of 125 and 100 is 25. There are 5 intervals of 25% in 125%, so I divided the number of cats into 5 intervals to find that 8 corresponds to 25%. There are four intervals of 25% in 100%, so there are four intervals of 8 in the whole quantity. 8∙4 = 32. There are 32 dogs in the animal shelter.
The number of animals combined is 32 + 40 = 72, so there are 72 animals in the animal shelter.

Question 10.
Angie scored 91 points on a test but only received a 65% grade on the test. How many points were possible on the test?
Answer:
The greatest common factor of 65 and 100 is 5. There are 13 intervals of 5% in 65%, so I divided 91 points into 13 intervals and found that 7 points corresponds to 5%. There are 20 intervals of 5% in 100%, so I multiplied 7 points times 20, which is 140 points. There were 140 points possible on Angie’s test.

For Problems 11–17, find the answer using any appropriate method.
Question 11.
Robbie owns 15% more movies than Rebecca, and Rebecca owns 10% more movies than Joshua. If Rebecca owns 220 movies, how many movies do Robbie and Joshua each have?
Answer:
Robbie owns 253 movies, and Joshua owns 200 movies.

Question 12.
20% of the seventh – grade students have math class in the morning. 16 2/3% of those students also have science class in the morning. If 30 seventh – grade students have math class in the morning but not science class, find how many seventh – grade students there are.
Answer:
There are 180 seventh – grade students.

Question 13.
The school bookstore ordered three – ring notebooks. They put 75% of the order in the warehouse and sold 80% of the rest in the first week of school. There are 25 notebooks left in the store to sell. How many three – ring notebooks did they originally order?
Answer:
The store originally ordered 500 three – ring notebooks.

Question 14.
In the first game of the year, the modified basketball team made 62.5% of their foul shot free throws. Matthew made all 6 of his free throws, which made up 25% of the team’s free throws. How many free throws did the team miss altogether?
Answer:
The team attempted 24 free throws, made 15 of them, and missed 9.

Question 15.
Aiden’s mom calculated that in the previous month, their family had used 40% of their monthly income for gasoline, and 63% of that gasoline was consumed by the family’s SUV. If the family’s SUV used $261.45 worth of gasoline last month, how much money was left after gasoline expenses?
Answer:
The amount of money spent on gasoline was $415; the monthly income was $1,037.50. The amount left over after gasoline expenses was $622.50.

Question 16.
Rectangle A is a scale drawing of Rectangle B and has 25% of its area. If Rectangle A has side lengths of 4 cm and 5 cm, what are the side lengths of Rectangle B?

Answer:
Area_A = length × width
Area_A = (5 cm)(4 cm)
Area_A = 20 cm²
The area of Rectangle A is 25% of the area of Rectangle B.
25% × 4 = 100%
20 × 4 = 80
So, the area of Rectangle B is 80 cm².
The value of the ratio of area A to area B is the square of the scale factor of the side lengths A:B.
The value of the ratio of area A:B is 20/80 = \(\frac{1}{4}\) , and \(\frac{1}{4}\) = (\(\frac{1}{2}\))², so the scale factor of the side lengths A:B is \(\frac{1}{2}\).
So, using the scale factor:
\(\frac{1}{2}\) (length_B) = 5 cm; length_B = 10 cm
\(\frac{1}{2}\) (width_B ) = 4 cm; width_B = 8 cm
The dimensions of Rectangle B are 8 cm and 10 cm.

Question 17.
Ted is a supervisor and spends 20% of his typical work day in meetings and 20% of that meeting time in his daily team meeting. If he starts each day at 7:30 a.m., and his daily team meeting is from 8:00 a.m. to 8:20 a.m., when does Ted’s typical work day end?
Answer:

20 minutes is \(\frac{1}{3}\) of an hour since \(\frac{20}{60}\) = \(\frac{1}{3}\).
Ted spends \(\frac{1}{3}\) hour in his daily team meeting, so \(\frac{1}{3}\) corresponds to 20% of his meeting time. There are 5 intervals of 20% in 100%, and 5(\(\frac{1}{3}\)) = \(\frac{5}{3}\), so Ted spends \(\frac{5}{3}\) hours in meetings.
\(\frac{5}{3}\) of an hour corresponds to 20% of Ted’s work day.

There are 5 intervals of 20% in 100%, and 5(\(\frac{5}{3}\)) = \(\frac{25}{3}\), so Ted spends \(\frac{25}{3}\) hours working. \(\frac{25}{3}\) hours = 8 \(\frac{1}{3}\) hours. Since \(\frac{1}{3}\) hour = 20 minutes, Ted works a total of 8 hours 20 minutes. If he starts at 7:30 a.m., he works 4 hours 30 minutes until 12:00 p.m., and since 8 \(\frac{1}{3}\) – 4 \(\frac{1}{2}\) = 3 \(\frac{5}{6}\), Ted works another 3 \(\frac{5}{6}\) hours after 12:00 p.m.

\(\frac{1}{6}\) hour = 10 minutes, and \(\frac{5}{6}\) hour = 50 minutes, so Ted works 3 hours 50 minutes after 12:00 p.m., which is 3:50 p.m. Therefore, Ted’s typical work day ends at 3:50 p.m.

Eureka Math Grade 7 Module 4 Lesson 5 Exit Ticket Answer Key

Question 1.
A tank that is 40% full contains 648 gallons of water. Use a double number line to find the maximum capacity of the water tank.
Answer:

I divided the percent line into intervals of 20% making five intervals of 20% in 100%. I know that I have to divide \(\frac{40}{2}\) to get 20, so I divided \(\frac{648}{2}\) to get 324 that corresponds with 20%. Since there are five 20% intervals in 100%, there are five 324 gallon intervals in the whole quantity, and 324∙5 = 1,620. The capacity of the tank is 1,620 gallons.

Question 2.
Loretta picks apples for her grandfather to make apple cider. She brings him her cart with 420 apples. Her grandfather smiles at her and says, “Thank you, Loretta. That is 35% of the apples that we need.”
Use mental math to find how many apples Loretta’s grandfather needs. Describe your method.
Answer:
420 is 35% of 1,200. 35 is not a factor of 100, but 35 and 100 have a common factor of 5. There are seven intervals of 5% in 35%, so I divided 420 apples into seven intervals; \(\frac{420}{7}\) = 60. There are 20 intervals of 5% in 100%, so I multiplied as follows:
60∙20
60∙2∙10
120∙10
1,200
Loretta’s grandfather needs a total of 1,200 apples to make apple cider.

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

Eureka Math Grade 7 Module 4 Lesson 4 Example Answer Key

Example 1: Finding a Percent Increase
Cassandra’s aunt said she will buy Cassandra another ring for her birthday. If Cassandra gets the ring for her birthday, what will be the percent increase in her ring collection?

Answer:
→ Looking back at our answers to the Opening Exercise, what percent is represented by 1 ring? If Cassandra gets the ring for her birthday, by what percent did her ring collection increase?
20% represents 1 ring, so her ring collection would increase by 20%.

→ Compare the number of new rings to the original total:
\(\frac{1}{5}\) = \(\frac{20}{100}\) = 0.20 = 20%

→ Use an algebraic equation to model this situation. The quantity is represented by the number of new rings.
Quantity = Percent × Whole. Let p represent the unknown percent.
1 = p∙5
\(\frac{1}{5}\) = p
\(\frac{1}{5}\) = \(\frac{20}{100}\) = 0.2 = 20%

Example 2: Percent Decrease
Ken said that he is going to reduce the number of calories that he eats during the day. Ken’s trainer asked him to start off small and reduce the number of calories by no more than 7%.
Ken estimated and consumed 2,200 calories per day instead of his normal 2,500 calories per day until his next visit with the trainer. Did Ken reduce his calorie intake by no more than 7%? Justify your answer.
Answer:
a. Ken reduced his daily calorie intake by 300 calories. Does 7% of 2,500 calories equal 300 calories?
Quantity = Percent × Whole

False, because 300≠175.

b. A 7% decrease means Ken would get 93% of his normal daily calorie intake since 100% – 7% = 93%. Ken consumed 2,200 calories, so does 93% of 2,500 equal 2,200?
Quantity = Percent × Whole

False. Because 2,200 ≠ 2,325, Ken’s estimation was wrong.

Example 3: Finding a Percent Increase or Decrease
Justin earned 8 badges in Scouts as of the Scout Master’s last report. Justin wants to complete 2 more badges so that he will have a total of 10 badges earned before the Scout Master’s next report.
a. If Justin completes the additional 2 badges, what will be the percent increase in badges?
Answer:
Quantity = Percent × Whole. Let p represent the unknown percent.
2 = p∙8
2(\(\frac{1}{8}\)) = p(\(\frac{1}{8}\))(8)
\(\frac{2}{8}\) = p
\(\frac{1}{4}\) = p
\(\frac{1}{4}\) = \(\frac{25}{100}\) = 25%
There would be a 25% increase in the number of badges.

b. Express the 10 badges as a percent of the 8 badges.
Answer:
8 badges is the whole, or 100%, and 2 badges represent 25% of the badges, so 10 badges represent
100% + 25% = 125% of the 8 badges.
Check:
10 = p∙8
10(\(\frac{1}{8}\)) = p(\(\frac{1}{8}\))(8)
\(\frac{10}{8}\) = p
\(\frac{5}{4}\) = p
\(\frac{5}{4}\) = \(\frac{125}{100}\) = 125%

c. Does 100% plus your answer in part (a) equal your answer in part (b)? Why or why not?
Answer:
Yes. My answer makes sense because 8 badges are the whole or 100%, and 2 badges represent 25% of the badges, so 10 badges represent 100% + 25%, or 125% of the 8 badges.

Example 4: Finding the Original Amount Given a Percent Increase or Decrease
The population of cats in a rural neighborhood has declined in the past year by roughly 30%. Residents hypothesize that this is due to wild coyotes preying on the cats. The current cat population in the neighborhood is estimated to be 12. Approximately how many cats were there originally?
Answer:
→ Do we know the part or the whole?
We know the part (how many cats are left), but we do not know the original whole.

→ Is this a percent increase or decrease problem? How do you know?
Percent decrease because the word declined means decreased.

→ If there was about a 30% decline in the cat population, then what percent of cats remain?
100% – 30% = 70%, so about 70% of the cats remain.

→ How do we write an equation to model this situation?
12 cats represent the quantity that is about 70% of the original number of cats. We are trying to find the whole, which equals the original number of cats. So, using Quantity = Percent × Whole and substituting the known values into the equation, we have 12 = 70%∙W, where W represents the original number of cats.
Quantity = Percent × Whole
12 = (\(\frac{7}{10}\))∙W
(12)(\(\frac{10}{7}\)) = (\(\frac{7}{10}\))(\(\frac{10}{7}\))∙W
\(\frac{120}{7}\) = W
W≈17.1≈17
There must have been 17 cats originally.

To find the original number of cats or the whole (100% of the cats), we need to add three more twelve sevenths to 12.
12 + 3(\(\frac{12}{7}\)) = \(\frac{84}{7}\) + \(\frac{36}{7}\) = \(\frac{120}{7}\)≈17
The decrease was given as approximately 30%, so there must have been 17 cats originally.

Example 5.
Lu’s math score on her achievement test in seventh grade was a 650. Her math teacher told her that her test level went up by 25% from her sixth grade test score level. What was Lu’s test score level in sixth grade?
Answer:
→ Does this represent a percent increase or decrease? How do you know?
Percent increase because the word up means increase.

→ Using the equation Quantity = Percent × Whole, what information do we know?
We know Lu’s test score level in seventh grade after the change, which is the quantity, and we know the percent. But we do not know the whole (her test score level from sixth grade).

→ If Lu’s sixth grade test score level represents the whole, then what percent represents the seventh grade level?
100% + 25% = 125%
→ How do we write an equation to model this situation? Let W represent Lu’s test score in sixth grade.
Quantity = Percent × Whole
650 = 125% × W
650 = 1.25W
650(\(\frac{1}{1.25}\)) = 1.25(\(\frac{1}{1.25}\))W
\(\frac{650}{1.25}\) = W
\(\frac{65,000}{125}\) = W
520 = W
Lu’s sixth grade test score level was 520.

Eureka Math Grade 7 Module 4 Lesson 4 Exercise Answer Key

Opening Exercise
Cassandra likes jewelry. She has 5 rings in her jewelry box.
a. In the box below, sketch Cassandra’s 5 rings.
Answer:

b. Draw a double number line diagram relating the number of rings as a percent of the whole set of rings.
Answer:

c. What percent is represented by the whole collection of rings? What percent of the collection does each ring represent?
Answer:
100%, 20%

Exercise 1.
a. Jon increased his trading card collection by 5 cards. He originally had 15 cards. What is the percent increase? Use the equation Quantity = Percent × Whole to arrive at your answer, and then justify your answer using a numeric or visual model.
Answer:
Quantity = Percent × Whole. Let p represent the unknown percent.
5 = p(15)
5(\(\frac{1}{15}\)) = p(15)(\(\frac{1}{15}\))
\(\frac{5}{15}\) = p
p = \(\frac{1}{3}\) = 0.3333…
0.3333 … = \(\frac{33}{100}\) + \(\frac{0.3333 \ldots}{100}\) = 33% + \(\frac{1}{3}\)% = 33\(\frac{1}{3}\)%

b. Suppose instead of increasing the collection by 5 cards, Jon increased his 15 – card collection by just 1 card. Will the percent increase be the same as when Cassandra’s ring collection increased by 1 ring (in Example 1)? Why or why not? Explain.
Answer:
No, it would not be the same because the part – to – whole relationship is different. Cassandra’s additional ring compared to the original whole collection was 1 to 5, which is equivalent to 20 to 100, which is 20%. Jon’s additional trading card compared to his original card collection is 1 to 15, which is less than 10%, since
\(\frac{1}{15}\)<\(\frac{1}{10}\), and \(\frac{1}{10}\) = 10%.

c. Based on your answer to part (b), how is displaying change as a percent useful?
Answer:
Representing change as a percent helps us to understand how large the change is compared to the whole.

Discussion
A sales representative is taking 10% off of your bill as an apology for any inconveniences.

Answer:

Exercise 2.
Skylar is answering the following math problem:
The value of an investment decreased by 10%. The original amount of the investment was $75.00. What is the current value of the investment?
a. Skylar said 10% of $75.00 is $7.50, and since the investment decreased by that amount, you have to subtract $7.50 from $75.00 to arrive at the final answer of $67.50. Create one algebraic equation that can be used to arrive at the final answer of $67.50. Solve the equation to prove it results in an answer of $67.50. Be prepared to explain your thought process to the class.
Answer:
Let F represent the final value of the investment.
The final value is 90% of the original investment, since 100% – 10% = 90%.
F = Percent × Whole
F = (0.90)(75)
F = 67.5
The final value of the investment is $67.50.

b. Skylar wanted to show the proportional relationship between the dollar value of the original investment, x, and its value after a 10% decrease, y. He creates the table of values shown below. Does it model the relationship? Explain. Then, provide a correct equation for the relationship Skylar wants to model.

Answer:
No. The table only shows the proportional relationship between the amount of the investment and the amount of the decrease, which is 10% of the amount of the investment. To show the relationship between the value of the investment before and after the 10% decrease, he needs to subtract each value currently in the y – column from each value in the x – column so that the y – column shows the following values: 67.5, 90, 180, 270, and 360. The correct equation is y = x – 0.10x, or y = 0.90x.

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

Question 1.
A store advertises 15% off an item that regularly sells for $300.
a. What is the sale price of the item?
Answer:
(0.85)300 = 255; the sale price is $255.

b. How is a 15% discount similar to a 15% decrease? Explain.
Answer:
In both cases, you are subtracting 15% of the whole from the whole, or finding 85% of the whole.

c. If 8% sales tax is charged on the sale price, what is the total with tax?
Answer:
(1.08)(255) = 275.40; the total with tax is $275.40.

d. How is 8% sales tax like an 8% increase? Explain.
Answer:
In both cases, you are adding 8% of the whole to the whole, or finding 108% of the whole.

Question 2.
An item that was selling for $72.00 is reduced to $60.00. Find the percent decrease in price. Round your answer to the nearest tenth.
Answer:
The whole is 72. 72 – 60 = 12. 12 is the part. Using Quantity = Percent × Whole, I get 12 = p × 72, where p represents the unknown percent, and working backward, I arrive at \(\frac{12}{72}\) = \(\frac{1}{6}\) = \(0.1 \overline{6}\) = p.
So, it is about a 16.7% decrease.

Question 3.
A baseball team had 80 players show up for tryouts last year and this year had 96 players show up for tryouts. Find the percent increase in players from last year to this year.
Answer:
The number of players that showed up last year is the whole; 16 players are the quantity of change since
96 – 80 = 16.
Quantity = Percent × Whole. Let p represent the unknown percent.
16 = p(80)
p = 0.2
0.2 = \(\frac{20}{100}\) = 20%
The number of players this year was a 20% increase from last year.

Question 4.
At a student council meeting, there was a total of 60 students present. Of those students, 35 were female.
a. By what percent is the number of females greater than the number of males?
Answer:
The number of males (60 – 35 = 25) at the meeting is the whole. The part (quantity) can be represented by the number of females (35) or how many more females there are than the number of males.
Quantity = Percent × Whole
35 = p(25)
p = 1.4
1.4 = 140%, which is 40% more than 100%. Therefore, there were 40% more females than males at the student council meeting.

b. By what percent is the number of males less than the number of females?
Answer:
The number of females (35) at the meeting is the whole. The part (quantity) can be represented by the number of males, or the number less of males than females (10).
Quantity = Percent × Whole
10 = p(35)
p≈0.29
0.29 = 29%
The number of males at the meeting is approximately 29% less than the number of females.

c. Why is the percent increase and percent decrease in parts (a) and (b) different?
Answer:
The difference in the number of males and females is the same in each case, but the whole quantities in parts (a) and (b) are different.

Question 5.
Once each day, Darlene writes in her personal diary and records whether the sun is shining or not. When she looked back though her diary, she found that over a period of 600 days, the sun was shining 60% of the time. She kept recording for another 200 days and then found that the total number of sunny days dropped to 50%. How many of the final 200 days were sunny days?
Answer:
To find the number of sunny days in the first 600 days, the total number of days is the whole.
Quantity = Percent × Whole. Let s represent the number of sunny days.
s = 0.6(600)
s = 360
There were 360 sunny days in the first 600 days.
The total number of days that Darlene observed was 800 days because 600 + 200 = 800.
d = 0.5(800)
d = 400
There was a total of 400 sunny days out of the 800 days.
The number of sunny days in the final 200 days is the difference of 400 days and 360 days.
400 – 360 = 40, so there were 40 sunny days of the last 200 days.

Question 6.
Henry is considering purchasing a mountain bike. He likes two bikes: One costs $500, and the other costs $600. He tells his dad that the bike that is more expensive is 20% more than the cost of the other bike. Is he correct? Justify your answer.
Answer:
Yes. Quantity = Percent × Whole. After substituting in the values of the bikes and percent, I arrive at the following equation: 600 = 1.2(500), which is a true equation.

Question 7.
State two numbers such that the lesser number is 25% less than the greater number.
Answer:
Answers will vary. One solution is as follows: Greater number is 100; lesser number is 75.

Question 8.
State two numbers such that the greater number is 75% more than the lesser number.
Answer:
Answers will vary. One solution is as follows: Greater number is 175; lesser number is 100.

Question 9.
Explain the difference in your thought process for Problems 7 and 8. Can you use the same numbers for each problem? Why or why not?
Answer:
No. The whole is different in each problem. In Problem 7, the greater number is the whole. In Problem 8, the lesser number is the whole.

Question 10.
In each of the following expressions, c represents the original cost of an item.

a. Circle the expression(s) that represents 10% of the original cost. If more than one answer is correct, explain why the expressions you chose are equivalent.
b. Put a box around the expression(s) that represents the final cost of the item after a 10% decrease. If more than one is correct, explain why the expressions you chose are equivalent.
Answer:
c – 0.10c
1c – 0.10c Multiplicative identity property of 1
(1 – 0.10)c Distributive property (writing a sum or difference as a product)
0.90c
Therefore, c – 0.10c = 0.90c.

c. Create a word problem involving a percent decrease so that the answer can be represented by expression (ii).
Answer:
Answers will vary. The store’s cashier told me I would get a 10% discount on my purchase. How can I find the amount of the 10% discount?

d. Create a word problem involving a percent decrease so that the answer can be represented by expression (i).
Answer:
Answers will vary. An item is on sale for 10% off. If the original price of the item is c, what is the final price after the 10% discount?

e. Tyler wants to know if it matters if he represents a situation involving a 25% decrease as 0.25x or (1 – 0.25)x. In the space below, write an explanation that would help Tyler understand how the context of a word problem often determines how to represent the situation.
Answer:
If the word problem asks you to find the amount of the 25% decrease, then 0.25x would represent it. If the problem asks you to find the value after a 25% decrease, then (1 – 0.25)x would be a correct representation.

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

Question 1.
Erin wants to raise her math grade to a 95 to improve her chances of winning a math scholarship. Her math average for the last marking period was an 81. Erin decides she must raise her math average by 15% to meet her goal. Do you agree? Why or why not? Support your written answer by showing your math work.
Answer:
No, I do not agree. 15% of 81 is 12.15. 81 + 12.15 = 93.15, which is less than 95. I arrived at my answer using the equation below to find 15% of 81.
Quantity = Percent × Whole
Let G stand for the number of points Erin’s grade will increase by after a 15% increase from 81. The whole is 81, and the percent is 15%. First, I need to find 15% of 81 to arrive at the number of points represented by a 15% increase. Then, I will add that to 81 to see if it equals 95, which is Erin’s goal.
G = 0.15 × 81
G = 12.15
Adding the points onto her average: 81.00 + 12.15 = 93.15
Comparing it to her goal: 93.15 < 95

Engage NY Eureka Math 7th Grade Module 4 Lesson 3 Answer Key

Eureka Math Grade 7 Module 4 Lesson 3 Example Answer Key

Example
a. The members of a club are making friendship bracelets to sell to raise money. Anna and Emily made 54 bracelets over the weekend. They need to produce 300 bracelets by the end of the week. What percent of the bracelets were they able to produce over the weekend?
Answer:

300 → 100%
1 → \(\frac{100}{300}\)%
54 → 54 ∙ \(\frac{100}{300}\)%
54 → 54 ∙ \(\frac{1}{3}\)%
54 → 18%
Anna and Emily were able to produce 18% of the total number of bracelets over the weekend

Quantity = Percent × Whole
Let p represent the unknown percent.
54 = p(300)
\(\frac{1}{300}\) (54) = \(\frac{1}{300}\) (300)p
\(\frac{54}{300}\) = 1p
\(\frac{18}{100}\) = p
\(\frac{18}{100}\) = 0.18 = 18%
Anna and Emily were able to produce 18% of the total bracelets over the weekend.

b. Anna produced 32 of the 54 bracelets produced by Emily and Anna over the weekend. Write the number of bracelets that Emily produced as a percent of those that Anna produced.
Answer:
Arithmetic Method:
32 → 100%
1 → \(\frac{100}{32}\)%
22 → 22 ∙ \(\frac{100}{32}\)%
22 → 100 ∙ \(\frac{22}{32}\)%
22 → 100 ∙ 0.6875%
22 → 68.75%

Algebraic Method:
Quantity = Percent × Whole
Let p represent the unknown percent.
22 = p(32)
\(\frac{1}{32}\) (22) = \(\frac{1}{32}\) (32)p
\(\frac{22}{32}\) = 1p
0.6875 = p
0.6875 = 68.75%

22 bracelets are 68.75% of the number of bracelets that Anna produced. Emily produced 22 bracelets; therefore, she produced 68.75% of the number of bracelets that Anna produced.

c. Write the number of bracelets that Anna produced as a percent of those that Emily produced.
Answer:
Arithmetic Method:
22 → 100%
1 → \(\frac{100}{22}\)%
32 → 32 ∙ \(\frac{100}{22}\)%
32 → 100 ∙ \(\frac{32}{22}\)%
32 → 100 ∙ \(\frac{16}{11}\)%
32 → \(\frac{1600}{11}\)%
32 → 145 \(\frac{5}{11}\)%

Algebraic Method:
Quantity = Percent × Whole
Let p represent the unknown percent.
32 = p(22)
\(\frac{1}{22}\) (32) = \(\frac{1}{22}\) (22)p
\(\frac{32}{22}\) = 1p
\(\frac{16}{11}\) = p
1 \(\frac{5}{11}\) = p
1 \(\frac{5}{11}\) = 1 \(\frac{5}{11}\) × 100% = 145 \(\frac{5}{11}\)%

32 bracelets are 145 \(\frac{5}{11}\)% of the number of bracelets that Emily produced. Anna produced 32 bracelets over the weekend, so Anna produced 145 \(\frac{5}{11}\)% of the number of bracelets that Emily produced.

Eureka Math Grade 7 Module 4 Lesson 3 Exercise Answer Key

Opening Exercise
If each 10 × 10 unit square represents one whole, then what percent is represented by the shaded region?

In the model above, 25% represents a quantity of 10 students. How many students does the shaded region represent?
Answer:
If 25% represents 10 students, then 1% represents \(\frac{10}{25}\), or \(\frac{2}{5}\), of a student. The shaded region covers 125 square units, or 125%, so since \(\frac{2}{5}\) ∙ 125 = 50, the shaded region represents 50 students.

Exercise 2.
There are 750 students in the seventh – grade class and 625 students in the eighth – grade class at Kent Middle School.
a. What percent is the seventh – grade class of the eighth – grade class at Kent Middle School?
The number of eighth graders is the whole amount. Let p represent the percent of seventh graders compared to eighth graders.
Quantity = Percent × Whole
Let p represent the unknown percent.
750 = p(625)
750(\(\frac{1}{625}\)) = p(625)(\(\frac{1}{625}\))
1.2 = p
1.2 = 120%
The number of seventh graders is 120% of the number of eighth graders.
There are 20% more seventh graders than eighth graders.
Alternate solution: There are 125 more seventh graders. 125 = p(625), p = 0.20. There are 20% more seventh graders than eighth graders.

b. The principal will have to increase the number of eighth – grade teachers next year if the seventh – grade enrollment exceeds 110% of the current eighth – grade enrollment. Will she need to increase the number of teachers? Explain your reasoning.
Answer:
The principal will have to increase the number of teachers next year. In part (a), we found out that the seventh grade enrollment was 120% of the number of eighth graders, which is greater than 110%.

Exercise 3.
At Kent Middle School, there are 104 students in the band and 80 students in the choir. What percent of the number of students in the choir is the number of students in the band?
Answer:
The number of students in the choir is the whole.
Quantity = Percent × Whole
Let p represent the unknown percent.
104 = p(80)
p = 1.3
1.3 = 130%
The number of students in the band is 130% of the number of students in the choir.

Exercise 4.
At Kent Middle School, breakfast costs $1.25 and lunch costs $3.75. What percent of the cost of lunch is the cost of breakfast?
Quantity = Percent × Whole
Let p represent the unknown percent.
1.25 = p(3.75)
1.25(\(\frac{1}{3.75}\)) = p(3.75)(\(\frac{1}{3.75}\))
p = \(\frac{1.25}{3.75}\)
p = \(\frac{1}{3}\)
\(\frac{1}{3}\) = \(\frac{1}{3}\) (100%) = 33 \(\frac{1}{3}\)%

The cost of breakfast is 33 \(\frac{1}{3}\)% of the cost of lunch.

Teacher may ask students what percent less than the cost of lunch is the cost of breakfast.
The cost of breakfast is 66\(\frac{2}{3}\)% less than the cost of lunch.

Teacher may ask what percent more is the cost of lunch than the cost of breakfast.
Let p represent the percent of lunch to breakfast.
3.75 = p(1.25)
3.75(\(\frac{1}{1.25}\)) = p(1.25)(\(\frac{1}{1.25}\))
p = \(\frac{3.75}{1.25}\) = 3 = 300%

The cost of lunch is 300% of the cost of breakfast.

Exercise 5.
Describe a real – world situation that could be modeled using the equation 398.4 = 0.83(x). Describe how the elements of the equation correspond with the real – world quantities in your problem. Then, solve your problem.
Answer:
Word problems will vary. Sample problem: A new tablet is on sale for 83% of its original sale price. The tablet is currently priced at $398.40. What was the original price of the tablet?

0.83 = \(\frac{83}{100}\) = 83%, so 0.83 represents the percent that corresponds with the current price. The current price ($398.40) is part of the original price; therefore, it is represented by 398.4. The original price is represented by x and is the whole quantity in this problem.
398.4 = 0.83x
\(\frac{1}{0.83}\) (398.4) = \(\frac{1}{0.83}\) (0.83)x
\(\frac{398.4}{0.83}\) = 1x
480 = x
The original price of the tablet was $480.00.

Eureka Math Grade 7 Module 4 Lesson 3 Problem Set Answer Key

Question 1.
Solve each problem using an equation.
a. 49.5 is what percent of 33?
Answer:
49.5 = p(33)
p = 1.5 = 150%

b. 72 is what percent of 180?
Answer:
72 = p(180)
p = 0.4 = 40%

c. What percent of 80 is 90?
Answer:
90 = p(80)
p = 1.125 = 112.5%

Question 2.
This year, Benny is 12 years old, and his mom is 48 years old.
a. What percent of his mom’s age is Benny’s age?
Answer:
Let p represent the percent of Benny’s age to his mom’s age.
12 = p(48)
p = 0.25 = 25%
Benny’s age is 25% of his mom’s age.

b. What percent of Benny’s age is his mom’s age?
Answer:
Let p represent the percent of his mom’s age to Benny’s age.
48 = p(12)
p = 4 = 400%
Benny’s mom’s age is 400% of Benny’s age.

c. In two years, what percent of his age will Benny’s mom’s age be at that time?
Answer:
In two years, Benny will be 14, and his mom will be 50.
14 → 100%
1 → (\(\frac{100}{14}\))%
50 → 50(\(\frac{100}{14}\)%
50 → 25(\(\frac{100}{7}\))%
50 → (\(\frac{2500}{7}\))%
50 → 357 \(\frac{1}{7}\)%
His mom’s age will be 357 \(\frac{1}{7}\)% of Benny’s age at that time.

d. In 10 years, what percent will Benny’s mom’s age be of his age?
Answer:
In 10 years, Benny will be 22 years old, and his mom will be 58 years old.
22 → 100%
1 → \(\frac{100}{22}\)%
58 → 58(\(\frac{100}{22}\))%
58 → 29(\(\frac{100}{11}\))%
58 → \(\frac{2900}{11}\)%
58 → 263 \(\frac{7}{11}\)%
In 10 years, Benny’s mom’s age will be 263 \(\frac{7}{11}\)% of Benny’s age at that time.

e. In how many years will Benny be 50% of his mom’s age?
Answer:
Benny will be 50% of his mom’s age when she is 200% of his age (or twice his age). Benny and his mom are always 36 years apart. When Benny is 36, his mom will be 72, and he will be 50% of her age. So, in 24 years, Benny will be 50% of his mom’s age.

d. As Benny and his mom get older, Benny thinks that the percent of difference between their ages will decrease as well. Do you agree or disagree? Explain your reasoning.
Answer:
Student responses will vary. Some students might argue that they are not getting closer since they are always 36 years apart. However, if you compare the percents, you can see that Benny‘s age is getting closer to 100% of his mom’s age, even though their ages are not getting any closer.

Question 3.
This year, Benny is 12 years old. His brother Lenny’s age is 175% of Benny’s age. How old is Lenny?
Answer:
Let L represent Lenny’s age. Benny’s age is the whole.
L = 1.75(12)
L = 21
Lenny is 21 years old.

Question 4.
When Benny’s sister Penny is 24, Benny’s age will be 125% of her age.
a. How old will Benny be then?
Answer:
Let b represent Benny’s age when Penny is 24.
b = 1.25(24)
b = 30
When Penny is 24, Benny will be 30.

b. If Benny is 12 years old now, how old is Penny now? Explain your reasoning.
Answer:
Penny is 6 years younger than Benny. If Benny is 12 now, then Penny is 6.

Question 5.
Benny’s age is currently 200% of his sister Jenny’s age. What percent of Benny’s age will Jenny’s age be in 4 years?
If Benny is 200% of Jenny’s age, then he is twice her age, and she is half of his age. Half of 12 is 6. Jenny is currently 6 years old. In 4 years, Answer:
Jenny will be 10 years old, and Benny will be 16 years old.
Quantity = Percent × Whole. Let p represent the unknown percent. Benny’s age is the whole.
10 = p(16)
p = \(\frac{10}{16}\)
p = \(\frac{5}{8}\)
p = 0.625 = 62.5%
In 4 years, Jenny will be 62.5% of Benny’s age.

Question 6.
At the animal shelter, there are 15 dogs, 12 cats, 3 snakes, and 5 parakeets.
a. What percent of the number of cats is the number of dogs?
Answer:
\(\frac{15}{12}\) = 1.25. That is 125%. The number of dogs is 125% the number of cats.

b. What percent of the number of cats is the number of snakes?
Answer:
\(\frac{3}{12}\) = \(\frac{1}{4}\) = 0.25. There are 25% as many snakes as cats.

c. What percent less parakeets are there than dogs?
Answer:
\(\frac{5}{15}\) = \(\frac{1}{3}\). That is 33 \(\frac{1}{3}\)%. There are 66 \(\frac{2}{3}\)% less parakeets than dogs.

d. Which animal has 80% of the number of another animal?
Answer:
\(\frac{12}{15}\) = \(\frac{4}{5}\) = \(\frac{8}{10}\) = 0.80. The number of cats is 80% the number of dogs.

e. Which animal makes up approximately 14% of the animals in the shelter?
Answer:
Quantity = Percent × Whole. The total number of animals is the whole.
q = 0.14(35)
q = 4.9
The quantity closest to 4.9 is 5, the number of parakeets.

Question 7.
Is 2 hours and 30 minutes more or less than 10% of a day? Explain your answer.
Answer:
2 hr.30 min. → 2.5 hr.; 24 hours is a whole day and represents the whole quantity in this problem.
10% of 24 hours is 2.4 hours.
2.5 > 2.4, so 2 hours and 30 minutes is more than 10% of a day.

Question 8.
A club’s membership increased from 25 to 30 members.
a. Express the new membership as a percent of the old membership.
Answer:
The old membership is the whole.
Quantity = Percent × Whole. Let p represent the unknown percent.
30 = p(25)
p = 1.2 = 120%
The new membership is 120% of the old membership.

b. Express the old membership as a percent of the new membership.
Answer:
The new membership is the whole.
30 → 100%
1 → \(\frac{100}{30}\)%
25 → 25 ∙ \(\frac{100}{30}\)%
25 → 5 ∙ 1\(\frac{100}{6}\)%
25 → \(\frac{500}{6}\)% = 83 \(\frac{1}{3}\)%
The old membership is 83 \(\frac{1}{3}\)% of the new membership.

Question 9.
The number of boys in a school is 120% the number of girls at the school.
a. Find the number of boys if there are 320 girls.
Answer:
The number of girls is the whole.
Quantity = Percent × Whole.
Let b represent the unknown number of boys at the school.
b = 1.2(320)
b = 384
If there are 320 girls, then there are 384 boys at the school.

b. Find the number of girls if there are 360 boys.
Answer:
The number of girls is still the whole.
Quantity = Percent × Whole.
Let g represent the unknown number of girls at the school.
360 = 1.2(g)
g = 300
If there are 360 boys at the school, then there are 300 girls.

Question 10.
The price of a bicycle was increased from $300 to $450.
a. What percent of the original price is the increased price?
Answer:
The original price is the whole.
Quantity = Percent × Whole. Let p represent the unknown percent.
450 = p(300)
p = 1.5
1.5 = \(\frac{150}{100}\) = 150%
The increased price is 150% of the original price.

b. What percent of the increased price is the original price?
Answer:
The increased price, $450, is the whole.
450 → 100%
1 → \(\frac{100}{450}\)%
300 → 300(\(\frac{100}{450}\))%
300 → 2(\(\frac{100}{3}\))%
300 → \(\frac{200}{3}\)%
300 → 66 \(\frac{2}{3}\)%
The original price is 66 \(\frac{2}{3}\)% of the increased price.

Question 11.
The population of Appleton is 175% of the population of Cherryton.
a. Find the population in Appleton if the population in Cherryton is 4,000 people.
Answer:
The population of Cherryton is the whole.
Quantity = Percent × Whole. Let a represent the unknown population of Appleton.
a = 1.75(4,000)
a = 7,000
If the population of Cherryton is 4,000 people, then the population of Appleton is 7,000 people.

b. Find the population in Cherryton if the population in Appleton is 10,500 people.
Answer:
The population of Cherryton is still the whole.
Quantity = Percent × Whole. Let c represent the unknown population of Cherryton.
10,500 = 1.75c
c = 10,500÷1.75
c = 6,000
If the population of Appleton is 10,500 people, then the population of Cherryton is 6,000 people.

Question 12.
A statistics class collected data regarding the number of boys and the number of girls in each classroom at their school during homeroom. Some of their results are shown in the table below.
a. Complete the blank cells of the table using your knowledge about percent.

Answer:

b. Using a coordinate plane and grid paper, locate and label the points representing the ordered pairs (x,y).
Answer:
See graph to the right.

c. Locate all points on the graph that would represent classrooms in which the number of girls y is 100% of the number of boys x. Describe the pattern that these points make.
Answer:
The points lie on a line that includes the origin; therefore, it is a proportional relationship.

d. Which points represent the classrooms in which the number of girls as a percent of the number of boys is greater than 100%? Which points represent the classrooms in which the number of girls as a percent of the number of boys is less than 100%? Describe the locations of the points in relation to the points in part (c).
Answer:
All points where y > x are above the line and represent classrooms where the number of girls is greater than 100% of the number of boys. All points where y < x are below the line and represent classrooms where the number of girls is less than 100% of the boys.

e. Find three ordered pairs from your table representing classrooms where the number of girls is the same percent of the number of boys. Do these points represent a proportional relationship? Explain your reasoning.
Answer:
There are two sets of points that satisfy this question:
{(3,6), (5,10), and (11,22)}: The points do represent a proportional relationship because there is a constant of proportionality k = \(\frac{y}{x}\) = 2.
{(4,2), (10,5), and (14,7)}: The points do represent a proportional relationship because there is a constant of proportionality k = \(\frac{y}{x}\) = \(\frac{1}{2}\).

f. Show the relationship(s) from part (e) on the graph, and label them with the corresponding equation(s).
Answer:

g. What is the constant of proportionality in your equation(s), and what does it tell us about the number of girls and the number of boys at each point on the graph that represents it? What does the constant of proportionality represent in the table in part (a)?
Answer:
In the equation y = 2x, the constant of proportionality is 2, and it tells us that the number of girls will be twice the number of boys, or 200% of the number of boys, as shown in the table in part (a).
In the equation y = 1/2 x, the constant of proportionality is 1/2, and it tells us that the number of girls will be half the number of boys, or 50% of the number of boys, as shown in the table in part (a).

Eureka Math Grade 7 Module 4 Lesson 3 Exit Ticket Answer Key

Solve each problem below using at least two different approaches.
Question 1.
Jenny’s great – grandmother is 90 years old. Jenny is 12 years old. What percent of Jenny’s great – grandmother’s age is Jenny’s age?
Answer:
Algebraic Solution:
Quantity = Percent × Whole. Let p represent the unknown percent.
Jenny’s great – grandmother’s age is the whole.
12 = p(90)
12 ∙ \(\frac{1}{90}\) = p(90) ∙ \(\frac{1}{90}\)
2 ∙ \(\frac{1}{15}\) = p(1)
\(\frac{2}{15}\) = p
\(\frac{2}{15}\) = \(\frac{2}{15}\) (100%) = 13 \(\frac{1}{3}\)%
Jenny’s age is 13 1/3% of her great – grandmother’s age.

Numeric Solution:
90 → 100%
1 → \(\frac{100}{90}\)%
12 → (12 ∙ \(\frac{100}{90}\))%
12 → (100 ∙ \(\frac{12}{90}\))%
12 → 100(\(\frac{2}{15}\))%
12 → 20(\(\frac{2}{3}\))%
12 → (\(\frac{40}{3}\))%
12 → 13 \(\frac{1}{3}\)%

Alternative Numeric Solution:
90 → 100%
9 → 10%
3 → \(\frac{10}{3}\)%
12 → 4(\(\frac{10}{3}\))%
12 → (\(\frac{40}{3}\))%
12 → 13 \(\frac{1}{3}\)%

Question 2.
Jenny’s mom is 36 years old. What percent of Jenny’s mother’s age is Jenny’s great – grandmother’s age?
Answer:
Quantity = Percent × Whole.
Let p represent the unknown percent. Jenny’s mother’s age is the whole.
90 = p(36)
90 ∙ \(\frac{1}{36}\) = p(36) ∙ \(\frac{1}{36}\)
5 ∙ \(\frac{1}{2}\) = p(1)
2.5 = p
2.5 = 250%
Jenny’s great grandmother’s age is 250% of Jenny’s mother’s age.

Eureka Math Grade 7 Module 4 Lesson 3 Part, Whole, or Percent—Round 1 Answer Key

Directions: Find each missing value.

Answer:

Eureka Math Grade 7 Module 4 Lesson 3 Part, Whole, or Percent—Round 2 Answer Key

Directions: Find each missing value.

Answer:

Engage NY Eureka Math 7th Grade Module 4 Lesson 2 Answer Key

Eureka Math Grade 7 Module 4 Lesson 2 Example Answer Key

Example 2: A Numeric Approach to Finding a Part, Given a Percent of the Whole
In Ty’s English class, 70% of the students completed an essay by the due date. There are 30 students in Ty’s English class. How many completed the essay by the due date?
Answer:
Whole → 100%
30 → 100%
\(\frac{30}{100}\) → 1%
70 ∙ \(\frac{3}{100}\) → 21
21 → 70%
70% of 30 is 21, so 21 of the students in Ty’s English class completed their essays on time.

Example 3: An Algebraic Approach to Finding a Part, Given a Percent of the Whole
A bag of candy contains 300 pieces of which 28% are red. How many pieces are red?
Which quantity represents the whole?
Answer:
The total number of candies in the bag, 300, is the whole because the number of red candies is being compared to it.

Which of the terms in the percent equation is unknown? Define a letter (variable) to represent the unknown quantity.
Answer:
We do not know the part, which is the number of red candies in the bag. Let r represent the number of red candies in the bag.

Write an expression using the percent and the whole to represent the number of pieces of red candy.
Answer:
\(\frac{28}{100}\) ∙ (300), or 0.28∙(300), is the amount of red candy since the number of red candies is 28% of the 300 pieces of candy in the bag.

Write and solve an equation to find the unknown quantity.
Part = Percent × Whole
r = \(\frac{28}{100}\) ∙ (300)
r 28 ∙ 3
r = 84
There are 84 red pieces of candy in the bag.

Example 4: Comparing Part of a Whole to the Whole with the Percent Formula
Zoey inflated 24 balloons for decorations at the middle school dance. If Zoey inflated 15% of the total number of balloons inflated for the dance, how many balloons are there total? Solve the problem using the percent formula, and verify your answer using a visual model.
Answer:
Part = Percent×Whole
The part is 24 balloons, and the percent is 15%, so let t represent the unknown total number of balloons.
24 = \(\frac{15}{100}\) t If a = b, then ac = bc.
\(\frac{100}{15}\) (24) = \(\frac{100}{15}\) (\(\frac{15}{100}\))t Multiplicative inverse
\(\frac{2400}{15}\) = 1t Multiplicative identity property of 1 and equivalent fractions
160 = t
The total number of balloons to be inflated for the dance was 160 balloons.

15% → 24
1% → \(\frac{24}{15}\) We want the quantity that corresponds with 100%, so first we find 1%.*
100% → \(\frac{24}{15}\) ∙ 100
100% → \(\frac{24}{3}\) ∙ 20
100% → 160
Student may also find 5% as is shown in the tape diagram below.
The total number of balloons to be inflated for the dance was 160 balloons.

Example 5: Finding the Percent Given a Part of the Whole and the Whole
Haley is making admission tickets to the middle school dance. So far she has made 112 tickets, and her plan is to make 320 tickets. What percent of the admission tickets has Haley produced so far? Solve the problem using the percent formula, and verify your answer using a visual model.
Answer:
Part = Percent×Whole
The part is 112 tickets, and the whole is 320 tickets, so let p represent the unknown percent.
112 = p(320) If” a = b”,then” ac = bc.
112 ∙ \(\frac{1}{320}\) = p(320) ∙ \(\frac{1}{320}\) “Multiplicative inverse
\(\frac{112}{320}\) = p(1) Multiplicative identity property of 1
\(\frac{7}{20}\) = p
0.35 = p
0.35 = \(\frac{35}{100}\) = 35%, so Haley has made 35% of the tickets for the dance.

We need to know the percent that corresponds with 112, so first we find the percent that corresponds with 1 ticket.
320 → 100%
1 → (\(\frac{100}{320}\))%
112 → 112 ∙ (\(\frac{100}{320}\))%
112 → 112 ∙ (\(\frac{5}{16}\))%
112 → 7∙(5)%
112 → 35%
Haley has made 35% of the tickets for the dance.

Eureka Math Grade 7 Module 4 Lesson 2 Exercise Answer Key

Opening Exercise
a. What is the whole unit in each scenario?

Answer:

b. Read each problem, and complete the table to record what you know.

Answer:

Exercise 1.
In Ty’s art class, 12% of the Flag Day art projects received a perfect score. There were 25 art projects turned in by Ty’s class. How many of the art projects earned a perfect score? (Identify the whole.)
Answer:

The whole is the number of art projects turned in by Ty’s class, 25.
\(\frac{25}{100}\) = 0.25; 0.25∙12 = 3; 12% of 25 is 3, so 3 art projects in Ty’s class received a perfect score.

Exercise 2
A bag of candy contains 300 pieces of which 28% are red. How many pieces are not red?
a. Write an equation to represent the number of pieces that are not red, n.
Answer:
Part = Percent×Whole
n = (100% – 28%)(300)

b. Use your equation to find the number of pieces of candy that are not red.
Answer:
If 28% of the candies are red, then the difference of 100% and 28% must be candies that are not red.
n = (100% – 28%)(300)
n = (72%)(300)
n = \(\frac{72}{100}\)(300)
n = 72∙3
n = 216
There are 216 pieces of candy in the bag that are not red.

c. Jah – Lil told his math teacher that he could use the answer from Example 3 and mental math to find the number of pieces of candy that are not red. Explain what Jah – Lil meant by that.
Answer:
He meant that once you know there are 84 red pieces of candy in a bag that contains 300 pieces of candy total, you just subtract 84 from 300 to know that 216 pieces of candy are not red.

Eureka Math Grade 7 Module 4 Lesson 2 Problem Set Answer Key

Question 1.
Represent each situation using an equation. Check your answer with a visual model or numeric method.
a. What number is 40% of 90?
Answer:
n = 0.40(90)
n = 36

b. What number is 45% of 90?
Answer:
n = 0.45(90)
n = 40.5

c. 27 is 30% of what number?
Answer:
27 = 0.3n
\(\frac{27}{0.3}\) = 1n
90 = n

d. 18 is 30% of what number?
Answer:
0.30n = 18
1n = \(\frac{18}{0.3}\)
n = 60

e. 25.5 is what percent of 85?
Answer:
25.5 = p(85)
\(\frac{25.5}{85}\) = 1p
0.3 = p
0.3 = \(\frac{30}{100}\) = 30%

f. 21 is what percent of 60?
21 = p(60)
\(\frac{21}{60}\) = 1p
0.35 = p
0.35 = \(\frac{35}{100}\) = 35%

Question 2.
40% of the students on a field trip love the museum. If there are 20 students on the field trip, how many love the museum?
Answer:
Let s represent the number of students who love the museum.
s = 0.40(20)
s = 8
Therefore, 8 students love the museum.

Question 3.
Maya spent 40% of her savings to pay for a bicycle that cost her $85.
a. How much money was in her savings to begin with?
Answer:
Let s represent the unknown amount of money in Maya’s savings.
85 = 0.4s
212.5 = s
Maya originally had $212.50 in her savings.

b. How much money does she have left in her savings after buying the bicycle?
Answer:
$212.50 – $85.00 = $127.50
She has $127.50 left in her savings after buying the bicycle.

Question 4.
Curtis threw 15 darts at a dartboard. 40% of his darts hit the bull’s – eye. How many darts did not hit the bull’s – eye?
Answer:
Let d represent the number of darts that hit the bull’s – eye.
d = 0.4(15)
d = 6
6 darts hit the bull’s – eye. 15 – 6 = 9
Therefore, 9 darts did not hit the bull’s – eye.

Question 5.
A tool set is on sale for $424.15. The original price of the tool set was $499.00. What percent of the original price is the sale price?
Answer:
Let p represent the unknown percent.
424.15 = p(499)
0.85 = p
The sale price is 85% of the original price.

Question 6.
Matthew scored a total of 168 points in basketball this season. He scored 147 of those points in the regular season and the rest were scored in his only playoff game. What percent of his total points did he score in the playoff game?
Answer:
Matthew scored 21 points during the playoff game because 168 – 147 = 21.
Let p represent the unknown percent.
21 = p(168)
0.125 = p
The points that Matthew scored in the playoff game were 12.5% of his total points scored in basketball this year.

Question 7.
Brad put 10 crickets in his pet lizard’s cage. After one day, Brad’s lizard had eaten 20% of the crickets he had put in the cage. By the end of the next day, the lizard had eaten 25% of the remaining crickets. How many crickets were left in the cage at the end of the second day?
Answer:
Let n represent the number of crickets eaten.
Day 1:
n = 0.2(10)
n = 2
At the end of the first day, Brad’s lizard had eaten 2 of the crickets.
Day 2:
n = 0.25(10 – 2)
n = 0.25(8)
n = 2
At the end of the second day, Brad’s lizard had eaten a total of 4 crickets, leaving 6 crickets in the cage.

Question 8.
A furnace used 40% of the fuel in its tank in the month of March and then used 25% of the remaining fuel in the month of April. At the beginning of March, there were 240 gallons of fuel in the tank. How much fuel (in gallons) was left at the end of April?
Answer:
March:
n = 0.4(240)
n = 96
Therefore, 96 gallons were used during the month of March, which means 144 gallons remain.
April:
n = 0.25(144)
n = 36
Therefore, 36 gallons were used during the month of April, which means 108 gallons remain.
There were 144 gallons of fuel remaining in the tank at the end of March and 108 gallons of fuel remaining at the end of April.

Question 9.
In Lewis County, there were 2,277 student athletes competing in spring sports in 2014. That was 110% of the number from 2013, which was 90% of the number from the year before. How many student athletes signed up for a spring sport in 2012?
Answer:
2013:
2,277 = 1.10a
2,070 = a
Therefore, 2,070 student athletes competed in spring sports in 2013.
2012:
2,070 = 0.9a
2,300 = a
Therefore, 2,300 student athletes competed in spring sports in 2012.
There were 2,070 students competing in spring sports in 2013 and 2,300 students in 2012.

Question 10.
Write a real – world word problem that could be modeled by the equation below. Identify the elements of the percent equation and where they appear in your word problem, and then solve the problem.
57.5 = p(250)
Answer:
Answers will vary. Greig is buying sliced almonds for a baking project. According to the scale, his bag contains 57.5 grams of almonds. Greig needs 250 grams of sliced almonds for his project. What percent of his total weight of almonds does Greig currently have?

The quantity 57.5 represents the part of the almonds that Greig currently has on the scale, the quantity 250 represents the 250 grams of almonds that he plans to purchase, and the variable p represents the unknown percent of the whole quantity that corresponds to the quantity 57.5.
57.5 = p(250)
\(\frac{1}{250}\) (57.5) = p(\(\frac{1}{250}\))(250)
\(\frac{57.5}{250}\) = p(1)
0.23 = p
0.23 = \(\frac{23}{100}\) = 23%
Greig currently has 23% of the total weight of almonds that he plans to buy.

Eureka Math Grade 7 Module 4 Lesson 2 Exit Ticket Answer Key

Question 1.
On a recent survey, 60% of those surveyed indicated that they preferred walking to running.
a. If 540 people preferred walking, how many people were surveyed?
Answer:
Let n represent the number of people surveyed.
0.60n is the number of people who preferred walking.
Since 540 people preferred walking,
0.60n = 540
n = \(\frac{540}{0.6}\) = \(\frac{5,400}{6}\) = 900
Therefore, 900 people were surveyed.

b. How many people preferred running?
Answer:
Subtract 540 from 900.
900 – 540 = 360
Therefore, 360 people preferred running.

Question 2.
Which is greater: 25% of 15 or 15% of 25? Explain your reasoning using algebraic representations or visual models.
Answer:
They are the same.
0.25×15 = \(\frac{25}{100}\) × 15 = 3.75
0.15×25 = \(\frac{15}{100}\) × 25 = 3.75
Also, you can see they are the same without actually computing the product because of any order, any grouping of multiplication.
\(\frac{25}{100}\) × 15 = 25 × \(\frac{1}{100}\) × 15 = 25 × \(\frac{15}{100}\)

Engage NY Eureka Math 7th Grade Module 4 Lesson 1 Answer Key

Eureka Math Grade 7 Module 4 Lesson 1 Example Answer Key

Example 1.
Use the definition of the word percent to write each percent as a fraction and then as a decimal.

Answer:

Example 2.
Fill in the chart by converting between fractions, decimals, and percents. Show your work in the space below.

Answer:

350% as a fraction: 350% = \(\frac{350}{100}\) = \(\frac{35}{10}\) = \(\frac{7}{2}\) = 3\(\frac{1}{2}\)
350% as a decimal: 350% = \(\frac{350}{100}\) = 3.50

Eureka Math Grade 7 Module 4 Lesson 1 Exercise Answer Key

Opening Exercise 1: Matching
Match the percents with the correct sentence clues.

Answer:

Opening Exercise 2.
Color in the grids to represent the following fractions:
a. \(\frac{30}{100}\)

Answer:

b. \(\frac{3}{100}\)

Answer:

c. \(\frac{\frac{1}{3}}{100}\)

Answer:

Eureka Math Grade 7 Module 4 Lesson 1 Problem Set Answer Key

Question 1.
Create a model to represent the following percents.
a. 90%
Answer:

b. 0.9%
Answer:

c. 900%
Answer:

c. \(\frac{9}{10}\)%
Answer:

Question 2.
Benjamin believes that \(\frac{1}{2}\)% is equivalent to 50%. Is he correct? Why or why not?
Answer:
Benjamin is not correct because \(\frac{1}{2}\)% is equivalent to 0.50%, which is equal to (\(\frac{\frac{1}{2}}{100}\)). The second percent is equivalent to \(\frac{50}{100}\). These percents are not equivalent.

Question 3.
Order the following from least to greatest:
100%, \(\frac{1}{100}\), 0.001%, \(\frac{1}{10}\), 0.001, 1.1, 10, and \(\frac{10,000}{100}\)
Answer:
0.001%, 0.001, \(\frac{1}{100}\), \(\frac{1}{10}\), 100%, 1.1, 10, and \(\frac{10,000}{100}\)

Question 4.
Fill in the chart by converting between fractions, decimals, and percents. Show work in the space below.

Answer:

Eureka Math Grade 7 Module 4 Lesson 1 Exit Ticket Answer Key

Question 1.
Fill in the chart converting between fractions, decimals, and percents. Show work in the space provided.

Answer:

Question 2.
Using the values from the chart in Problem 1, which is the least and which is the greatest? Explain how you arrived at your answers.
Answer:
The least of the values is \(\frac{2}{5}\)%, and the greatest is 1.125. To determine which value is the least and which is the greatest, compare all three values in decimal form, fraction form, or percents. When comparing the three decimals, 0.125, 1.125, and 0.004, one can note that 0.004 is the smallest value, so \(\frac{2}{5}\)% is the least of the values and 1.125 is the greatest.

Eureka Math Grade 7 Module 4 Lesson 1 Fractions, Decimals, and Percents—Round 1 Answer Key

Directions: Write each number in the alternate form indicated.

Answer:

Eureka Math Grade 7 Module 4 Lesson 1 Fractions, Decimals, and Percents—Round 2 Answer Key

Directions: Write each number in the alternate form indicated.

Answer: