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

Eureka Math Grade 7 Module 2 Lesson 4 Example Answer Key

Example 1:
Rule for Adding Integers with Same Signs
a. Represent the sum of 3+5 using arrows on the number line.

Answer:

3 + 5 = 8

i. How long is the arrow that represents 3?
Answer:
3 units

b. What direction does it point?
Answer:
Right/up

iii. How long is the arrow that represents 5?
Answer:
5 units

iv. What direction does it point?
Answer:
Right/up

v. What is the sum?
Answer:
8

vi. If you were to represent the sum using an arrow, how long would the arrow be, and what direction would it point?
Answer:
The arrow would be 8 units long and point to the right (or up on a vertical number line).

vii. What is the relationship between the arrow representing the number on the number line and the absolute value of the number?
Answer:
The length of an arrow representing a number is equal to the absolute value of the number.

vii. Do you think that adding two positive numbers will always give you a greater positive number? Why?
Answer:
Yes, because the absolute values are positive, so the sum will be a greater positive. On a number line, adding them would move you farther away from 0 (to the right or above) on a number line.

From part (a), use the same questions to elicit feedback.
→ In the Integer Game, I would combine -3 and -5 to give me -8.

b. Represent the sum of -3 + (-5) using arrows that represent -3 and -5 on the number line.

Answer:

-3 + (-5) = -8

i. How long is the arrow that represents -3?
Answer:
3 units

ii. What direction does it point?
Answer:
Left/down

iii. How long is the arrow that represents -5?
Answer:
5 units

iv. What direction does it point?
Answer:
Left/down

iv. What is the sum?
Answer:
-8

vi. If you were to represent the sum using an arrow, how long would the arrow be, and what direction would it point?
Answer:
The arrow would be 8 units long and point to the left (or down on a vertical number line).

vii. Do you think that adding two negative numbers will always give you a smaller negative number? Why?
Answer:
Yes, because the absolute values of negative numbers are positive, so the sum will be a greater positive. However, the opposite of a greater positive is a smaller negative. On a number line, adding two negative numbers would move you farther away from 0 (to the left or below) on a number line.

c. What do both examples have in common?
Answer:
The length of the arrow representing the sum of two numbers with the same sign is the same as the sum of the absolute values of both numbers.

The teacher writes the rule for adding integers with the same sign.

RULE: Add rational numbers with the same sign by adding the absolute values and using the common sign.

Example 2.
Rule for Adding Opposite Signs
a. Represent 5 + (-3) using arrows on the number line.

Answer:
5 + (-3) = 2

i. How long is the arrow that represents 5?
Answer:
5 units

ii. What direction does it point?
Answer:
Right/up

iii. How long is the arrow that represents -3?
Answer:
3 units

iv. What direction does it point?
Answer:
Left/down

v. Which arrow is longer?
Answer:
The arrow that is five units long, or the first arrow.

vi. What is the sum? If you were to represent the sum using an arrow, how long would the arrow be, and what direction would it point?
Answer:
5 + (-3) = 2
The arrow would be 2 units long and point right/up.

b. Represent the 4 + (-7) using arrows on the number line.

Answer:
4 + (-7) = -3

i. In the two examples above, what is the relationship between the length of the arrow representing the sum and the lengths of the arrows representing the two addends?
Answer:
The length of the arrow representing the sum is equal to the difference of the absolute values of the lengths of both arrows representing the two addends.

ii. What is the relationship between the direction of the arrow representing the sum and the direction of the arrows representing the two addends?
Answer:
The direction of the arrow representing the sum has the same direction as the arrow of the addend with the greater absolute value.

iii. Write a rule that will give the length and direction of the arrow representing the sum of two values that have opposite signs.
Answer:
The length of the arrow of the sum is the difference of the absolute values of the two addends. The direction of the arrow of the sum is the same as the direction of the longer arrow.

The teacher writes the rule for adding integers with opposite signs.

RULE: Add rational numbers with opposite signs by subtracting the absolute values and using the sign of the integer with the greater absolute value

Example 3.
Applying Integer Addition Rules to Rational Numbers
Find the sum of 6 + (-2\(\frac{1}{4}\)). The addition of rational numbers follows the same rules of addition for integers.
a. Find the absolute values of the numbers.
Answer:
|6| = 6
|-2\(\frac{1}{4}\)| = 2 \(\frac{1}{4}\)

b. Subtract the absolute values.
6 – 2\(\frac{1}{4}\) = 6 – \(\frac{9}{4}\) = \(\frac{24}{4}\) – \(\frac{9}{4}\) = \(\frac{15}{4}\) = 3\(\frac{3}{4}\)
Answer:

c. The answer will take the sign of the number that has the greater absolute value.
Answer:
Since 6 has the greater absolute value (arrow length), my answer will be positive 3 \(\frac{3}{4}\) .

Eureka Math Grade 7 Module 2 Lesson 3 Exercise Answer Key

Exercise 2.
a. Decide whether the sum will be positive or negative without actually calculating the sum.
i. -4 + (-2) ___
Answer:
Negative

ii. 5 + 9 ___
Answer:
Positive

iii. -6 + (-3) ____
Answer:
Negative

iv. -1 + (-11) ___
Answer:
Negative

v. 3 + 5 + 7 ___
Answer:
Positive

vi. -20 + (-15) __
Answer:
Negative

b. Find the sum.

i. 15 + 7
Answer:
22

ii. -4 + (-16)
Answer:
-20

iii. -18 + (-64)
Answer:
-82

iv. -205 + (-123)
Answer:
-328

Exercise 3.
Circle the integer with the greater absolute value. Decide whether the sum will be positive or negative without actually calculating the sum.

i. -1 + 2
Answer:

ii. 5 + (-9)
Answer:

iii. -6 + 3
Answer:

iv. -11 + 1
Answer:

b. Find the sum.

i. -10 + 7
Answer:
-3

ii. 8 + (-16)
Answer:
-8

iii. -12 + (65)
Answer:
53

iv. 105 + (-126)
Answer:
-21

Exercise 4.
Solve the following problems. Show your work.
a. Find the sum of
Answer:
|-18| = 18
|7| = 7
18 – 7 = 11
-11

b. If the temperature outside was 73 degrees at 5:00 p.m., but it fell 19 degrees by 10:00 p.m., what is the temperature at 10:00 p.m.? Write an equation and solve.
Answer:

c. Write an addition sentence, and find the sum using the diagram below.

Answer:

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

Question 1.
Write an addition problem that has a sum of -4 \(\frac{4}{5}\) and
a. The two addends have the same sign.
Answer:
Answers will vary. -1 \(\frac{4}{5}\) + (-3) = -4 \(\frac{4}{5}\) .

b. The two addends have different signs.
Answer:
Answers will vary. 1.8 + (-6.6) = -4.8.

Question 2.
In the Integer Game, what card would you need to draw to get a score of 0 if you have a -16, -35, and 18 in your hand?
Answer:
-16 + (-35) + 18 = -33, so I would need to draw a 33 because 33 is the additive inverse of -33.
-33 + 33 = 0.

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

Question 1.
Find the sum. Show your work to justify your answer.
a. 4 + 17
Answer:
4 + 17 = 21

b. -6 + (-12)
Answer:
-6 + (-12) = -18

c. 2.2 + (-3.7)
Answer:
2.2 + (-3.7) = -1.5

d. -3 + (-5) + 8
Answer:
-3 + (-5) + 8 = -8 + 8 = 0

e. \(\frac{1}{3}\) +(-2 \(\frac{1}{4}\))
Answer:
\(\frac{1}{3}\) +(-2 \(\frac{1}{4}\)) = \(\frac{1}{3}\) + (-\(\frac{9}{4}\) ) = \(\frac{4}{12}\) + (-\(\frac{27}{12}\) )= – \(\frac{23}{12}\) = -1\(\frac{11}{12}\)

Question 2.
Which of these story problems describes the sum 19+(-12)? Check all that apply. Show your work to justify your answer.
___ Jared’s dad paid him $19 for raking the leaves from the yard on Wednesday. Jared spent $12 at the movie theater on Friday. How much money does Jared have left?
Answer:
X

___ Jared owed his brother $19 for raking the leaves while Jared was sick. Jared’s dad gave him $12 for doing his chores for the week. How much money does Jared have now?

___ Jared’s grandmother gave him $19 for his birthday. He bought $8 worth of candy and spent another $4 on a new comic book. How much money does Jared have left over?
Answer:
X

Question 3.
Use the diagram below to complete each part.

Answer:

a. Label each arrow with the number the arrow represents.

b. How long is each arrow? What direction does each arrow point?

Answer:

c. Write an equation that represents the sum of the numbers. Find the sum.
Answer:
5 + (-3) + (-7) = -5

Question 4.
Jennifer and Katie were playing the Integer Game in class. Their hands are represented below.

a. What is the value of each of their hands? Show your work to support your answer.
Answer:
Jennifer’s hand has a value of -3 because 5 + (-8) = -3. Katie’s hand has a value of -2 because
-9 + 7 = -2.

b. If Jennifer drew two more cards, is it possible for the value of her hand not to change? Explain why or why not.
Answer:
It is possible for her hand not to change. Jennifer could get any two cards that are the exact opposites such as (-2) and 2. Numbers that are exact opposites are called additive inverses, and they sum to 0. Adding the number 0 to anything will not change the value.

c. If Katie wanted to win the game by getting a score of 0, what card would she need? Explain.
Answer:
Katie would need to draw a 2 because the additive inverse of -2 is 2. -2 + 2 = 0.

d. If Jennifer drew -1 and -2, what would be her new score? Show your work to support your answer.
Answer:
Jennifer’s new score would be -6 because -3 + (-1) + (-2) = -6.

Engage NY Eureka Math 7th Grade Module 2 Lesson 15 Answer Key

Eureka Math Grade 7 Module 2 Lesson 15 Exercise Answer Key

Exercise 1.
a. In the space below, create a word problem that involves integer multiplication. Write an equation to model the situation.
Answer:
Answers will vary.
Both times we went to the fair, I borrowed $3 from my older brother. -$3 × 2 = -$6

b. Now change the word problem by replacing the integers with non-integer rational numbers (fractions or decimals), and write the new equation.
Answer:
Answers will vary.
Both times we went to the fair, I borrowed $3.50 from my older brother. -$3.50×2=-$7.00

c. Was the process used to solve the second problem different from the process used to solve the first? Explain.
Answer:
No, the process was the same. Both times I had a positive number multiplied by a negative number, so the product is a negative number. The process, multiplication, is represented as repeated addition:
-$3.50+(-$3.50)= -$7.00.

→ Was the process for solving the second problem different from the process you used to solve the problem when it contained only integers?
→ No
Students record the rules in Exercise 1, part (d) of their student materials.

d. The Rules for Multiplying Rational Numbers are the same as the Rules for Multiplying Integers:
Answer:
1. Multiply the absolute values of the two rational numbers.
2. If the two numbers (factors) have the same sign, their product is positive.
3. If the two numbers (factors) have opposite signs, their product is negative.

Exercise 2.
a. In one year, Melinda’s parents spend $2,640.90 on cable and internet service. If they spend the same amount each month, what is the resulting monthly change in the family’s income?
Answer:
-2,640.90 ÷ 12 = -220.08
The average monthly change to their income is about -$220.08.

→ Are the rules for dividing rational numbers the same as the rules for dividing integers?
→ Yes

Students record the rules in Exercise 2, part (b) of their student materials.

b. The Rules for Dividing Rational Numbers are the same as the Rules for Dividing Integers:
1. Divide the absolute values of the two rational numbers.
2. If the two numbers (dividend and divisor) have the same sign, their quotient is positive.
3. If the two numbers (dividend and divisor) have opposite signs, their quotient is negative.

Exercise 3.
Use the fundraiser chart to help answer the questions that follow.

Answer:

Jeremy is selling plants for the school’s fundraiser, and listed above is a chart from his fundraiser order form. Use the information in the chart to answer the following questions. Show your work, and represent the answer as a rational number; then, explain your answer in the context of the situation.
a. If Tamara Jones writes a check to pay for the plants, what is the resulting change in her checking account balance?
Answer:
-4.25 × 2 = -8.50
Numerical Answer:
Answer:
-8.50
Explanation:
Answer:
Tamara Jones will need to deduct $8.50 from her checking account balance.

b. Mr. Clark wants to pay for his order with a $20 bill, but Jeremy does not have change. Jeremy tells Mr. Clark he will give him the change later. How will this affect the total amount of money Jeremy collects? Explain. What rational number represents the change that must be made to the money Jeremy collects?
Answer:
2.25 × 5 = 11.25 20.00 – 11.25 = 8.75
Numerical Answer:
Answer:
-8.75
Explanation:
Answer:
Jeremy collects too much money. He owes Mr. Clark $8.75. The adjustment Jeremy needs to make is -$8.75.

c. Jeremy’s sister, Susie, borrowed the money from their mom to pay for her order. Their mother has agreed to deduct an equal amount of money from Susie’s allowance each week for the next five weeks to repay the loan. What is the weekly change in Susie’s allowance?
Answer:
-2.50 ÷ 5 = – 0.50
Numerical Answer:
Answer:
-0.50
Explanation:
Answer:
Susie will lose $0.50 of her allowance each week.

d. Jeremy’s grandparents want to change their order. They want to order three daisies and one geranium, instead of four daisies. How does this change affect the amount of their order? Explain how you arrived at your answer.
Answer:
Original Order: 3.75 × 4 = 15.00
New Order: 3.75 × 3 + 2.25 = 11.25 + 2.25 = 13.50
15.00-13.50=1.50
Numerical Answer: 1.50
Explanation: Jeremy’s grandparents will get back $1.50 since the change in their order made it cheaper.
I got my answer by first calculating the cost of the original order. Second, I calculated the cost of the new order. Finally, I subtracted the two order totals to determine the difference in cost.

e. Jeremy approaches three people who do not want to buy any plants; however, they wish to donate some money for the fundraiser when Jeremy delivers the plants one week later. If the people promise to donate a total of $14.40, what will be the average cash donation?
Answer:
14.40 ÷ 3 = 4.80
Numerical Answer: 4.80
Explanation: The average cash donation will be $4.80 per person.

f. Jeremy spends one week collecting orders. If 22 people purchase plants totaling $270, what is the average amount of Jeremy’s sale?
Answer:
270 ÷ 22 ≈ 12.272
Numerical Answer: 12.27
Explanation: The average sale is about $12.27.

Eureka Math Grade 7 Module 2 Lesson 15 Problem Set Answer Key

Question 1.
At lunch time, Benjamin often borrows money from his friends to buy snacks in the school cafeteria. Benjamin borrowed $0.75 from his friend Clyde five days last week to buy ice cream bars. Represent the amount Benjamin borrowed as the product of two rational numbers; then, determine how much Benjamin owed his friend last week.
Answer:
5 (-0.75)=-3.75
Benjamin owed Clyde $3.75.

Question 2.
Monica regularly records her favorite television show. Each episode of the show requires 3.5% of the total capacity of her video recorder. Her recorder currently has 62% of its total memory free. If Monica records all five episodes this week, how much space will be left on her video recorder?
Answer:
62 + 5(-3.5) = 62 + (-17.5) = 44.5
Monica’s recorder will have 44.5% of its total memory free.

For Problems 3–5, find at least two possible sets of values that will work for each problem.

Question 3.
Fill in the blanks with two rational numbers (other than 1 and -1). ____ × (-\(\frac{1}{2}\)) × ____ = -20
What must be true about the relationship between the two numbers you chose?
Answer:
Answers may vary. Two possible solutions are 10 and 4 or -10 and -4. The two numbers must be factors of 40, and they must both have the same sign.

Question 4.
Fill in the blanks with two rational numbers (other than 1 and -1). -5.6 × 100÷ 80 × ____ × ____ =700
What must be true about the relationship between the two numbers you chose?
Answer:
Answers may vary. Two possible solutions are -50 and 2 or 25 and -4. The two numbers must be factors of -100, and they must have opposite signs.

Question 5.
Fill in the blanks with two rational numbers. ____ × ____ = -0.75
What must be true about the relationship between the two numbers you chose?
Answer:
Answers may vary. Two possible solutions are -3 and 0.25 or 0.5 and -1.5. The two numbers must be factors of -0.75, and they must have opposite signs.

For Problems 6–8, create word problems that can be represented by each expression, and then represent each product or quotient as a single rational number.

Question 6.
8×(-0.25)
Answer:
Answers may vary.
Example: Stacey borrowed a quarter from her mother every time she went to the grocery store so that she could buy a gumball from the gumball machine. Over the summer, Stacey went to the grocery store with her mom eight times. What rational number represents the dollar amount change in her mother’s money due to the purchase of gumballs?
Answer: -2

Question 7.
-6÷(1\(\frac{1}{3}\))
Answer:
Answers may vary.
Example: There was a loss of $6 on my investment over one-and-a-third months. Based on this, what was the investment’s average dollar amount change per month?
Answer: -4.50

Question 8.
–\(\frac{1}{2}\) × 12
Answer:
Answers may vary.
Example: I discarded exactly half of my card-point total in the Integer Game. If my card-point total was 12 before I discarded, which rational number represents the change to my hand’s total card-point total?
Answer: -6

Eureka Math Grade 7 Module 2 Lesson 15 Exit Ticket Answer Key

Harrison made up a game for his math project. It is similar to the Integer Game; however, in addition to integers, there are cards that contain other rational numbers such as -0.5 and -0.25. Write a multiplication or division equation to represent each problem below. Show all related work.

Question 1.
Harrison discards three -0.25 cards from his hand. How does this affect the overall point value of his hand? Write an equation to model this situation.
Answer:
-3 (-0.25) = 0.75
The point value of Harrison’s hand will increase by 0.75 points.

Question 2.
Ezra and Benji are playing the game with Harrison. After Ezra doubles his hand’s value, he has a total of -14.5 points. What was his hand’s value before he doubled it?
Answer:
-14.5 ÷ 2 = -7.25
Before Ezra doubled his hand, his hand had a point value of -7.25.

Question 3.
Benji has four -0.5 cards. What is his total score?
Answer:
4 × (-0.5) = -2.0
Benji’s total score is -2.0 points.

Eureka Math Grade 7 Module 2 Lesson 15 Integer Multiplication Round 1 Answer Key

Directions: Determine the product of the integers, and write it in the column to the right.

Answer:

Eureka Math Grade 7 Module 2 Lesson 15 Integer Multiplication Round 2 Answer Key

Answer:

Engage NY Eureka Math 7th Grade Module 2 Lesson 13 Answer Key

Eureka Math Grade 7 Module 2 Lesson 13 Example Answer Key

Example 1.
Representations of Rational Numbers in the Real World
Following the Opening Exercise and class discussion, describe why we need to know how to represent rational numbers in different ways.
Answer:
Different situations in the real world require different representations of rational numbers. Because of common usage in life outside of the classroom, we may automatically know that a quarter of a dollar is the same as 25 cents, or a quarter, but for people who are used to measuring money in only decimals, a quarter of a dollar might not make much sense.

Example 2.
Using Place Values to Write (Terminating) Decimals as Equivalent Fractions
a. What is the value of the number 2.25? How can this number be written as a fraction or mixed number?
Answer:
Two and twenty-five hundredths or 2\(\frac{25}{100}\)

→ How do we rewrite this fraction (or any fraction) in its simplest form?
→ If a factor is common to both the numerator and denominator of a fraction, the fraction can be simplified, resulting in a fraction whose numerator and denominator only have a common factor of 1 (the numerator and denominator are relatively prime).

b. Rewrite the fraction in its simplest form showing all steps that you use.
Answer:
Answer:
\(\frac{25}{100}\) = \(\frac{25}{4×25}\) = \(\frac{1}{4}\) → 2\(\frac{25}{100}\) = 2\(\frac{1}{4}\)

c. What is the value of the number 2.025? How can this number be written as a mixed number?
Answer:
Two and twenty-five thousandths, or 2\(\frac{25}{1,000}\)

d. Rewrite the fraction in its simplest form showing all steps that you use.
Answer:
\(\frac{25}{1,000}\) = \(\frac{25}{(100×10)}\)
\(\frac{25}{4×25×10}\) = \(\frac{1}{40}\) → 2\(\frac{25}{1,000}\) = 2\(\frac{1}{40}\)

Example 3.
Converting Fractions to Decimals—Fractions with Denominators Having Factors of only 2 or 5
a. What are decimals?
Answer:
Decimals specify points on the number line by repeatedly subdividing intervals into tenths. If a unit is divided into ten equal-sized pieces, one piece would be one-tenth of that unit.

b. Use the meaning of decimal to relate decimal place values.
Answer:
Each place value in a decimal is \(\frac{1}{10}\) of the value of the place to its left. This means that the denominators of the fractions that represent each decimal place value must be powers of ten.

c. Write the number \(\frac{3}{100}\) as a decimal. Describe your process.
Answer:
The decimal form is 0.03. The fraction includes a power of ten, 100, as its denominator. The value of the second decimal place is \(\frac{1}{100}\), so \(\frac{3}{100}\) in decimal form is 0.03.

→ How could we obtain an equivalent fraction to \(\frac{3}{20}\) with a power of ten in the denominator?
→ If there was another factor of 5 in the denominator, then we would have an equal number of 2’s and 5’s resulting in a power of ten. If we multiply the fraction by \(\frac{5}{5}\) (or 1), we get an equivalent fraction with a power of ten in its denominator.

d. Write the number \(\frac{3}{20}\) as a decimal. Describe your process.
Answer:
The fractional form is \(\frac{3}{20}\) = \(\frac{3}{2^{2} \times 5}\). The denominator lacks a factor of 5 to be a power of ten. To arrive at the decimal form, I multiply the fractional form by \(\frac{5}{5}\) to arrive at \(\frac{3}{2^{2} \times 5}\) × \(\frac{5}{5}\) = \(\frac{3 \times 5}{2^{2} \times 5^{2}}\) = \(\frac{15}{100}\), and \(\frac{15}{100}\) = 0.15.

e. Write the number \(\frac{10}{25}\) as a decimal. Describe your process.
Answer:
The fractional form is \(\frac{10}{25}\) = \(\frac{2×5}{5×5}\) ; and, since \(\frac{5}{5}\) = 1, then \(\frac{2×5}{5×5}\) = \(\frac{2}{5}\). The denominator lacks a factor of 2 to be a power of ten. To arrive at the decimal form, I multiply the fractional form by \(\frac{2}{2}\) to arrive at \(\frac{2}{5}\) × \(\frac{2}{2}\) = \(\frac{4}{10}\), and \(\frac{4}{10}\) = 0.4.

f. Write the number \(\frac{8}{40}\) as a decimal. Describe your process.
Answer:
The fractional form is \(\frac{8}{40}\) = \(\frac{2^{3}}{2^{3} \times 5}\). There are factors of 2 in the numerator and denominator that will cancel. If I leave one factor of two in the denominator, it will be 10 (a power of ten).
\(\frac{2^{3}}{2^{3} \times 5}\) = \(\frac{2^{2} \times 2}{2^{2} \times 2 \times 5}\) = \(\frac{2}{2 \times 5}\) = \(\frac{2}{10}\) = 0.2

Eureka Math Grade 7 Module 2 Lesson 13 Exercise Answer Key

Exercise 1.
Use place value to convert each terminating decimal to a fraction. Then rewrite each fraction in its simplest form.
a. 0.218
Answer:
\(\frac{218}{1,000}\) = \(\frac{109×2}{500×2}\) = \(\frac{109}{100}\) → 0.218 = \(\frac{109}{500}\)

b. 0.16
Answer:
\(\frac{16}{100}\) = \(\frac{4×4}{4×25}\) = \(\frac{4}{25}\) → 0.16 = \(\frac{4}{25}\)

c. 2.72
Answer:
\(\frac{72}{100}\) = \(\frac{4×18}{4×25}\) = \(\frac{18}{25}\) → 2.72 = 2\(\frac{18}{25}\)

d. 0.0005
Answer:
\(\frac{5}{10,000}\) = \(\frac{5×1}{5×2,000}\) = \(\frac{1}{2,000}\) → 0.0005 = \(\frac{1}{2,000}\)

→ What do you notice about the denominators of fractions that represent each decimal place?
→ The denominators are all powers of 10.
→ What are the prime factors of 10? 100? 1,000?

→ What prime factors make up the powers of 10?
→ The powers of 10 contain only the factors 2 and 5, and, in each case, the number of factors of 2 and 5 are equal to the number of factors of 10.
→ How can the prime factorization of the powers of ten be used to write fractions in decimal form?
→ Find an equivalent fraction whose denominator is a power of ten, then write the decimal representation using place values.

Exercise 2.
Convert each fraction to a decimal using an equivalent fraction.
a. \(\frac{3}{16}\) =
Answer:
\(\frac{3}{16}\) = \(\frac{3}{2×4}\) → \(\frac{3 \times 5^{4}}{2^{4} \times 5^{4}}\) = \(\frac{1,875}{10,000}\) → \(\frac{1875}{10,000}\) = 0.1875

b. \(\frac{7}{5}\)=
\(\frac{7}{5}\) → \(\frac{7×2}{5×2}\) = \(\frac{14}{10}\) → \(\frac{14}{10}\) = 1 \(\frac{4}{10}\) = 1.4

c. \(\frac{11}{32}\)=
\(\frac{11}{32}\) = \(\frac{11}{2^{5}}\) → \(\frac{11 \times 5^{5}}{2^{5} \times 5^{5}}\) = \(\frac{34,375}{100,000}\) → \(\frac{34375}{100,000}\) = 0.34375

d. \(\frac{35}{50}\)=
\(\frac{35}{50}\) = \(\frac{5 \times 7}{5^{2} \times 2}\) → \(\frac{7}{5×2}\) = \(\frac{7}{10}\) → \(\frac{7}{10}\) = 0.7

Eureka Math Grade 7 Module 2 Lesson 13 Problem Set Answer Key

Question 1.
Convert each terminating decimal to a fraction in its simplest form.
a. 0.4
Answer:
0.4 = \(\frac{2}{5}\)

b. 0.16
Answer:
0.16 = \(\frac{4}{25}\)

c. 0.625
Answer:
0.625 = \(\frac{5}{8}\)

d. 0.08
Answer:
0.08 = \(\frac{2}{25}\)

e. 0.012
Answer:
0.012 = \(\frac{3}{250}\)

Question 2.
Convert each fraction or mixed number to a decimal using an equivalent fraction.
a. \(\frac{4}{5}\)
Answer:
\(\frac{4}{5}\) = 0.8

b. \(\frac{3}{40}\)
Answer:
\(\frac{3}{40}\) = 0.075

c. \(\frac{8}{200}\)
Answer:
\(\frac{8}{200}\) = 0.04

d. 3\(\frac{5}{16}\)
Answer:
3 \(\frac{5}{16}\) = 3.3125

Question 3.
Tanja is converting a fraction into a decimal by finding an equivalent fraction that has a power of 10 in the denominator. Sara looks at the last step in Tanja’s work (shown below) and says that she cannot go any further. Is Sara correct? If she is, explain why. If Sara is incorrect, complete the remaining steps.
\(\frac{72}{480}\) = \(\frac{2^{3} \cdot 3^{2}}{2^{5} \cdot 3 \cdot 5}\)
Answer:
Tanja can finish the conversion since there is a factor pair of 3’s in the numerator and denominator that can be divided out with a quotient of 1.
Remaining Steps:
\(\frac{72}{480}\) = \(\frac{2^{3} \cdot 3^{2}}{2^{5} \cdot 3 \cdot 5}\) = \(\frac{3}{2^{2} \cdot 5}\)(\(\frac{5}{5}\)) = \(\frac{3 \cdot 5}{2^{2} \cdot 5^{2}}\) = \(\frac{15}{100}\)
Answer:0.5

Eureka Math Grade 7 Module 2 Lesson 13 Exit Ticket Answer Key

Question 1.
Write 3.0035 as a fraction. Explain your process.
Answer:
The right-most decimal place is the ten-thousandths place, so the number in fractional form would be 3\(\frac{35}{10,000}\). There are common factors of 5 in the numerator and denominator, and dividing both by these results in the fraction 3\(\frac{7}{2,000}\).

Question 2.
This week is just one of 40 weeks that you spend in the classroom this school year. Convert the fraction \(\frac{1}{40}\) to decimal form.
Answer:
\(\frac{1}{40}\) = \(\frac{1}{2^{3} \times 5}\)×\(\frac{5^{2}}{5^{2}}\) = \(\frac{5^{2}}{2^{3} \times 5^{3}}\) = \(\frac{25}{1,000}\) = 0.025

Engage NY Eureka Math 7th Grade Module 2 Lesson 11 Answer Key

Eureka Math Grade 7 Module 2 Lesson 11 Example Answer Key

Example 1.
Extending Whole Number Multiplication to the Integers
Part A: Complete quadrants I and IV of the table below to show how sets of matching integer cards will affect a player’s score in the Integer Game. For example, three 2’s would increase a player’s score by 0+2+2+2=6 points.

Answer:

a. What patterns do you see in the right half of the table?
Answer:
The products in quadrant I are positive and the products in quadrant IV are negative. When looking at the absolute values of the products, quadrants I and IV are a reflection of each other with respect to the middle row.

b. Enter the missing integers in the left side of the middle row, and describe what they represent.
Answer:
The numbers represent how many matching cards are being discarded or removed

Part B: Students complete quadrant II of the table.
Students describe, using an Integer Game scenario, what quadrant II of the table represents and record this in the student materials.

Part B: Complete quadrant II of the table.

Students answer the following questions:

c. What relationships or patterns do you notice between the products (values) in quadrant II and the products (values) in quadrant I?
Answer:
The products in quadrant II are all negative values. Looking at the absolute values of the products, quadrant I and II are a reflection of each other with respect to the center column.

d. What relationships or patterns do you notice between the products (values) in quadrant II and the products (values) in quadrant IV?
Answer:
The products in quadrants II and IV are all negative values. Each product of integers in quadrant II is equal to the product of their opposites in quadrant IV.

e. Use what you know about the products (values) in quadrants I, II, and IV to describe what quadrant III will look like when its products (values) are entered.
Answer:
The reflection symmetry of quadrant I to quadrants II and IV suggests that there should be similar relationships between quadrant II, III, and IV. The number patterns in quadrants II and IV also suggest that the products in quadrant III are positive values.

Part C: Discuss the following question. Then instruct students to complete the final quadrant of the table.
→ In the Integer Game, what happens to a player’s score when he removes a matching set of cards with negative values from his hand?
→ His score increases because the negative cards that cause the score to decrease are removed.
Students describe, using an Integer Game scenario, what quadrant III of the table represents and complete the quadrant in the student materials.

Part C: Complete quadrant III of the table.
Refer to the completed table to help you answer the following questions:

Students refer to the completed table to answer parts (f) and (g).

f. Is it possible to know the sign of a product of two integers just by knowing in which quadrant each integer is located? Explain.
Answer:
Yes, it is possible to know the sign of a product of two integers just by knowing each integer’s quadrant because the signs of the values in each of the quadrants are consistent. Two quadrants contain positive values, and the other two quadrants contain negative values.

g. Which quadrants contain which values? Describe an Integer Game scenario represented in each quadrant.
Answer:
Quadrants I and III contain all positive values. Picking up three 4’s is represented in quadrant I and increases your score. Removing three (-4)’s is represented in quadrant III and also increases your score. Quadrants II and IV contain all negative values. Picking up three (-4)’s is represented in quadrant IV and decreases your score. Removing three 4’s is represented in quadrant II and also decreases your score.

Eureka Math Grade 7 Module 2 Lesson 11 Exercise Answer Key

Exercise 1.
Multiplication of Integers in the Real World
Generate real-world situations that can be modeled by each of the following multiplication problems. Use the Integer Game as a resource.

a. -3 × 5
Answer:
I lost three $5 bills, and now I have -$15.

b. -6 × (-3)
Answer:
I removed six (-3)’s from my hand in the Integer Game, and my score increased 18 points.

c. 4 × (-7)
Answer:
If I lose 7 lb. per month for 4 months, my weight will change -28 lb. total.

Eureka Math Grade 7 Module 2 Lesson 11 Problem Set Answer Key

Question 1.
Complete the problems below. Then, answer the question that follows.

Which row shows the same pattern as the outlined column? Are the problems similar or different? Explain.
Answer:

The row outlined shows the same pattern as the outlined column. The problems have the same answers, but the signs of the integer factors are switched. For example, 3×(-1)=-3×1. This shows that the product of two integers with opposite signs is equal to the product of their opposites.

Question 2.
Explain why (-4) × (-5) = 20. Use patterns, an example from the Integer Game, or the properties of operations to support your reasoning.
Answer:
Answers may vary. Losing four -5 cards will increase a score in the Integer Game by 20 because a negative value decreases a score, but the score increases when it is removed.

Question 3.
Each time that Samantha rides the commuter train, she spends $4 for her fare. Write an integer that represents the change in Samantha’s money from riding the commuter train to and from work for 13 days. Explain your reasoning.
Answer:
If Samantha rides to and from work for 13 days, then she rides the train a total of 26 times. The cost of each ride can be represented by -4. So, the change to Samantha’s money is represented by -4 × 26 = -104. The change to Samantha’s money after 13 days of riding the train to and from work is -$104.

Question 4.
Write a real-world problem that can be modeled by 4 × (-7).
Answer:
Answers will vary. Every day, Alec loses 7 pounds of air pressure in a tire on his car. At that rate, what is the change in air pressure in his tire after 4 days?

Challenge:

Question 5.
Use properties to explain why for each integer a, -a = -1 × a. (Hint: What does (1 + (-1))×a equal? What is the additive inverse of a?)
Answer:
0 × a = 0 Zero product property
(1 + (-1)) × a = 0 Substitution
a + (-1 × a) = 0 Distributive property
Since a and (-1 × a) have a sum of zero, they must be additive inverses. By definition, the additive inverse of a is -a, so (-1 × a) = -a.

Eureka Math Grade 7 Module 2 Lesson 11 Exit Ticket Answer Key

Question 1.
Create a real-life example that can be modeled by the expression -2 × 4, and then state the product.
Answer:
Answers will vary. Tobi wants to lose 2 lb. each week for four weeks. Write an integer to represent Tobi’s weight change after four weeks. Tobi’s weight changes by –8 lb. after four weeks.

Question 2.
Two integers are multiplied, and their product is a positive number. What must be true about the two integers?
Answer:
Both integers must be positive numbers, or both integers must be negative numbers.

Engage NY Eureka Math 7th Grade Module 2 Mid Module Assessment Answer Key

Eureka Math Grade 7 Module 2 Mid Module Assessment Task Answer Key

Question 1.
Diamond used a number line to add. She started counting at 10, and then she counted until she was on the number -4 on the number line.

a. If Diamond is modeling addition, what number did she add to 10? Use the number line below to model your answer.

Answer:

10 + -14 = -4
Diamond added -14 to 10.

b. Write a real-world story problem that would fit this situation.
Answer:
Diamond had $10 and put it in the bank. She forgot about the monthly bank fee of $14.
Now her account has a balance of -$4.

c. Use absolute value to express the distance between 10 and -4.
Answer:
|10 – (-4)|
|10 + 4|
|14|
14
The distance between 10 and -4 is 14.

Question 2.
What value of a will make the equation a true statement? Explain how you arrived at your solution.
(-\(\frac{3}{4}\) + \(\frac{4}{3}\))
Answer:
(-\(\frac{3}{4}\) + \(\frac{4}{3}\))+a=0
(-\(\frac{3}{4}\) + \(\frac{4}{3}\) )
(-\(\frac{9}{12}\) + \(\frac{16}{12}\))
\(\frac{7}{12}\) + a = 0
a = –\(\frac{7}{12}\)
The value of a has to be –\(\frac{7}{12}\) because that’s the additive inverse of \(\frac{7}{12}\).

Question 3.
Every month, Ms. Thomas pays her car loan through automatic payments (withdrawals) from her savings account. She pays the same amount on her car loan each month. At the end of the year, her savings account balance changed by -$2,931 from payments made on her car loan.

a. What is the change in Ms. Thomas’ savings account balance each month due to her car payment?
Answer:

Her monthly payment is $244.25, so her account balance changes each month by -$244.25 when her payment is made.

b. Describe the total change to Ms. Thomas’ savings account balance after making six monthly payments on her car loan. Model your answer using a number sentence.
Answer:

6 × (-244.25) = -1,465.50
Ms. Thomas’ car loan changed her savings account balance by -$1,465.50 after “6” monthly payments.

Question 4.
Jesse and Miya are playing the Integer Card Game. The cards in Jesse’s hand are shown below:

a. What is the total score of Jesse’s hand? Support your answer by showing your work.
Answer:
3 + (-5 ) + 9 + (- 6)
(-2) + 3
1
Jesse’s score is 1.

b. Jesse picks up two more cards, but they do not affect his overall point total. State the value of each of the two cards, and tell why they do not affect his overall point total.
Answer:
3 + (-5) + 9 + (-6)
(-2) + 3
1
Jesse’s score is 1.

b. Jesse picks up two more cards, but they do not affect his overall point total. State the value of each of the two cards, and tell why they do not affect his overall point total.
Answer:
The values of the two cards must be opposites, such as -2 and 2, because opposites combine to get 0. 0 will not change the score in his hand.

c. Complete Jesse’s new hand to make this total score equal zero. What must be the value of the ? card? Explain how you arrived at your answer.

Answer:

The two given cards total -2. To get a sum of zero, I have to combine -2 with its opposite 2 because additive inverses (opposites) combine to get 0.

Question 5.
Michael’s father bought him a 16-foot board to cut into shelves for his bedroom. Michael plans to cut the board into 11 equal size lengths for his shelves.

a. The saw blade that Michael will use to cut the board will change the length of the board by -0.125 inches for each cut. How will this affect the total length of the board?
Answer:

The board will be cut in 10 places.

The 10 cuts take away 1.25 inches of the total length of the board. The usable length of the board is 1.25 inches shorter than 16 feet.

b. After making his cuts, what will the exact length of each shelf be?
Answer:
The board begins at 192 inches long.
192 – 1.25 = 190.75. The length of the board that can be used with the blade widths removed is 190.75 inches.

Continue the long division, and there will be a repeating remainder of 1. Therefore, the lengths of the shelves should be exactly \(17.34 \overline{09}\) inches.

Question 6.
Bryan and Jeanette were playing the Integer Card Game like the one you played in class. They were practicing adding and subtracting integers. Jeanette had a score of -10. Bryan took away one of Jeanette’s cards. He showed it to her. It was a -8. Jeanette recalculated her score to be -2, but Bryan disagreed. He said that her score should be -18 instead. Read their conversation and answer the question below.
“No Jeanette, removing a negative card means the same thing as subtracting a positive. So, negative 10 minus negative 8 is negative 18.”
“It does not! Removing a negative card is the same as adding the same positive card. My score will go up. Negative 10 minus negative 8 is negative 2.”
Based on their disagreement, who, if anyone, is right? Explain.
Answer:
Jeanette is correct that removing a negative is the same as adding the same positive card. Having a negative card in your hand decreases your score. If you remove that negative card, your score is no longer decreased by the card, so your score goes up.

Question 7.
The table below shows the temperature changes Monday morning in Bedford, New York over a 4-hour period after a cold front came through.

a. If the beginning temperature was -13°F at 5:00 a.m., what was the temperature at 9:00 a.m.?
Change in Temperature
Answer:

-13 + (-3) + (-2) + (-6) + 7
(-16) + (-8) + 7
-24 + 7
-17°F
The temperature at 9:00 a.m. was -17°F.

b. The same cold front hit Hartford, Connecticut the next morning. The temperature dropped by 7° each hour from 5:00 a.m. to 9:00 a.m. What was the beginning temperature at 5:00 a.m. if the temperature at 9:00 a.m. was -10°F?
Answer:
-10 + 4(7)
-10 + 28
18° F
The beginning temperature at 5:00 a.m. was 18°F.

c. In answering part (b), Josiah and Kate used different methods. Josiah said his method involved multiplication, while Kate said she did not use multiplication. Both students arrived at the correct answer. How is this possible? Explain.
Answer:
The temperature change was the same for each hour, so Josiah multiplied the 7-degree drop by 4 hours. Kate added the 7-degree drop 4 times. Kate used repeated addition, which is the same as multiplication.

Engage NY Eureka Math 7th Grade Module 2 Lesson 10 Answer Key

Eureka Math Grade 7 Module 2 Lesson 10 Example Answer Key

The numbers used throughout this example are a sample response. Students’ answers depend on the cards they pick.
Part A: Students pick four cards. Instruct students to record the values of their cards on the images in Part A. One of the four card images has a ★ beneath it. The ★ is used to indicate which of the four cards to copy (or multiply) in Part B.

Example 1.
Product of a Positive Integer and a Negative Integer

Part B: Instruct students to copy the value of the card with the ★ beneath it from Part A on each card with a ★ beneath it in Part B. The three remaining card values from Part A are entered in the three remaining card images in Part B. Students now have a total of six integer cards.

Use your cards from Part B to answer the questions below.

a. Write a product that describes the three matching cards.
Answer:
3 × (-5)

b. Write an expression that represents how each of the ★ cards changes your score.
Answer:
(-5) + (-5) + (-5)

c. Write an equation that relates these two expressions.
Answer:
3 × (-5) = (-5) + (-5) + (-5)

d. Write an integer that represents the total change to your score by the three ★ cards.
Answer:
-15

e. Write an equation that relates the product and how it affects your score.
Answer:
3 × (-5) = -15

Part C: Instruct students to pick three new cards and record the values of their cards on the images in Part C. The teacher chooses the fourth card and instructs the class to place a ★ beneath it to indicate which card will be cloned (multiplied) in Part D.

Part D: Instruct students to record the value of the card with the ★ beneath it from Part C on each image with a ★ beneath it in Part D. Also, rewrite the values of the three remaining cards on the other three images. Students now have a total of eight integer cards.

Use your cards from Part D to answer the questions below.

f. Write a product that describes the five matching cards.
Answer:
5 × 4

g. Write an expression that represents how each of the ★ cards changes your score.
Answer:
4 + 4 + 4 + 4 + 4

h. Write an equation that relates these two expressions.
Answer:
5 × 4 = 4 + 4 + 4 + 4 + 4

i. Write an integer that represents the total change to your score by the five ★ cards.
Answer:
20

j. Write an equation that relates the product and how it affects your score.
Answer:
5 × 4 = 20

Students write conclusions using their own words in the student materials.

k. Use the expression 5 × 4 to relate the multiplication of a positive valued card to addition.
Answer:
Multiplying a positive integer card is repeated addition of the positive integer card and increases your score.
5 × 4 = 4 + 4 + 4 + 4 + 4 = 20

i. Use the expression 3 × (-5) to relate the multiplication of a negative valued card to addition.
Answer:
Multiplying a negative integer card is repeated addition of the negative integer card and decreases your score.
3 × (-5) = (-5) + (-5) + (-5) = -15

Example 2.
Product of a Negative Integer and a Positive Integer
a. If all of the 4’s from the playing hand on the right are discarded, how will the score be affected? Model this using a product in an equation.

Answer:
The score decreases by 4 three consecutive times for a total decrease of 12 points. The equation is -3 × 4 = -12.

b. What three matching cards could be added to those pictured to get the same change in score? Model this using a product in an equation.
Answer:
To get the same change in score, you would add three (-4)’s. The equation is 3 × (-4) = -12.

c. Seeing how each play affects the score, relate the products that you used to model them. What do you conclude about multiplying integers with opposite signs?
Answer:
(-3) × 4 = 3 × (-4); adding a value multiple times has the same effect as removing the opposite value the same number of times.

Example 3.
Product of Two Negative Integers

a. If the matching cards from the playing hand on the right are discarded, how will this hand’s score be affected? Model this using a product in an equation.
Answer:
Removing -2 from the set of cards will cause the score to increase by 2. Removing all four of the (-2)’s causes the score to increase by two, four consecutive times for a total increase of 8; -4 × (-2) = 8.

b. What four matching cards could be added to those pictured to get the same change in score? Model this using a product in an equation.
Answer:
An increase of 8 could come from adding four 2’s to the cards shown; 4 × 2 = 8.

c. Seeing how each play affects the score, relate the products that you used to model them. What do you conclude about multiplying integers with the same sign?
Answer:
-4 × (-2) = 4 × 2; adding a value multiple times has the same effect as removing the opposite value the same number of times.

d. Using the conclusions from Examples 2 and 3, what can we conclude about multiplying integers? Write a few examples.
Answer:
The product of two integers is equal to the product of their opposites; removing two (-4)’s is the same as adding two 4’s; adding three (-5)’s is the same as removing three 5’s.
Examples: (-4) × (-5) = 4 × 5; (-2) × 7 = 2 × (-7); 6 × (-4) = (-6) × 4
Removing two (-4)’s is the same as adding two 4’s; adding three (-5)’s is the same as removing three 5’s.

Eureka Math Grade 7 Module 2 Lesson 10 Problem Set Answer Key

Question 1.
Describe sets of two or more matching integer cards that satisfy the criteria in each part below:
a. Cards increase the score by eight points.
Answer:
Picking up: eight 1’s, four 2’s, or two 4’s
OR
Removing: eight (-1)’s, four (-2)’s, or two (-4)’s

b. Cards decrease the score by 9 points.
Answer:
Picking up: nine (-1)’s or three (-3)’s
OR
Removing: nine 1’s or three 3’s

c. Removing cards that increase the score by 10 points.
Answer:
Ten (-1)’s, five (-2)’s, or two (-5)’s

d. Positive cards that decrease the score by 18 points.
Answer:
Removing eighteen 1’s, nine 2’s, six 3’s, three 6’s, or two 9’s.

Question 2.
You have the integer cards shown at the right when your teacher tells you to choose a card to multiply four times. If your goal is to get your score as close to zero as possible, which card would you choose? Explain how your choice changes your score.

Answer:
The best choice to multiply is the -3. The cards currently have a score of one. The new score with the -3 multiplied by 4 is -8. The scores where the other cards are multiplied by 4 are 10, -11, and 16, which are all further from zero.

Question 3.
Sherry is playing the Integer Game and is given a chance to discard a set of matching cards. Sherry determines that if she discards one set of cards, her score will increase by 12. If she discards another set, then her score will decrease by eight. If her matching cards make up all six cards in her hand, what cards are in Sherry’s hand? Are there any other possibilities?
Answer:
There are two possibilities:
2, 2, 2, 2, -6, -6
OR
-3, -3, -3, -3, 4, 4

Eureka Math Grade 7 Module 2 Lesson 10 Exit Ticket Answer Key

Question 1.
Natalie is playing the Integer Game and only shows you the four cards shown below. She tells you that the rest of her cards have the same values on them and match one of these four cards.

a. If all of the matching cards will increase her score by 18, what are the matching cards?
Answer:
If there were nine 2 cards, then 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2=18
9 × 2 = 18
If there were six 3 cards, then 3 + 3 + 3 + 3 + 3 + 3=18
6 × 3 = 18

b. If all of the matching cards will decrease her score by 12, what are the matching cards?
Answer:
If there were two -6 cards, then (-6) + (-6)=-12
2×(-6)=-12

Question 2.
A hand of six integer cards has one matching set of two or more cards. If the matching set of cards is removed from the hand, the score of the hand will increase by six. What are the possible values of these matching cards? Explain. Write an equation using multiplication showing how the matching cards yield an increase in score of six.
Answer:
If the matching cards are taken away from the playing hand and the score of the hand increases, then the matching cards must have negative values. The playing hand only has six cards, so the number of matching cards is limited to six. Taking away the following matching sets would increase the score by six:
Taking away one set of two -3 cards can be represented by -(-3)-(-3)
3 + 3 =6
3 × 2 = 6 or (-3) × (-2) = 6
Taking away one set of three -2 cards can be represented by -(-2) – (-2) – (-2)
2 + 2 + 2 =6
2 × 3 = 6 or (-2) × (-3) = 6
Taking away one set of six -1 cards can be represented by -(-1)-(-1)-(-1)-(-1)-(-1)-(-1)
1 + 1 + 1 + 1 + 1 + 1 = 6
1 × 6 = 6 or (-1) × (-6) = 6

Engage NY Eureka Math 7th Grade Module 2 Lesson 6 Answer Key

Eureka Math Grade 7 Module 2 Lesson 6 Example Answer Key

Example 1.
Formula for the Distance Between Two Rational Numbers
Find the distance between -3 and 2.
Step 1: Start on an endpoint.
Step 2: Count the number of units from the endpoint you started on to the other endpoint.

Answer:

Using a formula, _____
Answer:
| -3 – 2 | = | -3 + -2 | = | -5 | = 5 OR | 2 – (-3) | = | 2 + 3 | = | 5 | = 5

For two rational numbers p and q, the distance between p and q is |p – q|.

Example 2.
Change in Elevation vs. Distance
Distance is positive. Change in elevation or temperature may be positive or negative depending on whether it is increasing or decreasing (going up or down).
a. A hiker starts hiking at the beginning of a trail at a point which is 200 feet below sea level. He hikes to a location on the trail that is 580 feet above sea level and stops for lunch.
i. What is the vertical distance between 200 feet below sea level and 580 feet above sea level?
Answer:
| -200 – 580 | = | -200 + (-580) | = | -780 | = 780
The vertical distance is 780 feet.

ii. How should we interpret 780 feet in the context of this problem?
Answer:
The hiker hiked 780 feet from a point below sea level to a point above sea level.

b. After lunch, the hiker hiked back down the trail from the point of elevation, which is 580 feet above sea level, to the beginning of the trail, which is 200 feet below sea level.

Answer:

i. What is the vertical distance between 580 feet above sea level and 200 feet below sea level?
Answer:
| 580 – (-200) | = | 580 + 200 | = | 780 | = 780
The vertical distance is 780 feet.

ii. What is the change in elevation?
Answer:
780 feet

Eureka Math Grade 7 Module 2 Lesson 6 Exercise Answer Key

Exercise 1.
Use the number line to answer each of the following.

Answer:

Exercise 2.
Use the number line to answer each of the following questions.
a. What is the distance between 0 and -8?

Answer:

| 0 – (-8) | = | 0 + 8| = |8| = 8

b. What is the distance between -2 and –1\(\frac{1}{2}\)?

Answer:

| – 2 – (-1 \(\frac{1}{2}\) )|=| -2 + 1 \(\frac{1}{2}\) |=|-\(\frac{1}{2}\) | = \(\frac{1}{2}\)

c. What is the distance between -6 and -10?

Answer:

| -6 – (-10) | = | -6 + 10 | = | 4 | = 4

Exercise 3.
The distance between a negative number and a positive number is 12\(\frac{1}{2}\). What are the numbers?
Answer:
Answers will vary; a possible solution is -9\(\frac{1}{2}\) and 3. |-9\(\frac{1}{2}\) – 3 | = | -9\(\frac{1}{2}\) + (- 3) | = |-12\(\frac{1}{2}\)| = 12\(\frac{1}{2}\)

| -9\(\frac{1}{2}\) | = 9\(\frac{1}{2}\) | 3 | = 3

Exercise 4.
Use the distance formula to find each answer. Support your answer using a number line diagram.
a. Find the distance between -7 and -4.
Answer:
| -7 – (-4) | = | -7 + 4 | = | -3 | = 3
The distance between -7 and -4 is 3 units.

b. Find the change in temperature if the temperature rises from -18°F to 15°F (use a vertical number line).
Answer:
| 15 – (-18) | = | 15 + 18 | = | 33 | = 33
The change in temperature is 33°F.

c. Would your answer for part (b) be different if the temperature dropped from 15°F to -18°F? Explain.
Answer:
Yes. The distance between 15 and -18 on a number line is the same as the distance between -18 and 15 because the endpoints are the same, but the temperature would change by -33°F because it is going from a higher value to a lower value.

d. Beryl is the first person to finish a 5K race and is standing 15 feet beyond the finish line. Another runner, Jeremy, is currently trying to finish the race and has approximately 14 feet before he reaches the finish line. What is the minimum possible distance between Beryl and Jeremy?
Answer:
| -14 – 15 | = | -14 + (-15) | = | -29 | = 29
The minimum distance between Beryl and Jeremy is 29 feet.

e. What is the change in elevation from 140 feet above sea level to 40 feet below sea level? Explain.
Answer:
I used the distance formula: | 140 – (-40) | = 180 and a vertical number line to find the distance between the two elevations. However, the elevation is going from a higher elevation to a lower one, so the change would be negative. Therefore, the change in elevation is -180 feet.

Eureka Math Grade 7 Module 2 Lesson 6 Problem Set Answer Key

Question 1.
| -19 – 12 |
Answer:
| -19 – 12 | = | -19 + (-12) | = |-31| = 31

Question 2.
| 19 – (-12) |
Answer:
| 19 – (-12) | = | 19 + 12 | = | 31 | = 31

Question 3.
| 10 – (-43) |
Answer:
| 10 – (-43) | = | 10 + 43 | = | 53 | = 53

Question 4.
| – 10 – 43 |
Answer:
| – 10 – 43 | = | -10 + (-43) | = | -53 | = 53

Question 5.
| -1 – (-16) |
Answer:
| -1 – (-16) | = | -1 + 16 |= | 15 | = 15

Question 6.
| 1 – 16 |
Answer:
| 1 – 16 | = | 1 + (-16) | = | -15 | = 15

Question 7.
| 0 – (-9) |
Answer:
| 0 – (-9) | = | 0 + 9 | = | 9 | = 9

Question 8.
| 0 – 9|
Answer:
| 0 – 9| = | 0 + (-9) | = | -9 | = 9

Question 9.
| -14.5 – 13 |
Answer:
| -14.5 – 13 | = | -14.5 + (-13) | = | -27.5 | = 27.5

Question 10.
| 14.5 – (-13) |
Answer:
| 14.5 – (-13) | = | 14.5 + 13 | = |27.5| = 27.5

Question 11.
Describe any patterns you see in the answers to the problems in the left- and right-hand columns. Why do you think this pattern exists?
Answer:
Each problem in the right-hand column has the same answer as the problem across from it in the left-hand column. That is because you are finding the distance between the opposite numbers as compared to the first column. The difference between the opposite numbers is opposite the difference between the original numbers. The absolute values of opposite numbers are the same.

Eureka Math Grade 7 Module 2 Lesson 6 Exit Ticket Answer Key

Two Grade 7 students, Monique and Matt, both solved the following math problem:
If the temperature drops from 7°F to -17°F, by how much did the temperature decrease?
The students came up with different answers. Monique said the answer is 24°F, and Matt said the answer is 10°F. Who is correct? Explain, and support your written response with the use of a formula and a vertical number line diagram.
Answer:
Monique is correct. If you use the distance formula, you take the absolute value of the difference between 7 and -17 and that equals 24. Using a number line diagram, you can count the number of units between 7 and -17 to get 24.
| 7 – (-17) | = | 7 + 17 | = | 24 | = 24. There was a 24°F drop in the temperature.

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

Eureka Math Grade 7 Module 2 Lesson 3 Example Answer Key

Example 1.
Counting On to Express the Sum as Absolute Value on a Number Line

Horizontal Number Line Model of Counting Up

Horizontal Number Line Model of Counting Down

Vertical Number Line Model of Counting Up

Vertical Number Line Model of Counting Down

Counting up -4 is the same as the opposite of counting up 4 and also means counting down 4.

a. For each example above, what is the distance between 2 and the sum?
Answer:
4 units

b. Does the sum lie to the right or left of 2 on a horizontal number line? Above or below on a vertical number line?
Answer:
Horizontal: On the first model, the sum lies to the right of 2. On the second model, it lies to the left of 2.
Vertical: On the first model, the sum lies above 2. On the second model, it lies below 2.

c. Given the expression 54+81, determine, without finding the sum, the distance between 54 and the sum. Explain.
Answer:
The distance will be 81 units. When the q-value is positive, the sum will be to the right of (or above) the p-value the same number of units as the q-value.

d. Is the sum to the right or left of 54 on the horizontal number line? Above or below on a vertical number line?
Answer:
The sum is to the right of 54 on a horizontal number line and above 54 on a vertical number line.

e. Given the expression 14 + (-3), determine, without finding the sum, the distance between 14 and the sum. Explain.
Answer:
The distance will be 3 units. When the q-value is negative, the sum will be to the left of (or below) the p-value the same number of units as the q-value.

f. Is the sum to the right or left of 14 on the number line? Above or below on a vertical number line?
Answer:
The sum is to the left of 14 on a horizontal number line and below 14 on a vertical number line.

Eureka Math Grade 7 Module 2 Lesson 3 Exercise Answer Key

Exercise 2
Work with a partner to create a horizontal number line model to represent each of the following expressions. What is the sum?

a. -5 + 3
Answer:
-5 + 3 = -2. The sum is 3 units to the right of -5.

b. -6 + (-2)
Answer:
-6 + (-2) = -8. The sum is 2 units to the left of -6.

c. 7 + (-8)
Answer:
7 + (-8) = -1. The sum is 8 units to the left of 7.

Exercise 3.
Writing an Equation Using Verbal Descriptions
Write an equation, and using the number line, create an arrow diagram given the following information:
The sum of 6 and a number is 15 units to the left of 6 on the number line.
Answer:

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

Question 1.
Refer to the diagram to the right.
a. Write an equation for the diagram below.
Answer:
-5 + (-4) = -9

b. Find the sum.
Answer:
-9

c. Describe the sum in terms of the distance from the first addend. Explain.
Answer:
The sum is 4 units below -5 because |-4| = 4. I counted down from -5 four times and stopped at -9.

d. What integers do the arrows represent?
Answer:
The arrows represent the integers -4 and -5.

Question 2.
Jenna and Jay are playing the Integer Game. Below are the two cards they selected.
a. How do the models for these two addition problems differ on a number line? How are they the same?

Answer:
The p-values are the same. They are both 3, so the heads of the first arrows will be at the same point on the number line. The sums will both be five units from this point but in opposite directions.

b. If the order of the cards changed, how do the models for these two addition problems differ on a number line? How are they the same?

Answer:
The first addends are different, so the head of the first arrow in each model will be at different points on the number line. The sums are both three units to the right of the first addend.

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

Question 1.
Below is a table showing the change in temperature from morning to afternoon for one week.
a. Use the vertical number line to help you complete the table. As an example, the first row is completed for you.

Answer:

b. Do you agree or disagree with the following statement: “A rise of -7°C” means “a fall of 7°C?” Explain. (Note: No one would ever say, “A rise of -7 degrees”; however, mathematically speaking, it is an equivalent phrase.)
Answer:
Sample response: I agree with this statement because a rise of -7 is the opposite of a rise
of 7. The opposite of a rise of 7 is a fall of 7.

For Problems 2–3, refer to the Integer Game.

Question 2.
Terry selected two cards. The sum of her cards is -10.
a. Can both cards be positive? Explain why or why not.
Answer:
No. In order for the sum to be -10, at least one of the addends would have to be negative. If both cards are positive, then Terry would count up twice going to the right. Negative integers are to the left of 0.

b. Can one of the cards be positive and the other be negative? Explain why or why not.
Answer:
Yes. Since both cards cannot be positive, this means that one can be positive and the other negative.
She could have -11 and 1 or -12 and 2. The card with the greatest absolute value would have to be negative.

c. Can both cards be negative? Explain why or why not.
Answer:
Yes, both cards could be negative. She could have -8 and -2. On a number line, the sum of two negative integers will be to the left of 0.

Question 3.
When playing the Integer Game, the first two cards you selected were -8 and -10.
a. What is the value of your hand? Write an equation to justify your answer.
Answer:
-8 + (-10) = -18

b. For part (a), what is the distance of the sum from -8? Does the sum lie to the right or left of -8 on the number line?
Answer:
The distance is 10 units from -8, and it lies to the left of -8 on the number line.

c. If you discarded the -10 and then selected a 10, what would be the value of your hand? Write an equation to justify your answer.
Answer:
The value of the hand would be 2. -8+10=2.

Question 4.
Given the expression 67+(-35), can you determine, without finding the sum, the distance between 67 and the sum? Is the sum to the right or left of 67 on the number line?
Answer:
The distance would be 35 units from 67. The sum is to the left of 67 on the number line.

Question 5.
Use the information given below to write an equation. Then create an arrow diagram of this equation on the number line provided below.

Answer:
The sum of -4 and a number is 12 units to the right of -4 on a number line.
-4 + 12 = 8

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

Eureka Math Grade 7 Module 2 Lesson 2 Example Answer Key

Example 1:
Modeling Addition on the Number Line
Complete the steps to find the sum of -2+3 by filling in the blanks. Model the equation using straight arrows called vectors on the number line below.
a. Place the tail of the arrow on __ .
Answer:
0

b. Draw the arrow 2 units to the left of 0, and stop at __. The direction of the arrow is to the ___ left since you are counting down from 0.
Answer:
-2, left

c. Start the next arrow at the end of the first arrow, or at __ .
Answer:
-2

d. Draw the second arrow __ units to the right since you are counting up from -2.
Answer:
3

e. Stop at __ .
Answer:
1

f. Circle the number at which the second arrow ends to indicate the ending value.
Answer:

g. Repeat the process from parts (a)–(f) for the expression 3+(-2).
Answer:

h. What can you say about the sum of -2+3 and 3+(-2)? Does order matter when adding numbers? Why or why not?
Answer:
-2 + 3 is the same as 3 + (-2) because they both equal 1. The order does not matter when adding numbers because addition is commutative.

Example 2:
Expressing Absolute Value as the Length of an Arrow on the Real Number Line
a. How does absolute value determine the arrow length for -2? Use the number line provided to support your answer.
Answer:
|-2|=2, so the arrow is 2 units long. Because -2 is a negative number, the arrow points to the left.

b. How does the absolute value determine the arrow length for 3? Use the number line provided to support your answer.
Answer:
|3| = 3, so the arrow is 3 units long. Because 3 is positive, the arrow points to the right.

c. Describe how the absolute value helps you represent -10 on a number line.
Answer:
The absolute value can help me because it tells me how long my arrow should be when starting at 0 on the real number line. The |-10| = 10, so my arrow will be 10 units in length.

Example 3.
Finding Sums on a Real Number Line Model
Find the sum of the integers represented in the diagram below.

a. Write an equation to express the sum.
Answer:
5 + (-2) + 3 = 6

b. What three cards are represented in this model? How did you know?
Answer:
The cards are 5, -2, and 3 because the arrows show their lengths.

c. In what ways does this model differ from the ones we used in Lesson 1?
Answer:
In Lesson 1, a movement of 5 units was shown with 5 separate hops. In this lesson, 5 units are shown as one total movement with a straight arrow. Both represent the same total movement.

d. Can you make a connection between the sum of 6 and where the third arrow ends on the number line?
Answer:
The final position of the third arrow is 6. This means that the sum is 6.

e. Would the sum change if we changed the order in which we add the numbers, for example, (-2)+3+5?
Answer:
No, because addition is commutative. Order does not matter.

f. Would the diagram change? If so, how?
Answer:
Yes, the first arrow would start at 0 and point left 2 units. The second arrow would start at -2 and point right 3 units. The third arrow would start at 1 and point 5 units right but still end on 6.

Eureka Math Grade 7 Module 2 Lesson 2 Exercise Answer Key

Exercise 1:
Real-World Introduction to Integer Addition
Answer the questions below.
a. Suppose you received $10 from your grandmother for your birthday. You spent $4 on snacks. Using addition, how would you write an equation to represent this situation?
Answer:
10 + (-4) = 6

b. How would you model your equation on a number line to show your answer?
Answer:

Exercise 2
Create a number line model to represent each of the expressions below.
a. -6 + 4
Answer:

b. 3 + (-8)
Answer:

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

Jessica made the addition model below of the expression (-5)+(-2)+3.
a. Do the arrows correctly represent the numbers that Jessica is using in her expression?
Answer:
No. Jessica started her first arrow at -5 instead of 0. Negative numbers should be shown as counting down, so the arrow should have started at 0 and pointed left, ending on -5. The other arrows are drawn correctly, but they are in the wrong places because the starting arrow is in the wrong place.

b. Jessica used the number line diagram above to conclude that the sum of the three numbers is 1. Is she correct?
Answer:
Jessica is incorrect.

c. If she is incorrect, find the sum, and draw the correct model.
Answer:
The sum should be -4. -5 + (-2) + 3 = -4.

d. Write a real-world situation that would represent the sum.
Answer:
Answers will vary. A football team lost 5 yards on the first play. On the second play, the team lost another 2 yards. Then, the team gained 3 yards. After three plays, the team had a total yardage of -4 yards.

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

Represent Problems 1–3 using both a number line diagram and an equation.

Question 1.
David and Victoria are playing the Integer Card Game. David drew three cards, -6, 12, and -4. What is the sum of the cards in his hand? Model your answer on the number line below.
Answer:
(-6) + 12 + (-4) = 2

Question 2.
In the Integer Card Game, you drew the cards, 2, 8, and -11. Your partner gave you a 7 from his hand.
a. What is your total? Model your answer on the number line below.
Answer:
2 + 8 + (-11) + 7 = 6

b. What card(s) would you need to get your score back to zero? Explain. Use and explain the term additive inverse in your answer.
Answer:
You would need any combination of cards that sum to -6 because the additive inverse of 6 is -6.
6 + (-6) = 0

Question 3.
If a football player gains 40 yards on a play, but on the next play, he loses 10 yards, what would his total yards be for the game if he ran for another 60 yards? What did you count by to label the units on your number line?
Answer:
90 yards because 40+(-10)+60=90. Student answers may vary, but they should not choose to count by 1.

Question 4.
Find the sums.
a. -2 + 9
Answer:
7

b. -8 + -8
Answer:
-16

c. -4 + (-6) + 10
Answer:
0

d. 5 + 7 + (-11)
Answer:
1

Question 5.
Mark an integer between 1 and 5 on a number line, and label it point Z. Then, locate and label each of the following points by finding the sums.
Answer:
Answers will vary. Sample student response below.

a. Point A: Z+5
Answer:
Point A: 3+5=8

b. Point B: Z+(-3)
Answer:
Point B: 3+(-3)=0

c. Point C: (-4) + (-2) + Z
Answer:
Point C: (-4) + (-2) + 3 = -3

d. Point D: -3 + Z + 1
Answer:
Point D: -3 + 3 + 1 = 1

Question 6.
Write a story problem that would model the sum of the arrows in the number diagram below.

Answer:
Answers will vary. Jill got on an elevator and went to the 9^th floor. She accidently pressed the down button and went back to the lobby. She pressed the button for the 5^th floor and got off the elevator.

Question 7.
Do the arrows correctly represent the equation 4 + (-7) + 5 = 2? If not, draw a correct model below.

Answer:
No, the arrows are incorrect. The correct model is shown.

Engage NY Eureka Math 7th Grade Module 1 Lesson 21 Answer Key

Eureka Math Grade 7 Module 1 Lesson 21 Exploratory Challenge Answer Key

Exploratory Challenge: A New Scale Factor
The school plans to publish your work on the dream classroom in the next newsletter. Unfortunately, in order to fit the drawing on the page in the magazine, it must be \(\frac{1}{4}\) its current length. Create a new drawing (SD2) in which all of the lengths are \(\frac{1}{4}\) those in the original scale drawing (SD1) from Lesson 20.

Answer:

An example is included for students unable to create SD1 at the end of Lesson 20. Pose the following questions:
→ Would the new scale create a larger or smaller scale drawing as compared to the original drawing?
→ It would be smaller because \(\frac{1}{4}\) is smaller than one.
→ How would you use the scale factor between SD1 to SD2 to calculate the new scale drawing lengths without having to get the actual measurement first?
→ Take the original scale drawing lengths and multiply them by \(\frac{1}{4}\) to find the new scale lengths.
→ Once students have finished creating SD2, ask students to prove to the architect that SD2 is actually a scale drawing of the original room.
→ How can we go about proving that the new scale drawing (SD2) is actually a scale drawing of the original room?
→ The scale lengths of SD2 have to be proportional to the actual lengths. We need to find the constant of proportionality, the scale factor.
How do we find the new scale factor?
→ Divide one of the new scale lengths by its corresponding actual length.
→ If the actual measurement was not known, how could we find it?
→ Calculate the actual length by using the scale factor on the original drawing. Multiply the scale length of the original drawing by the original scale factor.

Eureka Math Grade 7 Module 1 Lesson 21 Exercise Answer Key

The picture shows an enlargement or reduction of a scale drawing of a trapezoid.

Using the scale factor written on the card you chose, draw your new scale drawing with correctly calculated measurements.

Answers may vary depending on the card. One sample response could be, 1 cm, 1\(\frac{5}{6}\) cm, 2 \(\frac{1}{3}\) cm, 1 \(\frac{2}{3}\) cm.

a. What is the scale factor between the original scale drawing and the one you drew?
Answer:
\(\frac{1}{3}\)

b. The longest base length of the actual trapezoid is 10 cm. What is the scale factor between the original scale drawing and the actual trapezoid?
Answer:
\(\frac{7}{10}\)

c. What is the scale factor between the new scale drawing you drew and the actual trapezoid?
Answer:

Eureka Math Grade 7 Module 1 Lesson 21 Problem Set Answer Key

Question 1.
Jake reads the following problem: If the original scale factor for a scale drawing of a square swimming pool is \(\frac{1}{90}\), and the length of the original drawing measured to be 8 inches, what is the length on the new scale drawing if the scale factor of the new scale drawing length to actual length is \(\frac{1}{144}\)?
He works out the problem:
8 inches ÷ \(\frac{1}{90}\) = 720 inches
720 inches × \(\frac{1}{144}\) = 5 inches
Is he correct? Explain why or why not.
Answer:
Jake is correct. He took the original scale drawing length and divided by the original scale factor to get the actual length, 720 inches. To get the new scale drawing length, he takes the actual length, 720, and multiplies by the new scale factor, \(\frac{1}{144}\), to get 5 inches.

Question 2.
What is the scale factor of the new scale drawing to the original scale drawing (SD2 to SD1)?
Answer:

Question 3.
Using the scale, if the length of the pool measures 10 cm on the new scale drawing:
a. Using the scale factor from Problem 1, \(\frac{1}{144}\), find the actual length of the pool in meters.
Answer:
14.40 m

b. What is the surface area of the floor of the actual pool? Rounded to the nearest tenth.
Answer:
14.4 m × 14.4 m
207.36 m² ≈ 207.4 m²

c. If the pool has a constant depth of 1.5 meters, what is the volume of the pool? Rounded to the nearest tenth.
Answer:
14.4 m × 14.4 m × 1.5 m
311.04 m³ ≈ 311.0 m³

d. If 1 cubic meter of water is equal to 264.2 gallons, how much water will the pool contain when completely filled? Rounded to the nearest unit.
Answer:
311.0 m³ × \(\frac{264.2 \text { gallons }}{1 \mathrm{~m}^{3}}\)
82,166.2 gallons

Question 4.
Complete a new scale drawing of your dream room from the Problem Set in Lesson 20 by either reducing by \(\frac{1}{4}\) or enlarging it by 4.
Answer:
Scale drawings will vary.

SD1 Example for students who were unable to create their own from Lesson 20

SCALE FACTOR:
\(\frac{1}{120}\)

Engage NY Eureka Math 7th Grade Module 2 Lesson 1 Answer Key

Eureka Math Grade 7 Module 2 Lesson 1 Example Answer Key

Example 2.
Counting Up and Counting Down on the Number Line
Use the number line below to practice counting up and counting down.
→ Counting up starting at 0 corresponds to positive ___ numbers.
Answer:
positive

→ Counting down starting at 0 corresponds to ___ numbers.
negative

Answer:

a. Where do you begin when locating a number on the number line?
Answer:
Start at 0.

b. What do you call the distance between a number and 0 on a number line?
Answer:
The absolute value

c. What is the relationship between 7 and -7?
Answer:
Answers will vary. 7 and -7 both have the same absolute values. They are both the same distance from zero, 0, but in opposite directions; therefore, 7 and -7 are opposites.

Example 3:
Using the Integer Game and the Number Line
What is the sum of the card values shown? Use the counting on method on the provided number line to justify your answer.

a. What is the final position on the number line? ___
Answer:
4

b. What card or combination of cards would you need to get back to 0? ___
Answer:
-4 or -1 and -3

Eureka Math Grade 7 Module 2 Lesson 1 Exercise Answer Key

Exercise 1.
Positive and Negative Numbers Review
With your partner, use the graphic organizer below to record what you know about positive and negative numbers. Add or remove statements during the whole-class discussion.

Exercise 2.
The Additive Inverse
Use the number line to answer each of the following questions.

a. How far is 7 from 0 and in which direction? ___
Answer:
7 units to the right

b. What is the opposite of 7? -7
Answer:
-7

c. How far is -7 from 0 and in which direction?
Answer:
7 units to the left

d. Thinking back to our previous work, explain how you would use the counting on method to represent the following: While playing the Integer Game, the first card selected is 7, and the second card selected is -7.
Answer:
I would start at 0 and count up 7 by moving to the right. Then, I would start counting back down from 7 to 0.

e. What does this tell us about the sum of 7 and its opposite, -7?
Answer:
The sum of 7 and -7 equals 0.
7 + (-7) = 0

f. Look at the curved arrows you drew for 7 and -7. What relationship exists between these two arrows that would support your claim about the sum of 7 and -7?
Answer:
The arrows are both the same distance from 0. They are just pointing in opposite directions.

g. Do you think this will hold true for the sum of any number and its opposite? Why?
Answer:
I think this will be true for the sum of any number and its opposite because when you start at 0 on the number line and move in one direction, moving in the opposite direction the same number of times will always take you back to zero.

Eureka Math Grade 7 Module 2 Lesson 1 Exit Ticket Answer Key

Question 1.
Your hand starts with the 7 card. Find three different pairs that would complete your hand and result in a value of zero.

Answer:
Answers will vary. (-3 and -4), (-5 and -2), (-10 and 3)

Question 2.
Write an equation to model the sum of the situation below.
A hydrogen atom has a zero charge because it has one negatively charged electron and one positively charged proton.
Answer:
(-1) + 1 = 0 or 1 + (-1) = 0

Question 3.
Write an equation for each diagram below. How are these equations alike? How are they different? What is it about the diagrams that lead to these similarities and differences?
Diagram A:

Diagram B:

Answer:
A: 4 + (-4)=0
B: -4 + 4 = 0
Answers will vary. Both equations are adding 4 and -4. The order of the numbers is different. The direction of A shows counting up 4 and then counting down 4. The direction of B shows counting down 4 and then counting up 4.
Students may also mention that both diagrams demonstrate a sum of zero, adding opposites, or adding additive inverses.

Eureka Math Grade 7 Module 2 Lesson 1 Problem Set Answer Key

The Problem Set provides practice with real-world situations involving the additive inverse such as temperature and money. Students also explore more scenarios from the Integer Game to provide a solid foundation for Lesson 2.

For Problems 1 and 2, refer to the Integer Game.

Question 1.
You have two cards with a sum of (-12) in your hand.
a. What two cards could you have?
Answer:
Answers will vary. (-6 and -6)

b. You add two more cards to your hand, but the total sum of the cards remains the same, (-12). Give some different examples of two cards you could choose.
Answer:
Answers will vary, but numbers must be opposites. (-2 and 2) and (4 and -4)

Question 2.
Choose one card value and its additive inverse. Choose from the list below to write a real-world story problem that would model their sum.
a. Elevation: above and below sea level
Answer:
Answers will vary. (A scuba diver is 20 feet below sea level. He had to rise 20 feet in order to get back on the boat.)

b. Money: credits and debits, deposits and withdrawals
Answer:
Answers will vary. (The bank charges a fee of $5 for replacing a lost debit card. If you make a deposit of $5, what would be the sum of the fee and the deposit?)

c. Temperature: above and below 0 degrees
Answer:
Answers will vary. (The temperature of one room is 5 degrees above 0. The temperature of another room is 5 degrees below zero. What is the sum of both temperatures?)

d. Football: loss and gain of yards
Answer:
Answers will vary. (A football player gained 25 yards on the first play. On the second play, he lost 25 yards. What is his net yardage after both plays?)

Question 3.
On the number line below, the numbers h and k are the same distance from 0. Write an equation to express the value of h + k. Explain.

Answer:
h + k = 0 because their absolute values are equal, but their directions are opposite. k is the additive inverse of h, and h is the additive inverse of k because they are the same distance from 0. Therefore, the sum of k and h is 0, because additive inverses have a sum of 0.

Question 4.
During a football game, Kevin gained five yards on the first play. Then he lost seven yards on the second play. How many yards does Kevin need on the next play to get the team back to where they were when they started? Show your work.
Answer:
He has to gain 2 yards.
5 + (-7) + 2 = 0, 5 + (-7) = -2, and -2 + 2 = 0.

QueWrite 5.
Write an addition number sentence that corresponds to the arrows below.

Answer:
10 + (-5) + (-5) = 0

Engage NY Eureka Math 7th Grade Module 2 End of Module Assessment Answer Key

Eureka Math Grade 7 Module 2 End of Module Assessment Task Answer Key

Question 1.
The water level in Ricky Lake changes at an average of –\(\frac{7}{16}\) inch every 3 years.

a. Based on the rate above, how much will the water level change after one year? Show your calculations and model your answer on the vertical number line, using 0 as the original water level.

Answer:

b. How much would the water level change over a 7-year period?
Answer:
Distance = rate × time
= – \(\frac{7}{48}\) × 7
= –\(\frac{49}{48}\)
= -1\(\frac{1}{48}\)
The water level drops 1\(\frac{1}{48}\) inches over a 7-year period.

c. When written in decimal form, is your answer to part (b) a repeating decimal or a terminating decimal? Justify your answer using long division.
Answer:

-1\(\frac{1}{48}\) written in decimal form is a repeating decimal because when converted using long division, the remainder repeats after the hundred-thousandths place.

Question 2.
Kay’s mother taught her how to make handmade ornaments to sell at a craft fair. Kay rented a table at the fair for $30 and set up her work station. Each ornament that she makes costs approximately $2.50 for materials. She sells each ornament for $6.00.

a. If x represents the number of ornaments sold at the craft fair, which of the following expressions would represent Kay’s profit? (Circle all choices that apply.)

A. -30 + 6x – 2.50x
B. 6x – 30 – 2.50x
C. 6x – 30
D. 4.50x-30
E. 3.50x-30
Answer:

b. Kay does not want to lose money on her business. Her mother told her she needs to sell enough ornaments to at least cover her expenses (costs for materials and table rental). Kay figures that if she sells 8 ornaments, she covers her expenses and does not lose any money. Do you agree? Explain and show work to support your answer.
Answer:
3.50x – 30
3.50(8) – 30
(24 + 4) – 30
28 – 30
-2
I disagree with Kay because selling 8 ornaments covers most of her costs but still leaves her $2 in debt.

c. Kay feels that if she earns a profit of $40.00 at this craft fair, her business will be successful enough for her to branch out to other craft fairs. How many ornaments does she have to sell to earn a $40.00 profit? Write and solve an equation; then explain how the steps and operations used in your algebraic solution compare to an arithmetic solution.
Answer:
3.50x – 30 = 40
3.50x – 30 + 30 = 40 + 30
3.50x + 0 = 70
3.50x = 70
3.50x (\(\frac{1}{3.50}\))= 70(\(\frac{1}{3.50}\))
1x = 20
To find the answer arithmetically, I would have to combine the $40 profit and $30 rental fee, then divide that sum ($70) by the $3.50 that she earns per ornament after costs.
Kay must sell 20 ornaments.

Question 3.
Travis received a letter from his bank saying that his checking account balance fell below zero. His account transaction log is shown below.

a. On which line did Travis make a mathematical error? Explain Travis’s mistake.
Answer:
On line 4, Travis subtracted $188 from $180 and got a positive answer. The difference should be -$8.00.

b. The bank charged Travis a $20 fee because his balance dropped below 0. He knows that he currently has an outstanding charge for $7.85 that he has not recorded yet. How much money will Travis have to deposit into his account so that the outstanding charge does not create another bank fee? Explain.
Answer:
Starting at Line 3:
180.00 – 188.00 + 20.00 – 5.95
-8.00 + 20.00 – 5.95
12.00 – 5.95
6.05
Travis’s actual balance should be $6.05.
6.05 + (- 20.00) overdraft fee
|-20.00| – | 6.05 |
– 13.95
– 13.95 + (-7.85) outstanding charge
| -13.95 | + | -7.85 |
– 21.80
To get his account back to 0, Travis needs to deposit $21.80 or more to avoid another overdraft fee.

Question 4.
The length of a rectangular envelope is 2\(\frac{1}{2}\) times its width. A plastic band surrounds the front and back of the envelope to secure it as shown in the picture. The plastic band is 39\(\frac{3}{8}\) inches long. Find the length and width of the envelope.

Answer:
The length of the plastic band is equivalent to the perimeter of the envelope.
Width: w
Length = 2\(\frac{1}{2}\) × width
w + w + (2\(\frac{1}{2}\)w) + (2\(\frac{1}{2}\)w) = 39\(\frac{3}{8}\)
2w + 5w = 39\(\frac{3}{8}\)
7w = 39\(\frac{3}{8}\)
7w = \(\frac{315}{8}\)
(\(\frac{1}{7}\))(7w) = (\(\frac{315}{8}\))(\(\frac{1}{7}\))
w = \(\frac{45}{8}\) = 5\(\frac{5}{8}\)

Length = 2\(\frac{1}{2}\) w
Length = 2\(\frac{1}{2}\) (5\(\frac{5}{8}\))
Length =\(\frac{5}{2}\) × \(\frac{45}{8}\)
Length = \(\frac{225}{16}\) = 14\(\frac{1}{16}\)

The length of the envelope is 14\(\frac{1}{16}\) inches and the width is 5\(\frac{5}{8}\) inches.

Question 5.
Juan and Mary are playing the Integer Card Game. The cards in their hands are shown below:

a. What are the scores in each of their hands?
Juan’s Score: 4
3 + 4 + 9 + (-12)
7 + 9 +(-12)
16 + (-12)
4
Mary’s Score: 4
-2 + 3 + 1 + 2
1 + 1 + 2
4

b. Lydia says that if Juan and Mary both take away their 3’s, Juan’s score will be higher than Mary’s. Marcus argues and says that Juan and Mary’s scores will be equal. Are either of them right? Explain.
Answer:
If both Juan and Mary lay down their 3’s, then both of their totals will be decreased by 3. Since both of their totals are 4, laying down a 3 would make both scores 1. Juan’s score and Mary’s score would be equal, so Marcus is correct.

c. Juan picks up another set of cards that is exactly like each card in his hand. Which of the following would make Mary’s score equal to Juan’s? Place a check mark ✓ by all that apply.

Explain why your selections will make Juan and Mary’s scores equal.
Answer:

Juan’s total doubles because every card in his hand doubled, so his total is 8. Each choice I selected would add 4 to Mary’s total to make it 8.

Engage NY Eureka Math 7th Grade Module 1 Lesson 22 Answer Key

Eureka Math Grade 7 Module 1 Lesson 22 Exploratory Challenge Answer Key

Using the new scale drawing of your dream room, list the similarities and differences between this drawing and the original drawing completed for Lesson 20.
Similarities
Differences
Answer:
Similarities
– Same room shape
– Placement of furniture
– Space between furniture
– Drawing of the original room
– Proportional

Differences
– One is bigger than the other
– Different scale factors

Original Scale Factor: ___
Answer:
\(\frac{1}{20}\)
New scale Factor: ___
Answer:
\(\frac{1}{30}\)

What is the relationship between these scale factors?
Answer:
\(\frac{1}{4}\)
Key Idea:
Two different scale drawings of the same top-view of a room are also scale drawings of each other. In other words, a scale drawing of a different scale can also be considered a scale drawing of the original scale drawing.

Eureka Math Grade 7 Module 1 Lesson 22 Example Answer Key

Example 1:
Building a Bench
To surprise her mother, Taylor helped her father build a bench for the front porch. Taylor’s father had the instructions with drawings, but Taylor wanted to have her own copy. She enlarged her copy to make it easier to read. Using the following diagram, fill in the missing information. To complete the first row of the table, write the scale factor of the bench to the bench, the bench to the original diagram, and the bench to Taylor’s diagram. Complete the remaining rows similarly.

The pictures below show the diagram of the bench shown on the original instructions and the diagram of the bench shown on Taylor’s enlarged copy of the instruction.

Answer:

Eureka Math Grade 7 Module 1 Lesson 22 Exercise Answer Key

Exercise 1.
Carmen and Jackie were driving separately to a concert. Jackie printed a map of the directions on a piece of paper before the drive, and Carmen took a picture of Jackie’s map on her phone. Carmen’s map had a scale factor of \(\frac{1}{563,270}\). Using the pictures, what is the scale of Carmen’s map to Jackie’s map? What was the scale factor of Jackie’s printed map to the actual distance?

Answer:

Exercise 2.
Ronald received a special toy train set for his birthday. In the picture of the train on the package, the boxcar has the following dimensions: length is 4\(\frac{5}{16}\) inches; width is 1\(\frac{1}{8}\) inches; height is 1\(\frac{5}{8}\) inches. The toy boxcar that Ronald received has dimensions l is 17.25 inches; w is 4.5 inches; h is 6.5 inches. If the actual boxcar is 50 feet long:
a. Find the scale factor of the picture on the package to the toy set.
Answer:

b. Find the scale factor of the picture on the package to the actual boxcar.
Answer:

c. Use these two scale factors to find the scale factor between the toy set and the actual boxcar.
Answer:

d. What is the width and height of the actual boxcar?
Answer:

Eureka Math Grade 7 Module 1 Lesson 22 Problem Set Answer Key

Question 1.
For the scale drawing, the actual lengths are labeled onto the scale drawing. Measure the lengths, in centimeters, of the scale drawing with a ruler, and draw a new scale drawing with a scale factor (SD2 to SD1) of \(\frac{1}{2}\).

Answer:

Question 2.
Compute the scale factor of the new scale drawing (SD2) to the first scale drawing (SD1) using the information from the given scale drawings.
a.
Answer:
\(\frac{1}{48}\)

b.
Answer:
36

c.
Answer:
\(\frac{5}{4}\)

Eureka Math Grade 7 Module 1 Lesson 22 Exit Ticket Answer Key

The school is building a new wheelchair ramp for one of the remodeled bathrooms. The original drawing was created by the contractor, but the principal drew another scale drawing to see the size of the ramp relative to the walkways surrounding it. Find the missing values on the table.

Answer:

Engage NY Eureka Math 7th Grade Module 1 Lesson 19 Answer Key

Eureka Math Grade 7 Module 1 Lesson 19 Example Answer Key

Exploring Area Relationships

Use the diagrams below to find the scale factor and then find the area of each figure.
Example 1

Scale factor: _________
Actual Area = ______________
Scale Drawing Area = ________________
Value of the Ratio of the Scale Drawing Area to the Actual Area: _________
Answer:
Scale factor: 2
Actual Area = 12 square units
Scale Drawing Area = 48 square units
Value of the Ratio of the Scale Drawing Area to the Actual Area:
\(\frac{48}{12}\) = 4

Example 2.

Scale factor: _________
Actual Area = ______________
Scale Drawing Area = ________________
Value of the Ratio of the Scale Drawing Area to the Actual Area: _________
Answer:
Scale factor: \(\frac{1}{3}\)
Actual Area = 54 square units
Scale Drawing Area = 6 square units
Value of the Ratio of Scale Drawing Area to Actual Area:
\(\frac{6}{54}\) = \(\frac{1}{9}\)

Example 3.

Scale factor: \(\frac{4}{3}\)
Actual Area = 27 square units
Scale Drawing Area = 48 square units
Value of the Ratio of Scale Drawing Area to Actual Area:
\(\frac{48}{27}\) = \(\frac{16}{9}\)

Guide students through completing the results statements on the student materials.

Results:
What do you notice about the ratio of the areas in Examples 1–3? Complete the statements below.
When the scale factor of the sides was 2, then the value of the ratio of the areas was __ .
Answer:
4
When the scale factor of the sides was \(\frac{1}{3}\), then the value of the ratio of the areas was __.
Answer:
\(\frac{1}{9}\)

When the scale factor of the sides was \(\frac{4}{3}\), then the value of the ratio of the areas was ___.
Answer:
\(\frac{16}{9}\)

Based on these observations, what conclusion can you draw about scale factor and area?
Answer:
The ratio of the areas is the scale factor multiplied by itself or squared.

If the scale factor is r, then the ratio of the areas is ___.
Answer:
r² to 1.

→ Why do you think this is? Why do you think it is squared (opposed to cubed or something else)?
→ When you are comparing areas, you are dealing with two dimensions instead of comparing one linear measurement to another.
→ How might you use this information in working with scale drawings?
→ In working with scale drawings, you could take the scale factor, r, and calculate r² to determine the relationship between the area of the scale drawing and the area of the actual picture. Given a blueprint for a room, the scale drawing dimensions could be used to find the scale drawing area and could then be applied to determine the actual area. The actual dimensions would not be needed.
→ Suppose a rectangle has an area of 12 square meters. If the rectangle is enlarged by a scale factor of three, what is the area of the enlarged rectangle based on Examples 1–3? Look and think carefully!
→ If the scale factor is 3, then the ratio of scale drawing area to actual area is 3² to 1², or 9 to 1. So, if its area is 12 square meters before it is enlarged to scale, then the enlarged rectangle will have an area of 12∙(\(\frac{9}{1}\)), or 12∙9, resulting in an area of 108 square meters.

Example 4:
They Said Yes!
The Student Government liked your half-court basketball plan. They have asked you to calculate the actual area of the court so that they can estimate the cost of the project.
Based on your drawing below, what will the area of the planned half-court be?
Scale Drawing: 1 inch on the drawing corresponds to 15 feet of actual length

Does the actual area you found reflect the results we found from Examples 1–3? Explain how you know.
Answer:
Method 1: Use the measurements we found in yesterday’s lesson to calculate the area of the half-court.
Actual area = 25 feet × 30 feet =750 square feet
Method 2: Apply the newly discovered Ratio of Area relationship.

Note to teachers: This can be applied to the given scale with no unit conversions (shown on left) or to the scale factor (shown on right). Both options are included here as possible student work and would provide for a rich discussion of why they both work and what method is preferred. See guiding questions below.
Answer:
Using Scale:
The Value of the Ratio of Areas: (\(\frac{15}{1}\))² = 225
Scale Drawing Area = 2 in.× 1\(\frac{2}{3}\) in.
= \(\frac{10}{3}\) square inches
Let x = scale drawing area, and let y = actual area.
y = kx
y = 225(\(\frac{10}{3}\))
y = \(\frac{225}{1}\).\(\frac{10}{3}\)
y = 750
The actual area using the given scale is 750 square feet.

Using Scale Factor:
The Value of the Ratio of Areas (\(\frac{180}{1}\))² = 32400
Scale Drawing Area = 2 in.×1\(\frac{2}{3}\) in.
= \(\frac{10}{3}\) square inches
Let x = scale drawing area, and let y = actual area.
y = kx
y = 32400(\(\frac{10}{3}\))
y = \(\frac{324000}{3}\)
y = 108000
The actual area is 108,000 square inches, or
108000 square inches × \(\frac{1 square feet}{144 square feet}\) = 750 square feet.
Ask students to share how they found their answer. Use guiding questions to find all three options as noted above.
→ What method do you prefer?
→ Is there a time you would choose one method over the other?
→ If we do not already know the actual dimensions, it might be faster to use Method 1 (ratio of areas). If we are re-carpeting a room based upon a scale drawing, we could just take the dimensions from the scale drawing, calculate area, and then apply the ratio of areas to find the actual amount of carpet we need to buy.
Guide students to complete the follow-up question in their student materials.

Does the actual area you found reflect the results we found from Examples 1–3? Explain how you know.
Answer:
Yes, the scale of 1 inch to 15 feet has a scale factor of 180, so the ratio of area should be (180)², or 32,400.
The drawing area is (2)(1\(\frac{2}{3}\)), or \(\frac{10}{3}\) square inches.
The actual area is 25 feet by 30 feet, or 750 square feet, or 108,000 square inches.
The value of the ratio of the areas is \(\frac{108,000}{\frac{10}{3}}\), or \(\frac{324,000}{10}\), or 32,400

It would be more efficient to apply this understanding to the scale, eliminating the need to convert units.
If we use the scale of \(\frac{15}{1}\), then the ratio of area is \(\frac{225}{1}\).
The drawing area is (2)(1\(\frac{2}{3}\)), or \(\frac{10}{3}\) square inches.
The actual area is 25 feet by 30 feet, or 750 square feet.
The ratio of area is \(\frac{750}{\frac{10}{3}}\), \(\frac{2250}{10}\), or \(\frac{225}{1}\).

Eureka Math Grade 7 Module 1 Lesson 19 Exercise Answer Key

Allow time for students to answer independently and then share results.

Question 1.
The triangle depicted by the drawing has an actual area of 36 square units. What is the scale of the drawing? (Note: Each square on the grid has a length of 1 unit.)

Answer:

Scale Drawing Area: \(\frac{1}{2}\) ∙ 6∙ 3 = 9 square units. Ratio of Scale Drawing Area to Actual Area: \(\frac{9}{36}\) = r²
Therefore, r (scale factor) = \(\frac{3}{6}\) since \(\frac{3}{6}\) ∙ \(\frac{3}{6}\) = \(\frac{9}{36}\). The scale factor is \(\frac{1}{2}\) . The scale is 1 unit of drawing length represents 2 units of actual length.

For Exercise 2, allow students time to measure the drawings of the apartments using a ruler and then compare measurements with a partner. Students then continue to complete parts (a)–(f) with a partner. Allow students time to share responses. Sample answers to questions are given below.

Question 2.
Use the scale drawings of two different apartments to answer the questions. Use a ruler to measure.

Answer:

a. Find the scale drawing area for both apartments, and then use it to find the actual area of both apartments.
Answer:

b. Which apartment has closets with more square footage? Justify your thinking.
Answer:

The suburban apartment has greater square footage in the closet floors.

c. Which apartment has the largest bathroom? Justify your thinking.
Answer:

The city apartment has the largest bathroom.

d. A one-year lease for the suburban apartment costs $750 per month. A one-year lease for the city apartment costs $925. Which apartment offers the greater value in terms of the cost per square foot?
Answer:
The suburban cost per square foot is \(\frac{750}{720}\), or approximately $1.04 per square foot. The city cost per square foot is \(\frac{925}{768}\), or approximately $1.20 per square foot. The suburban apartment offers a greater value (cheaper cost per square foot), $1.04 versus $1.20.

Eureka Math Grade 7 Module 1 Lesson 19 Exit Ticket Answer Key

A 1-inch length in the scale drawing below corresponds to a length of 12 feet in the actual room.

Question 1.
Describe how the scale or the scale factor can be used to determine the area of the actual dining room.
Answer:
The scale drawing will need to be enlarged to get the area or dimensions of the actual dining room. Calculate the area of the scale drawing, and then multiply by the square of the scale (or scale factor) to determine the actual area.

Question 2.
Find the actual area of the dining room.
Answer:
Scale drawing area of dining room: (1\(\frac{1}{2}\) in.×\(\frac{3}{4}\) in.) + (\(\frac{3}{4}\) in. × \(\frac{1}{2}\)in.) = \(\frac{12}{8}\) in² or 1\(\frac{1}{2}\) in²
Actual area of the dining room: \(\frac{12}{8}\) ft.× 144 ft. = 216 ft²
Or similar work completing conversions and using scale factor

Question 3.
Can a rectangular table that is 7 ft. long and 4 ft. wide fit into the narrower section of the dining room? Explain your answer.
Answer:
The narrower section of the dining room measures \(\frac{3}{4}\) by \(\frac{1}{2}\) in the drawing, or 9 feet by 6 feet in the actual room. Yes, the table will fit; however, it will only allow for 1 additional foot around all sides of the table for movement or chairs.

Engage NY Eureka Math 7th Grade Module 1 Lesson 18 Answer Key

Eureka Math Grade 7 Module 1 Lesson 18 Example Answer Key

Example 1.
Basketball at Recess?
Vincent proposes an idea to the Student Government to install a basketball hoop along with a court marked with all the shooting lines and boundary lines at his school for students to use at recess. He presents a plan to install a half-court design as shown below. After checking with the school administration, he is told it will be approved if it fits on the empty lot that measures 25 feet by 75 feet on the school property. Will the lot be big enough for the court he planned? Explain.

Answer:

Scale Factor: 1 inch corresponds to (15∙12) inches, or 180 inches, so the scale factor is 180. Let k = 180, x represent the scale drawing lengths in inches, and y represent the actual court lengths in inches. The y-values must be converted from feet to inches.
To find actual length: y = 180x
y = 180(2)
y = 360 inches, or 30 feet
To find actual width: y = 180x
y = 180(1 \(\frac{2}{3}\))
y = \(\frac{180}{1}\)∙\(\frac{5}{3}\)
y = 300 inches, or 25 feet
The actual court measures 25 feet by 30 feet. Yes, the lot is big enough for the court Vincent planned. The court will take up the entire width of the lot.

Example 2
The diagram shown represents a garden. The scale is 1 centimeter for every 20 meters. Each square in the drawing measures 1 cm by 1 cm. Find the actual length and width of the garden based upon the given drawing.

Answer:

Method 1:
Using the given scale: 1 cm of scale drawing length corresponds to 20 m of actual length.
k = 0 Drawing length to actual length
To find the actual length: y = 20x Where x represents the the scale drawing measurements in centimeters, and y represents the actual measurement in meters.
y = 20(8) Substitute the scale drawing length in place of x.
y = 160
The actual length is 80 m.
To find actual width: Divide the actual length by 2 since its drawing width is half the length.
The actual width is 80 m.

Method 2:
Use the scale factor: 1 cm of scale drawing length corresponds to 2000 cm of actual length.
k = 2000 Drawing length to actual length (in same units)
To find actual length: y = 2000x Where x represents the drawing measurement in centimeters, and
y represents the actual measurement in centimeters.
y = 2000(8) Substitute the scale drawing length in place of x.
y = 16000
The actual length is 16,000 cm, or 160 m.

To find actual width: y = 2000x
y = 2000(4) Substitute the scale drawing width in place of x.
y = 8000

Example 3
A graphic designer is creating an advertisement for a tablet. She needs to enlarge the picture given here so that 0.25 inches on the scale picture corresponds to 1 inch on the actual advertisement. What will be the length and width of the tablet on the advertisement?

Answer:

Using an Equation:
Find the constant of proportionality, k: k = 4
k = 4 (Scale factor since units of measure are the same; it is an enlargement.)
To find Actual Length: y = 4x Where x represents the picture measurement, and y represents the
actual advertisement measurement.
y = 4(1 \(\frac{1}{4}\)) Substitute the picture length in place of x.
y = 5
To find Actual Width: y = 4x
y = 4(1\(\frac{1}{8}\)) Substitute the picture width in place of y.
y = 4\(\frac{1}{2}\)
The tablet will be 5 inches by 4\(\frac{1}{2}\) inches on the actual advertisement.

Eureka Math Grade 7 Module 1 Lesson 18 Exercise Answer Key

Students from the high school are going to perform one of the acts from their upcoming musical at the atrium in the mall. The students want to bring some of the set with them so that the audience can get a better feel for the whole production. The backdrop that they want to bring has panels that measure 10 feet by 10 feet. The students are not sure if they will be able to fit these panels through the entrance of the mall since the panels need to be transported flat (horizontal). They obtain a copy of the mall floor plan, shown below, from the city planning office. Use this diagram to decide if the panels will fit through the entrance. Use a ruler to measure.

Answer the following questions.

a. Find the actual distance of the mall entrance, and determine whether the set panels will fit.
Answer:
Step 1: Relationship between lengths in drawing and lengths in actual
Scale:  or the value of the ratio \(\frac{36}{1}\) feet to inches
Scale factor calculations: inches to inches

= 432, an enlargement
Step 2: Find the actual distance of the entrance.
Using the given scale: \(\frac{3}{8}\)∙\(\frac{36}{1}\) = 13\(\frac{1}{2}\)
The actual distance of the entrance is 13 \(\frac{1}{2}\) feet wide.
OR
Using the scale factor: \(\frac{3}{8}\)∙\(\frac{432}{1}\) = 162
The actual distance of the entrance is 162 inches, or 13\(\frac{1}{2}\) feet, wide.
Yes, the set panels, which are 10 ft. ×10 ft. , will fit (lying flat) through the mall entrance.

b. What is the scale factor? What does it tell us?
Answer:
The scale factor is 432. Each length on the scale drawing is \(\frac{1}{432}\) of the actual length. The actual lengths are 432 times larger than the lengths in the scale drawing.

Eureka Math Grade 7 Module 1 Lesson 18 Problem Set Answer Key

Question 1.
A toy company is redesigning its packaging for model cars. The graphic design team needs to take the old image shown below and resize it so that \(\frac{1}{2}\) inch on the old packaging represents \(\frac{1}{3}\) inch on the new package. Find the length of the image on the new package.

Answer:
\(\frac{4}{3}\) inches; the scale \(\frac{1}{2}\) to \(\frac{1}{3}\) and the length of the original figure is 2, which is 4 halves, so in the scale drawing the length will be 4 thirds.

Question 2.
The city of St. Louis is creating a welcome sign on a billboard for visitors to see as they enter the city. The following picture needs to be enlarged so that \(\frac{1}{2}\) inch represents 7 feet on the actual billboard. Will it fit on a billboard that measures 14 feet in height?

Answer:
Yes, the drawing measures 1 inch in height, which corresponds to 14 feet on the actual billboard.

Question 3.
Your mom is repainting your younger brother’s room. She is going to project the image shown below onto his wall so that she can paint an enlarged version as a mural. Use a ruler to determine the length of the image of the train. Then determine how long the mural will be if the projector uses a scale where 1 inch of the image represents 2 \(\frac{1}{2}\) feet on the wall.

Answer:
The scale drawing measures 2 inches, so the image will measure 2 × 2.5, or 5 feet long, on the wall.

Question 4.
A model of a skyscraper is made so that 1 inch represents 75 feet. What is the height of the actual building if the height of the model is 18 \(\frac{3}{5}\) inches?
Answer:
1,395 feet

Question 5.
The portrait company that takes little league baseball team photos is offering an option where a portrait of your baseball pose can be enlarged to be used as a wall decal (sticker). Your height in the portrait measures 3 \(\frac{1}{2}\) inches. If the company uses a scale where 1 inch on the portrait represents 20 inches on the wall decal, find the height on the wall decal. Your actual height is 55 inches. If you stand next to the wall decal, will it be larger or smaller than you?
Answer:
Your height on the wall decal is 70 inches. The wall decal will be larger than your actual height (when you stand next to it).

Question 6.
The sponsor of a 5K run/walk for charity wishes to create a stamp of its billboard to commemorate the event. If the sponsor uses a scale where 1 inch represents 4 feet, and the billboard is a rectangle with a width of 14 feet and a length of 48 feet, what will be the shape and size of the stamp?
Answer:
The stamp will be a rectangle measuring 3 \(\frac{1}{2}\) inches by 12 inches.

Question 7.
Danielle is creating a scale drawing of her room. The rectangular room measures 20 \(\frac{1}{2}\) ft. by 25 ft. If her drawing uses the scale where 1 inch represents 2 feet of the actual room, will her drawing fit on an 8 \(\frac{1}{2}\) in. by 11 in. piece of paper?
Answer:
No, the drawing would be 10\(\frac{1}{4}\) inches by 12 \(\frac{1}{2}\) inches, which is larger than the piece of paper.

Question 8.
A model of an apartment is shown below where \(\frac{1}{4}\) inch represents 4 feet in the actual apartment. Use a ruler to measure the drawing and find the actual length and width of the bedroom.

Answer:
Ruler measurements: 1\(\frac{1}{8}\) inches by \(\frac{9}{16}\) inches.
The actual length would be 18 feet, and the actual width would be 9 feet.

Eureka Math Grade 7 Module 1 Lesson 18 Exit Ticket Answer Key

A drawing of a surfboard in a catalog shows its length as 8\(\frac{4}{9}\) inches. Find the actual length of the surfboard if \(\frac{1}{2}\) inch length on the drawing corresponds to \(\frac{3}{8}\) foot of actual length.
Answer:

y = kx
y = \(\frac{3}{4}\) x
=8 \(\frac{4}{9}\) ∙\(\frac{3}{4}\)
= \(\frac{76}{9}\) ∙\(\frac{3}{4}\)
= \(\frac{19}{3}\) ∙\(\frac{1}{1}\)
The actual surfboard measures 6 \(\frac{1}{3}\) feet long.

Note: Students could also use an equation where y represents the scale drawing, and x represents the actual measurement, in which case, k would equal \(\frac{4}{3}\) .