Eureka Math Grade 8 Module 7 Lesson 6 Answer Key

Engage NY Eureka Math 8th Grade Module 7 Lesson 6 Answer Key

Eureka Math Grade 8 Module 7 Lesson 6 Example Answer Key

Example 1.
Consider the fraction \(\frac{5}{8}\). Write an equivalent form of this fraction with a denominator that is a power of 10, and write the decimal expansion of this fraction.
Answer:
Consider the fraction \(\frac{5}{8}\). Is it equivalent to one with a denominator that is a power of 10? How do you know?
Yes. The fraction 5/8 has denominator 8 and so has factors that are products of 2’s only.
Write \(\frac{5}{8}\)as an equivalent fraction with a denominator that is a power of 10.

We have \(\frac{5}{8}\) = \(\frac{5}{2 \times 2 \times 2} = \frac{5 \times 5 \times 5 \times 5}{2 \times 2 \times 2 \times 5 \times 5 \times 5} = \frac{625}{10 \times 10 \times 10} = \frac{625}{10^{3}}\)
What is \(\frac{5}{8}\) as a finite decimal?
\(\frac{5}{8}\) = \(\frac{625}{1000}\) = 0.625

Example 2.
Consider the fraction \(\frac{17}{125}\). Is it equal to a finite or an infinite decimal? How do you know?
Answer:
→ Let’s consider the fraction \(\frac{17}{125}\). We want the decimal value of this number. Will it be a finite or an infinite decimal? How do you know?
→ We know that the fraction \(\frac{17}{125}\) is equal to a finite decimal because the denominator 125 is a product of 5’s, specifically, 5³, and so we can write the fraction as one with a denominator that is a power of 10.

→ What will we need to multiply 5³ by to obtain a power of 10?
We will need to multiply by 2³. Then, 5³ × 2³ = (5 × 2)³ = 〖10〗³.
→ Write \(\frac{17}{125}\) or its equivalent 17/5³ as a finite decimal.
\(\frac{17}{125}\) = \(\frac{17}{5^{3}} = \frac{17 \times 2^{3}}{5^{3} \times 2^{3}} = \frac{17 \times 8}{(5 \times 2)^{3}} = \frac{136}{10^{3}}\) = 0.136
(If the above two points are too challenging for some students, have them write out:
\(\frac{17}{125}\) = \(\frac{17}{5 \times 5 \times 5} = \frac{17 \times 2 \times 2 \times 2}{5 \times 5 \times 5 \times 2 \times 2 \times 2}\) = \(\frac{136}{1000}\) = 0.136.)

Example 3.
Will the decimal expansion of \(\frac{7}{80}\) be finite or infinite? If it is finite, find it.
Answer:
→ Will \(\frac{7}{80}\) have a finite or infinite decimal expansion?
We know that the fraction \(\frac{7}{80}\) is equal to a finite decimal because the denominator 80 is a product of 2’s and 5’s. Specifically, 2⁴ × 5. This means the fraction is equivalent to one with a denominator that is a power of 10.

→ What will we need to multiply 2⁴ × 5 by so that it is equal to (2 × 5)ⁿ = 10ⁿ for some n?
We will need to multiply by 5³ so that 2⁴ × 5⁴ = (2 × 5)⁴ = 10⁴.

→ Begin with\(\frac{7}{80}\) or \(\frac{7}{2^{4} \times 5}\). Use what you know about equivalent fractions to rewrite \(\frac{7}{80}\) in the form \(\frac{k}{10^{n}}\) and then write the decimal form of the fraction.
\(\frac{7}{80}\) = \(\frac{7}{2^{4} \times 5} = \frac{7 \times 5^{3}}{2^{4} \times 5 \times 5^{3}} = \frac{7 \times 125}{(2 \times 5)^{4}} = \frac{875}{10^{4}}\) = 0.0875

Example 4.
Will the decimal expansion of \(\frac{3}{160}\) be finite or infinite? If it is finite, find it.
Answer:
→ Will \(\frac{3}{160}\) have a finite or infinite decimal expansion?
We know that the fraction \(\frac{3}{160}\) is equal to a finite decimal because the denominator 160 is a product of 2’s and 5’s. Specifically, 2⁵ × 5. This means the fraction is equivalent to one with a denominator that is a power of 10.

→ What will we need to multiply 2⁵ × 5 by so that it is equal to (2 × 5)ⁿ = 10ⁿ for some n?
We will need to multiply by 5⁴ so that 2⁵ × 5⁵ = (2 × 5)⁵ = 10⁵.
Begin with \(\frac{3}{160}\) or \(\frac{3}{2^{5} \times 5}\). Use what you know about equivalent fractions to rewrite \(\frac{3}{160}\) in the form \(\frac{k}{10^{n}}\) and then write the decimal form of the fraction.
\(\frac{3}{160}\) = \(\frac{3}{2^{5} \times 5}=\frac{3 \times 5^{4}}{2^{5} \times 5 \times 5^{4}}=\frac{3 \times 625}{(2 \times 5)^{5}}=\frac{1875}{10^{5}}\) = 0.01875

Eureka Math Grade 8 Module 7 Lesson 6 Exercise Answer Key

Opening Exercise
a. Use long division to determine the decimal expansion of \(\frac{54}{20}\).
Answer:
\(\frac{54}{20}\) = 2.7

b. Use long division to determine the decimal expansion of \(\frac{7}{8}\).
Answer:
\(\frac{7}{8}\) = 0.875

c. Use long division to determine the decimal expansion of \(\frac{8}{9}\).
Answer:
\(\frac{8}{9}\) = 0.8888 “…”

d. Use long division to determine the decimal expansion of \(\frac{22}{7}\).
Answer:
\(\frac{22}{7}\) = 3.142857 “…”

e. What do you notice about the decimal expansions of parts (a) and (b) compared to the decimal expansions of parts (c) and (d)?
Answer:
The decimal expansions of parts (a) and (b) ended. That is, when I did the long division, I was able to stop after a few steps. That was different from the work I had to do in parts (c) and (d). In part (c), I noticed that the same number kept coming up in the steps of the division, but it kept going on. In part (d), when I did the long division, it did not end. I stopped dividing after I found a few decimal digits of the decimal expansion.

Exercises 1–5
You may use a calculator, but show your steps for each problem.

Exercise 1.
Consider the fraction \(\frac{3}{8}\) .
a. Write the denominator as a product of 2’s and/or 5’s. Explain why this way of rewriting the denominator helps to find the decimal representation of \(\frac{3}{8}\) .
Answer:
The denominator is equal to 2³. It is helpful to know that 8 = 2³ because it shows how many factors of 5 will be needed to multiply the numerator and denominator by so that an equivalent fraction is produced with a denominator that is a multiple of 10. When the denominator is a multiple of 10, the fraction can easily be written as a decimal using what I know about place value.

b. Find the decimal representation of \(\frac{3}{8}\). Explain why your answer is reasonable.
Answer:
\(\frac{3}{8}\) = \(\frac{3}{2^{3}}=\frac{3 \times 5^{3}}{2^{3} \times 5^{3}}=\frac{375}{10^{3}}\) = 0.375
The answer is reasonable because the decimal value, 0.375, is less than \(\frac{1}{2}\), just like the fraction \(\frac{3}{8}\).

Exercise 2.
Find the first four places of the decimal expansion of the fraction \(\frac{43}{64}\).
Answer:
The denominator is equal to 2⁶.
\(\frac{43}{64}\) = \(\frac{43}{2^{6}}=\frac{43 \times 5^{6}}{2^{6} \times 5^{6}}=\frac{671875}{10^{6}}\) = 0.671875
The decimal expansion to the first four decimal places is 0.6718.

Exercise 3.
Find the first four places of the decimal expansion of the fraction \(\frac{29}{125}\).
Answer:
The denominator is equal to 5³.
\(\frac{29}{125}\) = \(\frac{29}{5^{3}} = \frac{29 \times 2^{3}}{5^{3} \times 2^{3}} = \frac{232}{10^{3}}\) = 0.232
The decimal expansion to the first four decimal places is 0.2320.

Exercise 4.
Find the first four decimal places of the decimal expansion of the fraction \(\frac{19}{34}\).
Answer:
Using long division, the decimal expansion to the first four places is 0.5588….

Exercise 5.
Identify the type of decimal expansion for each of the numbers in Exercises 1–4 as finite or infinite. Explain why their decimal expansion is such.
We know that the number \(\frac{7}{8}\) had a finite decimal expansion because the denominator 8 is a product of 2’s and so is equivalent to a fraction with a denominator that is a power of 10. We know that the number \(\frac{43}{64}\) had a finite decimal expansion because the denominator 64 is a product of 2’s and so is equivalent to a fraction with a denominator that is a power of 10. We know that the number \(\frac{29}{125}\) had a finite decimal expansion because the denominator 125 is a product of 5’s and so is equivalent to a fraction with a denominator that is a power of 10. We know that the number \(\frac{19}{34}\) had an infinite decimal expansion because the denominator was not a product of 2’s or 5’s; it had a factor of 17 and so is not equivalent to a fraction with a denominator that is a power of 10.

Exercises 6–8
You may use a calculator, but show your steps for each problem.

Exercise 6.
Convert the fraction \(\frac{37}{40}\) to a decimal.
a. Write the denominator as a product of 2’s and/or 5’s. Explain why this way of rewriting the denominator helps to find the decimal representation of \(\frac{37}{40}\).
Answer:
The denominator is equal to 2³ × 5. It is helpful to know that 40 is equal to 2³ × 5 because it shows by how many factors of 5 the numerator and denominator will need to be multiplied to produce an equivalent fraction with a denominator that is a power of 10. When the denominator is a power of 10, the fraction can easily be written as a decimal using what I know about place value.

b. Find the decimal representation of \(\frac{37}{40}\). Explain why your answer is reasonable.
Answer:
\(\frac{37}{40}\) = \(\frac{37}{2^{3} \times 5} = \frac{37 \times 5^{2}}{2^{3} \times 5 \times 5^{2}} = \frac{925}{10^{3}}\) = 0.925
The answer is reasonable because the decimal value, 0.925, is less than 1, just like the fraction 37/40. Also, it is reasonable and correct because the fraction \(\frac{925}{1000}\) = \(\frac{37}{40}\); therefore, it has the decimal expansion 0.925.

Exercise 7.
Convert the fraction \(\frac{3}{250}\) to a decimal.
Answer:
The denominator is equal to 2 × 5³.
\(\frac{3}{250}\) = \(\frac{3}{2 \times 5^{3}} = \frac{3 \times 2^{2}}{2 \times 2^{2} \times 5^{3}} = \frac{12}{10^{3}}\) = 0.012

Exercise 8.
Convert the fraction \(\frac{7}{1250}\) to a decimal.
Answer:
The denominator is equal to 2 × 5⁴.
\(\frac{7}{1250}\) = \(\frac{7}{2 \times 5^{4}} = \frac{7 \times 2^{3}}{2 \times 2^{3} \times 5^{4}} = \frac{56}{10^{4}}\) = 0.0056

Eureka Math Grade 8 Module 7 Lesson 6 Problem Set Answer Key

Convert each fraction given to a finite decimal, if possible. If the fraction cannot be written as a finite decimal, then state how you know. You may use a calculator, but show your steps for each problem.
Question 1.
\(\frac{2}{32}\)
The fraction \(\frac{2}{32}\) simplifies to \(\frac{1}{16}\).
The denominator is equal to 2⁴.
\(\frac{1}{16}\) = \(\frac{1}{2^{4}} = \frac{1 \times 5^{4}}{2^{4} \times 5^{4}} = \frac{625}{10^{4}}\) = 0.0625

Question 2.
\(\frac{99}{125}\)
Answer:
The denominator is equal to 5³.
\(\frac{99}{125}\) = \(\frac{99}{125}\) = 0.792

Question 3.
\(\frac{15}{128}\)
The denominator is equal to 2⁷.
\(\frac{15}{128}\) = \(\frac{15}{128}\) = 0.1171875

Question 4.
\(\frac{8}{15}\)
Answer:
The fraction \(\frac{8}{15}\) is not a finite decimal because the denominator is equal to 3 × 5. Since the denominator cannot be expressed as a product of 2’s and 5’s, then \(\frac{8}{15}\) is not a finite decimal.

Question 5.
\(\frac{3}{28}\)
Answer:
The fraction \(\frac{3}{28}\) is not a finite decimal because the denominator is equal to 2² × 7. Since the denominator cannot be expressed as a product of 2’s and 5’s, then \(\frac{3}{28}\) is not a finite decimal.

Question 6.
\(\frac{13}{400}\)
Answer:
The denominator is equal to 2⁴ × 5².
\(\frac{13}{400}\) = \(\frac{13}{2^{4} \times 5^{2}} = \frac{13 \times 5^{2}}{2^{4} \times 5^{2} \times 5^{2}} = \frac{325}{10^{4}}\) = 0.0325

Question 7.
\(\frac{5}{64}\)
Answer:
The denominator is equal to 2⁶.
\(\frac{5}{64}\) = \(\frac{5}{2^{6}} = \frac{5 \times 5^{6}}{2^{6} \times 5^{6}} = \frac{78125}{10^{6}}\) = 0.078125

Question 8.
\(\frac{15}{35}\)
Answer:
The fraction \(\frac{15}{35}\) reduces to \(\frac{3}{7}\). The denominator 7 cannot be expressed as a product of 2’s and 5’s. Therefore, \(\frac{3}{7}\) is not a finite decimal.

Question 9.
\(\frac{199}{250}\)
Answer:
The denominator is equal to 2 × 5³.
\(\frac{199}{250}\) = \(\frac{199}{2 \times 5^{3}} = \frac{199 \times 2^{2}}{2 \times 2^{2} \times 5^{3}} = \frac{796}{10^{3}}\) = 0.796

Question 10.
\(\frac{219}{625}\)
Answer:
The denominator is equal to 5⁴.
\(\frac{219}{625}\) = \(\frac{219}{5^{4}} = \frac{219 \times 2^{4}}{2^{4} \times 5^{4}} = \frac{3504}{10^{4}}\) = 0.3504

Eureka Math Grade 8 Module 7 Lesson 6 Exit Ticket Answer Key

Convert each fraction to a finite decimal if possible. If the fraction cannot be written as a finite decimal, then state how you know. You may use a calculator, but show your steps for each problem.
Question 1.
\(\frac{9}{16}\)
Answer:
The denominator is equal to 2⁴.
\(\frac{9}{16}\) = \(\frac{9}{2^{4}} = \frac{9 \times 5^{4}}{2^{4} \times 5^{4}} = \frac{9 \times 625}{10^{4}} = \frac{5625}{10^{4}}\) = 0.5625

Question 2.
\(\frac{8}{125}\)
Answer:
The denominator is equal to 5³.
\(\frac{8}{125}\) = \(\frac{8}{5^{3}} = \frac{8 \times 2^{3}}{5^{3} \times 2^{3}} = \frac{8 \times 8}{10^{3}} = \frac{64}{10^{3}}\) = 0.064

Question 3.
\(\frac{4}{15}\)
Answer:
The fraction \(\frac{4}{15}\) is not a finite decimal because the denominator is equal to 5 × 3. Since the denominator cannot be expressed as a product of 2’s and 5’s, then \(\frac{4}{15}\) is not a finite decimal.

Question 4.
\(\frac{1}{200}\)
Answer:
The denominator is equal to 2³ × 5².
\(\frac{1}{200}\) = \(\frac{1}{2^{3} \times 5^{2}} = \frac{1 \times 5}{2^{3} \times 5^{2} \times 5} = \frac{5}{2^{3} \times 5^{3}} = \frac{5}{10^{3}}\) = 0.005