## Engage NY Eureka Math 5th Grade Module 3 Lesson 6 Answer Key

### Eureka Math Grade 5 Module 3 Lesson 6 Problem Set Answer Key

Question 1.

For the following problems, draw a picture using the rectangular fraction model and write the answer. Simplify your answer, if possible.

a. 1\(\frac{1}{4}\) – \(\frac{1}{3}\) =

b. 1\(\frac{1}{5}\) – \(\frac{1}{3}\) =

c. 1\(\frac{3}{8}\) – \(\frac{1}{2}\) =

d. 1\(\frac{2}{5}\) – \(\frac{1}{2}\) =

e. 1\(\frac{2}{7}\) – \(\frac{1}{3}\) =

f. 1\(\frac{2}{3}\) – \(\frac{3}{5}\) =

Answer:

a.

1\(\frac{1}{4}\) – \(\frac{1}{3}\) = \(\frac{5}{4}\) – \(\frac{1}{3}\)

lcm of 4 and 3 is 12 .

\(\frac{15}{12}\) – \(\frac{4}{12}\) = \(\frac{11}{12}\).

Explanation :

The Rectangle is divided into 4 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{1}{4}\) .

1\(\frac{1}{4}\) andÂ \(\frac{1}{3}\) have lcm 12 so, the rectangle is divided into 12 parts by drawing 2 horizontal lines . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts represent the difference of the given fraction to its total parts .

b.

1\(\frac{1}{5}\) – \(\frac{1}{3}\) = \(\frac{6}{5}\) – \(\frac{1}{3}\)

lcm of 5 and 3 is 15 .

\(\frac{18}{15}\) – \(\frac{5}{15}\) = \(\frac{13}{15}\)

Explanation :

The Rectangle is divided into 5 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{1}{5}\) .

1\(\frac{1}{5}\) andÂ \(\frac{1}{3}\) have lcm 15 so, the rectangle is divided into 15 parts by drawing 2 horizontal lines . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts represent the difference of the given fraction to its total parts .

c.

1\(\frac{3}{8}\) – \(\frac{1}{2}\) = \(\frac{11}{8}\) – \(\frac{1}{2}\)

lcm of 8 and 2 is 8 .

\(\frac{11}{8}\) – \(\frac{4}{8}\) =\(\frac{7}{8}\) .

Explanation :

The Rectangle is divided into 8 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{3}{8}\) .

1\(\frac{3}{8}\) andÂ \(\frac{1}{2}\) have lcm 8 so, the rectangle is divided into 8 parts . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts represent the difference of the given fraction to its total parts .

d.

1\(\frac{2}{5}\) – \(\frac{1}{2}\) = \(\frac{7}{5}\) – \(\frac{1}{2}\)

lcm of 5 and 2 is 10.

\(\frac{14}{10}\) – \(\frac{5}{10}\) = \(\frac{9}{10}\)

Explanation :

The Rectangle is divided into 5 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{2}{5}\) .

1\(\frac{2}{5}\) andÂ \(\frac{1}{2}\) have lcm 10 so, the rectangle is divided into 10 parts by drawing 1 horizontal lines . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts represent the difference of the given fraction to its total parts .

e.

1\(\frac{2}{7}\) – \(\frac{1}{3}\) = \(\frac{9}{7}\) – \(\frac{1}{3}\)

lcm of 7 and 3 is 21 .

\(\frac{27}{21}\) – \(\frac{7}{21}\) = \(\frac{20}{21}\) .

Explanation :

The Rectangle is divided into 7 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{2}{7}\) .

1\(\frac{2}{7}\) andÂ \(\frac{1}{3}\) have lcm 21 so, the rectangle is divided into 21 parts by drawing 2 horizontal lines . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts represent the difference of the given fraction to its total parts .

f.

1\(\frac{2}{3}\) – \(\frac{3}{5}\) = \(\frac{5}{3}\) – \(\frac{3}{5}\)

lcm of 3 and 5 is 15 .

\(\frac{25}{15}\) – \(\frac{9}{15}\) = \(\frac{16}{15}\)

Explanation :

The Rectangle is divided into 5 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{2}{3}\) .

1\(\frac{2}{3}\) andÂ \(\frac{3}{5}\) have lcm 15 so, the rectangle is divided into 15 parts by drawing 4 horizontal lines . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts represent the difference of the given fraction to its total parts .

Question 2.

Jean-Luc jogged around the lake in 1\(\frac{1}{4}\) hour. William jogged the same distance in \(\frac{5}{6}\) hour. How much longer did Jean-Luc take than William in hours?

Answer:

Time taken to jog for Jean-Luc = 1\(\frac{1}{4}\) hour .

Time taken to jog for William = \(\frac{5}{6}\) hour .

Time taken by Jean-Luc than William = 1\(\frac{1}{4}\) – \(\frac{5}{6}\)Â = \(\frac{5}{4}\) – \(\frac{5}{6}\)Â = \(\frac{30}{24}\)Â – \(\frac{20}{24}\) = \(\frac{10}{24}\) = \(\frac{5}{12}\) hour .

Question 3.

Is it true that 1\(\frac{2}{5}\) – \(\frac{3}{4}\) = \(\frac{1}{4}\) + \(\frac{2}{5}\)? Prove your answer.

Answer:

Yes , 1\(\frac{2}{5}\) – \(\frac{3}{4}\) = \(\frac{1}{4}\) + \(\frac{2}{5}\)

Explanation :

1\(\frac{2}{5}\) – \(\frac{3}{4}\) = \(\frac{7}{5}\) – \(\frac{3}{4}\) = \(\frac{28}{20}\) – \(\frac{15}{20}\) = \(\frac{13}{20}\) .

\(\frac{1}{4}\) + \(\frac{2}{5}\) = \(\frac{5}{20}\) + \(\frac{8}{20}\) = \(\frac{13}{20}\)

Therefore, 1\(\frac{2}{5}\) – \(\frac{3}{4}\) = \(\frac{1}{4}\) + \(\frac{2}{5}\) = \(\frac{13}{20}\)Â .

### Eureka Math Grade 5 Module 3 Lesson 6 Exit Ticket Answer Key

For the following problems, draw a picture using the rectangular fraction model and write the answer. Simplify your answer, if possible.

a. 1\(\frac{1}{5}\) – \(\frac{1}{2}\) =

b. 1\(\frac{1}{3}\) – \(\frac{5}{6}\) =

Answer:

a.

1\(\frac{1}{5}\) – \(\frac{1}{2}\) = \(\frac{6}{5}\) – \(\frac{1}{2}\)

lcm of 5 and 2 is 10 .

\(\frac{12}{10}\) – \(\frac{5}{10}\) = \(\frac{7}{10}\)

Explanation :

The Rectangle is divided into 5 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{2}{5}\) .

1\(\frac{2}{5}\) andÂ \(\frac{1}{2}\) have lcm 10 so, the rectangle is divided into 10 parts by drawing 1 horizontal line . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts represent the difference of the given fraction to its total parts .

b.

1\(\frac{1}{3}\) – \(\frac{5}{6}\) = \(\frac{4}{3}\) – \(\frac{5}{6}\)

lcm of 3 and 6 isÂ 6 .

\(\frac{8}{6}\) – \(\frac{5}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Explanation :

The Rectangle is divided into 3 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{1}{3}\) .

1\(\frac{1}{3}\) andÂ \(\frac{5}{6}\) have lcm 6 so, the rectangle is divided into 6 parts by drawing 2 horizontal lines . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts represent the difference of the given fraction to its total parts .

### Eureka Math Grade 5 Module 3 Lesson 6 Homework Answer Key

Question 1.

For the following problems, draw a picture using the rectangular fraction model and write the answer. Simplify your answer, if possible.

a. 1 – \(\frac{5}{6}\) =

b. \(\frac{3}{2}\) – \(\frac{5}{6}\) =

c. \(\frac{4}{3}\) – \(\frac{5}{7}\) =

d. 1\(\frac{1}{8}\) – \(\frac{3}{5}\) =

e. 1\(\frac{2}{5}\) – \(\frac{3}{4}\) =

f. 1\(\frac{5}{6}\) – \(\frac{7}{8}\) =

g. \(\frac{9}{7}\) – \(\frac{3}{4}\) =

h. 1\(\frac{3}{12}\) – \(\frac{2}{3}\) =

Answer:

a.

1 – \(\frac{5}{6}\)

lcm of 1 and 6 is 6 .

\(\frac{6}{6}\) – \(\frac{5}{6}\) = \(\frac{1}{6}\) .

Explanation :

The Rectangle is divided into 6 parts using vertical lines and shaded to represent the fraction \(\frac{6}{6}\) .

1 andÂ \(\frac{5}{6}\) have lcm 6 . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts without x marks represent the difference of the given fraction to its total parts .

b.

\(\frac{3}{2}\) – \(\frac{5}{6}\)

lcm of 2 and 6 isÂ 6 .

\(\frac{9}{6}\) – \(\frac{5}{6}\) = \(\frac{4}{6}\) = \(\frac{2}{3}\)

Explanation :

The Rectangle is divided into 2 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{1}{2}\) .

1\(\frac{1}{2}\) andÂ \(\frac{5}{6}\) have lcm 6 so, the rectangle is divided into 6 parts by drawing 2 horizontal lines . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts without x marks represent the difference of the given fraction to its total parts .

c.

\(\frac{4}{3}\) – \(\frac{5}{7}\)

lcm of 3 and 7 is 21 .

\(\frac{28}{21}\) – \(\frac{15}{21}\) = \(\frac{13}{21}\)

Explanation :

The Rectangle is divided into 3 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{1}{3}\) .

1\(\frac{1}{3}\) andÂ \(\frac{5}{7}\) have lcm 21 so, the rectangle is divided into 21 parts by drawing 6 horizontal lines . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts without x marks represent the difference of the given fraction to its total parts .

d.

1\(\frac{1}{8}\) – \(\frac{3}{5}\) = \(\frac{9}{8}\) – \(\frac{3}{5}\)

lcm of 8 and 5 is 40

\(\frac{45}{40}\) – \(\frac{24}{40}\) = \(\frac{21}{40}\)

Explanation :

The Rectangle is divided into 8 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{1}{8}\) .

1\(\frac{1}{8}\) andÂ \(\frac{3}{5}\) have lcm 40 so, the rectangle is divided into 40 parts by drawing 4 horizontal lines . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts without x marks represent the difference of the given fraction to its total parts .

e.

1\(\frac{2}{5}\) – \(\frac{3}{4}\) = \(\frac{7}{5}\) – \(\frac{3}{4}\)

lcm of 5 and 4 is 20 .

\(\frac{28}{20}\) – \(\frac{15}{20}\) = \(\frac{13}{20}\) .

Explanation :

The Rectangle is divided into 5 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{2}{5}\) .

1\(\frac{2}{5}\) andÂ \(\frac{3}{4}\) have lcm 20 so, the rectangle is divided into 20 parts by drawing 3 horizontal lines . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts without x marks represent the difference of the given fraction to its total parts .

f.

1\(\frac{5}{6}\) – \(\frac{7}{8}\) = \(\frac{11}{6}\) – \(\frac{7}{8}\)

lcm of 6 and 8 is 24 .

\(\frac{44}{24}\) – \(\frac{21}{24}\) = \(\frac{23}{24}\) .

Explanation :

The Rectangle is divided into 6 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{5}{6}\) .

1\(\frac{5}{6}\) andÂ \(\frac{7}{8}\) have lcm 24 so, the rectangle is divided into 24 parts by drawing 3 horizontal lines . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts without x marks represent the difference of the given fraction to its total parts .

g.

\(\frac{9}{7}\) – \(\frac{3}{4}\)

lcm of 7 and 4 is 28 .

\(\frac{36}{28}\) – \(\frac{21}{28}\) = \(\frac{15}{28}\)

Explanation :

The Rectangle is divided into 7 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{2}{7}\) .

1\(\frac{2}{7}\) andÂ \(\frac{3}{4}\) have lcm 28 so, the rectangle is divided into 28 parts by drawing 3 horizontal lines . after making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts without x marks represent the difference of the given fractions to its total parts .

h.

1\(\frac{3}{12}\) – \(\frac{2}{3}\) = \(\frac{15}{12}\) – \(\frac{2}{3}\)

lcm of 12 and 3 is 12 .

\(\frac{15}{12}\) – \(\frac{8}{12}\) = \(\frac{7}{12}\)

Explanation :

The Rectangle is divided into 12 parts using vertical lines and 2 rectangles are drawn and shaded to represent the fraction 1\(\frac{3}{12}\) .

1\(\frac{3}{12}\) andÂ \(\frac{2}{3}\) have lcm 12 . After making lcm equal then subtraction is done and subtraction is done by showing x-mark on the shaded parts.

The left over shaded parts without x marks represent the difference of the given fractions to its total parts .

Question 2.

Sam had 1\(\frac{1}{2}\) m of rope. He cut off \(\frac{5}{8}\) m and used it for a project. How much rope does Sam have left?

Answer:

Length of Rope with Sam = 1\(\frac{1}{2}\) m

Length of rope used for project = \(\frac{5}{8}\) m

Length of Rope left = 1\(\frac{1}{2}\) – \(\frac{5}{8}\) = \(\frac{3}{2}\) – \(\frac{5}{8}\) =

\(\frac{12}{8}\) – \(\frac{5}{8}\) = \(\frac{7}{8}\) .

Therefore, Length of rope left with sam = \(\frac{7}{8}\) m .

Question 3.

Jackson had 1\(\frac{3}{8}\) kg of fertilizer. He used some to fertilize a flower bed, and he only had \(\frac{2}{3}\) kg left. How much fertilizer was used in the flower bed?

Answer:

Quantity of fertilizers with Jackson = 1\(\frac{3}{8}\) kg

Quantity of fertilizers left = \(\frac{2}{3}\) kg .

Quantity of Fertilizers used for flower bedÂ = 1\(\frac{3}{8}\) kgÂ – \(\frac{2}{3}\) kg = \(\frac{11}{8}\) – \(\frac{2}{3}\) = \(\frac{33}{24}\) – \(\frac{16}{24}\) = \(\frac{17}{24}\) .

Therefore, Quantity of Fertilizers used for flower bedÂ =\(\frac{17}{24}\)