## 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$$\frac{2}{3}$$ – $$\frac{3}{5}$$ =
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$$\frac{1}{4}$$ and  $$\frac{1}{3}$$ 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$$\frac{1}{5}$$ and  $$\frac{1}{3}$$ 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$$\frac{3}{8}$$ and  $$\frac{1}{2}$$ 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$$\frac{2}{5}$$ and  $$\frac{1}{2}$$ 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$$\frac{2}{7}$$ and  $$\frac{1}{3}$$ 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$$\frac{2}{3}$$ and  $$\frac{3}{5}$$ 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?
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$$\frac{1}{4}$$ – $$\frac{5}{6}$$  = $$\frac{5}{4}$$ – $$\frac{5}{6}$$  = $$\frac{30}{24}$$  – $$\frac{20}{24}$$ = $$\frac{10}{24}$$ = $$\frac{5}{12}$$ hour .

Question 3.
Is it true that 1$$\frac{2}{5}$$ – $$\frac{3}{4}$$ = $$\frac{1}{4}$$ + $$\frac{2}{5}$$? Prove your answer.
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$$\frac{2}{5}$$ – $$\frac{3}{4}$$ = $$\frac{1}{4}$$ + $$\frac{2}{5}$$ = $$\frac{13}{20}$$  .

### 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$$\frac{1}{3}$$ – $$\frac{5}{6}$$ =
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$$\frac{2}{5}$$ and  $$\frac{1}{2}$$ 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 is  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$$\frac{1}{3}$$ and  $$\frac{5}{6}$$ 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$$\frac{3}{12}$$ – $$\frac{2}{3}$$ =
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 and  $$\frac{5}{6}$$ 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 is  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$$\frac{1}{2}$$ and  $$\frac{5}{6}$$ 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$$\frac{1}{3}$$ and  $$\frac{5}{7}$$ 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$$\frac{1}{8}$$ and  $$\frac{3}{5}$$ 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$$\frac{2}{5}$$ and  $$\frac{3}{4}$$ 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$$\frac{5}{6}$$ and  $$\frac{7}{8}$$ 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$$\frac{2}{7}$$ and  $$\frac{3}{4}$$ 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$$\frac{3}{12}$$ and  $$\frac{2}{3}$$ 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?
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?
Quantity of fertilizers with Jackson = 1$$\frac{3}{8}$$ kg
Quantity of fertilizers left = $$\frac{2}{3}$$ kg .
Quantity of Fertilizers used for flower bed  = 1$$\frac{3}{8}$$ kg  – $$\frac{2}{3}$$ kg = $$\frac{11}{8}$$ – $$\frac{2}{3}$$ = $$\frac{33}{24}$$ – $$\frac{16}{24}$$ = $$\frac{17}{24}$$ .
Therefore, Quantity of Fertilizers used for flower bed  =$$\frac{17}{24}$$