## Engage NY Eureka Math 5th Grade Module 3 Lesson 11 Answer Key

### Eureka Math Grade 5 Module 3 Lesson 11 Problem Set Answer Key

Question 1.

Generate equivalent fractions to get like units. Then, subtract.

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

b. \(\frac{7}{10}\) – \(\frac{1}{3}\) =

c. \(\frac{7}{8}\) – \(\frac{3}{4}\) =

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

e. 1\(\frac{3}{10}\) – \(\frac{1}{6}\) =

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

g. 5\(\frac{6}{7}\) – 2\(\frac{2}{3}\) =

h. Draw a number line to show that your answer to (g) is reasonable.

Answer:

a. \(\frac{1}{2}\) – \(\frac{1}{3}\) = \(\frac{1}{6}\)

Explanation :

\(\frac{1}{2}\) – \(\frac{1}{3}\)

lcm of 2 and 3 is 6

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

b. \(\frac{7}{10}\) – \(\frac{1}{3}\) = \(\frac{11}{30}\)

Explanation :

\(\frac{7}{10}\) – \(\frac{1}{3}\)

lcm of 10 and 3 is 30 .

\(\frac{21}{30}\) – \(\frac{10}{30}\) = \(\frac{11}{30}\)

c. \(\frac{7}{8}\) – \(\frac{3}{4}\) = \(\frac{1}{8 }\)

Explanation :

\(\frac{7}{8}\) – \(\frac{3}{4}\)

lcm of 8 and 4 is 8 .

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

d. 1\(\frac{2}{5}\) – \(\frac{3}{8}\) = 1\(\frac{31}{40}\)

Explanation :

1\(\frac{2}{5}\) – \(\frac{3}{8}\) = \(\frac{7}{5}\) – \(\frac{3}{8}\)

lcm of 5 and 8 is 40 .

\(\frac{56}{40}\) – \(\frac{15}{40}\) = \(\frac{71}{40}\) = 1\(\frac{31}{40}\)

e. 1\(\frac{3}{10}\) – \(\frac{1}{6}\) = 1\(\frac{4}{30}\)

Explanation :

1\(\frac{3}{10}\) – \(\frac{1}{6}\) = \(\frac{13}{10}\) – \(\frac{1}{6}\)

lcm of 6 and 10 is 30.

\(\frac{39}{30}\) – \(\frac{5}{30}\) = \(\frac{34}{30}\) = 1\(\frac{4}{30}\)

f. 2\(\frac{1}{3}\) – 1\(\frac{1}{5}\) =1\(\frac{2}{15}\)

Explanation :

2\(\frac{1}{3}\) – 1\(\frac{1}{5}\) = \(\frac{7}{3}\) – \(\frac{6}{5}\)

lcm of 3 and 5 is 15 .

\(\frac{35}{15}\) – \(\frac{18}{15}\) = \(\frac{17}{15}\) = 1\(\frac{2}{15}\)

g. 5\(\frac{6}{7}\) – 2\(\frac{2}{3}\) = 3\(\frac{4}{21}\) .

Explanation :

5\(\frac{6}{7}\) – 2\(\frac{2}{3}\) = \(\frac{41}{7}\) – \(\frac{8}{3}\)

lcm of 7 and 3 is 21 .

\(\frac{123}{21}\) – \(\frac{56}{21}\) = \(\frac{67}{21}\) = 3\(\frac{4}{21}\) .

Question 2.

George says that, to subtract fractions with different denominators, you always have to multiply the denominators to find the common unit; for example:

\(\frac{3}{8}-\frac{1}{6}=\frac{18}{48}-\frac{8}{48}\)

Show George how he could have chosen a denominator smaller than 48, and solve the problem.

Answer:

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

lcm of 8 and 6 is 24 .

[late3x]\frac{9}{24}[/latex] – \(\frac{4}{24}\) = \(\frac{5}{24}\)

Multiplies of 8 and 6 are .

8 : 16, 24, 32, 40, 48 .

6 : 12, 18, 24, 30, 36, 48.

common multiple smaller than 48 is 24 .

Question 3.

Meiling has 1\(\frac{1}{4}\) liter of orange juice. She drinks \(\frac{1}{3}\) liter. How much orange juice does she have left? (Extension: If her brother then drinks twice as much as Meiling, how much is left?)

Answer:

Fraction of Quantity of Juice with Meiling = 1\(\frac{1}{4}\) = \(\frac{5}{4}\)

Fraction of Quantity of Juice drank by Meiling = \(\frac{1}{3}\)

Fraction of Quantity of Juice left = \(\frac{5}{4}\) – \(\frac{1}{3}\) = \(\frac{15}{12}\) – \(\frac{4}{12}\) = \(\frac{11}{12}\) .

Therefore , Fraction of Quantity of Juice left = \(\frac{11}{12}\) .

Question 4.

Harlan used 3\(\frac{1}{2}\) kg of sand to make a large hourglass. To make a smaller hourglass, he only used 1\(\frac{3}{7}\) kg of sand. How much more sand did it take to make the large hourglass than the smaller one?

Answer:

Fraction of Quantity of sand used for large hourglass = 3\(\frac{1}{2}\) kg = \(\frac{7}{2}\) kg

Fraction of Quantity of sand used for small hourglass = \(\frac{10}{7}\) kg

Fraction of Quantity of sand to make the large hourglass than the smaller one = \(\frac{7}{2}\) – \(\frac{10}{7}\) = \(\frac{49}{14}\) – \(\frac{20}{14}\) = \(\frac{29}{14}\) = 2\(\frac{1}{14}\) .

Therefore, Fraction of Quantity of sand to make the large hourglass than the smaller one = 2\(\frac{1}{14}\) .

### Eureka Math Grade 5 Module 3 Lesson 11 Exit Ticket Answer Key

Generate equivalent fractions to get like units. Then, subtract.

a. \(\frac{3}{4}\) – \(\frac{3}{10}\) =

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

Answer:

a. \(\frac{3}{4}\) – \(\frac{3}{10}\) = \(\frac{9}{20}\)

Explanation :

\(\frac{3}{4}\) – \(\frac{3}{10}\)

lcm of 4 and 10 are 20 .

\(\frac{15}{20}\) – \(\frac{6}{20}\) = \(\frac{9}{20}\)

b. 3\(\frac{1}{2}\) – 1\(\frac{1}{3}\) = 2\(\frac{1}{6}\)

Explanation :

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

lcm of 2 and 3 is 6

\(\frac{21}{6}\) – \(\frac{8}{6}\) = \(\frac{13}{6}\) = 2\(\frac{1}{6}\)

### Eureka Math Grade 5 Module 3 Lesson 11 Homework Answer Key

Question 1.

Generate equivalent fractions to get like units. Then, subtract.

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

b. \(\frac{7}{8}\) – \(\frac{1}{3}\) =

c. \(\frac{7}{10}\) – \(\frac{3}{5}\) =

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

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

f. 5\(\frac{6}{7}\) – 3\(\frac{2}{3}\) =

g. 15\(\frac{7}{8}\) – 5\(\frac{3}{4}\) =

h. 15\(\frac{5}{8}\) – 3\(\frac{1}{3}\) =

Answer:

a. \(\frac{1}{2}\) – \(\frac{1}{5}\) = \(\frac{3}{10}\)

Explanation :

\(\frac{1}{2}\) – \(\frac{1}{5}\)

lcm of 2 and 5 is 10 .

\(\frac{5}{10}\) – \(\frac{2}{10}\) = \(\frac{3}{10}\)

b. \(\frac{7}{8}\) – \(\frac{1}{3}\) = \(\frac{13}{24}\)

Explanation :

\(\frac{7}{8}\) – \(\frac{1}{3}\)

lcm of 8 and 3 is 24 .

\(\frac{21}{24}\) – \(\frac{8}{24}\) = \(\frac{13}{24}\)

c. \(\frac{7}{10}\) – \(\frac{3}{5}\) = \(\frac{1}{10}\)

Explanation :

\(\frac{7}{10}\) – \(\frac{3}{5}\)

lcm of 10 and 5 is 10 .

\(\frac{7}{10}\) – \(\frac{6}{10}\) = \(\frac{1}{10}\)

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

Explanation :

1\(\frac{5}{6}\) – \(\frac{2}{3}\) = \(\frac{11}{6}\) – \(\frac{2}{3}\)

lcm of 6 and 3 is 6

\(\frac{11}{6}\) – \(\frac{4}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

e. 2\(\frac{1}{4}\) – 1\(\frac{1}{5}\) = 1\(\frac{1}{20}\)

Explanation :

2\(\frac{1}{4}\) – 1\(\frac{1}{5}\) = \(\frac{9}{4}\) – \(\frac{6}{5}\)

lcm of 4 and 5 is 20 .

\(\frac{45}{20}\) – \(\frac{24}{20}\) = \(\frac{21}{20}\) = 1\(\frac{1}{20}\)

f. 5\(\frac{6}{7}\) – 3\(\frac{2}{3}\) = 2 \(\frac{4}{21}\)

Explanation :

5\(\frac{6}{7}\) – 3\(\frac{2}{3}\) = \(\frac{41}{7}\) – \(\frac{11}{3}\)

lcm of 7 and 3 is 21

\(\frac{123}{21}\) – \(\frac{77}{21}\) = \(\frac{46}{21}\) = 2 \(\frac{4}{21}\)

g. 15\(\frac{7}{8}\) – 5\(\frac{3}{4}\) = 10\(\frac{1}{8}\)

Explanation :

15\(\frac{7}{8}\) – 5\(\frac{3}{4}\) = \(\frac{127}{8}\) – \(\frac{23}{4}\)

lcm of 8 and 4 is 8 .

\(\frac{127}{8}\) – \(\frac{46}{8}\) = \(\frac{81}{8}\) = 10\(\frac{1}{8}\) .

h. 15\(\frac{5}{8}\) – 3\(\frac{1}{3}\) = 12 \(\frac{7}{24}\)

Explanation :

15\(\frac{5}{8}\) – 3\(\frac{1}{3}\) = \(\frac{125}{8}\) – \(\frac{10}{3}\)

lcm of 3 and 8 is 24 .

\(\frac{375}{24}\) – \(\frac{80}{24}\) = \(\frac{295}{24}\) =12 \(\frac{7}{24}\)

Question 2.

Sandy ate \(\frac{1}{6}\) of a candy bar. John ate \(\frac{3}{4}\) of it. How much more of the candy bar did John eat than Sandy?

Answer:

Fraction of candy ate by sandy = \(\frac{1}{6}\)

Fraction of candy ate by John = \(\frac{3}{4}\)

Fraction of the candy bar ate more by John eat than Sandy = \(\frac{3}{4}\) – \(\frac{1}{6}\) = \(\frac{9}{12}\) – \(\frac{2}{12}\) = \(\frac{7}{12}\)

Therefore, Fraction of the candy bar ate more by John eat than Sandy = \(\frac{7}{12}\) .

Question 3.

4\(\frac{1}{2}\) yards of cloth are needed to make a woman’s dress. 2\(\frac{2}{7}\) yards of cloth are needed to make a girl’s dress. How much more cloth is needed to make a woman’s dress than a girl’s dress?

Answer:

Fraction of cloth needed to make women’s dress = 4\(\frac{1}{2}\) yards = \(\frac{9}{2}\) yards

Fraction of cloth needed to make girl’s dress = 2\(\frac{2}{7}\) yards = \(\frac{16}{7}\) yards

Fraction of more cloth needed to make a woman’s dress than a girl’s dress = \(\frac{9}{2}\) – \(\frac{16}{7}\)

= \(\frac{63}{14}\) – \(\frac{32}{14}\) = \(\frac{31}{14}\) = 2\(\frac{3}{14}\) yards .

Therefore, Fraction of more cloth needed to make a woman’s dress than a girl’s dress = 2\(\frac{3}{14}\) yards

Question 4.

Bill reads \(\frac{1}{5}\) of a book on Monday. He reads \(\frac{2}{3}\) of the book on Tuesday. If he finishes reading the book on Wednesday, what fraction of the book did he read on Wednesday?

Answer:

Fraction of book read on Monday =\(\frac{1}{5}\)

Fraction of book read on Tuesday = \(\frac{2}{3}\)

Fraction of book read on both days = \(\frac{1}{5}\) + \(\frac{2}{3}\) = \(\frac{3}{15}\) + \(\frac{10}{15}\) = \(\frac{13}{15}\) .

Therefore, Fraction of book read on both days = \(\frac{13}{15}\)

Question 5.

Tank A has a capacity of 9.5 gallons. 6\(\frac{1}{3}\) gallons of the tank’s water are poured out. How many gallons of water are left in the tank?

Answer:

Fraction of Capacity of Tank A = 9.5 gallons

Fraction of Capacity of water poured out = 6\(\frac{1}{3}\) gallons = \(\frac{19}{3}\) gallons .

Fraction of Capacity of water left = 9.5 – \(\frac{19}{3}\) = \(\frac{95}{10}\) – \(\frac{19}{3}\) = \(\frac{285}{30}\) – \(\frac{190}{30}\) = \(\frac{95}{30}\) = \(\frac{19}{6}\) = 3\(\frac{1}{6}\)

Therefore, Fraction of Capacity of water left = 3\(\frac{1}{6}\) .