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\(\frac{1}{6}\) .