grade 5 module 3 lesson 11 homework

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Eureka Math Grade 5 Module 3 Lesson 11 Answer Key

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$\frac{1}{6}$ .