eureka math grade 5 module 4 lesson 8 homework answers

Eureka Math Grade 5 Answer Key | Engage NY Math 5th Grade Answer Key Solutions

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• Eureka Math Grade 5 Module 5 Answer Key
• Eureka Math Grade 5 Module 6 Answer Key

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Eureka Math Grade 5 Module 4 Lesson 15 Answer Key

Engage ny eureka math 5th grade module 4 lesson 15 answer key, eureka math grade 5 module 4 lesson 15 problem set answer key.

Question 1. Solve. Draw a rectangular fraction model to explain your thinking. Then, write a multiplication sentence. The first one is done for you.

b. \(\frac{3}{4}\) of \(\frac{4}{5}\) =

Answer: latex]\frac{3}{4}[/latex] of \(\frac{4}{5}\) = \(\frac{3}{5}\).

Explanation: Given that \(\frac{3}{4}\) of \(\frac{4}{5}\) which is \(\frac{3}{4}\) × \(\frac{4}{5}\) = \(\frac{3}{5}\)

c. \(\frac{2}{5}\) of \(\frac{2}{3}\)=

Answer: latex]\frac{2}{5}[/latex] of \(\frac{2}{3}\) = \(\frac{4}{15}\).

Explanation: Given that \(\frac{2}{5}\) of \(\frac{2}{3}\) which is \(\frac{2}{5}\) × \(\frac{2}{3}\) = \(\frac{4}{15}\)

d. \(\frac{4}{5}\) × \(\frac{2}{3}\) =

Answer: latex]\frac{4}{5}[/latex] of \(\frac{2}{3}\) = \(\frac{8}{15}\).

Explanation: Given that \(\frac{4}{5}\) of \(\frac{2}{3}\) which is \(\frac{4}{5}\) × \(\frac{2}{3}\) = \(\frac{8}{15}\)

e. \(\frac{3}{4}\) × \(\frac{2}{3}\)=

Answer: latex]\frac{3}{4}[/latex] of \(\frac{2}{3}\) = \(\frac{1}{2}\).

Explanation: Given that \(\frac{3}{4}\) of \(\frac{2}{3}\) which is \(\frac{3}{4}\) × \(\frac{2}{3}\) = \(\frac{1}{2}\)

Question 2. Multiply. Draw a rectangular fraction model if it helps you, or use the method in the example.

a. \(\frac{3}{4}\) × \(\frac{5}{6}\)

Answer: latex]\frac{3}{4}[/latex] of \(\frac{5}{6}\) = \(\frac{5}{8}\).

Explanation: Given that \(\frac{3}{4}\) of \(\frac{5}{6}\) which is \(\frac{3}{4}\) × \(\frac{5}{6}\) = \(\frac{5}{8}\).

b. \(\frac{4}{5}\) × \(\frac{5}{8}\)

Answer: latex]\frac{4}{5}[/latex] of \(\frac{5}{8}\) = \(\frac{1}{2}\).

Explanation: Given that \(\frac{4}{5}\) of \(\frac{5}{8}\) which is \(\frac{4}{5}\) × \(\frac{5}{8}\) = \(\frac{1}{2}\)

c. \(\frac{2}{3}\) × \(\frac{6}{7}\)

Answer: latex]\frac{2}{3}[/latex] of \(\frac{6}{7}\) = \(\frac{1}{7}\).

Explanation: Given that \(\frac{2}{3}\) of \(\frac{6}{7}\) which is \(\frac{2}{3}\) × \(\frac{6}{7}\) = \(\frac{1}{7}\)

d. \(\frac{4}{9}\) × \(\frac{3}{10}\)

Answer: latex]\frac{4}{9}[/latex] of \(\frac{3}{10}\) = \(\frac{2}{15}\).

Explanation: Given that \(\frac{4}{9}\) of \(\frac{3}{10}\) which is \(\frac{4}{9}\) × \(\frac{3}{10}\) = \(\frac{2}{15}\).

Question 3. Phillip’s family traveled \(\frac{3}{10}\) of the distance to his grandmother’s house on Saturday. They traveled \(\frac{4}{7}\) of the remaining distance on Sunday. What fraction of the total distance to his grandmother’s house was traveled on Sunday?

Answer: Philip’s family traveled on Sunday is \(\frac{2}{5}\).

Explanation: Given that Phillip’s family traveled \(\frac{3}{10}\) of the distance to his grandmother’s house on Saturday, so the remaining is 1 – \(\frac{3}{10}\) which is \(\frac{7}{10}\). So Philip’s family traveled on Sunday is \(\frac{4}{7}\) × \(\frac{7}{10}\) which is \(\frac{2}{5}\).

Question 4. Santino bought a \(\frac{3}{4}\)-pound bag of chocolate chips. He used \(\frac{2}{3}\) of the bag while baking. How many pounds of chocolate chips did he use while baking?

Answer: The number of pounds of chocolate chips did he use while baking is \(\frac{1}{2}\) lb.

Explanation: Given that Santino bought a \(\frac{3}{4}\)-pound bag of chocolate chips and he used \(\frac{2}{3}\) of the bag while baking. So the number of pounds of chocolate chips did he use while baking is \(\frac{3}{4}\) × \(\frac{2}{3}\) which is \(\frac{1}{2}\) lb.

Question 5. Farmer Dave harvested his corn. He stored \(\frac{5}{9}\) of his corn in one large silo and \(\frac{3}{4}\) of the remaining corn in a small silo. The rest was taken to market to be sold. a. What fraction of the corn was stored in the small silo?

Answer: The fraction of the corn was stored in the small silo \(\frac{1}{3}\).

Explanation: Given that Dave has stored \(\frac{5}{9}\) of his corn in one large silo. Let the total corn be ‘X’, and the amount of corn stored in the silo is \(\frac{5}{9}\)X. The amount of corn remaining is X – \(\frac{5}{9}\)X which is \(\frac{9X – 5X}{9}\) = \(\frac{4X}{9}\). Thus the amount of corn stored in the small silo is \(\frac{3}{4}\) × \(\frac{4}{9}\)X which is \(\frac{1}{3}\)X. Thus the fraction of the corn was stored in the small silo \(\frac{1}{3}\).

b. If he harvested 18 tons of corn, how many tons did he take to market?

Answer: The amount of corn taken to market is 9 tonnes.

Explanation: The amount of corn solid in the market is \(\frac{4X}{9}\) – \(\frac{X}{3}\) which is \(\frac{X}{9}\). Thus the amount of corn taken to market is 18 × \(\frac{1}{9}\) which is 9 tonnes.

Eureka Math Grade 5 Module 4 Lesson 15 Exit Ticket Answer Key

Question 1. Solve. Draw a rectangular fraction model to explain your thinking. Then, write a multiplication sentence.

a. \(\frac{2}{3}\) of \(\frac{3}{5}\) =

Answer: latex]\frac{2}{3}[/latex] of \(\frac{3}{5}\) = \(\frac{2}{5}\).

Explanation: Given that \(\frac{2}{3}\) of \(\frac{3}{5}\) which is \(\frac{2}{3}\) × \(\frac{3}{5}\) = \(\frac{2}{5}\).

b. \(\frac{4}{9}\) × \(\frac{3}{8}\) =

Answer: latex]\frac{4}{9}[/latex] of \(\frac{3}{8}\) = \(\frac{1}{6}\).

Explanation: Given that \(\frac{4}{9}\) of \(\frac{3}{8}\) which is \(\frac{4}{9}\) × \(\frac{3}{8}\) = \(\frac{1}{6}\).

Question 2. A newspaper’s cover page is \(\frac{3}{8}\) text, and photographs fill the rest. If \(\frac{2}{5}\) of the text is an article about endangered species, what fraction of the cover page is the article about endangered species?

Answer: The fraction of the cover page is the article about endangered species \(\frac{3}{20}\).

Explanation: Given that a newspaper’s cover page is \(\frac{3}{8}\) text, and photographs fill the rest, and if \(\frac{2}{5}\) of the text is an article about endangered species. So the fraction of the cover page is the article about endangered species \(\frac{3}{8}\) × \(\frac{2}{5}\) which is \(\frac{3}{20}\).

Eureka Math Grade 5 Module 4 Lesson 15 Homework Answer Key

Question 1. Solve. Draw a rectangular fraction model to explain your thinking. Then, write a multiplication sentence. a. \(\frac{2}{3}\) of \(\frac{3}{4}\) =

Answer: latex]\frac{2}{3}[/latex] of \(\frac{3}{4}\) = \(\frac{1}{2}\).

Explanation: Given that \(\frac{2}{3}\) of \(\frac{3}{4}\) which is \(\frac{2}{3}\) × \(\frac{3}{4}\) = \(\frac{1}{2}\).

b. \(\frac{2}{5}\) of \(\frac{3}{4}\) =

Answer: latex]\frac{2}{5}[/latex] of \(\frac{3}{4}\) = \(\frac{3}{10}\).

Explanation: Given that \(\frac{2}{5}\) of \(\frac{3}{4}\) which is \(\frac{2}{5}\) × \(\frac{3}{4}\) = \(\frac{3}{10}\).

c. \(\frac{2}{5}\) of \(\frac{4}{5}\) =

Answer: latex]\frac{2}{5}[/latex] of \(\frac{4}{5}\) = \(\frac{8}{25}\).

Explanation: Given that \(\frac{2}{5}\) of \(\frac{4}{5}\) which is \(\frac{2}{5}\) × \(\frac{4}{5}\) = \(\frac{8}{25}\).

d. \(\frac{4}{5}\) of \(\frac{3}{4}\) =

Answer: latex]\frac{4}{5}[/latex] of \(\frac{3}{4}\) = \(\frac{3}{5}\).

Explanation: Given that \(\frac{4}{5}\) of \(\frac{3}{4}\) which is \(\frac{4}{5}\) × \(\frac{3}{4}\) = \(\frac{3}{5}\).

Question 2. Multiply. Draw a rectangular fraction model if it helps you. a. \(\frac{5}{6}\) × \(\frac{3}{10}\)

Answer: latex]\frac{5}{6}[/latex] of \(\frac{3}{10}\) = \(\frac{1}{4}\).

Explanation: Given that \(\frac{5}{6}\) of \(\frac{3}{10}\) which is \(\frac{5}{6}\) × \(\frac{3}{10}\) = \(\frac{1}{4}\).

b. \(\frac{3}{4}\) × \(\frac{4}{5}\)

Explanation: Given that \(\frac{3}{4}\) of \(\frac{4}{5}\) which is \(\frac{3}{4}\) × \(\frac{4}{5}\) = \(\frac{3}{5}\).

c. \(\frac{5}{6}\) × \(\frac{5}{8}\)

d. \(\frac{3}{4}\) × \(\frac{5}{12}\)

e. \(\frac{8}{9}\) × \(\frac{2}{3}\)

Answer: latex]\frac{8}{9}[/latex] of \(\frac{2}{3}\) = \(\frac{16}{27}\).

Explanation: Given that \(\frac{8}{9}\) of \(\frac{2}{3}\) which is \(\frac{8}{9}\) × \(\frac{2}{3}\) = \(\frac{16}{27}\).

f. \(\frac{3}{7}\) × \(\frac{2}{9}\)

Answer: latex]\frac{3}{7}[/latex] of \(\frac{2}{9}\) = \(\frac{2}{21}\).

Explanation: Given that \(\frac{3}{7}\) of \(\frac{2}{9}\) which is \(\frac{3}{7}\) × \(\frac{2}{9}\) = \(\frac{2}{21}\).

Question 3. Every morning, Halle goes to school with a 1-liter bottle of water. She drinks \(\frac{1}{4}\) of the bottle before school starts and \(\frac{2}{3}\) of the rest before lunch. a. What fraction of the bottle does Halle drink after school starts but before lunch?

Answer: The fraction of the bottle does Halle drinks after school starts but before lunch is \(\frac{1}{2}\).

Explanation: Given that Halle goes to school with a 1-liter bottle of water and she drinks \(\frac{1}{4}\) of the bottle before school starts and \(\frac{2}{3}\) of the rest before lunch and the amount left after drinking before school starts are 1 – \(\frac{1}{4}\) which is \(\frac{3}{4}\) and the fraction of the bottle does Halle drinks after school starts but before lunch is \(\frac{2}{3}\) of Amount left = \(\frac{2}{3}\) × \(\frac{3}{4}\) = \(\frac{1}{2}\).

b. How many milliliters are left in the bottle at lunch?

Answer: The amount that left in the bottle at lunch is 250 milliliters.

Explanation: The amount that left in the bottle at lunch is 1 – (\(\frac{3}{4}\) + \(\frac{1}{2}\)) = \(\frac{1}{4}\), as we know that 1 litre is 1000 milliliters, so \(\frac{1}{4}\) litre is \(\frac{1}{4}\) × 1000 which is 250 milliliters.

Question 4. Moussa delivered \(\frac{3}{8}\) of the newspapers on his route in the first hour and \(\frac{4}{5}\) of the rest in the second hour. What fraction of the newspapers did Moussa deliver in the second hour?

Question 5. Rose bought some spinach. She used \(\frac{3}{5}\) of the spinach on a pan of spinach pie for a party and \(\frac{3}{4}\) of the remaining spinach for a pan for her family. She used the rest of the spinach to make a salad. a. What fraction of the spinach did she use to make the salad?

b. If Rose used 3 pounds of spinach to make the pan of spinach pie for the party, how many pounds of spinach did Rose use to make the salad?

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