Eureka Math Grade 5 Module 4 Lesson 15 Answer Key (2024)

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.

a. \(\frac{2}{3}\) of \(\frac{3}{5}\)
\(\frac{2}{3}\) × \(\frac{3}{5}\) = \(\frac{6}{15}\) = \(\frac{2}{5}\)
Eureka Math Grade 5 Module 4 Lesson 15 Answer Key (1)

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.

Eureka Math Grade 5 Module 4 Lesson 15 Answer Key (2)

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. \(\frac{3}{4}\) × \(\frac{4}{5}\)

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. \(\frac{5}{6}\) × \(\frac{5}{8}\)

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
\(\frac{4}{9}\) × \(\frac{3}{8}\) = \(\frac{1}{6}\).

d. \(\frac{3}{4}\) × \(\frac{5}{12}\)

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
\(\frac{4}{9}\) × \(\frac{3}{8}\) = \(\frac{1}{6}\).

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?

Eureka Math Grade 5 Module 4 Lesson 15 Answer Key (2024)

FAQs

What grade does Eureka math go up to? ›

Eureka Math® is a holistic Prekindergarten through Grade 12 curriculum that carefully sequences mathematical progressions in expertly crafted modules, making math a joy to teach and learn. We provide in-depth professional development, learning materials, and a community of support.

What are the four core components of a Eureka Math TEKS lesson? ›

A typical Eureka lesson is comprised of four critical components: fluency practice, concept development (including a problem set), application problem, and student debrief (including the Exit Ticket).

Is Eureka Math a curriculum? ›

An Elementary, Middle, And High School Math Curriculum. Eureka Math® is a math program designed to advance equity in the math classroom by helping students build enduring math knowledge.

Why was Eureka Math created? ›

Eureka was built around the core principle that students need to know more than just what works when solving a prob- lem—they need to understand why it works. The curriculum goes beyond facts and formulas, teaching students to think about math conceptually.

What is the hardest math in 5th grade? ›

Some of the hardest math problems for fifth graders involve multiplying: multiplying using square models, multiplying fractions and whole numbers using expanded form, and multiplying fractions using number lines.

What is the hardest math grade? ›

The hardest math class you can take in high school is typically AP Calculus BC or IB Math HL. These courses cover a wide range of advanced mathematical concepts, including calculus, trigonometry, and statistics. Students who take these courses must be able to think abstractly and solve complex problems.

Is Eureka Math good or bad? ›

Is Eureka Math a good curriculum? The answer to this question depends on the target audience. If you're a teacher in a public school who needs to cover State Standards and your goal is merely to prepare students for State tests, then Eureka may be a good curriculum for you.

Is Eureka Math still free? ›

Is Eureka Math free? Yes. Anyone can download the entire PK–12 Eureka Math curriculum, along with a variety of instructional materials and support resources, for free.

Does Khan Academy align with Eureka Math? ›

To access our aligned resources, go to the Courses dropdown menu in the top left corner of your screen and select See all Math. From the Math page you can view all Math courses including the courses aligned to the Eureka Math/EngageNY curriculum.

Is Zearn related to Eureka Math? ›

Zearn Math K–5 lessons follow the scope and sequence of Eureka Math/EngageNY. All Middle School materials align to Eureka Math/EngageNY on the unit level and may be reordered to directly follow the curriculum's scope and sequence.

What are the goals of Eureka Math? ›

Eureka Math is designed to support students in gaining a solid understanding of concepts, a high degree of procedural skill and fluency, and the ability to apply math to solve problems in and outside the classroom. There is also an intentional coherence linking topics and thinking across grades.

What is the highest level of math in 9th grade? ›

9th grade math usually focuses on Algebra I, but can include other advanced mathematics such as Geometry, Algebra II, Pre-Calculus or Trigonometry.

What grade does go math go up to? ›

Go Math! for Grades K–6 combines trusted content, practice, and games with user-friendly tools aimed at guiding every learner toward mastery.

What is advanced math in 8th grade called? ›

Almost every school district in the state offers an accelerated math option for selected students. These students take Algebra I in 8th grade. These students complete Algebra II, Geometry and Precalculus one year earlier than their peers.

What grade level is go math for? ›

Go Math! (K-6) on Ed is an easy-to-implement core curriculum with an effective instructional approach that includes robust differentiation and assessment resources that engage all levels of learners and support all levels of teachers, from novice to master.

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