# Into Math Grade 5 Module 9 Lesson 2 Answer Key Multiply Mixed Numbers

We included HMH Into Math Grade 5 Answer Key PDF Module 9 Lesson 2 Multiply Mixed Numbers to make students experts in learning maths.

## HMH Into Math Grade 5 Module 9 Lesson 2 Answer Key Multiply Mixed Numbers

I Can solve real world problems involving multiplication of mixed numbers by writing an equation to model the problem.

Arun paints on a large rectangular canvas that measures 2$$\frac{1}{4}$$ feet by 4$$\frac{1}{3}$$ feet. He wants to enter his painting into an art contest. The contest rules say that the area of all canvases must be less than 4$$\frac{1}{3}$$ square feet. Arun cannot enter the contest as his painting is too large.

Explanation:
The measurements of Arun’s canva are 8+1+$$\frac{2}{3}$$+$$\frac{1}{12}$$ which is 9+$$\frac{8}{12}$$+$$\frac{1}{12}$$ = 9$$\frac{9}{12}$$ sq ft. So Arun cannot enter the contest as his painting is too large.
Based on these rules, can Arun enter his painting into the contest? Explain with words or drawings.  Turn and Talk How would your answer change if Arun’s canvas measured $$\frac{5}{6}$$ foot by 4$$\frac{1}{3}$$ feet? Justify your reasoning.

Build Understanding

1. Tam draws a logo for her company’s website. The logo is a rectangle that is 1$$\frac{1}{3}$$ inches wide and 1$$\frac{3}{4}$$ inches long. What is the area of the logo? A. Each unit square shown represents a square with a side length of 1 inch. Use this area model to represent the area of Tam’s logo.

The area is $$\frac{7}{3}$$ sq inches.

Explanation:
Given that the logo is a rectangle that is 1$$\frac{1}{3}$$ inches wide and 1$$\frac{3}{4}$$ inches long. So the area will be 1$$\frac{1}{3}$$ × 1$$\frac{3}{4}$$ which is $$\frac{4}{3}$$ × $$\frac{7}{4}$$ = $$\frac{7}{3}$$ sq inches.

B. Why does this area model show four unit squares? 2$$\frac{4}{12}$$ sq in.

Explanation:
The measurements are 1+$$\frac{1}{3}$$+$$\frac{3}{4}$$+$$\frac{3}{12}$$ which is 1+$$\frac{4+9+3}{12}$$ = 1 + $$\frac{16}{12}$$
= 2$$\frac{4}{12}$$ sq in. C. How did you show the width of 1$$\frac{1}{3}$$ inches?
__________________________

The width of 1$$\frac{1}{3}$$ inches is $$\frac{4}{3}$$ inches.

D. How did you show the length of 1$$\frac{3}{4}$$ inches?
__________________________

The width of 1$$\frac{3}{4}$$ inches is $$\frac{7}{3}$$ inches.

E. How many equal-sized parts represent the area of the logo?
___________________
___________________

The area of each equal-sized part is $$\frac{7}{3}$$ sq inches.
F. What is the area of each equal-sized part? How do you know?
___________________
___________________

The area of each equal-sized part is $$\frac{7}{3}$$ sq inches.
G. What is the area of the logo? Justify your reasoning.
___________________
___________________

The area of the logo is $$\frac{7}{3}$$ sq inches.

Step It Out

2. Tam makes another logo. The area of the logo is represented by the purple shading. The area is $$\frac{27}{10}$$ sq in.

Explanation:
The area of the logo is 1$$\frac{4}{5}$$ × 1$$\frac{1}{2}$$ which is $$\frac{9}{5}$$ × $$\frac{3}{2}$$ = $$\frac{27}{10}$$ sq in.

A. How many equal-sized parts represent the area of the logo?
____________________

The equal-sized parts represent the area of the logo is $$\frac{27}{10}$$ sq in.

B. Find the area of one equal-sized part.
____________________
C. What is the area of the logo? Explain your reasoning. Write an equation to model the problem.
____________________

The area of the logo is $$\frac{27}{10}$$ sq in.

Explanation:
The equation to model the problem is 1$$\frac{4}{5}$$ × 1$$\frac{1}{2}$$ which is $$\frac{9}{5}$$ × $$\frac{3}{2}$$ = $$\frac{27}{10}$$ sq in.

D. Rename 1$$\frac{1}{2}$$ and 1$$\frac{4}{5}$$ as fractions greater than 1.
____________________

$$\frac{27}{10}$$ which is greater than 1.

Explanation:
The fraction greater than 1 is $$\frac{9}{5}$$ × $$\frac{3}{2}$$ = $$\frac{27}{10}$$ which is greater than 1.

E. Use your answers from Part D to write an equation to model the area of the logo.
____________________

The equation is $$\frac{9}{5}$$ × $$\frac{3}{2}$$ = $$\frac{27}{10}$$.

Turn and Talk How are the equations that model the area of the logo in Parts C and E related? How do they connect to the visual model?

Check Understanding Math Board

Question 1.
Jonah’s new poster has a width of 1$$\frac{1}{4}$$ yards and a length of 1$$\frac{1}{3}$$ yards. Use this area model to represent the area of the poster. What is the area of the poster? Write an equation using fractions greater than 1 to model the problem. The area is 1$$\frac{2}{3}$$ sq yards.

Explanation:
Given that Jonah’s new poster has a width of 1$$\frac{1}{4}$$ yards and a length of 1$$\frac{1}{3}$$ yards. So the area is 1$$\frac{1}{4}$$ × 1$$\frac{1}{3}$$ which is $$\frac{5}{4}$$ × $$\frac{4}{3}$$ = $$\frac{5}{3}$$ sq yards.

Question 2.
Use Structure The flagpole at a park is 3$$\frac{1}{3}$$ yards tall. The flagpole at a museum is 1$$\frac{1}{2}$$ times as tall as the height of the flagpole at the park. • Explain how to write an equation to model the height of the museum’s flagpole.
• What is the height of the museum’s flagpole?

The equation is 3$$\frac{1}{3}$$ × 1$$\frac{1}{2}$$ = 5 sq yards.

Explanation:
Given that the flagpole at a park is 3$$\frac{1}{3}$$ yards tall and the flagpole at a museum is 1$$\frac{1}{2}$$ times as tall as the height of the flagpole at the park. So the equation will be 3$$\frac{1}{3}$$ × 1$$\frac{1}{2}$$ which is $$\frac{10}{3}$$ × $$\frac{3}{2}$$ = 5 sq yards.

Question 3.
Model with Mathematics Debbie measures the length and the width of a cell phone. The length is 5$$\frac{3}{5}$$ inches, and the width is 2$$\frac{4}{5}$$ inches. What is the area of the front of the cell phone? Write an equation using fractions greater than 1 to model the problem.
5$$\frac{3}{5}$$ × 2$$\frac{4}{5}$$ = 15$$\frac{17}{25}$$ sq in.

Explanation:
Given that the length is 5$$\frac{3}{5}$$ inches, and the width is 2$$\frac{4}{5}$$ inches. So the equation is 5$$\frac{3}{5}$$ × 2$$\frac{4}{5}$$ which is $$\frac{28}{5}$$ × $$\frac{14}{5}$$ = $$\frac{392}{25}$$ sq in.

Multiply.

Question 4.
2$$\frac{1}{5}$$ × 3$$\frac{1}{2}$$ = ___________________
2$$\frac{1}{5}$$ × 3$$\frac{1}{2}$$ = 7$$\frac{7}{10}$$.

Explanation:
The product of 2$$\frac{1}{5}$$ × 3$$\frac{1}{2}$$ which is $$\frac{11}{5}$$ × $$\frac{7}{2}$$ = $$\frac{77}{10}$$
= 7$$\frac{7}{10}$$.

Question 5.
____ = 5$$\frac{1}{2}$$ × 1$$\frac{1}{3}$$
5$$\frac{1}{2}$$ × 1$$\frac{1}{3}$$ = 7$$\frac{1}{3}$$.

Explanation:
The product of 5$$\frac{1}{2}$$ × 1$$\frac{1}{3}$$ which is $$\frac{11}{2}$$ × $$\frac{4}{3}$$ = $$\frac{44}{6}$$
= 7$$\frac{1}{3}$$.

Question 6.
2$$\frac{1}{4}$$ × 1$$\frac{2}{3}$$ = ___________________
2$$\frac{1}{4}$$ × 1$$\frac{2}{3}$$ = 3$$\frac{3}{4}$$.

Explanation:
The product of 2$$\frac{1}{4}$$ × 1$$\frac{2}{3}$$ which is $$\frac{9}{4}$$ × $$\frac{5}{3}$$ = $$\frac{45}{12}$$
= 3$$\frac{3}{4}$$.

Question 7.
STEM An early computer connection called USB 1.0 transfers about 1$$\frac{1}{2}$$ megabytes of data each second. A later connection called USB 2.0 transfers data 40$$\frac{1}{2}$$ times as fast as the speed of the USB 1.0. How many megabytes can a USB 2.0 connection transfer each second? _______
The megabytes can a USB 2.0 connection transfer each second will be 60$$\frac{3}{4}$$ mb.
Given that about 1$$\frac{1}{2}$$ megabytes of data each second and later connection called USB 2.0 transfers data 40$$\frac{1}{2}$$ times as fast as the speed of the USB 1.0. So the megabytes can a USB 2.0 connection transfer each second will be 1$$\frac{1}{2}$$ × 40$$\frac{1}{2}$$ which is $$\frac{3}{2}$$ × $$\frac{81}{2}$$ = $$\frac{243}{4}$$ = 60$$\frac{3}{4}$$ mb.