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.
Spark Your Learning
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.
Answer:
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.
Answer:
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.
Answer:
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?
Answer:
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?
__________________________
Answer:
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?
__________________________
Answer:
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?
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Answer:
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?
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Answer:
The area of each equal-sized part is \(\frac{7}{3}\) sq inches.
G. What is the area of the logo? Justify your reasoning.
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Answer:
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.
Answer:
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?
____________________
Answer:
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.
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C. What is the area of the logo? Explain your reasoning. Write an equation to model the problem.
____________________
Answer:
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.
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Answer:
\(\frac{27}{10}\) 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.
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Answer:
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.
Answer:
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.
On Your Own
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?
Answer:
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.
Answer:
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\(\frac{1}{2}\) = ___________________
Answer:
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\(\frac{1}{3}\)
Answer:
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\(\frac{2}{3}\) = ___________________
Answer:
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? _______
Answer:
The megabytes can a USB 2.0 connection transfer each second will be 60\(\frac{3}{4}\) mb.
Explanation:
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.
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