Into Math Grade 5 Module 17 Lesson 3 Answer Key Assess Reasonableness of Quotients

We included HMH Into Math Grade 5 Answer Key PDF Module 17 Lesson 3 Assess Reasonableness of Quotients to make students experts in learning maths.

HMH Into Math Grade 5 Module 17 Lesson 3 Answer Key Assess Reasonableness of Quotients

I Can estimate the quotient of a decimal division problem by using compatible numbers.

Spark Your Learning
A food store received a shipment of the spice turmeric. The store manager is trying to decide whether to sell the turmeric in 3 equal bags, 4 equal bags, 5 equal bags, or 6 equal bags. She wants the turmeric in each bag to weigh between 0.3 pound and 0.4 pound. Into how many bags should she separate the turmeric?
HMH Into Math Grade 5 Module 17 Lesson 3 Answer Key Assess Reasonableness of Quotients 1
Answer:
6 bags she should separate the turmeric and sell.

Explanation:
Total weight of turmeric bag = 2.3 lb.
Number of 3 equal bags = 2.3 ÷ 3 = 0.767.
Number of 4 equal bags = 2.3 ÷ 4 = 0.575.
Number of 5 equal bags = 2.3 ÷ 5 = 0.46.
Number of 6 equal bags = 2.3 ÷ 6 = 0.38.
Weight of the turmeric in each bag she wants between 0.3 pound and 0.4 pound.
=> Weight of each bag between 0.3 and 0.4 = 0.38.

Turn and Talk Will the weight of the turmeric in each bag be closer to 0.3 pound or 0.4 pound? Explain.
Answer:
Yes, he weight of the turmeric in each bag be closer to 0.3 pound or 0.4 pound.

Explanation:
Total weight of turmeric bag = 2.3 lb.
Number of 3 equal bags = 2.3 ÷ 3 = 0.767 lb.
Number of 4 equal bags = 2.3 ÷ 4 = 0.575 lb.
Number of 5 equal bags = 2.3 ÷ 5 = 0.46 lb.
Number of 6 equal bags = 2.3 ÷ 6 = 0.38 lb.
Weight of the turmeric in each bag she wants between 0.3 pound and 0.4 pound.
=> Weight of each bag between 0.3 and 0.4 = 0.38pounds.

Build Understanding
Question 1.
A cook bought an envelope of the spice saffron. She wants to store equal amounts of the saffron in 8 glass jars. About how much saffron will the cook put in each jar?
HMH Into Math Grade 5 Module 17 Lesson 3 Answer Key Assess Reasonableness of Quotients 2
A. What is an expression you could use to model the problem?
Answer:
Amount of saffron the cook puts in each jar = 0.2375 oz.

Explanation:
Number of glass jars she wants to store equal the saffron = 8.
Weight of the saffron envelope = 1.9 oz.
Amount of saffron the cook puts in each jar = Weight of the saffron envelope ÷ Number of glass jars she wants to store equal the saffron
= 1.9  ÷ 8
= 0.2375 oz.

B. A compatible number is close to the dividend and can be divided evenly by the divisor. How can you use a compatible number to make your estimate?
Answer:
We can choose a number that is close to the dividend. And easy to divide by the divisor. Compatible numbers are numbers that are easy to compute with mental math.

Explanation:
Compatible numbers are numbers that are close to the numbers they’re replacing that divide evenly into each other. The quotient is the result you get when you divide.
We can choose a number that is close to the dividend. And easy to divide by the divisor. Compatible numbers are numbers that are easy to compute with mental math.

C. Do you think that your estimate is more than or less than the actual amount of saffron that will be put into each jar? Why?
Answer:
Yes, my estimate is more than the actual amount of saffron that will be put into each jar because my estimate value is more and which shows the weight also increases in jars.

Explanation:
Estimation for saffron weight of my choice number is going to be bigger than actual weight of it.

 

D. What is a different compatible number that you could have used to estimate? What is the estimate?
Answer:
An estimate is a rough or approximate calculation.
My Estimation for saffron weight = 2.4 oz because its close to 1.9 oz  and can be divided by 8 evenly.

Explanation:
In mathematics, compatible numbers are the numbers that are easy to add, subtract, multiply, or divide mentally. Compatible numbers are close in value to the actual numbers that make estimating the answer and computing problems easier.
My Estimation for saffron weight = 2.4 oz because its close to 1.9 oz and can be divided by 8 evenly.

E. Is this estimate more than or less than the actual amount? How do you know?
Answer:
My estimate is more than the actual amount.

Explanation:
My Estimation for saffron weight = 2.4 oz because its close to 1.9 and can be divided by 8 evenly.
It is more than the 1.9 oz.

F. The cook thinks that there will be about 0.23 oz of saffron in each jar. Is this amount reasonable? Explain.
Answer:
Yes, is amount reasonable because amount of saffron the cook puts in each jar = 0.2375 oz, which is about 0.23.

Explanation:
Number of glass jars she wants to store equal the saffron = 8.
Weight of the saffron envelope = 1.9 oz.
Amount of saffron the cook puts in each jar = Weight of the saffron envelope ÷ Number of glass jars she wants to store equal the saffron
= 1.9  ÷ 8
= 0.2375 oz.

 

Turn and Talk Would 0.28 ounce be a more reasonable estimate than 0.23 ounce? Explain.
Answer:
No, 0.23 oz is reasonable than 0.28 oz because it is easy to calculate and to estimate.

Explanation:
Number of glass jars she wants to store equal the saffron = 8.
Weight of the saffron envelope = 1.9 oz.
Amount of saffron the cook puts in each jar = Weight of the saffron envelope ÷ Number of glass jars she wants to store equal the saffron
= 1.9  ÷ 8
= 0.2375 oz.
Comparison of 0.28 ounce be a more reasonable estimate than 0.23 ounce:
=> 1.9 × 0.28 = 0.532 oz.
=> 1.9 × 0.23 = 0.437 oz.

Step It Out
Question 2.
Twenty-three friends decide to order some food. The friends decide to share the total cost evenly. About how much will each friend pay for the food?
HMH Into Math Grade 5 Module 17 Lesson 3 Answer Key Assess Reasonableness of Quotients 3
A. Write a division expression for the problem.
Answer:
Amount of money each friend pays for the food = $6.47.

Explanation:
Number of friends decide to order some food = 23.
Total cost of food = $148.80.
The friends decide to share the total cost evenly.
=> Amount of money each friend pays for the food = Total cost of food ÷ Number of friends decide to order some food
= $148.80 ÷ 23
= $6.47.

B. Estimate the quotient. To make numbers compatible, you may have to adjust both the divisor and the dividend.
To what number could you adjust the divisor?
Answer:
161 number is the adjust for  the divisor.

Explanation:
=> Amount of money each friend pays for the food = Total cost of food ÷ Number of friends decide to order some food
= $148.80 ÷ 23
= $6.47.
Compatible number = 161 ÷ 23 = 7.

To what number could you adjust the dividend so that it can be divided evenly by your adjusted divisor?
Answer:
To adjust the dividend, we need to check the nearest next number which is divisible by divisor.
=> 161 ÷ 23 = 7.

Explanation:
By checking the factors of 23:
148.80 ÷ 23 = 6.47. So, next whole number after 6 is 7.
=> 23 × 7 = 161.

C. What is your estimate?
Answer:
My estimation for $148.80 = $161.

Explanation:
My estimation for $148.80 = $161.

D. About how much will each friend pay for the food?
Answer:
Amount of money each friend pays for the food = $6.47.

Explanation:
Number of friends decide to order some food = 23.
Total cost of food = $148.80.
The friends decide to share the total cost evenly.
=> Amount of money each friend pays for the food = Total cost of food ÷ Number of friends decide to order some food
= $148.80 ÷ 23
= $6.47.

 

Turn and Talk When do you not adjust the divisor to make numbers compatible?
Answer:
Well, we don’t adjust the divisor to make numbers compatible if they dividend is exactly divisible by the divisor.

Explanation:
Compatible numbers are numbers that are close to the numbers they’re replacing that divide evenly into each other. The quotient is the result you get when you divide.

Check Understanding
Question 1.
A ribbon is 0.42 meter long. Sean cuts the ribbon into 9 equal pieces. Is 0.05 meter a reasonable estimate for the length of each piece of ribbon? Explain.
Answer:
No, 0.05 meter is not a reasonable estimate for the length of each piece of ribbon because Length of each piece = 0.47 meter which is more than 0.05.

Explanation:
Length of the ribbon = 0.42 meter.
Number of pieces Sean cuts the ribbon = 9.
Length of each piece = Length of the ribbon ÷ Number of pieces Sean cuts the ribbon
= 0.42 ÷ 9
= 0.47 meter.

 

Estimate the quotient by using compatible numbers.
Question 2.
158.2 ÷ 7 Dividend: ____________ Divisor: ____________ Quotient: ____________
Answer:
Dividend: 158.2      Divisor: 7            Quotient: 22.6.

Explanation:
158.2 ÷ 7 = 22.6.
Compatible number = 161 ÷ 7 = 23.

Question 3.
183.75 ÷ 18 Dividend: ____________ Divisor: ____________ Quotient: ____________
Answer:
Dividend: 183.75          Divisor: 18            Quotient: 10.2

Explanation:
183.75 ÷ 18 = 10.2.
Compatible number = 198 ÷ 18 = 11.

On Your Own
Question 4.
Model with Mathematics Kathy pours equal amounts of the water from the bucket into each of 7 bottles. About how much water goes into each bottle?
HMH Into Math Grade 5 Module 17 Lesson 3 Answer Key Assess Reasonableness of Quotients 4
Write a division expression for the problem.
Answer:
Number of liters of water goes into each bottle = 0.92.

Explanation:
Number of bottles Kathy pours equal amounts of the water from the bucket = 7.
Weight of bucket of water = 6.5 Liters.
Number of liters of water goes into each bottle = Weight of bucket of water ÷ Number of bottles Kathy pours equal amounts of the water from the bucket
= 6.5 ÷ 7
= 0.92.

Rename the dividend as a number of tenths, and adjust it to be a compatible number with 7.
Answer:
6.5 + 0.5 = 7.

Explanation:
Rename the dividend as a number of tenths.
=> 6.5 + 0.5 = 7.

Estimate the quotient. About how much water goes into each bottle?
Answer:
Number of liters of water goes into each bottle = Weight of bucket of water ÷ Number of bottles Kathy pours equal amounts of the water from the bucket
= 6.5 ÷ 7
= 0.92.

Explanation:
Estimation of quotient for 6.5 liters = around 0.95 liter.

Question 5.
Terence collects 0.61 pound of sand along a river. He wants to distribute the sand equally among 8 buckets. Is it reasonable for Terence to put 0.08 pound of sand in each bucket? Explain.
Answer:
No, its not reasonable for Terence to put 0.08 pound of sand in each bucket because sand won’t be enough to put in all buckets equally as actually Number of pounds of sand goes into each bucket = 0.07625.

Explanation:
Weight of sand along a river Terence collects = 0.61 pound.
Number of buckets he wants to distribute the sand equally = 8.
Number of pounds of sand goes into each bucket = Weight of sand along a river Terence collects ÷ Number of buckets he wants to distribute the sand equally
= 0.61 pound ÷ 8
= 0.07625.

Question 6.
Financial Literacy Ms. Smythe buys a bike on credit for $434.95. To avoid paying interest, she agrees to make 12 equal monthly payments for the bike. About how much will Ms. Smythe pay each month?
Answer:
Amount of money she pays each month = $36.25.

Explanation:
Amount of bike Ms. Smythe buys on credit = $434.95.
Number of equal monthly payments for the bike she agrees to pay = 12.
Amount of money she pays each month = Amount of bike Ms. Smythe buys on credit ÷ Number of equal monthly payments for the bike she agrees to pay
= $434.95 ÷ 12
= $36.25.

Estimate the quotient by using compatible numbers.
Question 7.
21.8 ÷ 6 Dividend: ____________ Divisor: ____________ Quotient: ____________
Answer:
Dividend: 21.8          Divisor: 6           Quotient: 3.633.

Explanation:
Dividend ÷ Divisor = Quotient.
21.8 ÷ 6 = 3.633.
Compatible numbers: 24 ÷ 6 = 4.

Question 8.
141.19 ÷ 32 Dividend: ____________ Divisor: ____________ Quotient: ____________
Answer:
Dividend: 141.19       Divisor: 32         Quotient: 4.4122.

Explanation:
Dividend ÷ Divisor = Quotient.
141.19 ÷ 32 = 4.4122.
Compatible numbers: 160 ÷ 32 = 5.

I’m in a Learning Mindset!
What types of decisions did I make when estimating decimal quotients?
Answer:
It is important for the students to know how to estimate because sometimes they may not need to find an exact answer for a problem.

Explanation:
It is important for the students to know how to estimate because sometimes they may not need to find an exact answer for a problem. Also, estimation is an excellent way for the students to check their quotients to see if it is reasonable.

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