Module 2 * Lesson 20 - Mrs. Neubecker's 5th Grade

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Module 2 – Lesson 20

Objective: Divide twoand three-digit dividends by two-digit divisors with single-digit quotients and make connections to a written method.

Fluency Practice – Group Count by

Multi-Digit Numbers

• Count by 21 forward and backward, changing directions occasionally.

– 21, 42, 63, 84, 105, 126, 147, 168, 189, 210, 231,

252

• Count by 43 forward and backward, changing directions occasionally.

– 43, 86, 129, 172, 215, 258, 301, 344, 387, 430,

473

Fluency Practice – Estimate & Divide

• 607 ÷ 19

– 600 ÷ 20 = 30

• 123 ÷ 24

– 120 ÷ 20 = 6

• 891 ÷ 96

– 900 ÷ 90 = 10

• 5,482 ÷ 62

– 5,400 ÷ 60 = 90 (This is the best option for the estimate that will be closer to the actual answer. You don’t always have to follow the rounding rules in estimating.)

– 6,000 ÷ 60 = 100 (This will work, but will not be super close to the real answer.)

Fluency Practice – Divide by Multiples of Ten with Remainders

• (Solve using the standard algorithm)

– 73 ÷ 50 = 1 r 23

– 70 ÷ 30 = 2 r 10

– 432 ÷ 70 = 6 r 12

– 157 ÷ 30 =

5 r 7

1

50 73

-50 r 23

23

2

30 70

-60 r 10

10

70 432

-420

6 r 12

12

5

30 157

-150

7 r 7

Application Problem

• Billy has 2.4 m of ribbon for crafts. He wants to share it evenly with 12 friends. How many centimeters of ribbon would 7 friends get?

– 1 st step convert 2.4 meters to centimeters

• 2.4 x 100 = 240 cm

– 2 nd step divide 240 cm by 12 (there are 12 friends)

• 240 ÷ 12 ( You could think of (24 x 10) ÷ 12 = 24 ÷ 12 x 10 = 24

÷12 = 2 x 10 = 20

– 20 cm per friend

– 3 rd step multiply 20 by 7 (because we want to know how many 7 friends would get)

• 20 x 7 = 140 cm

– Final step – statement of solution

• 7 of the friends would get 140 cm of ribbon.

Concept Development – Problem 1

• 72 ÷ 21

– 1 st – find a multiple of 20 close to 72 that makes this division easy. (Think of estimating first to help get close to the actual answer in the division process.)

• 60 ÷ 20 = 3

– Why chose 60 instead of 80?

• Because 80 ÷ 20 = 4 which is too big.

– What does our estimate of 3 tell us?

• That there is about 3 twenty-ones in 72.

• Where should we record our answer?

– In the ones place, because we are going into 72, not 7.

Concept Development – Problem 1

• 72 ÷ 21

– What is our next step?

• Multiple 21 x 3

– What is 21 x 3

• 63

– What is the next step?

• Subtract 63 from 72

– What is the answer of 72-63?

• 9

– What is the next step?

• Check to see if our difference is smaller than the divisor (21).

Concept Development – Problem 1

• 72 ÷ 21

– Is 9 smaller than 21?

• Yes

– What is the next step?

• Check to see if there are any more numbers to bring down.

– Is there any more numbers to bring down?

• No

– What is the quotient (answer to a division problem)?

• 3 r 9

– How did estimating help us?

• It gave us a place to start with in our division process.

Concept Development – Problem 1

• 72 ÷ 21

– Below is the long division steps we just did.

3

21 72

-63 r 9

9

• How can we check our work?

– Take the quotient and times it by the divisor and then add the remainder.

• What is (21 x 3) + 9

– 63 + 9 = 72

• Does the answer match the dividend?

– Yes

• If you answered yes then you solved the problem correctly.

Concept Development – Problem 2

• 94 ÷ 43

– What would we estimate our dividend and divisor to? Why?

• 80 ÷ 40 = 2, because if we went to 120 ÷ 40 that would be too much.

– What number should we try first in the steps to division? Why?

• 2 because in our estimate we decided on 80 ÷ 40 = 2.

– Now complete the steps to solving the problem.

Use the standard algorithm.

Concept Development – Problem 2

• 94 ÷ 43 = 2 r 8

43

2

94

-86

8 r 8

• Now check your work.

– (43 x 2) + 8 = 86 + 8 = 94

Concept Development – Problem 3

• 84 ÷ 23 =

• Solve using the standard algorithm and estimating (to help get started). Show all your work.

• What would be the estimate?

– 80 ÷ 20 (estimate) = 4

• (note if 4 is to big then try 1 group less, which is 3.)

23

3

84

-69

15 r 15

23 x 3

69

+15

84

Concept Development – Problem 4

• 57 ÷ 29

– Solve using the standard algorithm and estimating.

– What is your answer and was the estimate too large?

– 1 r 28, yes it was too large.

Exit Ticket

• Divide, then check with multiplication

– 78 ÷ 21

– 89 ÷ 37

End of Lesson Activities

• Debrief

• Problem Set

• Homework

• Exit Ticket

Problem Set

• Divide, then check with multiplication. The first one is done for you.

• 1 a.

17 r 14

17

65

-51

14 x 3

51

+14

65

• 1 b. 49 ÷ 21 1 c. 78 ÷ 39 1 d. 84 ÷ 32 1 e. 77 ÷ 25 1 f. 68 ÷ 12

• 2. When dividing 82 by 43, Linda estimated the quotient to be 2. Examine

Linda’s work and explain what she needs to do next. On the right, show how you would solve the problem.

2

Your work

2

43 82

40 80

43 82

-86

Linda’s estimation: ??

Linda’s work:

Problem Set

• A number divided by 43 has a quotient of 3 with

28 as a remainder. Find the number. Show your work.

• Write another division problem that has a quotient of 3 and a remainder of 28.

• Mrs. Silverstein sold 91 cupcakes at a food fair.

The cupcakes were sold in boxes of “a baker’s dozen” which is 13. She sold all the cupcakes at

$15 per box. How much money did she receive?

Homework

• Divide, then check with multiplication. The first one is done for you.

• 1 a. 72 ÷ 32 r 10

31

2

72

-62

31 x 2 = 62

62 + 10 = 72

10

• 1 b. 89 ÷ 21 1 c. 94 ÷ 33 1 d. 67 ÷ 19 1 e. 79 ÷ 25 1 f. 83 ÷ 21

• 2. A 189 square-foot rectangular office has a length of 21 feet. What is the width of the office?

• 3. While preparing for a morning conference, Principal Corsetti is laying out 15 dozen bagels on square plates. Each plate can hold 14 bagels.

– A. How many plates of bagels will Mr. Corsetti have?

– B. How many more bagels would be needed to fill the final plate with bagels?

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