lesson 20: i can divide two and three digit dividends by two digit divisors with single digit...
TRANSCRIPT
5th Grade Module 2 – Lesson 20
Lesson 20: I can divide two and three digit dividends by two digit
divisors with single digit quotients and make connections
to a written method.
5th Grade Module 2 – Lesson 20
Estimate and Divide
607 ÷ 19 Say the divisor rounded to the nearest ten…÷ 20
Name the multiple of 20 that’s closest
to 607…600
Estimate the value…
30
5th Grade Module 2 – Lesson 20
Estimate and Divide
123 ÷ 24 Say the divisor rounded to the nearest ten…÷ 20
Name the multiple of 20 that’s closest
to 123…120
Estimate the value…
6
5th Grade Module 2 – Lesson 20
Estimate and Divide
891 ÷ 96 Say the divisor rounded to the nearest ten…÷ 10
0
Name the multiple of 100 that’s closest to
891…900
Estimate the value…
9
5th Grade Module 2 – Lesson 20
Estimate and Divide
5,482 ÷ 62 Say the divisor rounded to the nearest ten…÷ 60
Name the multiple of 60 that’s closest
to 5,482…5,400
Estimate the value…
90
5th Grade Module 2 – Lesson 20
Application Problem
Bill has 2.4 m of ribbons for crafts. He wants to share it evenly with 12 friends. How many centimeters of ribbon
would 7 friends get?
Solution on next slide…
5th Grade Module 2 – Lesson 20
Solution
2.4 m x 100 = 240 cm240 ÷ 12 =
24 tens ÷12 =2 tens =
20 cm per friend
Think of 2.4 m as 240 centimeters
5th Grade Module 2 – Lesson 20
Solution
20 cm per friend x 7 =
140 cm for 7 friends
5th Grade Module 2 – Lesson 20
Concept Development
72÷ 21
What is our whole? 72
Find a multiple of 20 close to 72 that makes this
division easy.60
Let’s use rounding to find an estimated quotient!
≈ 60 ÷ 20= 6 ÷ 2
= 3
Let’s solve using the standard algorithm on
the next page.
5th Grade Module 2 – Lesson 20
Let’s Try That One Again Using the Standard Algorithm!
72 ÷ 21 = 3 21 72 - 63
9
R 9
Check: 21 x 3 = 63
(add the remainder!) 63 + 9 = 72
5th Grade Module 2 – Lesson 20
Concept Development
94 ÷ 43
Round the divisor 40
Find a multiple of the divisor that makes the
division easy.80
Let’s use rounding to find an estimated quotient!
94 ÷ 43 ≈ 80 ÷ 40
= 8 ÷ 4 = 2
Let’s solve using the standard algorithm on
the next page.
5th Grade Module 2 – Lesson 20
Let’s Try That One Again Using the Standard Algorithm!
94 ÷ 43 = 2 43 94 - 86
8
R 8
Check: 43 x 2 = 86
(add the remainder!) 86 + 8 = 94
5th Grade Module 2 – Lesson 20
Concept Development
84 ÷ 23
Round the divisor to find a multiple of the divisor that
makes the division easy.
20
60
Let’s use rounding to find an estimated quotient!
84 ÷ 23 ≈ 60 ÷ 20
= 6 ÷ 2 = 3
Let’s solve using the standard algorithm on
the next page.
5th Grade Module 2 – Lesson 20
Let’s Try That One Again Using the Standard Algorithm!
84 ÷ 23 = 3 23 84 - 69
15
R 15
Check: 23 x 3 = 69
(add the remainder!) 69 + 15 = 84
5th Grade Module 2 – Lesson 20
Work independently to find the quotient. Remember to estimate, divide and check.
Check your work with your partner.
Solution (use ink tools):
57 ÷ 29
5th Grade Module 2 – Lesson 20
Get Ready to Complete theProblem Set on Your Own!
Complete Pages 2.F.20 & 2.F.21You will have 15 minutes to work.
Try your Best!
5th Grade Module 2 – Lesson 20
LET’S DEBRIEF• What pattern did you notice between 1(c) and 1(f)?
• Did your initial estimates work for every example in Problem (1)?Why or why not? What happened in 1(d)?
• In Problem 2, what would you tell Linda in order to help her solve the problem? What lesson does Linda need to learn?
• Explain your thought process as you set up and began to solve Problems 3 and 4.
• What is the importance of estimation when dividing with two-digit divisors?
5th Grade Module 2 – Lesson 20
EXITTICKETPage 2.F.22