 # 5th Grade Module 2 – Lesson 20

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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 Estimate the value… Say the divisor rounded to the nearest ten… Name the multiple of 20 that’s closest to 607… ÷ 20 600 30 5th Grade Module 2 – Lesson 20

Estimate and Divide 123 ÷ 24 Say the divisor rounded to the nearest ten… Name the multiple of 20 that’s closest to 123… Estimate the value… ÷ 20 120 6 5th Grade Module 2 – Lesson 20

Estimate and Divide Say the divisor rounded to the nearest ten… Estimate the value… Name the multiple of 100 that’s closest to 891… 891 ÷ 96 ÷ 100 900 9 5th Grade Module 2 – Lesson 20

Estimate and Divide Say the divisor rounded to the nearest ten… Estimate the value… Name the multiple of 60 that’s closest to 5,482… 5,482 ÷ 62 ÷ 60 5,400 90 5th Grade Module 2 – Lesson 20

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 cm 240 ÷ 12 = 24 tens ÷12 = 2 tens =
Think of 2.4 m as 240 centimeters 2.4 m x 100 = 240 cm 240 ÷ 12 = 24 tens ÷12 = 2 tens = 20 cm per friend 5th Grade Module 2 – Lesson 20

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 72 What is our whole? Let’s use rounding to find an estimated quotient! 60 Find a multiple of 20 close to 72 that makes this division easy. ≈ 60 ÷ 20 = 6 ÷ 2 Let’s solve using the standard algorithm on the next page. = 3 5th Grade Module 2 – Lesson 20

Let’s Try That One Again Using the Standard Algorithm!
72 ÷ 21 = R 9 - 63 9 Check: 21 x 3 = 63 (add the remainder!) = 72 5th Grade Module 2 – Lesson 20

Concept Development 94 ÷ 43 40 Round the divisor Let’s use rounding to find an estimated quotient! 80 Find a multiple of the divisor that makes the division easy. 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 = R 8 - 86 8 Check: 43 x 2 = 86 (add the remainder!) = 94 5th Grade Module 2 – Lesson 20

Let’s use rounding to find an estimated quotient!
Concept Development 84 ÷ 23 20 Round the divisor to find a multiple of the divisor that makes the division easy. Let’s use rounding to find an estimated quotient! 60 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 = R 15 - 69 15 Check: 23 x 3 = 69 (add the remainder!) = 84 5th Grade Module 2 – Lesson 20

57 ÷ 29 Work independently to find the quotient.
Remember to estimate, divide and check. Check your work with your partner. Solution (use ink tools): 5th Grade Module 2 – Lesson 20

Complete Pages 2.F.20 & 2.F.21 You will have 15 minutes to work. Try your Best! 5th Grade Module 2 – Lesson 20

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

5th Grade Module 2 – Lesson 20
EXIT TICKET Page 2.F.22 5th Grade Module 2 – Lesson 20