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Lob: Dividing using the chunking method. “Rather than take away small amounts, we take away chunks !”. 48. ÷. 6. =. 8. 48. 6. 6. 6. 6. 6. 6. 6. 6. 0. 6. 12. 18. 36. 42. 30. 24. 6-6. 12-6. 18-6. 24-6. 30-6. 36-6. 42-6. 48-6. How many 6’s have been subtracted?.
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Lob: Dividing using the chunking method. “Rather than take away small amounts, we take away chunks!”
48 ÷ 6 = 8 48 6 6 6 6 6 6 6 6 0 6 12 18 36 42 30 24 6-6 12-6 18-6 24-6 30-6 36-6 42-6 48-6 How many 6’s have been subtracted? That means, there are 8 sixes in 48!
48 ÷ 6 = 8 48 6 6 6 6 6 6 6 6 0 6 12 18 36 42 30 24 6-6 12-6 18-6 24-6 30-6 36-6 42-6 48-6 How many 6’s have been subtracted? That means, there are 8 sixes in 48!
42 ÷ 7 = 6 42 7 7 7 7 7 7 0 7 28 35 21 14 7-7 14-7 21-7 28-7 35-7 42-7 How many 7’s have been subtracted? That means, there are 6 sevens in 42!
63 ÷ 9 = 7 63 9 9 9 9 9 9 9 0 9 18 45 54 36 27 9-9 18-9 27-9 36-9 45-9 54-9 63-9 How many 9’s have been subtracted? That means, there are 7 nines in 63!
Thumbs up if you understand, how to use the number line and down if you don’t. Now we can go onto “chunking”. Rather than take away small amounts we can take away CHUNKS!
Q. 84 divided by 4 = • We know we can write 84÷ 4. this means “How many 4’sare there in 84?” 2. On a number line using small jumps of 4 will take a lot of time. 3. We need to think bigger. Change the small ‘chunks’ of 4+4+4+4+4+4+4+4+4+4, (ten times) into larger chunks such as 4X 10 (40). This makes the chunks easier and quicker. 4 x 10 = 40 4 x 10 = 40 4 x 1 = 4 84 0 4 44 • 10 chunks of 4 (40) • Another 10chunks of 4 (40) • Finally 1chunk of 4 (4) 84 ÷ 4 = 21
Q. 96 divided by 6 = • We know we can write 96÷ 6. this means “How many 6’sare there in 96?” 2. On a number line using small jumps of 6 will take a lot of time. 3. We need to think bigger. Change the small ‘chunks’ of 6+6+6+6+6+6+6+6+6+6, (ten times) into larger chunks such as 6 X 10 (60). This makes the chunks easier and quicker. 6 x 6 = 36 6 x 10 = 60 96 0 36 • 10 chunks of 6 (60) • Finally 6chunks of 6 (36) 96 ÷ 6 = 16
Q. 108 divided by 6 = • We know we can write 108÷ 6. this means “How many 6’sare there in 108?” 2. On a number line using small jumps of 6 will take a lot of time. 3. We need to think bigger. Change the small ‘chunks’ of 6+6+6+6+6+6+6+6+6+6, (ten times) into larger chunks such as 6 X 10 (60). This makes the chunks easier and quicker. 6 x 10 = 60 6 x 8 = 48 108 0 48 108 ÷ 6 = 18 • 10 chunks of 6 (60) • Finally 8chunks of 6 (48)
What’s the biggest divisor chunks (2, 5, 6 and 3) you can remove from their dividend? Work it out! Don’t forget to use a number line! • 12 ÷ 2 = • 45 ÷ 3 = • 78 ÷ 6 = • 54 ÷ 3 = • Challenge calculations- 240 ÷ 15 = • 161 ÷ 5 = • 135 ÷ 4 = 6 15 13 18 16 32 r 1 33 r 3 REMEMBER-Dividend÷Divisor=Quotient
Q.If a box holds 28 chocolates, how many boxes can be filled with 350 chocolates? A.12boxes can be filled 10plus2chunksof28equals12 chunks. • We know we can write 350 ÷ 28. this means “How many 28’s are there in 350?” • On a number line using small jumps of 28 will take a lot of time. _____________________________________________________________ 0 +28 TEN TIMES 280 336 350 3. We need to think bigger. Change the small ‘chunks’ of 28+28+28, (ten times) into larger chunks such as 28 X 10 (280). This makes the chunks easier and quicker. 28X 10 28 X 2 14 Left 4. 10 chunks of 28 (280) 5. 2 chunks of 28( 56) 6. There are not enough chocolates to fill a further box as there are 350- 336 (14) left.
What’s the biggest divisor chunks (2, 5, 6 and 3) you can remove from their dividend? Work it out! Don’t forget to use a number line! • 12 ÷ 2 = • 108 ÷ 5 = • 60 ÷ 6 = • 54 ÷ 3 = • Challenge calculations- 240 ÷ 15 = • 161 ÷ 5 = • 135 ÷ 4 = 6 21 r 2 10 18 16 32 r 1 33 r 3 REMEMBER-Dividend÷Divisor=Quotient
Next step!Beginning a more formal method. The number line is turned vertically, and “chunks” of a number can be taken away from the starting number until you get to 0. 72 ÷ 5 = 10x5 -50 4x5 -20 -2 14 r 2 72 • 2 • 0 ( 10 lots of 5 ) ( 22 left ) 2 2 - 2 0 ( 4 lots of 5 ) 22 ( 2remaining) 2 So there are 14 lots of 5 and 2 left over. 0