Basic Civil Engineering first year Notes- Chapter 4 Building.pptx
Dividing by Chunking (4 digit by 2 digit)
1. Dividing a 4 Digit Number by a 2 Digit Number For Example: 1937 ÷ 13 By Chunking For more maths help & free games related to this, visit: www.makemymathsbetter.com
2. When “chunking” I think of the question as referring to a certain number of sweets (in this case 1937) that needs to be shared between a certain number of children (in this case 13). I then ask myself, how many sweets can I give to each child? 1 9 3 7 13
3. In this example, I’ll start by imagining that I’m giving each child 100 sweets. This is 13 x 100 = 1300 13 1 9 3 7
4. In this example, I’ll start by imagining that I’m giving each child 100 sweets. This is 13 x 100 = 1300 This amount is placed under the 1937 13 x 100 1 9 3 7 13 1 3 0 0
5. To find the number of sweets remaining, I then subtract 1300 from 1937 13 x 100 1 9 3 7 13 1 3 0 0
6. To find the number of sweets remaining, I then subtract 1300 from 1937 Which is 637 13 x 100 1 9 3 7 13 1 3 0 0 - 6 3 7
7. From the remaining 637 “sweets”, I then imagine that I am giving 40 to each child. This is 13 x 40 = 520 13 x 100 1 9 3 7 13 1 3 0 0 - 6 3 7
8. From the remaining 637 “sweets”, I then imagine that I am giving 40 to each child. This is 13 x 40 = 520 This amount is placed under the 637 13 x 100 1 9 3 7 13 1 3 0 0 - 6 3 7 5 2 0 13 x 40
9. To find the number of sweets remaining, I then subtract 520 from 637 13 x 100 1 9 3 7 13 1 3 0 0 - 6 3 7 5 2 0 13 x 40
10. To find the number of sweets remaining, I then subtract 520 from 637 Which is 117 13 x 100 1 9 3 7 13 1 3 0 0 - 6 3 7 5 2 0 13 x 40 - 1 1 7
11. From the remaining 117 “sweets”, I then imagine that I am giving 9 to each child. This is 13 x 9 = 117 13 x 100 1 9 3 7 13 1 3 0 0 - 6 3 7 5 2 0 13 x 40 - 1 1 7
12. From the remaining 117 “sweets”, I then imagine that I am giving 9 to each child. This is 13 x 9 = 117 This amount is placed under the 117 13 x 100 1 9 3 7 13 1 3 0 0 - 6 3 7 5 2 0 13 x 40 - 1 1 7 1 1 7 13 x 9
13. To find the number of sweets remaining, I then subtract 117 from 117 13 x 100 1 9 3 7 13 1 3 0 0 - 6 3 7 5 2 0 13 x 40 - 1 1 7 1 1 7 13 x 9
14. To find the number of sweets remaining, I then subtract 117 from 117 Which leaves zero 13 x 100 1 9 3 7 13 1 3 0 0 - 6 3 7 5 2 0 13 x 40 - 1 1 7 1 1 7 13 x 9 0 -
15. Altogether, each child has been given 100 + 40 + 9 sweets 13 x 100 1 9 3 7 13 1 3 0 0 - 6 3 7 5 2 0 13 x 40 - 1 1 7 1 1 7 13 x 9 0 -
16. Altogether, each child has been given 100 + 40 + 9 sweets So, 1937 ÷ 13 = 149 13 x 100 1 9 3 7 13 1 3 0 0 - 6 3 7 5 2 0 13 x 40 - 1 1 7 1 1 7 13 x 9 0 -
17. 13 x 100 1 9 3 7 13 1 3 0 0 - 6 3 7 5 2 0 13 x 40 - 1 1 7 1 1 7 13 x 9 0 - For more maths help & free games related to this, visit: www.makemymathsbetter.com