Simon and Howard had 45 pencils together. Simon had 13 more pencils than Howard. Both of them sold some of their pencils. Howard sold 40% less pencils than what Simon sold. In the end, Howard had half as many pencils left as Simon. How many pencils had Simon left?
| |
Simon
(1) |
Howard
(2) |
Make p the same
(2)x2 = (3) |
Comparing Simon
and Howard at first |
13 more |
|
|
| Before |
29 |
16 |
32 |
| Change |
- 5 u |
- 3 u |
- 6 u |
| After |
2 p |
1 p |
2 p |
Number of pencils that Howard had at first
= (45 - 13) ÷ 2
= 32 ÷ 2
= 16
Number of pencils that Simon had at first
= 16 + 13
= 29
Number of pencils that Howard sold in percent
= 100% - 40%
= 60%
60% =
60100 =
35 Make p the same.
(1) = (3)
29 - 5 u = 32 - 6 u
6 u - 5 u = 32 - 29
1 u = 3
1 u = 3 ÷ 1 = 3
Number of pencils that Simon had left
= 29 - 5 u
= 29 - 5 x 3
= 29 - 15
= 14
Answer(s): 14
