Perry and Jenson had 60 pencils together. Jenson had 4 less pencils than Perry. Both of them sold some of their pencils. Jenson sold 25% more pencils than what Perry sold. In the end, Jenson had half as many pencils left as Perry. How many pencils had Jenson left?
| |
Perry
(1) |
Jenson
(2) |
Make p the same
(2)x2 = (3) |
Comparing Perry
and Jenson at first |
4 more |
|
|
| Before |
32 |
28 |
56 |
| Change |
- 4 u |
- 5 u |
- 10 u |
| After |
2 p |
1 p |
2 p |
Number of pencils that Jenson had at first
= (60 - 4) ÷ 2
= 56 ÷ 2
= 28
Number of pencils that Perry had at first
= 28 + 4
= 32
Number of pencils that Jenson sold in percent
= 100% + 25%
= 125%
125% =
125100 =
54 Make p the same.
(1) = (3)
32 - 4 u = 56 - 10 u
10 u - 4 u = 56 - 32
6 u = 24
1 u = 24 ÷ 6 = 4
Number of pencils that Jenson had left
= 28 - 5 u
= 28 - 5 x 4
= 28 - 20
= 8
Answer(s): 8
