Cole and Peter had 117 pens together. Peter had 7 less pens than Cole. Both of them sold some of their pens. Peter sold 10% more pens than what Cole sold. In the end, Peter had half as many pens left as Cole. How many pens had Peter left?
| |
Cole
(1) |
Peter
(2) |
Make p the same
(2)x2 = (3) |
Comparing Cole
and Peter at first |
7 more |
|
|
| Before |
62 |
55 |
110 |
| Change |
- 10 u |
- 11 u |
- 22 u |
| After |
2 p |
1 p |
2 p |
Number of pens that Peter had at first
= (117 - 7) ÷ 2
= 110 ÷ 2
= 55
Number of pens that Cole had at first
= 55 + 7
= 62
Number of pens that Peter sold in percent
= 100% + 10%
= 110%
110% =
110100 =
1110 Make p the same.
(1) = (3)
62 - 10 u = 110 - 22 u
22 u - 10 u = 110 - 62
12 u = 48
1 u = 48 ÷ 12 = 4
Number of pens that Peter had left
= 55 - 11 u
= 55 - 11 x 4
= 55 - 44
= 11
Answer(s): 11
