Sam and Pierre had 102 markers together. Pierre had 2 less markers than Sam. Both of them sold some of their markers. Pierre sold 10% more markers than what Sam sold. In the end, Pierre had half as many markers left as Sam. How many markers had Pierre left?
| |
Sam
(1) |
Pierre
(2) |
Make p the same
(2)x2 = (3) |
Comparing Sam
and Pierre at first |
2 more |
|
|
| Before |
52 |
50 |
100 |
| Change |
- 10 u |
- 11 u |
- 22 u |
| After |
2 p |
1 p |
2 p |
Number of markers that Pierre had at first
= (102 - 2) ÷ 2
= 100 ÷ 2
= 50
Number of markers that Sam had at first
= 50 + 2
= 52
Number of markers that Pierre sold in percent
= 100% + 10%
= 110%
110% =
110100 =
1110 Make p the same.
(1) = (3)
52 - 10 u = 100 - 22 u
22 u - 10 u = 100 - 52
12 u = 48
1 u = 48 ÷ 12 = 4
Number of markers that Pierre had left
= 50 - 11 u
= 50 - 11 x 4
= 50 - 44
= 6
Answer(s): 6
