Pierre had 180 more coins than Brandon. Pierre gave 20% of his coins to Brandon. Brandon in return gave 50% of his coins to Pierre. In the end, Pierre had 549 more coins than Brandon. How many coins did Pierre have at first?
| |
Pierre |
Brandon |
Comparing Pierre
and Peter at first |
180 more |
|
| Before |
5 u + 180 |
5 u |
| Change 1 |
- 1 u - 36 |
+ 1 u + 36 |
| After 1 |
4 u + 144 |
6 u + 36 |
| Change 2 |
+ 3 u + 18 |
- 3 u - 18 |
| After 2 |
7 u + 162 |
3 u + 18 |
20% =
20100 =
1520% x 180
=
20100 x 180
= 36
50% x 36
=
50100 x 36
= 18
50% x 6 u
=
50100 x 6 u
= 3 u
Pierre had 549 more coins than Brandon in the end. If another 549 coins are given to Brandon, both will have the same number of coins.
7 u + 162 = 3 u + 18 + 549
7 u - 3 u = 18 + 549 - 162
15 u = 405
1 u = 405 ÷ 15 = 27
Number of Pierre's coins at first
= 5 u + 180
= 5 x 27 + 180
= 135 + 180
= 315
Answer: 315
