John had 200 more coins than Ian. John gave 25% of his coins to Ian. Ian in return gave 60% of his coins to John. In the end, Ian had 880 less coins than John. How many coins did John have at first?
| |
John |
Ian |
Comparing John
and Peter at first |
200 more |
|
| Before |
4 u + 200 |
4 u |
| Change 1 |
- 1 u - 50 |
+ 1 u + 50 |
| After 1 |
3 u + 150 |
5 u + 50 |
| Change 2 |
+ 3 u + 30 |
- 3 u - 30 |
| After 2 |
6 u + 180 |
2 u + 20 |
25% =
25100 =
1425% x 200
=
25100 x 200
= 50
60% x 50
=
60100 x 50
= 30
60% x 5 u
=
60100 x 5 u
= 3 u
Ian had 880 less coins than John in the end. If another 880 coins are given to Ian, both will have the same number of coins.
6 u + 180 = 2 u + 20 + 880
6 u - 2 u = 20 + 880 - 180
4 u = 720
1 u = 720 ÷ 4 = 40
Number of Ian's coins at first
= 4 u
= 4 x 40
= 160
Answer: 160
