Seth, Lee and Asher had a total of 137 coins. The ratio of the number of coins that Lee had to the number of coins that Asher had was 6 : 5. After Seth and Lee each gave away half of their coins, the 3 children had 96 coins left. How many coins did Lee and Asher have in the end?
| |
Seth |
Lee |
Asher |
Total |
| Before |
2 p |
6 u |
5 u |
137 |
| Change |
- 1 p |
- 3 u |
|
- 41 |
| After |
1 p |
3 u |
5 u |
96 |
Total number of coins that Seth and Lee gave away
= 137 - 96
= 41
1 p + 3 u = 41 --- (1)
1 p + 3 u + 5 u = 96
1 p + 8 u = 96 --- (2)
(2) - (1)
(1 p + 8 u) - (1 p + 3 u) = 96 - 41
8 u - 3 u = 55
5 u = 55
1 u = 55 ÷ 5 = 11
Total number of coins that Lee and Asher had in the end
= 3 u + 5 u
= 8 u
= 8 x 11
= 88
Answer(s): 88
