Sean, Nick and Oliver had a total of 141 coins. The ratio of the number of coins that Nick had to the number of coins that Oliver had was 6 : 7. After Sean and Nick each gave away half of their coins, the 3 children had 102 coins left. How many coins did Nick and Oliver have at first?
| |
Sean |
Nick |
Oliver |
Total |
| Before |
2 p |
6 u |
7 u |
141 |
| Change |
- 1 p |
- 3 u |
|
- 39 |
| After |
1 p |
3 u |
7 u |
102 |
Total number of coins that Sean and Nick gave away
= 141 - 102
= 39
1 p + 3 u = 39 --- (1)
1 p + 3 u + 7 u = 102
1 p + 10 u = 102 --- (2)
(2) - (1)
(1 p + 10 u) - (1 p + 3 u) = 102 - 39
10 u - 3 u = 63
7 u = 63
1 u = 63 ÷ 7 = 9
Total number of coins that Nick and Oliver had at first
= 6 u + 7 u
= 13 u
= 13 x 9
= 117
Answer(s): 117
