Perry, Bobby and Ryan had a total of 98 coins. The ratio of the number of coins that Bobby had to the number of coins that Ryan had was 6 : 5. After Perry and Bobby each gave away half of their coins, the 3 children had 69 coins left. How many coins did Bobby and Ryan have at first?
| |
Perry |
Bobby |
Ryan |
Total |
| Before |
2 p |
6 u |
5 u |
98 |
| Change |
- 1 p |
- 3 u |
|
- 29 |
| After |
1 p |
3 u |
5 u |
69 |
Total number of coins that Perry and Bobby gave away
= 98 - 69
= 29
1 p + 3 u = 29 --- (1)
1 p + 3 u + 5 u = 69
1 p + 8 u = 69 --- (2)
(2) - (1)
(1 p + 8 u) - (1 p + 3 u) = 69 - 29
8 u - 3 u = 40
5 u = 40
1 u = 40 ÷ 5 = 8
Total number of coins that Bobby and Ryan had at first
= 6 u + 5 u
= 11 u
= 11 x 8
= 88
Answer(s): 88
