Oscar and Bobby had some coins in the ratio of 5 : 2. Bobby and Sean had some coins in the ratio of 3 : 2. Oscar sold 24 coins to Sean and ended up with half the number of coins Bobby had. How many coins did Sean have in the end?
| Oscar |
Bobby |
Sean |
| 5x3 |
2x3 |
|
| |
3x2 |
2x2 |
| 15 u |
6 u |
4 u |
The number of coins that Bobby had at first is repeated. Make the number of coins that Bobby had at first the same. LCM of 2 and 3 is 6.
| |
Oscar |
Bobby |
Sean |
| Before |
15 u |
6 u |
4 u |
| Change |
- 24 |
|
+ 24 |
| After |
3 u |
6 u |
4 u + 24 |
Number of coins that Oscar had in the end
= 6 u ÷ 2
= 3 u
Number of coins that Oscar sold to Sean
= 15 u - 3 u
= 12 u
12 u = 24
1 u = 24 ÷ 12 = 2
Number of coins that Sean had in the end
= 4 u + 24
= 4 x 2 + 24
= 8 + 24
= 32
Answer(s): 32
