Elijah and Bryan had some coins in the ratio of 2 : 1. Bryan and Tim had some coins in the ratio of 6 : 5. Elijah sold 99 coins to Tim and ended up with half the number of coins Bryan had. How many coins did Tim have in the end?
| Elijah |
Bryan |
Tim |
| 2x6 |
1x6 |
|
| |
6x1 |
5x1 |
| 12 u |
6 u |
5 u |
The number of coins that Bryan had at first is repeated. Make the number of coins that Bryan had at first the same. LCM of 1 and 6 is 6.
| |
Elijah |
Bryan |
Tim |
| Before |
12 u |
6 u |
5 u |
| Change |
- 99 |
|
+ 99 |
| After |
3 u |
6 u |
5 u + 99 |
Number of coins that Elijah had in the end
= 6 u ÷ 2
= 3 u
Number of coins that Elijah sold to Tim
= 12 u - 3 u
= 9 u
9 u = 99
1 u = 99 ÷ 9 = 11
Number of coins that Tim had in the end
= 5 u + 99
= 5 x 11 + 99
= 55 + 99
= 154
Answer(s): 154
