Charlie and Xavier collected some coins. If Charlie gave Xavier 54 coins, both would have an equal number of coins. If Xavier gave Charlie 44 coins, Charlie would have 5 times as many coins as Xavier. Find the number of coins each of them had.
- Charlie?
- Xavier?
| Case 1 |
Case 2 |
| Charlie |
Xavier |
Charlie |
Xavier |
| 3 u + 54 |
3 u - 54 |
5 u - 44 |
1 u + 44 |
| - 54 |
+ 54 |
+ 44 |
- 44 |
1x3 =
3 u |
1x3 =
3 u |
5x1 =
5 u |
1x1 =
1 u |
(a)
The total number of stamps remains unchanged in both cases. Make the total number of coins the same. LCM of 2 and 6 is 6.
The number of coins that Xavier had at first is the same in both cases.
3 u - 54 = 1 u + 44
3 u - 1 u = 54 + 44
2 u = 98
1 u = 98 ÷ 2 = 49
Number of coins that Charlie had
= 3 u + 54
= 3 x 49 + 54
= 201
(b)
Number of coins that Xavier had
= 3 u - 54
= 3 x 49 - 54
= 93
Answer(s): (a) 201; (b) 93
