Xavier and David collected some coins. If Xavier gave David 59 coins, both would have an equal number of coins. If David gave Xavier 33 coins, Xavier would have 5 times as many coins as David. Find the number of coins each of them had.
- Xavier?
- David?
| Case 1 |
Case 2 |
| Xavier |
David |
Xavier |
David |
| 3 u + 59 |
3 u - 59 |
5 u - 33 |
1 u + 33 |
| - 59 |
+ 59 |
+ 33 |
- 33 |
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 David had at first is the same in both cases.
3 u - 59 = 1 u + 33
3 u - 1 u = 59 + 33
2 u = 92
1 u = 92 ÷ 2 = 46
Number of coins that Xavier had
= 3 u + 59
= 3 x 46 + 59
= 197
(b)
Number of coins that David had
= 3 u - 59
= 3 x 46 - 59
= 79
Answer(s): (a) 197; (b) 79
