Eric and Fred collected some stickers. If Eric gave Fred 55 stickers, both would have an equal number of stickers. If Fred gave Eric 41 stickers, Eric would have 5 times as many stickers as Fred. Find the number of stickers each of them had.
- Eric?
- Fred?
| Case 1 |
Case 2 |
| Eric |
Fred |
Eric |
Fred |
| 3 u + 55 |
3 u - 55 |
5 u - 41 |
1 u + 41 |
| - 55 |
+ 55 |
+ 41 |
- 41 |
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 stickers the same. LCM of 2 and 6 is 6.
The number of stickers that Fred had at first is the same in both cases.
3 u - 55 = 1 u + 41
3 u - 1 u = 55 + 41
2 u = 96
1 u = 96 ÷ 2 = 48
Number of stickers that Eric had
= 3 u + 55
= 3 x 48 + 55
= 199
(b)
Number of stickers that Fred had
= 3 u - 55
= 3 x 48 - 55
= 89
Answer(s): (a) 199; (b) 89
