Pierre and Bobby had an equal number of sweets at first. Pierre and Bobby ate 34 and 7 sweets respectively. In the end, Bobby has four times as many sweets as Pierre. How many sweets did each of them has at first?
| |
Pierre |
Bobby |
| Before |
1 u + 34 |
4 u + 7 |
| Change |
- 34 |
- 7 |
| After |
1 u |
4 u |
Pierre and Bobby had an equal number of sweets at first.
4 u + 7 = 1 u + 34
4 u - 1 u = 34 - 7
3 u = 27
1 u = 27 ÷ 3 = 9
Number of sweets that each of them had at first
= 1 u + 34
= 9 + 34
= 43
Answer(s): 43
