Zane and David shared equally a bunch of longans. After Zane had eaten 45 of his longans and David had eaten 33 of his, David had 4 times as many longans left as Zane. How many longans had each of them at first?
| |
Zane |
David |
| Before |
1 u + 45 |
4 u + 33 |
| Change |
- 45 |
- 33 |
| After |
1 u |
4 u |
Number of longans each had at first was the same.
4 u + 33 = 1 u + 45
4 u - 1 u = 45 - 33
3 u = 12
1 u = 12 ÷ 3 = 4
Number of longans that they had each at first
= 1 u + 45
= 4 + 45
= 49
Answer(s): 49
