Min and Mary have some erasers. If Min gives 40 erasers to Mary, Mary will have 5 times the number of erasers as Min. If Mary gives 25 erasers to Min, Min will have
35 as many erasers as Mary. How many erasers does Min have at first?
|
Case 1 |
Case 2 |
|
Min |
Mary |
Min |
Mary |
Before |
4 u + 40 |
20 u - 40 |
9 u - 25 |
15 u + 25 |
Change |
- 40 |
+ 40 |
+ 25 |
- 25 |
After |
1x4 = 4 u |
5x4 = 20 u |
3x3 = 9 u |
5x3 = 15 u |
The total number of erasers in both cases remains unchanged. Make the total number of erasers in both cases the same. LCM of 6 and 8 is 24.
Number of erasers that Min had at first is the same in both cases.
9 u - 25 = 4 u + 40
9 u - 4 u = 40 + 25
5 u = 65
1 u = 13
Number of erasers that Min had at first
= 4 u + 40
= 4 x 13 + 40
= 52 + 40
= 92
Answer(s): 92