Kathy and Min have some wafers each. If Kathy gives Min 12 wafers, Kathy will have
34 as many wafers as Min. If Kathy gives Min 2 wafers, they will have an equal number of wafers. How many wafers does Min have at first?
| |
Case 1 |
Case 2 |
| |
Kathy |
Min |
Kathy |
Min |
| Before |
3 u + 12 |
4 u - 12 |
3.5 u + 2 |
3.5 u - 2 |
| Change |
- 12 |
+ 12 |
- 2 |
+ 2 |
| After |
3 u |
4 u |
3.5 u |
3.5 u |
Total number of wafers that Kathy and Min have
= 3 u + 4 u
= 7 u
Number of wafers that Kathy and Min each has in the end is the same.
Number of wafers that Kathy and Min each has in the end in Case 2
= 7 u ÷ 2
= 3.5 u
Number of wafers that Kathy had at first is the same in Case 1 and Case 2.
3.5 u + 2 = 3 u + 12
3.5 u - 3 u = 12 - 2
0.5 u = 10
1 u = 10 ÷ 0.5 = 20
Number of wafers that Min has
= 4 u - 12
= 4 x 20 - 12
= 80 - 12
= 68
Answer(s): 68
