Kathy and Yen have some tarts each. If Kathy gives Yen 21 tarts, Kathy will have
38 as many tarts as Yen. If Kathy gives Yen 19 tarts, they will have an equal number of tarts. How many tarts does Kathy have at first?
| |
Case 1 |
Case 2 |
| |
Kathy |
Yen |
Kathy |
Yen |
| Before |
3 u + 21 |
8 u - 21 |
5.5 u + 19 |
5.5 u - 19 |
| Change |
- 21 |
+ 21 |
- 19 |
+ 19 |
| After |
3 u |
8 u |
5.5 u |
5.5 u |
Total number of tarts that Kathy and Yen have
= 3 u + 8 u
= 11 u
Number of tarts that Kathy and Yen each has in the end is the same.
Number of tarts that Kathy and Yen each has in the end in Case 2
= 11 u ÷ 2
= 5.5 u
Number of tarts that Kathy had at first is the same in Case 1 and Case 2.
5.5 u + 19 = 3 u + 21
5.5 u - 3 u = 21 - 19
0.5 u = 2
1 u = 2 ÷ 0.5 = 4
Number of tarts that Kathy has
= 3 u + 21
= 3 x 4 + 21
= 12 + 21
= 33
Answer(s): 33
