Esther and Hilda have some coins. If Esther gives 39 coins to Hilda, Hilda will have 5 times the number of coins as Esther. If Hilda gives 31 coins to Esther, Esther will have
25 as many coins as Hilda. How many coins does Esther have at first?
|
Case 1 |
Case 2 |
|
Esther |
Hilda |
Esther |
Hilda |
Before |
7 u + 39 |
35 u - 39 |
12 u - 31 |
30 u + 31 |
Change |
- 39 |
+ 39 |
+ 31 |
- 31 |
After |
1x7 = 7 u |
5x7 = 35 u |
2x6 = 12 u |
5x6 = 30 u |
The total number of coins in both cases remains unchanged. Make the total number of coins in both cases the same. LCM of 6 and 7 is 42.
Number of coins that Esther had at first is the same in both cases.
12 u - 31 = 7 u + 39
12 u - 7 u = 39 + 31
5 u = 70
1 u = 14
Number of coins that Esther had at first
= 7 u + 39
= 7 x 14 + 39
= 98 + 39
= 137
Answer(s): 137