Irene and Joelle have some coins. If Irene gives 16 coins to Joelle, Joelle will have 4 times the number of coins as Irene. If Joelle gives 82 coins to Irene, Irene will have
35 as many coins as Joelle. How many coins does Irene have at first?
|
Case 1 |
Case 2 |
|
Irene |
Joelle |
Irene |
Joelle |
Before |
8 u + 16 |
32 u - 16 |
15 u - 82 |
25 u + 82 |
Change |
- 16 |
+ 16 |
+ 82 |
- 82 |
After |
1x8 = 8 u |
4x8 = 32 u |
3x5 = 15 u |
5x5 = 25 u |
The total number of coins in both cases remains unchanged. Make the total number of coins in both cases the same. LCM of 5 and 8 is 40.
Number of coins that Irene had at first is the same in both cases.
15 u - 82 = 8 u + 16
15 u - 8 u = 16 + 82
7 u = 98
1 u = 14
Number of coins that Irene had at first
= 8 u + 16
= 8 x 14 + 16
= 112 + 16
= 128
Answer(s): 128