Marion and Erika have some coins. If Marion gives 20 coins to Erika, Erika will have 5 times the number of coins as Marion. If Erika gives 64 coins to Marion, Marion will have
23 as many coins as Erika. How many coins does Marion have at first?
|
Case 1 |
Case 2 |
|
Marion |
Erika |
Marion |
Erika |
Before |
5 u + 20 |
25 u - 20 |
12 u - 64 |
18 u + 64 |
Change |
- 20 |
+ 20 |
+ 64 |
- 64 |
After |
1x5 = 5 u |
5x5 = 25 u |
2x6 = 12 u |
3x6 = 18 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 5 is 30.
Number of coins that Marion had at first is the same in both cases.
12 u - 64 = 5 u + 20
12 u - 5 u = 20 + 64
7 u = 84
1 u = 12
Number of coins that Marion had at first
= 5 u + 20
= 5 x 12 + 20
= 60 + 20
= 80
Answer(s): 80