Three bags, X, Y and Z contained a certain number of coins. Bag Z contained
310 as many coins as X and Y combined. There were 85 more coins in Bag X than in Bag Z. Bag Y contained 131 more coins than Bag Z. How many coins were in Bag Y?
Bag X |
Bag Y |
Bag Z |
Total |
10 u |
3 u |
13 u |
3 u + 85 |
3 u + 131 |
|
|
The total number of coins in Bag X and Bag Y is repeated.
3 u + 85 + 3 u + 131 = 10 u
6 u + 216 = 10 u
10 u - 6 u = 216
4 u = 216
1 u = 216 ÷ 4 = 54
Number of coins in Bag Y
= 3 u + 131
= 3 x 54 + 131
= 293
Answer(s): 293