Xavier, Ben and John shared a bag of coins. Ben took 27 as many coins as John. Xavier took twice as many coins as the total Ben and John took. After Xavier had given 50 to Ben and 4 to John, Ben gave 8 to John. In the end, all three of them had the same number of coins. Find the difference between the number of coins that Xavier and Ben had at first.
| |
Ben |
John |
Xavier |
Total |
| Before |
2 u |
7 u |
18 u |
27 u |
| Change 1 |
+ 50 |
|
- 50 |
|
| Change 2 |
|
+ 4 |
- 4 |
|
| Change 3 |
- 8 |
+ 8 |
|
|
| After |
1x9 =
9 u |
1x9 =
9 u |
1x9 =
9 u |
3x9 =
27 u |
Total number of coins that Ben and John had at first
= 2 u + 7 u
= 9 u
Number of coins that Xavier had at first
= 2 x 9 u
= 18 u
The total number of coins remains unchanged. Make the total number of coins the same. LCM of 3 and 27 is 27.
Number of coins that John received from Ben and Xavier
= 9 u - 7 u
= 2 u
2 u = 4 + 8
2 u = 12
1 u = 12 ÷ 2 = 6
Difference between the number coins that Xavier and Ben had at first
= 18 u - 2 u
= 16 u
= 16 x 6
= 96
Answer(s): 96
