Ben, Simon and Perry shared a bag of coins. Simon took 27 as many coins as Perry. Ben took twice as many coins as the total Simon and Perry took. After Ben had given 59 to Simon and 13 to Perry, Simon gave 3 to Perry. In the end, all three of them had the same number of coins. Find the difference between the number of coins that Ben and Simon had at first.
| |
Simon |
Perry |
Ben |
Total |
| Before |
2 u |
7 u |
18 u |
27 u |
| Change 1 |
+ 59 |
|
- 59 |
|
| Change 2 |
|
+ 13 |
- 13 |
|
| Change 3 |
- 3 |
+ 3 |
|
|
| After |
1x9 =
9 u |
1x9 =
9 u |
1x9 =
9 u |
3x9 =
27 u |
Total number of coins that Simon and Perry had at first
= 2 u + 7 u
= 9 u
Number of coins that Ben 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 Perry received from Simon and Ben
= 9 u - 7 u
= 2 u
2 u = 13 + 3
2 u = 16
1 u = 16 ÷ 2 = 8
Difference between the number coins that Ben and Simon had at first
= 18 u - 2 u
= 16 u
= 16 x 8
= 128
Answer(s): 128
