Charlie, Ken and Vaidev shared a bag of coins. Ken took 27 as many coins as Vaidev. Charlie took twice as many coins as the total Ken and Vaidev took. After Charlie had given 25 to Ken and 2 to Vaidev, Ken gave 4 to Vaidev. In the end, all three of them had the same number of coins. Find the difference between the number of coins that Charlie and Ken had at first.
| |
Ken |
Vaidev |
Charlie |
Total |
| Before |
2 u |
7 u |
18 u |
27 u |
| Change 1 |
+ 25 |
|
- 25 |
|
| Change 2 |
|
+ 2 |
- 2 |
|
| Change 3 |
- 4 |
+ 4 |
|
|
| After |
1x9 =
9 u |
1x9 =
9 u |
1x9 =
9 u |
3x9 =
27 u |
Total number of coins that Ken and Vaidev had at first
= 2 u + 7 u
= 9 u
Number of coins that Charlie 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 Vaidev received from Ken and Charlie
= 9 u - 7 u
= 2 u
2 u = 2 + 4
2 u = 6
1 u = 6 ÷ 2 = 3
Difference between the number coins that Charlie and Ken had at first
= 18 u - 2 u
= 16 u
= 16 x 3
= 48
Answer(s): 48
