Liam, Jack and Pierre shared a bag of coins. Jack took 37 as many coins as Pierre. Liam took twice as many coins as the total Jack and Pierre took. After Liam had given 72 to Jack and 18 to Pierre, Jack gave 9 to Pierre. In the end, all three of them had the same number of coins. Find the difference between the number of coins that Liam and Jack had at first.
| |
Jack |
Pierre |
Liam |
Total |
| Before |
3 u |
7 u |
20 u |
30 u |
| Change 1 |
+ 72 |
|
- 72 |
|
| Change 2 |
|
+ 18 |
- 18 |
|
| Change 3 |
- 9 |
+ 9 |
|
|
| After |
1x10 =
10 u |
1x10 =
10 u |
1x10 =
10 u |
3x10 =
30 u |
Total number of coins that Jack and Pierre had at first
= 3 u + 7 u
= 10 u
Number of coins that Liam had at first
= 2 x 10 u
= 20 u
The total number of coins remains unchanged. Make the total number of coins the same. LCM of 3 and 30 is 30.
Number of coins that Pierre received from Jack and Liam
= 10 u - 7 u
= 3 u
3 u = 18 + 9
3 u = 27
1 u = 27 ÷ 3 = 9
Difference between the number coins that Liam and Jack had at first
= 20 u - 3 u
= 17 u
= 17 x 9
= 153
Answer(s): 153
