Luis, Paul and Caden shared a bag of coins. Paul took 35 as many coins as Caden. Luis took twice as many coins as the total Paul and Caden took. After Luis had given 17 to Paul and 7 to Caden, Paul gave 2 to Caden. In the end, all three of them had the same number of coins. Find the difference between the number of coins that Luis and Paul had at first.
| |
Paul |
Caden |
Luis |
Total |
| Before |
3 u |
5 u |
16 u |
24 u |
| Change 1 |
+ 17 |
|
- 17 |
|
| Change 2 |
|
+ 7 |
- 7 |
|
| Change 3 |
- 2 |
+ 2 |
|
|
| After |
1x8 =
8 u |
1x8 =
8 u |
1x8 =
8 u |
3x8 =
24 u |
Total number of coins that Paul and Caden had at first
= 3 u + 5 u
= 8 u
Number of coins that Luis had at first
= 2 x 8 u
= 16 u
The total number of coins remains unchanged. Make the total number of coins the same. LCM of 3 and 24 is 24.
Number of coins that Caden received from Paul and Luis
= 8 u - 5 u
= 3 u
3 u = 7 + 2
3 u = 9
1 u = 9 ÷ 3 = 3
Difference between the number coins that Luis and Paul had at first
= 16 u - 3 u
= 13 u
= 13 x 3
= 39
Answer(s): 39
