Paul, Fred and Julian shared a bag of coins. Fred took 27 as many coins as Julian. Paul took four times as many coins as the total Fred and Julian took. After Paul had given 139 to Fred and 71 to Julian, Fred gave 9 to Julian. In the end, all three of them had the same number of coins. Find the difference between the number of coins that Paul and Fred had at first.
| |
Fred |
Julian |
Paul |
Total |
| Before |
2 u |
7 u |
36 u |
45 u |
| Change 1 |
+ 139 |
|
- 139 |
|
| Change 2 |
|
+ 71 |
- 71 |
|
| Change 3 |
- 9 |
+ 9 |
|
|
| After |
1x15 =
15 u |
1x15 =
15 u |
1x15 =
15 u |
3x15 =
45 u |
Total number of coins that Fred and Julian had at first
= 2 u + 7 u
= 15 u
Number of coins that Paul had at first
= 4 x 15 u
= 36 u
The total number of coins remains unchanged. Make the total number of coins the same. LCM of 3 and 45 is 45.
Number of coins that Julian received from Fred and Paul
= 15 u - 7 u
= 8 u
8 u = 71 + 9
8 u = 80
1 u = 80 ÷ 8 = 10
Difference between the number coins that Paul and Fred had at first
= 36 u - 2 u
= 34 u
= 34 x 10
= 340
Answer(s): 340
