Julian, Henry and Jeremy shared a bag of coins. Henry took 35 as many coins as Jeremy. Julian took twice as many coins as the total Henry and Jeremy took. After Julian had given 28 to Henry and 4 to Jeremy, Henry gave 8 to Jeremy. In the end, all three of them had the same number of coins. Find the difference between the number of coins that Julian and Henry had at first.
| |
Henry |
Jeremy |
Julian |
Total |
| Before |
3 u |
5 u |
16 u |
24 u |
| Change 1 |
+ 28 |
|
- 28 |
|
| Change 2 |
|
+ 4 |
- 4 |
|
| Change 3 |
- 8 |
+ 8 |
|
|
| After |
1x8 =
8 u |
1x8 =
8 u |
1x8 =
8 u |
3x8 =
24 u |
Total number of coins that Henry and Jeremy had at first
= 3 u + 5 u
= 8 u
Number of coins that Julian 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 Jeremy received from Henry and Julian
= 8 u - 5 u
= 3 u
3 u = 4 + 8
3 u = 12
1 u = 12 ÷ 3 = 4
Difference between the number coins that Julian and Henry had at first
= 16 u - 3 u
= 13 u
= 13 x 4
= 52
Answer(s): 52
