Fabian, Henry and Lee shared a bag of coins. Henry took 35 as many coins as Lee. Fabian took twice as many coins as the total Henry and Lee took. After Fabian had given 22 to Henry and 2 to Lee, Henry gave 7 to Lee. In the end, all three of them had the same number of coins. Find the difference between the number of coins that Fabian and Henry had at first.
| |
Henry |
Lee |
Fabian |
Total |
| Before |
3 u |
5 u |
16 u |
24 u |
| Change 1 |
+ 22 |
|
- 22 |
|
| Change 2 |
|
+ 2 |
- 2 |
|
| Change 3 |
- 7 |
+ 7 |
|
|
| After |
1x8 =
8 u |
1x8 =
8 u |
1x8 =
8 u |
3x8 =
24 u |
Total number of coins that Henry and Lee had at first
= 3 u + 5 u
= 8 u
Number of coins that Fabian 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 Lee received from Henry and Fabian
= 8 u - 5 u
= 3 u
3 u = 2 + 7
3 u = 9
1 u = 9 ÷ 3 = 3
Difference between the number coins that Fabian and Henry had at first
= 16 u - 3 u
= 13 u
= 13 x 3
= 39
Answer(s): 39
