Lee, Reggie and Gabriel shared a bag of coins. Reggie took 25 as many coins as Gabriel. Lee took twice as many coins as the total Reggie and Gabriel took. After Lee had given 22 to Reggie and 6 to Gabriel, Reggie gave 2 to Gabriel. In the end, all three of them had the same number of coins. Find the difference between the number of coins that Lee and Reggie had at first.
| |
Reggie |
Gabriel |
Lee |
Total |
| Before |
2 u |
5 u |
14 u |
21 u |
| Change 1 |
+ 22 |
|
- 22 |
|
| Change 2 |
|
+ 6 |
- 6 |
|
| Change 3 |
- 2 |
+ 2 |
|
|
| After |
1x7 =
7 u |
1x7 =
7 u |
1x7 =
7 u |
3x7 =
21 u |
Total number of coins that Reggie and Gabriel had at first
= 2 u + 5 u
= 7 u
Number of coins that Lee had at first
= 2 x 7 u
= 14 u
The total number of coins remains unchanged. Make the total number of coins the same. LCM of 3 and 21 is 21.
Number of coins that Gabriel received from Reggie and Lee
= 7 u - 5 u
= 2 u
2 u = 6 + 2
2 u = 8
1 u = 8 ÷ 2 = 4
Difference between the number coins that Lee and Reggie had at first
= 14 u - 2 u
= 12 u
= 12 x 4
= 48
Answer(s): 48
