Archie, Wesley and Elijah shared a bag of coins. Wesley took 27 as many coins as Elijah. Archie took thrice as many coins as the total Wesley and Elijah took. After Archie had given 57 to Wesley and 18 to Elijah, Wesley gave 7 to Elijah. In the end, all three of them had the same number of coins. Find the difference between the number of coins that Archie and Wesley had at first.
| |
Wesley |
Elijah |
Archie |
Total |
| Before |
2 u |
7 u |
27 u |
36 u |
| Change 1 |
+ 57 |
|
- 57 |
|
| Change 2 |
|
+ 18 |
- 18 |
|
| Change 3 |
- 7 |
+ 7 |
|
|
| After |
1x12 =
12 u |
1x12 =
12 u |
1x12 =
12 u |
3x12 =
36 u |
Total number of coins that Wesley and Elijah had at first
= 2 u + 7 u
= 12 u
Number of coins that Archie had at first
= 3 x 12 u
= 27 u
The total number of coins remains unchanged. Make the total number of coins the same. LCM of 3 and 36 is 36.
Number of coins that Elijah received from Wesley and Archie
= 12 u - 7 u
= 5 u
5 u = 18 + 7
5 u = 25
1 u = 25 ÷ 5 = 5
Difference between the number coins that Archie and Wesley had at first
= 27 u - 2 u
= 25 u
= 25 x 5
= 125
Answer(s): 125
