Albert, Wesley and Liam shared a bag of buttons. Wesley took 25 as many buttons as Liam. Albert took twice as many buttons as the total Wesley and Liam took. After Albert had given 12 to Wesley and 2 to Liam, Wesley gave 2 to Liam. In the end, all three of them had the same number of buttons. Find the difference between the number of buttons that Albert and Wesley had at first.
| |
Wesley |
Liam |
Albert |
Total |
| Before |
2 u |
5 u |
14 u |
21 u |
| Change 1 |
+ 12 |
|
- 12 |
|
| Change 2 |
|
+ 2 |
- 2 |
|
| Change 3 |
- 2 |
+ 2 |
|
|
| After |
1x7 =
7 u |
1x7 =
7 u |
1x7 =
7 u |
3x7 =
21 u |
Total number of buttons that Wesley and Liam had at first
= 2 u + 5 u
= 7 u
Number of buttons that Albert had at first
= 2 x 7 u
= 14 u
The total number of buttons remains unchanged. Make the total number of buttons the same. LCM of 3 and 21 is 21.
Number of buttons that Liam received from Wesley and Albert
= 7 u - 5 u
= 2 u
2 u = 2 + 2
2 u = 4
1 u = 4 ÷ 2 = 2
Difference between the number buttons that Albert and Wesley had at first
= 14 u - 2 u
= 12 u
= 12 x 2
= 24
Answer(s): 24
