John, Sean and Xavier shared a bag of buttons. Sean took 35 as many buttons as Xavier. John took twice as many buttons as the total Sean and Xavier took. After John had given 27 to Sean and 5 to Xavier, Sean gave 7 to Xavier. In the end, all three of them had the same number of buttons. Find the difference between the number of buttons that John and Sean had at first.
| |
Sean |
Xavier |
John |
Total |
| Before |
3 u |
5 u |
16 u |
24 u |
| Change 1 |
+ 27 |
|
- 27 |
|
| Change 2 |
|
+ 5 |
- 5 |
|
| Change 3 |
- 7 |
+ 7 |
|
|
| After |
1x8 =
8 u |
1x8 =
8 u |
1x8 =
8 u |
3x8 =
24 u |
Total number of buttons that Sean and Xavier had at first
= 3 u + 5 u
= 8 u
Number of buttons that John had at first
= 2 x 8 u
= 16 u
The total number of buttons remains unchanged. Make the total number of buttons the same. LCM of 3 and 24 is 24.
Number of buttons that Xavier received from Sean and John
= 8 u - 5 u
= 3 u
3 u = 5 + 7
3 u = 12
1 u = 12 ÷ 3 = 4
Difference between the number buttons that John and Sean had at first
= 16 u - 3 u
= 13 u
= 13 x 4
= 52
Answer(s): 52
