Andy, Xavier and Fred shared $212. During a shopping trip, Andy spent
37 of his share, Xavier spent
25 of his share and Fred spent
14 of his share. Given that all three boys each spent the same amount of money, how much more money do Andy and Xavier have than Fred at first?
| |
Andy |
Xavier |
Fred |
Total |
| Before |
7x2 =
14 u |
5x3 =
15 u |
4x6 =
24 u |
$212 |
| Change |
- 3x2 =
- 6 u |
- 2x3 =
- 6 u |
- 1x6 =
- 6 u |
|
| After |
4x2 =
8 u |
3x3 =
9 u |
3x6 =
18 u |
|
The amounts that Andy, Xavier and Fred spent is the same. Make the amounts that each boy spent the same. LCM of 3, 2 and 1 is 6.
Amounts that the 3 boys had at first
= 14 u + 15 u + 24 u
= 53 u
53 u = 212
1 u = 212 ÷ 53 = 4
Amount that Andy and Xavier have at first
= 14 u + 15 u
= 29 u
Amount that Andy and Xavier have more than Fred at first
= 29 u - 24 u
= 5 u
= 5 x 4
= $20
Answer(s): $20
