Paul, Xavier and Valen shared $260. During a shopping trip, Paul spent
310 of his share, Xavier spent
27 of his share and Valen spent
14 of his share. Given that all three boys each spent the same amount of money, how much more money do Paul and Xavier have than Valen at first?
| |
Paul |
Xavier |
Valen |
Total |
| Before |
10x2 =
20 u |
7x3 =
21 u |
4x6 =
24 u |
$260 |
| Change |
- 3x2 =
- 6 u |
- 2x3 =
- 6 u |
- 1x6 =
- 6 u |
|
| After |
7x2 =
14 u |
5x3 =
15 u |
3x6 =
18 u |
|
The amounts that Paul, Xavier and Valen 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
= 20 u + 21 u + 24 u
= 65 u
65 u = 260
1 u = 260 ÷ 65 = 4
Amount that Paul and Xavier have at first
= 20 u + 21 u
= 41 u
Amount that Paul and Xavier have more than Valen at first
= 41 u - 24 u
= 17 u
= 17 x 4
= $68
Answer(s): $68
