Xavier, Luke and George shared $294. During a shopping trip, Xavier spent
38 of his share, Luke spent
25 of his share and George spent
13 of his share. Given that all three boys each spent the same amount of money, how much more money do Xavier and Luke have than George at first?
| |
Xavier |
Luke |
George |
Total |
| Before |
8x2 =
16 u |
5x3 =
15 u |
3x6 =
18 u |
$294 |
| Change |
- 3x2 =
- 6 u |
- 2x3 =
- 6 u |
- 1x6 =
- 6 u |
|
| After |
5x2 =
10 u |
3x3 =
9 u |
2x6 =
12 u |
|
The amounts that Xavier, Luke and George 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
= 16 u + 15 u + 18 u
= 49 u
49 u = 294
1 u = 294 ÷ 49 = 6
Amount that Xavier and Luke have at first
= 16 u + 15 u
= 31 u
Amount that Xavier and Luke have more than George at first
= 31 u - 18 u
= 13 u
= 13 x 6
= $78
Answer(s): $78
