Luke, Luis and Sam shared $344. During a shopping trip, Luke spent
38 of his share, Luis spent
25 of his share and Sam spent
12 of his share. Given that all three boys each spent the same amount of money, how much more money do Luke and Luis have than Sam at first?
| |
Luke |
Luis |
Sam |
Total |
| Before |
8x2 =
16 u |
5x3 =
15 u |
2x6 =
12 u |
$344 |
| Change |
- 3x2 =
- 6 u |
- 2x3 =
- 6 u |
- 1x6 =
- 6 u |
|
| After |
5x2 =
10 u |
3x3 =
9 u |
1x6 =
6 u |
|
The amounts that Luke, Luis and Sam 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 + 12 u
= 43 u
43 u = 344
1 u = 344 ÷ 43 = 8
Amount that Luke and Luis have at first
= 16 u + 15 u
= 31 u
Amount that Luke and Luis have more than Sam at first
= 31 u - 12 u
= 19 u
= 19 x 8
= $152
Answer(s): $152
