Ahmad, Paul and Elijah shared $715. During a shopping trip, Ahmad spent
29 of his share, Paul spent
37 of his share and Elijah spent
14 of his share. Given that all three boys each spent the same amount of money, how much more money do Ahmad and Paul have than Elijah at first?
| |
Ahmad |
Paul |
Elijah |
Total |
| Before |
9x3 =
27 u |
7x2 =
14 u |
4x6 =
24 u |
$715 |
| Change |
- 2x3 =
- 6 u |
- 3x2 =
- 6 u |
- 1x6 =
- 6 u |
|
| After |
7x3 =
21 u |
4x2 =
8 u |
3x6 =
18 u |
|
The amounts that Ahmad, Paul and Elijah spent is the same. Make the amounts that each boy spent the same. LCM of 2, 3 and 1 is 6.
Amounts that the 3 boys had at first
= 27 u + 14 u + 24 u
= 65 u
65 u = 715
1 u = 715 ÷ 65 = 11
Amount that Ahmad and Paul have at first
= 27 u + 14 u
= 41 u
Amount that Ahmad and Paul have more than Elijah at first
= 41 u - 24 u
= 17 u
= 17 x 11
= $187
Answer(s): $187
