Cole, Ahmad and Riordan shared $94. During a shopping trip, Cole spent
37 of his share, Ahmad spent
25 of his share and Riordan spent
13 of his share. Given that all three boys each spent the same amount of money, how much more money do Cole and Ahmad have than Riordan at first?
| |
Cole |
Ahmad |
Riordan |
Total |
| Before |
7x2 =
14 u |
5x3 =
15 u |
3x6 =
18 u |
$94 |
| Change |
- 3x2 =
- 6 u |
- 2x3 =
- 6 u |
- 1x6 =
- 6 u |
|
| After |
4x2 =
8 u |
3x3 =
9 u |
2x6 =
12 u |
|
The amounts that Cole, Ahmad and Riordan 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 + 18 u
= 47 u
47 u = 94
1 u = 94 ÷ 47 = 2
Amount that Cole and Ahmad have at first
= 14 u + 15 u
= 29 u
Amount that Cole and Ahmad have more than Riordan at first
= 29 u - 18 u
= 11 u
= 11 x 2
= $22
Answer(s): $22
