Howard, Ian and Ken shared $715. During a shopping trip, Howard spent
310 of his share, Ian spent
27 of his share and Ken spent
14 of his share. Given that all three boys each spent the same amount of money, how much more money do Howard and Ian have than Ken at first?
| |
Howard |
Ian |
Ken |
Total |
| Before |
10x2 =
20 u |
7x3 =
21 u |
4x6 =
24 u |
$715 |
| Change |
- 3x2 =
- 6 u |
- 2x3 =
- 6 u |
- 1x6 =
- 6 u |
|
| After |
7x2 =
14 u |
5x3 =
15 u |
3x6 =
18 u |
|
The amounts that Howard, Ian and Ken 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 = 715
1 u = 715 ÷ 65 = 11
Amount that Howard and Ian have at first
= 20 u + 21 u
= 41 u
Amount that Howard and Ian have more than Ken at first
= 41 u - 24 u
= 17 u
= 17 x 11
= $187
Answer(s): $187
