Liam, Jenson and Howard shared $62. During a shopping trip, Liam spent
35 of his share, Jenson spent
23 of his share and Howard spent
12 of his share. Given that all three boys each spent the same amount of money, how much more money do Liam and Jenson have than Howard at first?
| |
Liam |
Jenson |
Howard |
Total |
| Before |
5x2 =
10 u |
3x3 =
9 u |
2x6 =
12 u |
$62 |
| Change |
- 3x2 =
- 6 u |
- 2x3 =
- 6 u |
- 1x6 =
- 6 u |
|
| After |
2x2 =
4 u |
1x3 =
3 u |
1x6 =
6 u |
|
The amounts that Liam, Jenson and Howard 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
= 10 u + 9 u + 12 u
= 31 u
31 u = 62
1 u = 62 ÷ 31 = 2
Amount that Liam and Jenson have at first
= 10 u + 9 u
= 19 u
Amount that Liam and Jenson have more than Howard at first
= 19 u - 12 u
= 7 u
= 7 x 2
= $14
Answer(s): $14
