Perry, Simon and Gabriel shared $235. During a shopping trip, Perry spent
27 of his share, Simon spent
34 of his share and Gabriel spent
13 of his share. Given that all three boys each spent the same amount of money, how much more money do Perry and Simon have than Gabriel at first?
| |
Perry |
Simon |
Gabriel |
Total |
| Before |
7x3 =
21 u |
4x2 =
8 u |
3x6 =
18 u |
$235 |
| Change |
- 2x3 =
- 6 u |
- 3x2 =
- 6 u |
- 1x6 =
- 6 u |
|
| After |
5x3 =
15 u |
1x2 =
2 u |
2x6 =
12 u |
|
The amounts that Perry, Simon and Gabriel 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
= 21 u + 8 u + 18 u
= 47 u
47 u = 235
1 u = 235 ÷ 47 = 5
Amount that Perry and Simon have at first
= 21 u + 8 u
= 29 u
Amount that Perry and Simon have more than Gabriel at first
= 29 u - 18 u
= 11 u
= 11 x 5
= $55
Answer(s): $55
