A ferry can carry a maximum of 35 adults or 45 children. A ticket for an adult cost $55 and a ticket for a child cost $35. A group of adults and children took 4 such ferries at maximum load on a ferry trip. They paid a total of $7420 for the tickets.
- How many adults were there?
- How many children were there?
Number of adults |
Amount paid for adults |
Number of children |
Amount paid for children |
Total amount paid |
140 |
140 x 55 = $7700 |
0 |
0 x 35 = $0 |
$7700 |
- 7 |
|
+ 9 |
|
- $70 |
133 |
133 x 55 = $7315 |
9 |
9 x 35 = $315 |
$7630 |
112 |
|
36 |
|
$7420 |
35 adults → 45 children
7 adults → 9 children
From every decrease of 7 adults, there is an increase of 9 children and the total amount paid will be reduced by $70.
If all 4 ferries are occupied by adults, total number of adults
= 4 x 35
= 140
Total difference
= 7700 - 7420
= $280
Small difference
= 7700 - 7630
= $70
Number of sets
= 280 ÷ 70
= 4
Total decrease in the number of adults
= 4 x 7
= 28
(a)
Number of adults
= 140 - 28
= 112
(b)
Number of children
= 4 x 9
= 36
Answer(s): (a) 112; (b) 36