A ferry can carry a maximum of 35 adults or 50 children. A ticket for an adult cost $55 and a ticket for a child cost $30. A group of adults and children took 3 such ferries at maximum load on a ferry trip. They paid a total of $5095 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 |
105 |
105 x 55 = $5775 |
0 |
0 x 30 = $0 |
$5775 |
- 7 |
|
+ 10 |
|
- $85 |
98 |
98 x 55 = $5390 |
10 |
10 x 30 = $300 |
$5690 |
49 |
|
80 |
|
$5095 |
35 adults → 50 children
7 adults → 10 children
From every decrease of 7 adults, there is an increase of 10 children and the total amount paid will be reduced by $85.
If all 3 ferries are occupied by adults, total number of adults
= 3 x 35
= 105
Total difference
= 5775 - 5095
= $680
Small difference
= 5775 - 5690
= $85
Number of sets
= 680 ÷ 85
= 8
Total decrease in the number of adults
= 8 x 7
= 56
(a)
Number of adults
= 105 - 56
= 49
(b)
Number of children
= 8 x 10
= 80
Answer(s): (a) 49; (b) 80