A ferry can carry a maximum of 30 adults or 45 children. A ticket for an adult cost $40 and a ticket for a child cost $25. A group of adults and children took 3 such ferries at maximum load on a ferry trip. They paid a total of $3580 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 |
90 |
90 x 40 = $3600 |
0 |
0 x 25 = $0 |
$3600 |
- 2 |
|
+ 3 |
|
- $5 |
88 |
88 x 40 = $3520 |
3 |
3 x 25 = $75 |
$3595 |
82 |
|
12 |
|
$3580 |
30 adults → 45 children
2 adults → 3 children
From every decrease of 2 adults, there is an increase of 3 children and the total amount paid will be reduced by $5.
If all 3 ferries are occupied by adults, total number of adults
= 3 x 30
= 90
Total difference
= 3600 - 3580
= $20
Small difference
= 3600 - 3595
= $5
Number of sets
= 20 ÷ 5
= 4
Total decrease in the number of adults
= 4 x 2
= 8
(a)
Number of adults
= 90 - 8
= 82
(b)
Number of children
= 4 x 3
= 12
Answer(s): (a) 82; (b) 12