A ferry can carry a maximum of 30 adults or 40 children. A ticket for an adult cost $40 and a ticket for a child cost $20. A group of adults and children took 4 such ferries at maximum load on a ferry trip. They paid a total of $4600 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 |
120 |
120 x 40 = $4800 |
0 |
0 x 20 = $0 |
$4800 |
- 3 |
|
+ 4 |
|
- $40 |
117 |
117 x 40 = $4680 |
4 |
4 x 20 = $80 |
$4760 |
105 |
|
20 |
|
$4600 |
30 adults → 40 children
3 adults → 4 children
From every decrease of 3 adults, there is an increase of 4 children and the total amount paid will be reduced by $40.
If all 4 ferries are occupied by adults, total number of adults
= 4 x 30
= 120
Total difference
= 4800 - 4600
= $200
Small difference
= 4800 - 4760
= $40
Number of sets
= 200 ÷ 40
= 5
Total decrease in the number of adults
= 5 x 3
= 15
(a)
Number of adults
= 120 - 15
= 105
(b)
Number of children
= 5 x 4
= 20
Answer(s): (a) 105; (b) 20