A ferry can carry a maximum of 30 adults or 50 children. A ticket for an adult cost $35 and a ticket for a child cost $20. A group of adults and children took 3 such ferries at maximum load on a ferry trip. They paid a total of $3100 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 35 =
$3150 |
0 |
0 x 20 =
$0 |
$3150 |
| - 3 |
|
+ 5 |
|
- $5 |
| 87 |
87 x 35 =
$3045 |
5 |
5 x 20 =
$100 |
$3145 |
| 60 |
|
50 |
|
$3100 |
30 adults → 50 children
3 adults → 5 children
From every decrease of 3 adults, there is an increase of 5 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
= 3150 - 3100
= $50
Small difference
= 3150 - 3145
= $5
Number of sets
= 50 ÷ 5
= 10
Total decrease in the number of adults
= 10 x 3
= 30
(a)
Number of adults
= 90 - 30
= 60
(b)
Number of children
= 10 x 5
= 50
Answer(s): (a) 60; (b) 50
