A ferry can carry a maximum of 30 adults or 40 children. A ticket for an adult cost $55 and a ticket for a child cost $40. A group of adults and children took 4 such ferries at maximum load on a ferry trip. They paid a total of $6580 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 55 =
$6600 |
0 |
0 x 40 =
$0 |
$6600 |
| - 3 |
|
+ 4 |
|
- $5 |
| 117 |
117 x 55 =
$6435 |
4 |
4 x 40 =
$160 |
$6595 |
| 108 |
|
16 |
|
$6580 |
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 $5.
If all 4 ferries are occupied by adults, total number of adults
= 4 x 30
= 120
Total difference
= 6600 - 6580
= $20
Small difference
= 6600 - 6595
= $5
Number of sets
= 20 ÷ 5
= 4
Total decrease in the number of adults
= 4 x 3
= 12
(a)
Number of adults
= 120 - 12
= 108
(b)
Number of children
= 4 x 4
= 16
Answer(s): (a) 108; (b) 16
