A ferry can carry a maximum of 30 adults or 45 children. A ticket for an adult cost $60 and a ticket for a child cost $35. A group of adults and children took 2 such ferries at maximum load on a ferry trip. They paid a total of $3495 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 |
| 60 |
60 x 60 =
$3600 |
0 |
0 x 35 =
$0 |
$3600 |
| - 2 |
|
+ 3 |
|
- $15 |
| 58 |
58 x 60 =
$3480 |
3 |
3 x 35 =
$105 |
$3585 |
| 46 |
|
21 |
|
$3495 |
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 $15.
If all 2 ferries are occupied by adults, total number of adults
= 2 x 30
= 60
Total difference
= 3600 - 3495
= $105
Small difference
= 3600 - 3585
= $15
Number of sets
= 105 ÷ 15
= 7
Total decrease in the number of adults
= 7 x 2
= 14
(a)
Number of adults
= 60 - 14
= 46
(b)
Number of children
= 7 x 3
= 21
Answer(s): (a) 46; (b) 21
