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 3 such ferries at maximum load on a ferry trip. They paid a total of $3240 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 20 = $0 |
$3600 |
- 3 |
|
+ 4 |
|
- $40 |
87 |
87 x 40 = $3480 |
4 |
4 x 20 = $80 |
$3560 |
63 |
|
36 |
|
$3240 |
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 3 ferries are occupied by adults, total number of adults
= 3 x 30
= 90
Total difference
= 3600 - 3240
= $360
Small difference
= 3600 - 3560
= $40
Number of sets
= 360 ÷ 40
= 9
Total decrease in the number of adults
= 9 x 3
= 27
(a)
Number of adults
= 90 - 27
= 63
(b)
Number of children
= 9 x 4
= 36
Answer(s): (a) 63; (b) 36