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 $30. A group of adults and children took 2 such ferries at maximum load on a ferry trip. The total amount paid for the children was $1590 less than the total amount paid for the adults.
- How many adults were there?
- How many children were there?
Number of adults |
Amount paid for adults |
Number of children |
Amount paid for children |
Difference in total amounts paid |
60 |
60 x 55 = 3300 |
0 |
0 x 30 = 0 |
3300 - 0 = 3300 |
- 3 |
|
+ 4 |
|
- 285 |
57 |
57 x 55 = 3135 |
4 |
4 x 30 = 120 |
3135 - 120 = 3015 |
42 |
|
24 |
|
1590 |
30 adults → 40 children
(÷10)3 adults → 4 children
From every decrease of 3 adults, there is an increase of 4 children and the difference in the total amounts paid between the adults and the children will be reduced by $285.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 30
= 60
Total difference
= 3300 - 1590
= $1710
Small difference
= 3300 - 3015
= $285
Number of sets
= 1710 ÷ 285
= 6
Total decrease in the number of adults
= 6 x 3
= 18
(a)
Number of adults
= 60 - 18
= 42
(b)
Number of children
= 6 x 4
= 24
Answer(s): (a) 42; (b) 24