A ferry can carry a maximum of 30 adults or 50 children. A ticket for an adult cost $55 and a ticket for a child cost $35. A group of adults and children took 3 such ferries at maximum load on a ferry trip. The total amount paid for the children was $1550 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 |
| 90 |
90 x 55 = 4950 |
0 |
0 x 35 = 0 |
4950 - 0 = 4950 |
| - 3 |
|
+ 5 |
|
- 340 |
| 87 |
87 x 55 = 4785 |
5 |
5 x 35 = 175 |
4785 - 175 = 4610 |
| 60 |
|
50 |
|
1550 |
30 adults → 50 children
(÷10)3 adults → 5 children
From every decrease of 3 adults, there is an increase of 5 children and the difference in the total amounts paid between the adults and the children will be reduced by $340.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 30
= 90
Total difference
= 4950 - 1550
= $3400
Small difference
= 4950 - 4610
= $340
Number of sets
= 3400 ÷ 340
= 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
