A ferry can carry a maximum of 20 adults or 50 children. A ticket for an adult cost $45 and a ticket for a child cost $25. 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 $550 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 45 = 2700 |
0 |
0 x 25 = 0 |
2700 - 0 = 2700 |
| - 2 |
|
+ 5 |
|
- 215 |
| 58 |
58 x 45 = 2610 |
5 |
5 x 25 = 125 |
2610 - 125 = 2485 |
| 40 |
|
50 |
|
550 |
20 adults → 50 children
(÷10)2 adults → 5 children
From every decrease of 2 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 $215.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 20
= 60
Total difference
= 2700 - 550
= $2150
Small difference
= 2700 - 2485
= $215
Number of sets
= 2150 ÷ 215
= 10
Total decrease in the number of adults
= 10 x 2
= 20
(a)
Number of adults
= 60 - 20
= 40
(b)
Number of children
= 10 x 5
= 50
Answer(s): (a) 40; (b) 50
