A ferry can carry a maximum of 30 adults or 40 children. A ticket for an adult cost $50 and a ticket for a child cost $25. A group of adults and children took 4 such ferries at maximum load on a ferry trip. The total amount paid for the children was $4250 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 |
| 120 |
120 x 50 = 6000 |
0 |
0 x 25 = 0 |
6000 - 0 = 6000 |
| - 3 |
|
+ 4 |
|
- 250 |
| 117 |
117 x 50 = 5850 |
4 |
4 x 25 = 100 |
5850 - 100 = 5750 |
| 99 |
|
28 |
|
4250 |
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 $250.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 30
= 120
Total difference
= 6000 - 4250
= $1750
Small difference
= 6000 - 5750
= $250
Number of sets
= 1750 ÷ 250
= 7
Total decrease in the number of adults
= 7 x 3
= 21
(a)
Number of adults
= 120 - 21
= 99
(b)
Number of children
= 7 x 4
= 28
Answer(s): (a) 99; (b) 28
