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 $40. 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 $2760 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 40 = 0 |
4950 - 0 = 4950 |
- 3 |
|
+ 5 |
|
- 365 |
87 |
87 x 55 = 4785 |
5 |
5 x 40 = 200 |
4785 - 200 = 4585 |
72 |
|
30 |
|
2760 |
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 $365.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 30
= 90
Total difference
= 4950 - 2760
= $2190
Small difference
= 4950 - 4585
= $365
Number of sets
= 2190 ÷ 365
= 6
Total decrease in the number of adults
= 6 x 3
= 18
(a)
Number of adults
= 90 - 18
= 72
(b)
Number of children
= 6 x 5
= 30
Answer(s): (a) 72; (b) 30