A ferry can carry a maximum of 30 adults or 40 children. A ticket for an adult cost $45 and a ticket for a child cost $30. 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 $2265 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 45 = 4050 |
0 |
0 x 30 = 0 |
4050 - 0 = 4050 |
- 3 |
|
+ 4 |
|
- 255 |
87 |
87 x 45 = 3915 |
4 |
4 x 30 = 120 |
3915 - 120 = 3795 |
69 |
|
28 |
|
2265 |
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 $255.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 30
= 90
Total difference
= 4050 - 2265
= $1785
Small difference
= 4050 - 3795
= $255
Number of sets
= 1785 ÷ 255
= 7
Total decrease in the number of adults
= 7 x 3
= 21
(a)
Number of adults
= 90 - 21
= 69
(b)
Number of children
= 7 x 4
= 28
Answer(s): (a) 69; (b) 28