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 $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 $1700 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 25 = 0 |
4050 - 0 = 4050 |
- 3 |
|
+ 4 |
|
- 235 |
87 |
87 x 45 = 3915 |
4 |
4 x 25 = 100 |
3915 - 100 = 3815 |
60 |
|
40 |
|
1700 |
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 $235.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 30
= 90
Total difference
= 4050 - 1700
= $2350
Small difference
= 4050 - 3815
= $235
Number of sets
= 2350 ÷ 235
= 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 4
= 40
Answer(s): (a) 60; (b) 40