A ferry can carry a maximum of 25 adults or 35 children. A ticket for an adult cost $45 and a ticket for a child cost $20. A group of adults and children took 2 such ferries at maximum load on a ferry trip. The total amount paid for the children was $60 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 |
50 |
50 x 45 = 2250 |
0 |
0 x 20 = 0 |
2250 - 0 = 2250 |
- 5 |
|
+ 7 |
|
- 365 |
45 |
45 x 45 = 2025 |
7 |
7 x 20 = 140 |
2025 - 140 = 1885 |
20 |
|
42 |
|
60 |
25 adults → 35 children
(÷5)5 adults → 7 children
From every decrease of 5 adults, there is an increase of 7 children and the difference in the total amounts paid between the adults and the children will be reduced by $365.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 25
= 50
Total difference
= 2250 - 60
= $2190
Small difference
= 2250 - 1885
= $365
Number of sets
= 2190 ÷ 365
= 6
Total decrease in the number of adults
= 6 x 5
= 30
(a)
Number of adults
= 50 - 30
= 20
(b)
Number of children
= 6 x 7
= 42
Answer(s): (a) 20; (b) 42