A ferry can carry a maximum of 20 adults or 50 children. A ticket for an adult cost $60 and a ticket for a child cost $40. 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 $2880 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 |
80 |
80 x 60 = 4800 |
0 |
0 x 40 = 0 |
4800 - 0 = 4800 |
- 2 |
|
+ 5 |
|
- 320 |
78 |
78 x 60 = 4680 |
5 |
5 x 40 = 200 |
4680 - 200 = 4480 |
68 |
|
30 |
|
2880 |
20 adults → 50 children
(÷10)2 adults → 5 children
From every decrease of 2 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 $320.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 20
= 80
Total difference
= 4800 - 2880
= $1920
Small difference
= 4800 - 4480
= $320
Number of sets
= 1920 ÷ 320
= 6
Total decrease in the number of adults
= 6 x 2
= 12
(a)
Number of adults
= 80 - 12
= 68
(b)
Number of children
= 6 x 5
= 30
Answer(s): (a) 68; (b) 30