A ferry can carry a maximum of 30 adults or 50 children. A ticket for an adult cost $35 and a ticket for a child cost $30. 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 $1080 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 |
60 |
60 x 35 = 2100 |
0 |
0 x 30 = 0 |
2100 - 0 = 2100 |
- 3 |
|
+ 5 |
|
- 255 |
57 |
57 x 35 = 1995 |
5 |
5 x 30 = 150 |
1995 - 150 = 1845 |
48 |
|
20 |
|
1080 |
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 $255.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 30
= 60
Total difference
= 2100 - 1080
= $1020
Small difference
= 2100 - 1845
= $255
Number of sets
= 1020 ÷ 255
= 4
Total decrease in the number of adults
= 4 x 3
= 12
(a)
Number of adults
= 60 - 12
= 48
(b)
Number of children
= 4 x 5
= 20
Answer(s): (a) 48; (b) 20