A ferry can carry a maximum of 30 adults or 50 children. A ticket for an adult cost $40 and a ticket for a child cost $25. 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 $2840 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 |
120 |
120 x 40 = 4800 |
0 |
0 x 25 = 0 |
4800 - 0 = 4800 |
- 3 |
|
+ 5 |
|
- 245 |
117 |
117 x 40 = 4680 |
5 |
5 x 25 = 125 |
4680 - 125 = 4555 |
96 |
|
40 |
|
2840 |
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 $245.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 30
= 120
Total difference
= 4800 - 2840
= $1960
Small difference
= 4800 - 4555
= $245
Number of sets
= 1960 ÷ 245
= 8
Total decrease in the number of adults
= 8 x 3
= 24
(a)
Number of adults
= 120 - 24
= 96
(b)
Number of children
= 8 x 5
= 40
Answer(s): (a) 96; (b) 40