A ferry can carry a maximum of 30 adults or 50 children. A ticket for an adult cost $45 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 $3320 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 45 = 5400 |
0 |
0 x 25 = 0 |
5400 - 0 = 5400 |
- 3 |
|
+ 5 |
|
- 260 |
117 |
117 x 45 = 5265 |
5 |
5 x 25 = 125 |
5265 - 125 = 5140 |
96 |
|
40 |
|
3320 |
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 $260.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 30
= 120
Total difference
= 5400 - 3320
= $2080
Small difference
= 5400 - 5140
= $260
Number of sets
= 2080 ÷ 260
= 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