A ferry can carry a maximum of 30 adults or 45 children. A ticket for an adult cost $35 and a ticket for a child cost $20. A group of adults and children took 3 such ferries at maximum load on a ferry trip. The total amount paid for the children was $2370 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 |
90 |
90 x 35 = 3150 |
0 |
0 x 20 = 0 |
3150 - 0 = 3150 |
- 2 |
|
+ 3 |
|
- 130 |
88 |
88 x 35 = 3080 |
3 |
3 x 20 = 60 |
3080 - 60 = 3020 |
78 |
|
18 |
|
2370 |
30 adults → 45 children
(÷15)2 adults → 3 children
From every decrease of 2 adults, there is an increase of 3 children and the difference in the total amounts paid between the adults and the children will be reduced by $130.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 30
= 90
Total difference
= 3150 - 2370
= $780
Small difference
= 3150 - 3020
= $130
Number of sets
= 780 ÷ 130
= 6
Total decrease in the number of adults
= 6 x 2
= 12
(a)
Number of adults
= 90 - 12
= 78
(b)
Number of children
= 6 x 3
= 18
Answer(s): (a) 78; (b) 18