A ferry can carry a maximum of 30 adults or 45 children. A ticket for an adult cost $25 and a ticket for a child cost $20. 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 $510 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 25 = 1500 |
0 |
0 x 20 = 0 |
1500 - 0 = 1500 |
- 2 |
|
+ 3 |
|
- 110 |
58 |
58 x 25 = 1450 |
3 |
3 x 20 = 60 |
1450 - 60 = 1390 |
42 |
|
27 |
|
510 |
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 $110.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 30
= 60
Total difference
= 1500 - 510
= $990
Small difference
= 1500 - 1390
= $110
Number of sets
= 990 ÷ 110
= 9
Total decrease in the number of adults
= 9 x 2
= 18
(a)
Number of adults
= 60 - 18
= 42
(b)
Number of children
= 9 x 3
= 27
Answer(s): (a) 42; (b) 27