A ferry can carry a maximum of 20 adults or 50 children. A ticket for an adult cost $25 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 $150 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 |
|
+ 5 |
|
- 150 |
58 |
58 x 25 = 1450 |
5 |
5 x 20 = 100 |
1450 - 100 = 1350 |
42 |
|
45 |
|
150 |
20 adults → 50 children
(÷10)2 adults → 5 children
From every decrease of 2 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 $150.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 20
= 60
Total difference
= 1500 - 150
= $1350
Small difference
= 1500 - 1350
= $150
Number of sets
= 1350 ÷ 150
= 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 5
= 45
Answer(s): (a) 42; (b) 45