A ferry can carry a maximum of 20 adults or 50 children. A ticket for an adult cost $50 and a ticket for a child cost $40. 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 $1600 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 |
80 |
80 x 50 = 4000 |
0 |
0 x 40 = 0 |
4000 - 0 = 4000 |
- 2 |
|
+ 5 |
|
- 300 |
78 |
78 x 50 = 3900 |
5 |
5 x 40 = 200 |
3900 - 200 = 3700 |
64 |
|
40 |
|
1600 |
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 $300.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 20
= 80
Total difference
= 4000 - 1600
= $2400
Small difference
= 4000 - 3700
= $300
Number of sets
= 2400 ÷ 300
= 8
Total decrease in the number of adults
= 8 x 2
= 16
(a)
Number of adults
= 80 - 16
= 64
(b)
Number of children
= 8 x 5
= 40
Answer(s): (a) 64; (b) 40