A ferry can carry a maximum of 30 adults or 45 children. A ticket for an adult cost $50 and a ticket for a child cost $35. 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 $4975 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 50 = 6000 |
0 |
0 x 35 = 0 |
6000 - 0 = 6000 |
- 2 |
|
+ 3 |
|
- 205 |
118 |
118 x 50 = 5900 |
3 |
3 x 35 = 105 |
5900 - 105 = 5795 |
110 |
|
15 |
|
4975 |
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 $205.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 30
= 120
Total difference
= 6000 - 4975
= $1025
Small difference
= 6000 - 5795
= $205
Number of sets
= 1025 ÷ 205
= 5
Total decrease in the number of adults
= 5 x 2
= 10
(a)
Number of adults
= 120 - 10
= 110
(b)
Number of children
= 5 x 3
= 15
Answer(s): (a) 110; (b) 15