A ferry can carry a maximum of 30 adults or 45 children. A ticket for an adult cost $40 and a ticket for a child cost $25. A group of adults and children took 4 such ferries at maximum load on a ferry trip. They paid a total of $4775 for the tickets.
- How many adults were there?
- How many children were there?
Number of adults |
Amount paid for adults |
Number of children |
Amount paid for children |
Total amount paid |
120 |
120 x 40 = $4800 |
0 |
0 x 25 = $0 |
$4800 |
- 2 |
|
+ 3 |
|
- $5 |
118 |
118 x 40 = $4720 |
3 |
3 x 25 = $75 |
$4795 |
110 |
|
15 |
|
$4775 |
30 adults → 45 children
2 adults → 3 children
From every decrease of 2 adults, there is an increase of 3 children and the total amount paid will be reduced by $5.
If all 4 ferries are occupied by adults, total number of adults
= 4 x 30
= 120
Total difference
= 4800 - 4775
= $25
Small difference
= 4800 - 4795
= $5
Number of sets
= 25 ÷ 5
= 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