A ferry can carry a maximum of 25 adults or 45 children. A ticket for an adult cost $55 and a ticket for a child cost $30. A group of adults and children took 2 such ferries at maximum load on a ferry trip. They paid a total of $2720 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 |
50 |
50 x 55 = $2750 |
0 |
0 x 30 = $0 |
$2750 |
- 5 |
|
+ 9 |
|
- $5 |
45 |
45 x 55 = $2475 |
9 |
9 x 30 = $270 |
$2745 |
20 |
|
54 |
|
$2720 |
25 adults → 45 children
5 adults → 9 children
From every decrease of 5 adults, there is an increase of 9 children and the total amount paid will be reduced by $5.
If all 2 ferries are occupied by adults, total number of adults
= 2 x 25
= 50
Total difference
= 2750 - 2720
= $30
Small difference
= 2750 - 2745
= $5
Number of sets
= 30 ÷ 5
= 6
Total decrease in the number of adults
= 6 x 5
= 30
(a)
Number of adults
= 50 - 30
= 20
(b)
Number of children
= 6 x 9
= 54
Answer(s): (a) 20; (b) 54