A ferry can carry a maximum of 35 adults or 45 children. A ticket for an adult cost $50 and a ticket for a child cost $30. A group of adults and children took 3 such ferries at maximum load on a ferry trip. They paid a total of $4930 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 |
105 |
105 x 50 = $5250 |
0 |
0 x 30 = $0 |
$5250 |
- 7 |
|
+ 9 |
|
- $80 |
98 |
98 x 50 = $4900 |
9 |
9 x 30 = $270 |
$5170 |
77 |
|
36 |
|
$4930 |
35 adults → 45 children
7 adults → 9 children
From every decrease of 7 adults, there is an increase of 9 children and the total amount paid will be reduced by $80.
If all 3 ferries are occupied by adults, total number of adults
= 3 x 35
= 105
Total difference
= 5250 - 4930
= $320
Small difference
= 5250 - 5170
= $80
Number of sets
= 320 ÷ 80
= 4
Total decrease in the number of adults
= 4 x 7
= 28
(a)
Number of adults
= 105 - 28
= 77
(b)
Number of children
= 4 x 9
= 36
Answer(s): (a) 77; (b) 36