A ferry can carry a maximum of 35 adults or 45 children. A ticket for an adult cost $35 and a ticket for a child cost $25. A group of adults and children took 3 such ferries at maximum load on a ferry trip. They paid a total of $3595 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 35 = $3675 |
0 |
0 x 25 = $0 |
$3675 |
- 7 |
|
+ 9 |
|
- $20 |
98 |
98 x 35 = $3430 |
9 |
9 x 25 = $225 |
$3655 |
77 |
|
36 |
|
$3595 |
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 $20.
If all 3 ferries are occupied by adults, total number of adults
= 3 x 35
= 105
Total difference
= 3675 - 3595
= $80
Small difference
= 3675 - 3655
= $20
Number of sets
= 80 ÷ 20
= 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