A ferry can carry a maximum of 35 adults or 50 children. A ticket for an adult cost $40 and a ticket for a child cost $20. A group of adults and children took 3 such ferries at maximum load on a ferry trip. They paid a total of $3800 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 40 = $4200 |
0 |
0 x 20 = $0 |
$4200 |
- 7 |
|
+ 10 |
|
- $80 |
98 |
98 x 40 = $3920 |
10 |
10 x 20 = $200 |
$4120 |
70 |
|
50 |
|
$3800 |
35 adults → 50 children
7 adults → 10 children
From every decrease of 7 adults, there is an increase of 10 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
= 4200 - 3800
= $400
Small difference
= 4200 - 4120
= $80
Number of sets
= 400 ÷ 80
= 5
Total decrease in the number of adults
= 5 x 7
= 35
(a)
Number of adults
= 105 - 35
= 70
(b)
Number of children
= 5 x 10
= 50
Answer(s): (a) 70; (b) 50