A ferry can carry a maximum of 35 adults or 50 children. A ticket for an adult cost $50 and a ticket for a child cost $30. A group of adults and children took 4 such ferries at maximum load on a ferry trip. They paid a total of $6500 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 |
140 |
140 x 50 = $7000 |
0 |
0 x 30 = $0 |
$7000 |
- 7 |
|
+ 10 |
|
- $50 |
133 |
133 x 50 = $6650 |
10 |
10 x 30 = $300 |
$6950 |
70 |
|
100 |
|
$6500 |
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 $50.
If all 4 ferries are occupied by adults, total number of adults
= 4 x 35
= 140
Total difference
= 7000 - 6500
= $500
Small difference
= 7000 - 6950
= $50
Number of sets
= 500 ÷ 50
= 10
Total decrease in the number of adults
= 10 x 7
= 70
(a)
Number of adults
= 140 - 70
= 70
(b)
Number of children
= 10 x 10
= 100
Answer(s): (a) 70; (b) 100