A ferry can carry a maximum of 35 adults or 50 children. A ticket for an adult cost $45 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 $4575 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 45 = $4725 |
0 |
0 x 30 = $0 |
$4725 |
- 7 |
|
+ 10 |
|
- $15 |
98 |
98 x 45 = $4410 |
10 |
10 x 30 = $300 |
$4710 |
35 |
|
100 |
|
$4575 |
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 $15.
If all 3 ferries are occupied by adults, total number of adults
= 3 x 35
= 105
Total difference
= 4725 - 4575
= $150
Small difference
= 4725 - 4710
= $15
Number of sets
= 150 ÷ 15
= 10
Total decrease in the number of adults
= 10 x 7
= 70
(a)
Number of adults
= 105 - 70
= 35
(b)
Number of children
= 10 x 10
= 100
Answer(s): (a) 35; (b) 100