A ferry can carry a maximum of 35 adults or 40 children. A ticket for an adult cost $55 and a ticket for a child cost $40. A group of adults and children took 3 such ferries at maximum load on a ferry trip. They paid a total of $5190 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 55 = $5775 |
0 |
0 x 40 = $0 |
$5775 |
- 7 |
|
+ 8 |
|
- $65 |
98 |
98 x 55 = $5390 |
8 |
8 x 40 = $320 |
$5710 |
42 |
|
72 |
|
$5190 |
35 adults → 40 children
7 adults → 8 children
From every decrease of 7 adults, there is an increase of 8 children and the total amount paid will be reduced by $65.
If all 3 ferries are occupied by adults, total number of adults
= 3 x 35
= 105
Total difference
= 5775 - 5190
= $585
Small difference
= 5775 - 5710
= $65
Number of sets
= 585 ÷ 65
= 9
Total decrease in the number of adults
= 9 x 7
= 63
(a)
Number of adults
= 105 - 63
= 42
(b)
Number of children
= 9 x 8
= 72
Answer(s): (a) 42; (b) 72