A ferry can carry a maximum of 35 adults or 45 children. A ticket for an adult cost $60 and a ticket for a child cost $40. A group of adults and children took 2 such ferries at maximum load on a ferry trip. They paid a total of $3960 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 |
70 |
70 x 60 = $4200 |
0 |
0 x 40 = $0 |
$4200 |
- 7 |
|
+ 9 |
|
- $60 |
63 |
63 x 60 = $3780 |
9 |
9 x 40 = $360 |
$4140 |
42 |
|
36 |
|
$3960 |
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 $60.
If all 2 ferries are occupied by adults, total number of adults
= 2 x 35
= 70
Total difference
= 4200 - 3960
= $240
Small difference
= 4200 - 4140
= $60
Number of sets
= 240 ÷ 60
= 4
Total decrease in the number of adults
= 4 x 7
= 28
(a)
Number of adults
= 70 - 28
= 42
(b)
Number of children
= 4 x 9
= 36
Answer(s): (a) 42; (b) 36