A ferry can carry a maximum of 35 adults or 45 children. A ticket for an adult cost $40 and a ticket for a child cost $20. A group of adults and children took 2 such ferries at maximum load on a ferry trip. They paid a total of $2200 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 40 = $2800 |
0 |
0 x 20 = $0 |
$2800 |
- 7 |
|
+ 9 |
|
- $100 |
63 |
63 x 40 = $2520 |
9 |
9 x 20 = $180 |
$2700 |
28 |
|
54 |
|
$2200 |
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 $100.
If all 2 ferries are occupied by adults, total number of adults
= 2 x 35
= 70
Total difference
= 2800 - 2200
= $600
Small difference
= 2800 - 2700
= $100
Number of sets
= 600 ÷ 100
= 6
Total decrease in the number of adults
= 6 x 7
= 42
(a)
Number of adults
= 70 - 42
= 28
(b)
Number of children
= 6 x 9
= 54
Answer(s): (a) 28; (b) 54