A ferry can carry a maximum of 30 adults or 40 children. A ticket for an adult cost $35 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 $1925 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 |
60 |
60 x 35 = $2100 |
0 |
0 x 20 = $0 |
$2100 |
- 3 |
|
+ 4 |
|
- $25 |
57 |
57 x 35 = $1995 |
4 |
4 x 20 = $80 |
$2075 |
39 |
|
28 |
|
$1925 |
30 adults → 40 children
3 adults → 4 children
From every decrease of 3 adults, there is an increase of 4 children and the total amount paid will be reduced by $25.
If all 2 ferries are occupied by adults, total number of adults
= 2 x 30
= 60
Total difference
= 2100 - 1925
= $175
Small difference
= 2100 - 2075
= $25
Number of sets
= 175 ÷ 25
= 7
Total decrease in the number of adults
= 7 x 3
= 21
(a)
Number of adults
= 60 - 21
= 39
(b)
Number of children
= 7 x 4
= 28
Answer(s): (a) 39; (b) 28