A ferry can carry a maximum of 35 adults or 40 children. A ticket for an adult cost $30 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 $1850 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 30 = $2100 |
0 |
0 x 20 = $0 |
$2100 |
- 7 |
|
+ 8 |
|
- $50 |
63 |
63 x 30 = $1890 |
8 |
8 x 20 = $160 |
$2050 |
35 |
|
40 |
|
$1850 |
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 $50.
If all 2 ferries are occupied by adults, total number of adults
= 2 x 35
= 70
Total difference
= 2100 - 1850
= $250
Small difference
= 2100 - 2050
= $50
Number of sets
= 250 ÷ 50
= 5
Total decrease in the number of adults
= 5 x 7
= 35
(a)
Number of adults
= 70 - 35
= 35
(b)
Number of children
= 5 x 8
= 40
Answer(s): (a) 35; (b) 40