A ferry can carry a maximum of 35 adults or 40 children. A ticket for an adult cost $45 and a ticket for a child cost $35. A group of adults and children took 3 such ferries at maximum load on a ferry trip. They paid a total of $4515 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 45 = $4725 |
0 |
0 x 35 = $0 |
$4725 |
- 7 |
|
+ 8 |
|
- $35 |
98 |
98 x 45 = $4410 |
8 |
8 x 35 = $280 |
$4690 |
63 |
|
48 |
|
$4515 |
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 $35.
If all 3 ferries are occupied by adults, total number of adults
= 3 x 35
= 105
Total difference
= 4725 - 4515
= $210
Small difference
= 4725 - 4690
= $35
Number of sets
= 210 ÷ 35
= 6
Total decrease in the number of adults
= 6 x 7
= 42
(a)
Number of adults
= 105 - 42
= 63
(b)
Number of children
= 6 x 8
= 48
Answer(s): (a) 63; (b) 48