A ferry can carry a maximum of 35 adults or 45 children. A ticket for an adult cost $55 and a ticket for a child cost $35. A group of adults and children took 2 such ferries at maximum load on a ferry trip. They paid a total of $3500 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 55 = $3850 |
0 |
0 x 35 = $0 |
$3850 |
- 7 |
|
+ 9 |
|
- $70 |
63 |
63 x 55 = $3465 |
9 |
9 x 35 = $315 |
$3780 |
35 |
|
45 |
|
$3500 |
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 $70.
If all 2 ferries are occupied by adults, total number of adults
= 2 x 35
= 70
Total difference
= 3850 - 3500
= $350
Small difference
= 3850 - 3780
= $70
Number of sets
= 350 ÷ 70
= 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 9
= 45
Answer(s): (a) 35; (b) 45