A ferry can carry a maximum of 25 adults or 35 children. A ticket for an adult cost $60 and a ticket for a child cost $35. A group of adults and children took 4 such ferries at maximum load on a ferry trip. The total amount paid for the children was $550 less than the total amount paid for the adults.
- How many adults were there?
- How many children were there?
Number of adults |
Amount paid for adults |
Number of children |
Amount paid for children |
Difference in total amounts paid |
100 |
100 x 60 = 6000 |
0 |
0 x 35 = 0 |
6000 - 0 = 6000 |
- 5 |
|
+ 7 |
|
- 545 |
95 |
95 x 60 = 5700 |
7 |
7 x 35 = 245 |
5700 - 245 = 5455 |
50 |
|
70 |
|
550 |
25 adults → 35 children
(÷5)5 adults → 7 children
From every decrease of 5 adults, there is an increase of 7 children and the difference in the total amounts paid between the adults and the children will be reduced by $545.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 25
= 100
Total difference
= 6000 - 550
= $5450
Small difference
= 6000 - 5455
= $545
Number of sets
= 5450 ÷ 545
= 10
Total decrease in the number of adults
= 10 x 5
= 50
(a)
Number of adults
= 100 - 50
= 50
(b)
Number of children
= 10 x 7
= 70
Answer(s): (a) 50; (b) 70