A ferry can carry a maximum of 25 adults or 40 children. A ticket for an adult cost $60 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. The total amount paid for the children was $1020 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 |
75 |
75 x 60 = 4500 |
0 |
0 x 35 = 0 |
4500 - 0 = 4500 |
- 5 |
|
+ 8 |
|
- 580 |
70 |
70 x 60 = 4200 |
8 |
8 x 35 = 280 |
4200 - 280 = 3920 |
45 |
|
48 |
|
1020 |
25 adults → 40 children
(÷5)5 adults → 8 children
From every decrease of 5 adults, there is an increase of 8 children and the difference in the total amounts paid between the adults and the children will be reduced by $580.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 25
= 75
Total difference
= 4500 - 1020
= $3480
Small difference
= 4500 - 3920
= $580
Number of sets
= 3480 ÷ 580
= 6
Total decrease in the number of adults
= 6 x 5
= 30
(a)
Number of adults
= 75 - 30
= 45
(b)
Number of children
= 6 x 8
= 48
Answer(s): (a) 45; (b) 48