A ferry can carry a maximum of 25 adults or 35 children. A ticket for an adult cost $65 and a ticket for a child cost $40. 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 $2265 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 65 = 6500 |
0 |
0 x 40 = 0 |
6500 - 0 = 6500 |
- 5 |
|
+ 7 |
|
- 605 |
95 |
95 x 65 = 6175 |
7 |
7 x 40 = 280 |
6175 - 280 = 5895 |
65 |
|
49 |
|
2265 |
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 $605.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 25
= 100
Total difference
= 6500 - 2265
= $4235
Small difference
= 6500 - 5895
= $605
Number of sets
= 4235 ÷ 605
= 7
Total decrease in the number of adults
= 7 x 5
= 35
(a)
Number of adults
= 100 - 35
= 65
(b)
Number of children
= 7 x 7
= 49
Answer(s): (a) 65; (b) 49