A ferry can carry a maximum of 20 adults or 50 children. A ticket for an adult cost $35 and a ticket for a child cost $25. 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 $1045 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 |
80 |
80 x 35 = 2800 |
0 |
0 x 25 = 0 |
2800 - 0 = 2800 |
- 2 |
|
+ 5 |
|
- 195 |
78 |
78 x 35 = 2730 |
5 |
5 x 25 = 125 |
2730 - 125 = 2605 |
62 |
|
45 |
|
1045 |
20 adults → 50 children
(÷10)2 adults → 5 children
From every decrease of 2 adults, there is an increase of 5 children and the difference in the total amounts paid between the adults and the children will be reduced by $195.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 20
= 80
Total difference
= 2800 - 1045
= $1755
Small difference
= 2800 - 2605
= $195
Number of sets
= 1755 ÷ 195
= 9
Total decrease in the number of adults
= 9 x 2
= 18
(a)
Number of adults
= 80 - 18
= 62
(b)
Number of children
= 9 x 5
= 45
Answer(s): (a) 62; (b) 45