A ferry can carry a maximum of 30 adults or 50 children. A ticket for an adult cost $50 and a ticket for a child cost $25. A group of adults and children took 2 such ferries at maximum load on a ferry trip. The total amount paid for the children was $1900 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 |
60 |
60 x 50 = 3000 |
0 |
0 x 25 = 0 |
3000 - 0 = 3000 |
- 3 |
|
+ 5 |
|
- 275 |
57 |
57 x 50 = 2850 |
5 |
5 x 25 = 125 |
2850 - 125 = 2725 |
48 |
|
20 |
|
1900 |
30 adults → 50 children
(÷10)3 adults → 5 children
From every decrease of 3 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 $275.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 30
= 60
Total difference
= 3000 - 1900
= $1100
Small difference
= 3000 - 2725
= $275
Number of sets
= 1100 ÷ 275
= 4
Total decrease in the number of adults
= 4 x 3
= 12
(a)
Number of adults
= 60 - 12
= 48
(b)
Number of children
= 4 x 5
= 20
Answer(s): (a) 48; (b) 20