A ferry can carry a maximum of 20 adults or 50 children. A ticket for an adult cost $50 and a ticket for a child cost $35. 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 $900 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 |
40 |
40 x 50 = 2000 |
0 |
0 x 35 = 0 |
2000 - 0 = 2000 |
- 2 |
|
+ 5 |
|
- 275 |
38 |
38 x 50 = 1900 |
5 |
5 x 35 = 175 |
1900 - 175 = 1725 |
32 |
|
20 |
|
900 |
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 $275.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 20
= 40
Total difference
= 2000 - 900
= $1100
Small difference
= 2000 - 1725
= $275
Number of sets
= 1100 ÷ 275
= 4
Total decrease in the number of adults
= 4 x 2
= 8
(a)
Number of adults
= 40 - 8
= 32
(b)
Number of children
= 4 x 5
= 20
Answer(s): (a) 32; (b) 20