A ferry can carry a maximum of 20 adults or 45 children. A ticket for an adult cost $50 and a ticket for a child cost $25. 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 $25 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 |
- 4 |
|
+ 9 |
|
- 425 |
56 |
56 x 50 = 2800 |
9 |
9 x 25 = 225 |
2800 - 225 = 2575 |
32 |
|
63 |
|
25 |
20 adults → 45 children
(÷5)4 adults → 9 children
From every decrease of 4 adults, there is an increase of 9 children and the difference in the total amounts paid between the adults and the children will be reduced by $425.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 20
= 60
Total difference
= 3000 - 25
= $2975
Small difference
= 3000 - 2575
= $425
Number of sets
= 2975 ÷ 425
= 7
Total decrease in the number of adults
= 7 x 4
= 28
(a)
Number of adults
= 60 - 28
= 32
(b)
Number of children
= 7 x 9
= 63
Answer(s): (a) 32; (b) 63