A ferry can carry a maximum of 20 adults or 35 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 $125 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 25 = 0 |
2000 - 0 = 2000 |
- 4 |
|
+ 7 |
|
- 375 |
36 |
36 x 50 = 1800 |
7 |
7 x 25 = 175 |
1800 - 175 = 1625 |
20 |
|
35 |
|
125 |
20 adults → 35 children
(÷5)4 adults → 7 children
From every decrease of 4 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 $375.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 20
= 40
Total difference
= 2000 - 125
= $1875
Small difference
= 2000 - 1625
= $375
Number of sets
= 1875 ÷ 375
= 5
Total decrease in the number of adults
= 5 x 4
= 20
(a)
Number of adults
= 40 - 20
= 20
(b)
Number of children
= 5 x 7
= 35
Answer(s): (a) 20; (b) 35