A ferry can carry a maximum of 25 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 $375 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 |
50 |
50 x 50 = 2500 |
0 |
0 x 25 = 0 |
2500 - 0 = 2500 |
- 5 |
|
+ 7 |
|
- 425 |
45 |
45 x 50 = 2250 |
7 |
7 x 25 = 175 |
2250 - 175 = 2075 |
25 |
|
35 |
|
375 |
25 adults → 35 children
(÷5)5 adults → 7 children
From every decrease of 5 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 $425.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 25
= 50
Total difference
= 2500 - 375
= $2125
Small difference
= 2500 - 2075
= $425
Number of sets
= 2125 ÷ 425
= 5
Total decrease in the number of adults
= 5 x 5
= 25
(a)
Number of adults
= 50 - 25
= 25
(b)
Number of children
= 5 x 7
= 35
Answer(s): (a) 25; (b) 35