A ferry can carry a maximum of 25 adults or 45 children. A ticket for an adult cost $25 and a ticket for a child cost $20. A group of adults and children took 4 such ferries at maximum load on a ferry trip. The total amount paid for the children was $670 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 |
100 |
100 x 25 = 2500 |
0 |
0 x 20 = 0 |
2500 - 0 = 2500 |
- 5 |
|
+ 9 |
|
- 305 |
95 |
95 x 25 = 2375 |
9 |
9 x 20 = 180 |
2375 - 180 = 2195 |
70 |
|
54 |
|
670 |
25 adults → 45 children
(÷5)5 adults → 9 children
From every decrease of 5 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 $305.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 25
= 100
Total difference
= 2500 - 670
= $1830
Small difference
= 2500 - 2195
= $305
Number of sets
= 1830 ÷ 305
= 6
Total decrease in the number of adults
= 6 x 5
= 30
(a)
Number of adults
= 100 - 30
= 70
(b)
Number of children
= 6 x 9
= 54
Answer(s): (a) 70; (b) 54