A ferry can carry a maximum of 20 adults or 50 children. A ticket for an adult cost $45 and a ticket for a child cost $20. 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 $660 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 45 = 1800 |
0 |
0 x 20 = 0 |
1800 - 0 = 1800 |
| - 2 |
|
+ 5 |
|
- 190 |
| 38 |
38 x 45 = 1710 |
5 |
5 x 20 = 100 |
1710 - 100 = 1610 |
| 28 |
|
30 |
|
660 |
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 $190.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 20
= 40
Total difference
= 1800 - 660
= $1140
Small difference
= 1800 - 1610
= $190
Number of sets
= 1140 ÷ 190
= 6
Total decrease in the number of adults
= 6 x 2
= 12
(a)
Number of adults
= 40 - 12
= 28
(b)
Number of children
= 6 x 5
= 30
Answer(s): (a) 28; (b) 30
