A ferry can carry a maximum of 30 adults or 40 children. A ticket for an adult cost $30 and a ticket for a child cost $25. 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 $2650 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 |
120 |
120 x 30 = 3600 |
0 |
0 x 25 = 0 |
3600 - 0 = 3600 |
- 3 |
|
+ 4 |
|
- 190 |
117 |
117 x 30 = 3510 |
4 |
4 x 25 = 100 |
3510 - 100 = 3410 |
105 |
|
20 |
|
2650 |
30 adults → 40 children
(÷10)3 adults → 4 children
From every decrease of 3 adults, there is an increase of 4 children and the difference in the total amounts paid between the adults and the children will be reduced by $190.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 30
= 120
Total difference
= 3600 - 2650
= $950
Small difference
= 3600 - 3410
= $190
Number of sets
= 950 ÷ 190
= 5
Total decrease in the number of adults
= 5 x 3
= 15
(a)
Number of adults
= 120 - 15
= 105
(b)
Number of children
= 5 x 4
= 20
Answer(s): (a) 105; (b) 20