A ferry can carry a maximum of 30 adults or 45 children. A ticket for an adult cost $40 and a ticket for a child cost $30. A group of adults and children took 3 such ferries at maximum load on a ferry trip. The total amount paid for the children was $1900 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 |
90 |
90 x 40 = 3600 |
0 |
0 x 30 = 0 |
3600 - 0 = 3600 |
- 2 |
|
+ 3 |
|
- 170 |
88 |
88 x 40 = 3520 |
3 |
3 x 30 = 90 |
3520 - 90 = 3430 |
70 |
|
30 |
|
1900 |
30 adults → 45 children
(÷15)2 adults → 3 children
From every decrease of 2 adults, there is an increase of 3 children and the difference in the total amounts paid between the adults and the children will be reduced by $170.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 30
= 90
Total difference
= 3600 - 1900
= $1700
Small difference
= 3600 - 3430
= $170
Number of sets
= 1700 ÷ 170
= 10
Total decrease in the number of adults
= 10 x 2
= 20
(a)
Number of adults
= 90 - 20
= 70
(b)
Number of children
= 10 x 3
= 30
Answer(s): (a) 70; (b) 30