A ferry can carry a maximum of 20 adults or 45 children. A ticket for an adult cost $55 and a ticket for a child cost $30. 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 $1950 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 |
80 |
80 x 55 = 4400 |
0 |
0 x 30 = 0 |
4400 - 0 = 4400 |
- 4 |
|
+ 9 |
|
- 490 |
76 |
76 x 55 = 4180 |
9 |
9 x 30 = 270 |
4180 - 270 = 3910 |
60 |
|
45 |
|
1950 |
20 adults → 45 children
(÷5)4 adults → 9 children
From every decrease of 4 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 $490.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 20
= 80
Total difference
= 4400 - 1950
= $2450
Small difference
= 4400 - 3910
= $490
Number of sets
= 2450 ÷ 490
= 5
Total decrease in the number of adults
= 5 x 4
= 20
(a)
Number of adults
= 80 - 20
= 60
(b)
Number of children
= 5 x 9
= 45
Answer(s): (a) 60; (b) 45