A ferry can carry a maximum of 25 adults or 45 children. A ticket for an adult cost $45 and a ticket for a child cost $40. 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 $1575 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 45 = 4500 |
0 |
0 x 40 = 0 |
4500 - 0 = 4500 |
- 5 |
|
+ 9 |
|
- 585 |
95 |
95 x 45 = 4275 |
9 |
9 x 40 = 360 |
4275 - 360 = 3915 |
75 |
|
45 |
|
1575 |
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 $585.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 25
= 100
Total difference
= 4500 - 1575
= $2925
Small difference
= 4500 - 3915
= $585
Number of sets
= 2925 ÷ 585
= 5
Total decrease in the number of adults
= 5 x 5
= 25
(a)
Number of adults
= 100 - 25
= 75
(b)
Number of children
= 5 x 9
= 45
Answer(s): (a) 75; (b) 45