A ferry can carry a maximum of 25 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 $1685 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 55 = 5500 |
0 |
0 x 30 = 0 |
5500 - 0 = 5500 |
- 5 |
|
+ 9 |
|
- 545 |
95 |
95 x 55 = 5225 |
9 |
9 x 30 = 270 |
5225 - 270 = 4955 |
65 |
|
63 |
|
1685 |
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 $545.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 25
= 100
Total difference
= 5500 - 1685
= $3815
Small difference
= 5500 - 4955
= $545
Number of sets
= 3815 ÷ 545
= 7
Total decrease in the number of adults
= 7 x 5
= 35
(a)
Number of adults
= 100 - 35
= 65
(b)
Number of children
= 7 x 9
= 63
Answer(s): (a) 65; (b) 63