A ferry can carry a maximum of 35 adults or 45 children. A ticket for an adult cost $65 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 $4210 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 |
140 |
140 x 65 = 9100 |
0 |
0 x 40 = 0 |
9100 - 0 = 9100 |
- 7 |
|
+ 9 |
|
- 815 |
133 |
133 x 65 = 8645 |
9 |
9 x 40 = 360 |
8645 - 360 = 8285 |
98 |
|
54 |
|
4210 |
35 adults → 45 children
(÷5)7 adults → 9 children
From every decrease of 7 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 $815.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 35
= 140
Total difference
= 9100 - 4210
= $4890
Small difference
= 9100 - 8285
= $815
Number of sets
= 4890 ÷ 815
= 6
Total decrease in the number of adults
= 6 x 7
= 42
(a)
Number of adults
= 140 - 42
= 98
(b)
Number of children
= 6 x 9
= 54
Answer(s): (a) 98; (b) 54