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 $35. 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 $135 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 |
75 |
75 x 45 = 3375 |
0 |
0 x 35 = 0 |
3375 - 0 = 3375 |
- 5 |
|
+ 9 |
|
- 540 |
70 |
70 x 45 = 3150 |
9 |
9 x 35 = 315 |
3150 - 315 = 2835 |
45 |
|
54 |
|
135 |
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 $540.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 25
= 75
Total difference
= 3375 - 135
= $3240
Small difference
= 3375 - 2835
= $540
Number of sets
= 3240 ÷ 540
= 6
Total decrease in the number of adults
= 6 x 5
= 30
(a)
Number of adults
= 75 - 30
= 45
(b)
Number of children
= 6 x 9
= 54
Answer(s): (a) 45; (b) 54