A ferry can carry a maximum of 20 adults or 35 children. A ticket for an adult cost $45 and a ticket for a child cost $30. 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 $360 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 |
60 |
60 x 45 = 2700 |
0 |
0 x 30 = 0 |
2700 - 0 = 2700 |
- 4 |
|
+ 7 |
|
- 390 |
56 |
56 x 45 = 2520 |
7 |
7 x 30 = 210 |
2520 - 210 = 2310 |
36 |
|
42 |
|
360 |
20 adults → 35 children
(÷5)4 adults → 7 children
From every decrease of 4 adults, there is an increase of 7 children and the difference in the total amounts paid between the adults and the children will be reduced by $390.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 20
= 60
Total difference
= 2700 - 360
= $2340
Small difference
= 2700 - 2310
= $390
Number of sets
= 2340 ÷ 390
= 6
Total decrease in the number of adults
= 6 x 4
= 24
(a)
Number of adults
= 60 - 24
= 36
(b)
Number of children
= 6 x 7
= 42
Answer(s): (a) 36; (b) 42