A ferry can carry a maximum of 20 adults or 35 children. A ticket for an adult cost $55 and a ticket for a child cost $35. A group of adults and children took 2 such ferries at maximum load on a ferry trip. The total amount paid for the children was $340 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 |
40 |
40 x 55 = 2200 |
0 |
0 x 35 = 0 |
2200 - 0 = 2200 |
- 4 |
|
+ 7 |
|
- 465 |
36 |
36 x 55 = 1980 |
7 |
7 x 35 = 245 |
1980 - 245 = 1735 |
24 |
|
28 |
|
340 |
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 $465.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 20
= 40
Total difference
= 2200 - 340
= $1860
Small difference
= 2200 - 1735
= $465
Number of sets
= 1860 ÷ 465
= 4
Total decrease in the number of adults
= 4 x 4
= 16
(a)
Number of adults
= 40 - 16
= 24
(b)
Number of children
= 4 x 7
= 28
Answer(s): (a) 24; (b) 28