A ferry can carry a maximum of 30 adults or 35 children. A ticket for an adult cost $30 and a ticket for a child cost $20. 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 $520 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 30 = 1800 |
0 |
0 x 20 = 0 |
1800 - 0 = 1800 |
- 6 |
|
+ 7 |
|
- 320 |
54 |
54 x 30 = 1620 |
7 |
7 x 20 = 140 |
1620 - 140 = 1480 |
36 |
|
28 |
|
520 |
30 adults → 35 children
(÷5)6 adults → 7 children
From every decrease of 6 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 $320.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 30
= 60
Total difference
= 1800 - 520
= $1280
Small difference
= 1800 - 1480
= $320
Number of sets
= 1280 ÷ 320
= 4
Total decrease in the number of adults
= 4 x 6
= 24
(a)
Number of adults
= 60 - 24
= 36
(b)
Number of children
= 4 x 7
= 28
Answer(s): (a) 36; (b) 28