A ferry can carry a maximum of 25 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 4 such ferries at maximum load on a ferry trip. The total amount paid for the children was $1260 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 |
100 |
100 x 30 = 3000 |
0 |
0 x 20 = 0 |
3000 - 0 = 3000 |
- 5 |
|
+ 7 |
|
- 290 |
95 |
95 x 30 = 2850 |
7 |
7 x 20 = 140 |
2850 - 140 = 2710 |
70 |
|
42 |
|
1260 |
25 adults → 35 children
(÷5)5 adults → 7 children
From every decrease of 5 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 $290.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 25
= 100
Total difference
= 3000 - 1260
= $1740
Small difference
= 3000 - 2710
= $290
Number of sets
= 1740 ÷ 290
= 6
Total decrease in the number of adults
= 6 x 5
= 30
(a)
Number of adults
= 100 - 30
= 70
(b)
Number of children
= 6 x 7
= 42
Answer(s): (a) 70; (b) 42