A ferry can carry a maximum of 35 adults or 40 children. A ticket for an adult cost $30 and a ticket for a child cost $25. 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 $50 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 |
70 |
70 x 30 = 2100 |
0 |
0 x 25 = 0 |
2100 - 0 = 2100 |
- 7 |
|
+ 8 |
|
- 410 |
63 |
63 x 30 = 1890 |
8 |
8 x 25 = 200 |
1890 - 200 = 1690 |
35 |
|
40 |
|
50 |
35 adults → 40 children
(÷5)7 adults → 8 children
From every decrease of 7 adults, there is an increase of 8 children and the difference in the total amounts paid between the adults and the children will be reduced by $410.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 35
= 70
Total difference
= 2100 - 50
= $2050
Small difference
= 2100 - 1690
= $410
Number of sets
= 2050 ÷ 410
= 5
Total decrease in the number of adults
= 5 x 7
= 35
(a)
Number of adults
= 70 - 35
= 35
(b)
Number of children
= 5 x 8
= 40
Answer(s): (a) 35; (b) 40