A ferry can carry a maximum of 35 adults or 50 children. A ticket for an adult cost $60 and a ticket for a child cost $40. 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 $100 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 60 = 4200 |
0 |
0 x 40 = 0 |
4200 - 0 = 4200 |
- 7 |
|
+ 10 |
|
- 820 |
63 |
63 x 60 = 3780 |
10 |
10 x 40 = 400 |
3780 - 400 = 3380 |
35 |
|
50 |
|
100 |
35 adults → 50 children
(÷5)7 adults → 10 children
From every decrease of 7 adults, there is an increase of 10 children and the difference in the total amounts paid between the adults and the children will be reduced by $820.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 35
= 70
Total difference
= 4200 - 100
= $4100
Small difference
= 4200 - 3380
= $820
Number of sets
= 4100 ÷ 820
= 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 10
= 50
Answer(s): (a) 35; (b) 50