A ferry can carry a maximum of 35 adults or 50 children. A ticket for an adult cost $65 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 $1130 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 65 = 4550 |
0 |
0 x 40 = 0 |
4550 - 0 = 4550 |
- 7 |
|
+ 10 |
|
- 855 |
63 |
63 x 65 = 4095 |
10 |
10 x 40 = 400 |
4095 - 400 = 3695 |
42 |
|
40 |
|
1130 |
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 $855.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 35
= 70
Total difference
= 4550 - 1130
= $3420
Small difference
= 4550 - 3695
= $855
Number of sets
= 3420 ÷ 855
= 4
Total decrease in the number of adults
= 4 x 7
= 28
(a)
Number of adults
= 70 - 28
= 42
(b)
Number of children
= 4 x 10
= 40
Answer(s): (a) 42; (b) 40