A ferry can carry a maximum of 35 adults or 45 children. A ticket for an adult cost $45 and a ticket for a child cost $30. 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 $3960 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 |
140 |
140 x 45 = 6300 |
0 |
0 x 30 = 0 |
6300 - 0 = 6300 |
- 7 |
|
+ 9 |
|
- 585 |
133 |
133 x 45 = 5985 |
9 |
9 x 30 = 270 |
5985 - 270 = 5715 |
112 |
|
36 |
|
3960 |
35 adults → 45 children
(÷5)7 adults → 9 children
From every decrease of 7 adults, there is an increase of 9 children and the difference in the total amounts paid between the adults and the children will be reduced by $585.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 35
= 140
Total difference
= 6300 - 3960
= $2340
Small difference
= 6300 - 5715
= $585
Number of sets
= 2340 ÷ 585
= 4
Total decrease in the number of adults
= 4 x 7
= 28
(a)
Number of adults
= 140 - 28
= 112
(b)
Number of children
= 4 x 9
= 36
Answer(s): (a) 112; (b) 36