A ferry can carry a maximum of 35 adults or 40 children. A ticket for an adult cost $45 and a ticket for a child cost $40. 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 $3125 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 40 = 0 |
6300 - 0 = 6300 |
- 7 |
|
+ 8 |
|
- 635 |
133 |
133 x 45 = 5985 |
8 |
8 x 40 = 320 |
5985 - 320 = 5665 |
105 |
|
40 |
|
3125 |
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 $635.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 35
= 140
Total difference
= 6300 - 3125
= $3175
Small difference
= 6300 - 5665
= $635
Number of sets
= 3175 ÷ 635
= 5
Total decrease in the number of adults
= 5 x 7
= 35
(a)
Number of adults
= 140 - 35
= 105
(b)
Number of children
= 5 x 8
= 40
Answer(s): (a) 105; (b) 40