A ferry can carry a maximum of 35 adults or 40 children. A ticket for an adult cost $35 and a ticket for a child cost $30. A group of adults and children took 3 such ferries at maximum load on a ferry trip. The total amount paid for the children was $1735 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 |
| 105 |
105 x 35 = 3675 |
0 |
0 x 30 = 0 |
3675 - 0 = 3675 |
| - 7 |
|
+ 8 |
|
- 485 |
| 98 |
98 x 35 = 3430 |
8 |
8 x 30 = 240 |
3430 - 240 = 3190 |
| 77 |
|
32 |
|
1735 |
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 $485.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 35
= 105
Total difference
= 3675 - 1735
= $1940
Small difference
= 3675 - 3190
= $485
Number of sets
= 1940 ÷ 485
= 4
Total decrease in the number of adults
= 4 x 7
= 28
(a)
Number of adults
= 105 - 28
= 77
(b)
Number of children
= 4 x 8
= 32
Answer(s): (a) 77; (b) 32
