A ferry can carry a maximum of 35 adults or 50 children. A ticket for an adult cost $35 and a ticket for a child cost $25. 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 $210 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 25 = 0 |
3675 - 0 = 3675 |
- 7 |
|
+ 10 |
|
- 495 |
98 |
98 x 35 = 3430 |
10 |
10 x 25 = 250 |
3430 - 250 = 3180 |
56 |
|
70 |
|
210 |
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 $495.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 35
= 105
Total difference
= 3675 - 210
= $3465
Small difference
= 3675 - 3180
= $495
Number of sets
= 3465 ÷ 495
= 7
Total decrease in the number of adults
= 7 x 7
= 49
(a)
Number of adults
= 105 - 49
= 56
(b)
Number of children
= 7 x 10
= 70
Answer(s): (a) 56; (b) 70