A ferry can carry a maximum of 30 adults or 50 children. A ticket for an adult cost $25 and a ticket for a child cost $20. 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 $1775 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 |
120 |
120 x 25 = 3000 |
0 |
0 x 20 = 0 |
3000 - 0 = 3000 |
- 3 |
|
+ 5 |
|
- 175 |
117 |
117 x 25 = 2925 |
5 |
5 x 20 = 100 |
2925 - 100 = 2825 |
99 |
|
35 |
|
1775 |
30 adults → 50 children
(÷10)3 adults → 5 children
From every decrease of 3 adults, there is an increase of 5 children and the difference in the total amounts paid between the adults and the children will be reduced by $175.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 30
= 120
Total difference
= 3000 - 1775
= $1225
Small difference
= 3000 - 2825
= $175
Number of sets
= 1225 ÷ 175
= 7
Total decrease in the number of adults
= 7 x 3
= 21
(a)
Number of adults
= 120 - 21
= 99
(b)
Number of children
= 7 x 5
= 35
Answer(s): (a) 99; (b) 35