Final answer:
To solve this problem, set up a system of equations using the given ratios of passengers and fares. Solve the equations to find the amounts collected from each class of passengers. The amount collected from I class passengers is closest to Rs. 1000.
Step-by-step explanation:
To solve this problem, we can set up a system of equations. Let's denote the number of passengers traveling by I class as x and the number of passengers traveling by II class as y. We are given the ratio between the two classes, which is 1:50. This can be written as x:y = 1:50.
We are also given the ratio of the fares between the two classes, which is 3:1. This can be written as the total amount collected from the passengers traveling in each class is in the same ratio. Let the amount collected from I class passengers be a and the amount collected from II class passengers be b. This gives us the equation:
a:b = 3:1
We are told that the total amount collected from all passengers is Rs. 1,325. This means that a + b = 1325.
We can use these two equations to solve for a and b. Let's first solve for a:
a:b = 3:1 ---> a/b = 3/1 ---> a = 3b
Substituting this into the equation a + b = 1325:
3b + b = 1325 ---> 4b = 1325 ---> b = 1325/4 ---> b = 331.25
So the amount collected from II class passengers is Rs. 331.25.
Now, we can find the amount collected from I class passengers:
a = 3b ---> a = 3(331.25) ---> a = 993.75
Therefore, the amount collected from I class passengers is Rs. 993.75, which is closest to Rs. 1000 (option D).