There are 25 rooms in a hotel. Each room can accommodate at the most three people. For each room, the single occupancy charge is Rs. 2000 per day, the double occupancy charge is Rs. 3000 per day, and the triple occupancy charge is Rs. 3500 per day.
If there are 55 people staying in the hotel today, what is the maximum possible revenue from room occupancy charges today?

There are 25 rooms in total and 55 people need to be occupied in the these 25 rooms.

The cost of a single, double and triple occupancy room is Rs. 2,000, Rs. 3,000 and Rs. 3,500 respectively. 

Let the number of single, double and triple occupancy rooms used are $$x$$, $$y$$ and $$z$$ respectively. 

We know the total number of rooms are 25.
$$x+y+z=25$$
$$4x+4y+4z=100$$  -----> Eq(1)

Further the total number of people are 55
$$x+2y+3z=55$$ 
$$2y+3z=55-x$$   -----> Eq(2)

The revenue(R) can be written as:
R=$$2000x+3000y+3500z$$
R=$$500(4x+6y+7z)$$

Use Eq(1):
R=$$500(100+2y+3z)$$

Use Eq(2):
R=$$500(100+55-x)$$
R=$$500(155-x)$$

We have to maximize the revenue R.
It implies we have to minimize $$x$$.
Minimum value $$x=0$$ gives $$y=20$$ and $$z=5$$ from Eq(1) and Eq(2).
Maximum Revenue $$=500*155=77500$$

Option C.