By walking at 3/4 of her usual speed, a lady reaches her office 30 minutes later than her usual time. Find the usual time taken by her to reach her office?

Let the speed of the lady be 4S and the time taken to reach the office be T. 

So, the distance can be calculated as 4S * T = 4ST.

In the second case, when she walks at 3/4 of her speed, the speed of the lady becomes 3S. The time taken is given as T + 30. 

So, the distance covered in this case is 3S * (T + 30) = 3ST + 90S.

We know that the distance to the office remains constant and must be the same in both cases. 

Equating them, we get,

4ST = 3ST + 90S

ST = 90S

T = 90 minutes.

So, the usual time taken by the lady is 90 minutes.

Hence, the correct answer is option C.