Walking at $$\frac{3}{4}$$ of his usual speed, a person reaches his office 18 minutes late than the usual time. His usual time in minutes is:

Let the distance between his home to his office $$=d$$ and his usual speed is $$=v$$

Now, usual time is taken by the man to reaching office $$(t)=\dfrac{d}{v}----------------(i)$$

As per the question,

 the new speed of man $$=\dfrac{3v}{4}$$

So, $$\Rightarrow 18+t=\dfrac{d}{\dfrac{3v}{4}}$$

$$\Rightarrow 18+t=\dfrac{4d}{3v}--------(ii)$$

From the equation (i) and (ii)

$$\Rightarrow 18+t=\dfrac{4t}{3}$$

$$\Rightarrow \dfrac{4t}{3}-t=18$$

$$\Rightarrow \dfrac{4t-3t}{3}=18$$

$$\Rightarrow t=18\times 3$$

$$\Rightarrow t=54$$min