Walking at $$\frac25$$​ of his normal speed, Abhijeet is 18 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and his office is:

Correct option is A

​Given:

Abhijeet is walking at $$\frac{2}{5}$$​ of his normal speed.

He is 18 minutes late in reaching his office.

Formula Used:

Time = $$\frac{\text{Distance}}{\text{Speed}}$$​​

Solution:

Let the normal speed be S and the distance be D

Now, 

​$$\frac{5D}{2S} - \frac{D}{S} = 18$$​​

​$$\frac{5D}{2S} - \frac{2D}{2S} = 18$$​​

​$$\frac{3D}{2S} = 18$$​​

3D = 36S

D = 12S

Now, the usual time  = $$\frac{D}{S} = \frac{12S}{S}$$​ = 12 minutes

Thus, the usual time taken by Abhijeet to reach his office is 12 minutes

Alternate Solution: ​

When distance is constant;

Speed $$\propto \frac{1}{time}$$ 

Speed changes  -     5         :    2 

so, time change -    2        :    5 

5 - 2 = 3 unit 

3 unit = 18 min.

then 2 unit = $$\frac{18}{3} \times 2 = 12 \ min.$$​​