Walking at speed of 5 km/hr,a man reaches his office 6 minutes late. Walking at 6 km/hr, he reaches there 2 minutes early. The distance of his office is:

Let ideal time to reach office is in = $$t$$ hours and let distance = $$d$$ km

When he walks at 5 km/hr, he reaches his office 6 minutes late

Using, distance = speed $$\times$$ time

=> $$d=5(t+\frac{6}{60})$$ ---------------(i)

Similarly, $$d=6(t-\frac{2}{60})$$ ---------------(ii)

Comparing equations (i) and (ii), we get :

=> $$5(t+\frac{6}{60})=6(t-\frac{2}{60})$$

=> $$5t+\frac{1}{2}=6t-\frac{1}{5}$$

=> $$6t-5t=\frac{1}{2}+\frac{1}{5}$$

=> $$t=\frac{(5+2)}{10}=\frac{7}{10}$$

Substituting it in equation (i), => $$d=5\times(\frac{7}{10}+\frac{6}{60})$$

= $$5\times\frac{48}{60}=\frac{48}{12}=4$$ km

=> Ans - (C)