If Anuj walks at the speed of 4 km/hr, then he reaches his school 6 minutes late but if he walks at the speed of 5 km/hr, then he reaches 6 minutes before the scheduled time. What is the distance (in km) of his school from his house?

Let distance between school and house = $$d$$ km and ideal time to reach school = $$t$$ hours

When he walks at a speed of 4 km/hr, he reaches school 6 minutes late, => using distance = speed $$\times$$ time

=> $$d=4\times(t+\frac{6}{60})$$ ---------------(i)

Similarly, $$d=5\times(t-\frac{6}{60})$$ --------------(ii)

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

=> $$4(t+\frac{6}{60})=5(t-\frac{6}{60})$$

=> $$4t+\frac{2}{5}=5t-\frac{1}{2}$$

=> $$5t-4t=\frac{2}{5}+\frac{1}{2}$$

=> $$t=\frac{(4+5)}{10}=\frac{9}{10}$$

Substituting it in equation (i), => $$d=4(\frac{9}{10}+\frac{6}{60})$$

= $$4(\frac{(54+6)}{60})=\frac{60}{15}=4$$ km

=> Ans - (A)