You arrive at your school 5 minutes late if you walk with a speed of 4 km/h, but you arrive 10 minutes before the scheduled time if you walk with a speed of 5 km/h. The distance of your school from your house (in km) is

Here the distance between home and school will remain same , let the distance be D

Let the original time taken be T hours

as in both cases the distance remain constant, we can use

$$\frac{S1}{S2}$$ = $$\frac{T2}{T1}$$

T1 = T + $$\frac{1}{12}$$ hr , S1 = 4 km/hr

T2 = T - $$\frac{1}{6}$$ hr , S2 = 5 km/hr

So, $$\frac{4}{5}$$ = $$\frac{(T-\frac{1}{6})}{(T+\frac{1}{12})}$$

T = $$\frac{7}{6}$$ hour

Distance = Speed x Time

Distance of School = 4 x $$\frac{15}{12}$$ = 5 km