A boy starts from his house and walks at a speed of 5 km/hr and reaches his school 3 minutes late. Next day he starts at the same time and increases his speed by 4 km/hr and reaches 3 minutes early. What is the distance (in km) between the school and 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 5 km/hr, he reaches school 3 minutes late, => using distance = speed $$\times$$ time

=> $$d=5\times(t+\frac{3}{60})$$ ---------------(i)

Similarly, $$d=9\times(t-\frac{3}{60})$$ --------------(ii)

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

=> $$5(t+\frac{3}{60})=9(t-\frac{3}{60})$$

=> $$5t+\frac{1}{4}=9t-\frac{9}{20}$$

=> $$9t-5t=\frac{1}{4}+\frac{9}{20}$$

=> $$4t=\frac{(5+9)}{20}=\frac{14}{20}$$

=> $$t=\frac{7}{10}\times\frac{1}{4}=\frac{7}{40}$$

Substituting it in equation (i), => $$d=5(\frac{7}{40}+\frac{3}{60})$$

= $$5(\frac{(21+6)}{120})=\frac{27}{24}=1.125$$ km

=> Ans - (C)