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 $$ imes$$ time => $$d=5 imes(t+frac{3}{60})$$ ---------------(i) Similarly, $$d=9 imes(t-frac{3}{60})$$ --------------(ii) Comparing, equations (i) and (ii), we get : => $$5(t+frac{3}{60})=9(t-frac{3}{60})$$ => $$5t+frac{1}{4}=9t-frac{9}{20}$$ => $$9t-5t=frac{1}{4}+frac{9}{20}$$ => $$4t=frac{(5+9)}{20}=frac{14}{20}$$ => $$t=frac{7}{10} imesfrac{1}{4}=frac{7}{40}$$ 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)