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)
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