Question 56

A person cycles from hostel to college at a speed of 36 kmph and reaches 7 minutes late. If he cycles at a speed of 45 kmph and he reaches early by 5 minutes. Find the distance between hostel and college.

Solution

Let ideal time taken = $$t$$ hours

Also, speed is inversely proportional to time.

=> $$\frac{36}{45}=\frac{t-\frac{5}{60}}{t+\frac{7}{60}}$$

=> $$4t+\frac{7}{15}=5t-\frac{5}{12}$$

=> $$5t-4t=\frac{7}{15}+\frac{5}{12}$$

=> $$t=\frac{28+25}{60}=\frac{53}{60}$$

$$\therefore$$ Distance = speed $$\times$$ time

= $$36\times(\frac{53}{60}+\frac{7}{60})$$

= $$36\times1=36$$ km

=> Ans - (C)

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