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A student starts from his house with a speed 2kmph and reaches his college 3 min late. Next day he increases his speed by 1kmph and reaches college 3 min earlier the distance between his house and college is

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1.6 km

2 km

0.6 km

3 km

answer is A.

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Detailed Solution

Lets assume the students needs time 't' to reach at correct time

In the first case,

$\frac{S}{2} = t + \frac{3}{60}$ ...(1)

in the second case,

$\frac{S}{3} = t - \frac{3}{60}$ ...(2)

From (1) and (2)

$\frac{S}{2} - \frac{S}{3} = \frac{6}{60}$

$\frac{S}{6} = \frac{6}{60} \Rightarrow S = 0.6 km$

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