If a man walks at the rate of 4 kmph , he misses a train by only 6 min . However, if he walks at the rate of 5 kmph he reaches the station 6 minutes before the arrival of the train. The distance covered by him to reach the station is
Let the total distance be ' D ' and time be ' T ' \[ 4\left(T+\frac{6}{60}\right)=5\left(T-\frac{6}{60}\right) \] then \[4 T+\frac{2}{5}=5 T-\frac{1}{2}\] \[T=\frac{9}{10} \] Hence distance 'D' \[=4\left(\frac{9}{10}+\frac{6}{60}\right)=4 \mathrm{~km}\] Choice (A)\[\]
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