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If I walk at 3 km/hr, I miss a train by 2 minutes. If however I walk at 4 km/hr, then I reach the station 2 minutes before the arrival of the train. How far do I walk to reach the station?
A. $\dfrac{3}{4}$ km
B. $\dfrac{4}{5}$ km
C. $\dfrac{5}{4}$ km
D. 1 km

Last updated date: 13th Jun 2024
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Hint: We will find the difference of time in both the speeds by given timings and then use the general formula for time related to distance and spend and equate them both.

Complete answer:
We are given that, if the person walks with the speed of 3 km / hr, he / she reaches late to the station by 2 minutes. We are also given that, if the person walks with the speed of 4 km / hr, he / she reaches the station 2 minutes early.
Therefore, the time difference in both the journeys is = (2 + 2) minutes = 4 minutes
Since, we have our speed in hours and the options of distance in hours as well. So, let us convert 4 minutes into hours as well.
Since 60 minutes = 1 hour
$ \Rightarrow $ 1 minute = $\dfrac{1}{{60}}$ hours
$ \Rightarrow $ 4 minute = $4 \times \dfrac{1}{{60}}$ hours
Simplifying the calculations on the right hand side, we will then obtain:-
$ \Rightarrow $ 1 minute = $\dfrac{1}{{15}}$ hours
Therefore, the time difference of both the journeys is $\dfrac{1}{{15}}$ hours. ……………….(1)
Now, we also know that $Dis\tan ce = Speed \times Time$
$ \Rightarrow Time = \dfrac{{Dis\tan ce}}{{Speed}}$
Assuming the distance to be x km, and putting in both the speeds to get difference of time, we will get:-$ \Rightarrow $
 Difference in time = $\dfrac{x}{3} - \dfrac{x}{4}$ ………………(2)
Equating the equations (1) and (2), we will then obtain:-
$ \Rightarrow \dfrac{x}{3} - \dfrac{x}{4} = \dfrac{1}{{15}}$
Taking the LCM on the left hand side, we will obtain:-
$ \Rightarrow \dfrac{{4x - 3x}}{{12}} = \dfrac{1}{{15}}$
Simplifying the numerator on the left hand side to obtain the following expression:-
$ \Rightarrow \dfrac{x}{{12}} = \dfrac{1}{{15}}$
Taking the 12 from denominator in left hand side to multiplication in right hand side to obtain the following:-
$ \Rightarrow x = \dfrac{{12}}{{15}}km$
Cutting off 3 by taking common from numerator and denominator, we will get:-
$ \Rightarrow x = \dfrac{4}{5}km$

Hence, the correct option is (B).

The students must note that they must commit to memory the following formula:-
$ \Rightarrow Dis\tan ce = Speed \times Time$
The students must note that they may forget to convert the time in hours from minutes but that will create mistakes in the final answer because on the other hand we take the time in hours and on this side, we have time in minutes.