Question

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

Answer 1

Hint:
Assuming x to the distance and using the relation ${\text{time = }}\dfrac{{{\text{distance}}}}{{{\text{speed}}}}$we get the time as $T = \left( {\dfrac{x}{3} + \dfrac{2}{{60}}} \right)$when I walk at a speed of 3 km/hr and $T = \left( {\dfrac{x}{4} - \dfrac{2}{{60}}} \right)$ when I walk at a speed of 4 km/hr and equating these both we get the value of x.

Complete step by step solution:
Let x be the distance which I walk to reach the station
We know that ${\text{time = }}\dfrac{{{\text{distance}}}}{{{\text{speed}}}}$
Therefore when I walk at a speed of 3 km/hr I reach the station 2 minutes late
$ \Rightarrow T = \left( {\dfrac{x}{3} + \dfrac{2}{{60}}} \right)$ ……….(1)
Here the extra 2 minutes is written as $\dfrac{2}{{60}}$because our calculation in done in terms of hours so we convert minutes to hours by dividing it by 60
Here we add the two minutes as I take two minutes extra (in addition )

 Therefore when I walk at a speed of 4 km/hr I reach the station 2 minutes early
$ \Rightarrow T = \left( {\dfrac{x}{4} - \dfrac{2}{{60}}} \right)$ ……….(2)
Here we subtract the two minutes as I reach 2 minutes earlier
Equating (1) and (2)
$
   \Rightarrow \left( {\dfrac{x}{3} + \dfrac{2}{{60}}} \right) = \left( {\dfrac{x}{4} - \dfrac{2}{{60}}} \right) \\
   \Rightarrow \dfrac{x}{3} - \dfrac{x}{4} = \dfrac{{ - 1}}{{30}} - \dfrac{1}{{30}} \\
   \Rightarrow \dfrac{{4x - 3x}}{{12}} = \dfrac{{ - 2}}{{30}} \\
   \Rightarrow \dfrac{x}{{12}} = \dfrac{{ - 1}}{{15}} \\
   \Rightarrow x = \dfrac{{ - 12}}{{15}} = \dfrac{{ - 4}}{5}km \\
 $
Since distance cannot be negative
$ \Rightarrow x = \dfrac{4}{5}km$

The correct option is c.

Note:
If distance travelled for each part of the journey,
Ie ${d_1} = {d_2} = .... = {d_n} = d$, then the average speed of the object is Harmonic Mean of speeds.
Let each distance be covered with speeds ${s_1},{s_2},...,{s_n}{\text{ in }}{t_1},{t_2},...,{t_n}$ times respectively.
Then ${t_1} = \dfrac{d}{{{s_1}}},{t_2} = \dfrac{d}{{{s_2}}},...,{t_n} = \dfrac{d}{{{s_n}}}$