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) ${\displaystyle \frac{3}{4}}km$
B) ${\displaystyle \frac{4}{5}}km$
C) ${\displaystyle \frac{5}{4}}km$
D) 1 km

Answer 1

Hint:
Assuming x to the distance and using the relation $\text{time = }{\displaystyle \frac{\text{distance}}{\text{speed}}}$we get the time as $T=\left({\displaystyle \frac{x}{3}}+{\displaystyle \frac{2}{60}}\right)$when I walk at a speed of 3 km/hr and $T=\left({\displaystyle \frac{x}{4}}-{\displaystyle \frac{2}{60}}\right)$ 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 $\text{time = }{\displaystyle \frac{\text{distance}}{\text{speed}}}$
Therefore when I walk at a speed of 3 km/hr I reach the station 2 minutes late
$\Rightarrow T=\left({\displaystyle \frac{x}{3}}+{\displaystyle \frac{2}{60}}\right)$ ……….(1)
Here the extra 2 minutes is written as ${\displaystyle \frac{2}{60}}$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
$\Rightarrow T=\left({\displaystyle \frac{x}{4}}-{\displaystyle \frac{2}{60}}\right)$ ……….(2)
Here we subtract the two minutes as I reach 2 minutes earlier
Equating (1) and (2)
$$\begin{gathered} \Rightarrow \left({\displaystyle \frac{x}{3}}+{\displaystyle \frac{2}{60}}\right)=\left({\displaystyle \frac{x}{4}}-{\displaystyle \frac{2}{60}}\right) \\ \Rightarrow {\displaystyle \frac{x}{3}}-{\displaystyle \frac{x}{4}}={\displaystyle \frac{-1}{30}}-{\displaystyle \frac{1}{30}} \\ \Rightarrow {\displaystyle \frac{4x-3x}{12}}={\displaystyle \frac{-2}{30}} \\ \Rightarrow {\displaystyle \frac{x}{12}}={\displaystyle \frac{-1}{15}} \\ \Rightarrow x={\displaystyle \frac{-12}{15}}={\displaystyle \frac{-4}{5}}km \end{gathered}$$
Since distance cannot be negative
$\Rightarrow x={\displaystyle \frac{4}{5}}km$

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 $t_1={\displaystyle \frac{d}{s_1}},t_2={\displaystyle \frac{d}{s_2}},...,t_n={\displaystyle \frac{d}{s_n}}$