Question

A man walks from his house to the Railway station to catch a train, which is running as per schedule. If he walks at $6\,{\rm{km/h}}$, he misses the train by $9$ minutes. However, if he walks at $7\,{\rm{km/h}}$, he reaches the station $6$ minutes before the departure of the train. The distance of his home to the Railway Station is:
A. $2\,{\rm{km}}$
B.$10.5\,{\rm{km}}$
C.$1.05\,{\rm{km}}$
D.$1.25\,{\rm{km}}$

Answer 1

Hint:
In the solution, we need to calculate the distances for the given two cases. For that we need to use the formula which relates distance, time and speed, $d = v \times t$ where $d$ is the distance, $t$ is the time and $v$ represent the speed. In both the cases, since the total distance he covered is the same, we need to compare both the expressions of the distances.

Complete step by step solution:
Let the distance between his home and the railway station be ‘$d$’. The time taken to reach the station is ‘$t$’. In this problem, whatever may be the speed of the person while walking, the distance between the home and the station will always be the same in both the cases.
For the 1st case:
Speed $\left( v \right) = 6\,{\rm{km/h}}$, time is $t + \dfrac{9}{{60}}$
Convert the time in hours by dividing by $60,$ to convert to hours.
Hence, the relation is:
$d = 6 \times \left( {t + \dfrac{9}{{60}}} \right)$ …… (1)
For the 2nd case:
Speed $\left( v \right) = 7{\rm{km/h}}$, time is $t - \dfrac{6}{{60}}$
Convert the time in hours by dividing by $60,$ to convert to hours.
Hence, the relation is:
$d = 7 \times \left( {t - \dfrac{6}{{60}}} \right)$ …… (2)

However, in both the cases, the distances will be same in both the cases, so equation (1) and (2) will be equal:

$\begin{array}{c} \Rightarrow 6 \times \left( {t + \dfrac{9}{{60}}} \right) = 7 \times \left( {t - \dfrac{6}{{60}}} \right)\\ \Rightarrow 6\left( {t + 0.15} \right) = 7\left( {t - 0.1} \right)\\ \Rightarrow 6t + 0.9 = 7t - 0.7\\ \Rightarrow 6t - 7t = - 0.7 - 0.9\\ \Rightarrow - t = - 1.6\\ \Rightarrow t = 1.6\,{\rm{h}}\end{array}$
Putting the value of $t = 1.6\,{\rm{h}}$in equation (2):
$\begin{array}{l} \Rightarrow d = 7 \times \left( {t - \dfrac{6}{{60}}} \right)\\ \Rightarrow d = 7 \times \left( {1.6 - 0.1} \right)\\ \Rightarrow d = 7 \times 1.5\\ \Rightarrow d = 10.5\,{\rm{km}}\end{array}$
Hence, the distance between his home and the railway station is $10.5\,{\rm{km}}$.

Note:
In this problem, we need to apply the formula which relates distance, time and speed, $d = v \times t$. Here we have to determine the distance from his home to the railway station. From the given information we can calculate the distances for both the cases. So equating the distance for both the cases, we can easily calculate the distance between his home and railway station.