Question

A train leaves the station 1 hour before the scheduled time. The driver decreases its speed by 50 km per hour. At the next station 300 km away the train reached on time. Find the original speed of the train.
A.100 km/hour
B.150 km/hour
C.125 km/hour
D.200 km/hour

Answer 1

Hint First we will assume that the normal speed of the train is represented by \[x\] km/hour and then use formula of speed, \[\dfrac{{{\text{Distance}}}}{{{\text{TIme}}}}\]. When the driver decreases the speed of the train, we have the new speed \[x - 50{\text{ km/hr}}\]. Comparing both the speeds to find the value of original speed.

Complete step-by-step answer:
We are given that a train leaves the station 1 hour before the scheduled time and the driver decreases its speed by 50 km per hour
Let us assume that the normal speed of the train is represented by \[x\] km/hour.
First, we will compute the normal time to cover 300 km by train.
\[\dfrac{{300}}{x}\]
Since the train takes 1 more hour, then we will find the time taken by the train from the old time.
\[ \Rightarrow 1 + \dfrac{{300}}{x} = \dfrac{{x + 300}}{x}\]
We know that the speed is calculated using the formula,\[\dfrac{{{\text{Distance}}}}{{{\text{TIme}}}}\].
Substituting the value of distance and time taken by the train in the above formula of speed to find the new speed of the train, we get
\[ \Rightarrow \dfrac{{300}}{{\dfrac{{x + 300}}{x}}} = \dfrac{{300x}}{{x + 300}}{\text{ km/hr}}\]
But since the driver decreases the speed of the train, we have the new speed \[x - 50{\text{ km/hr}}\].
Now we have,
\[ \Rightarrow \dfrac{{300x}}{{x + 300}} = x - 50\]
Cross-multiplying in above equation, we get
\[
   \Rightarrow 300x = \left( {x - 50} \right)\left( {x + 300} \right) \\
   \Rightarrow 300x = {x^2} - 50x + 300x - 15000 \\
   \Rightarrow 300x = {x^2} + 250x - 15000 \\
 \]
Subtracting the above equation by \[300x\] on both sides, we get
\[
   \Rightarrow 300x - 300x = {x^2} + 250x - 15000 - 300x \\
   \Rightarrow 0 = {x^2} - 50x - 15000 \\
   \Rightarrow {x^2} - 50x - 15000 = 0 \\
 \]
\[
   \Rightarrow {x^2} - 150x + 100x - 1500 = 0 \\
   \Rightarrow x\left( {x - 150} \right) + 100\left( {x - 150} \right) = 0 \\
   \Rightarrow \left( {x - 150} \right)\left( {x + 100} \right) = 0 \\
 \]
\[ \Rightarrow x - 150 = 0\] or \[x + 100 = 0\]
\[ \Rightarrow x = 150{\text{ or }} - 100\]
So we reject the negative value as the train always runs in the same direction.
\[x = 150{\text{ km/hr}}\]
So, the normal speed of the train is 150 km/hour.
Hence, option B is correct.

Note In solving these types of questions, students should know the basic concepts of speed, distance and time. Since speed and time are inversely proportional, when time increases, speed decreases and vice versa.