Question

A train travels 360km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.
A.45 km/hr
B.60 km/hr
C.40 km/hr
D.55 km/hr

Answer 1

Hint: Let us assume the speed of the train be $v$ km/hr, and time taken for the journey be $t$ hours. We can use the relationship for the uniform speed, that is ${\text{speed = }}\dfrac{{{\text{distance}}}}{{{\text{time}}}}$ to formulate equations according to the relations given in the question. The equations then can be solved to find the value of $v$ .

Complete step-by-step answer:
Since the speed of the train is uniform, we can say the speed of the train is a fixed value throughout the journey of 360 km.
Let us assume the uniform speed of the train be $v$ km/hr, and the time taken for the journey of 360 km with uniform speed $v$ km/hr be $t$ hours.
It is known that the uniform speed of an object is related to time by the relation ${\text{speed = }}\dfrac{{{\text{distance}}}}{{{\text{time}}}}$ .
Substituting the value $v$ for speed, 360 for distance and $t$ for time in the relation ${\text{speed = }}\dfrac{{{\text{distance}}}}{{{\text{time}}}}$ , we get
$v = \dfrac{{360}}{t}$
On rearranging the above equation, we get
$t = \dfrac{{360}}{v}$
It is given in the question that if the speed of the train had been 5 km/hr more, the time would have been reduced by 1 hour.
The speed increased by 5 km/hr more, can be given as $v + 5$ km/hr, as the original speed of the train was $v$ km/hr.
Similarly, the reduced time can be represented as $t - 1$ hours, as the original time for the journey with uniform speed $v$ km/hr was $t$ hours.
Thus, substituting the value the value $v + 5$ for speed, 360 for distance and $t - 1$ for time in the relation ${\text{speed = }}\dfrac{{{\text{distance}}}}{{{\text{time}}}}$ , we get
$v + 5 = \dfrac{{360}}{{t - 1}}$
Substituting the value $\dfrac{{360}}{v}$ for $t$ in the above equation, we get
$v + 5 = \dfrac{{360}}{{\dfrac{{360}}{v} - 1}}$
Solving the above equation for the value of $v$, we get
$
  v + 5 = \dfrac{{360v}}{{360 - v}} \\
   \Rightarrow \left( {v + 5} \right)\left( {360 - v} \right) = 360v \\
   \Rightarrow 360v + 1800 - 5v - {v^2} = 360v \\
   \Rightarrow {v^2} + 5v - 1800 = 0 \\
   \Rightarrow {v^2} + 45v - 40v - 1800 = 0 \\
   \Rightarrow v\left( {v + 45} \right) - 40\left( {v + 45} \right) = 0 \\
   \Rightarrow \left( {v + 45} \right)\left( {v - 40} \right) = 0 \\
   \Rightarrow v = - 45,40 \\
$
Since the speed of the train is a physical quantity , it cannot be negative.
Thus the speed of the train is 40 km/hr.

Note: The speed an object and time taken for a journey cannot be negative. We can additionally solve for the time of the journey by the formula $t = \dfrac{{360}}{v}$. Since the equation ${v^2} + 5v - 1800 = 0$ is a quadratic equation, only the physically possible values must be taken upon solving the equation.