Question

A train travelling at a uniform speed of 360km, would have taken 48 minutes less to travel the same distance if its speed were 5km/hr more. Find the original speed of the train in km.hr.

Answer 1

Hint: In this question let the original speed of the train be V km/hr and the time taken to complete the journey be t hours. Use the constraints of questions along with the relationship between distance, speed and time to formulate mathematical equations.
i.e. Distance = speed $ \times $ time.

Complete step-by-step answer:
Let the speed of the train be V km/hr.
And the time it takes to cover a distance be (t) hr.
Now it is given that the train covers a distance (d) = 360 km with some speed.
Now it is given that the speed is increased by 5 km/hr. it will cover the same distance in 48 minutes less.
Now as we know 60 min = 1 hour.
Therefore 48 min = (48/60) hour
Therefore new speed = (V + 5) km/hr.
And the new time = (t – (48/60)) hr.

Now the distance is the same in both the cases.
And we know that the distance, speed and time is related as
Distance = speed $ \times $ time.
Therefore
$ \Rightarrow 360 = V.t$ Km............................. (1)
And
\[ \Rightarrow 360 = \left( {V + 5} \right)\left( {t - \dfrac{{48}}{{60}}} \right)\] Km.............................. (2)
So equate these two equation we have,
\[ \Rightarrow \left( {V + 5} \right)\left( {t - \dfrac{{48}}{{60}}} \right) = V.t\]

Now simplify the above equation we have,
\[ \Rightarrow V.t - \dfrac{{48}}{{60}}V + 5t - \dfrac{{5 \times 48}}{{60}} = V.t\]
\[ \Rightarrow - \dfrac{4}{5}V + 5t - 4 = 0\]
Now from equation (1) \[t = \dfrac{{360}}{V}\] so substitute this value in above equation we have,
\[ \Rightarrow - \dfrac{4}{5}V + 5 \times \dfrac{{360}}{V} - 4 = 0\]

Now simplify the above equation we have,
\[ \Rightarrow - 4{V^2} + 9000 - 20V = 0\]
Divide by -4 throughout we have,
\[ \Rightarrow {V^2} + 5V - 2250 = 0\]

Now factorize this equation we have,
\[ \Rightarrow {V^2} + 50V - 45V - 2250 = 0\]
$ \Rightarrow V\left( {V + 50} \right) - 45\left( {V + 50} \right) = 0$
$ \Rightarrow \left( {V + 50} \right)\left( {V - 45} \right) = 0$
$ \Rightarrow V = 45, - 50$

Negative speed is not possible.
So the speed of the train is 45 km/hr.
So this is the required answer.

Note: It is important to notice here that the negative value of speed is not considered because speed is a scalar quantity and it is not vector like velocity. So speed has only magnitude and not direction and the magnitude can’t be negative. The quadratic equation could have been solved directly using the Dharacharya formula which is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.