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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.

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 uniform speed.

Now it is given that the speed is increased by 5 km/hr. it will cover the same distance in 1 hour less.

Therefore new speed = (V + 5) km/hr.

And the new time = (t – 1) 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

\[ \Rightarrow 360 = \left( {V + 5} \right)\left( {t - 1} \right)\] Km.............................. (2)

So equate these two equation we have,

\[ \Rightarrow \left( {V + 5} \right)\left( {t - 1} \right) = V.t\]

Now simplify the above equation we have,

\[ \Rightarrow V.t - V + 5t - 5 = V.t\]

\[ \Rightarrow - V + 5t - 5 = 0\]

Now from equation (1) \[t = \dfrac{{360}}{V}\] so substitute this value in above equation we have,

\[ \Rightarrow - V + 5 \times \dfrac{{360}}{V} - 5 = 0\]

Now simplify the above equation we have,

\[ \Rightarrow - {V^2} + 1800 - 5V = 0\]

Divide by -1 throughout we have,

\[ \Rightarrow {V^2} + 5V - 1800 = 0\]

Now factorize this equation we have,

\[ \Rightarrow {V^2} + 45V - 40V - 1800 = 0\]

$ \Rightarrow V\left( {V + 45} \right) - 40\left( {V + 45} \right) = 0$

$ \Rightarrow \left( {V + 45} \right)\left( {V - 40} \right) = 0$

$ \Rightarrow V = 40, - 45$

Negative speed is not possible.

So the speed of the train is 40 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 as well.

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 uniform speed.

Now it is given that the speed is increased by 5 km/hr. it will cover the same distance in 1 hour less.

Therefore new speed = (V + 5) km/hr.

And the new time = (t – 1) 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

\[ \Rightarrow 360 = \left( {V + 5} \right)\left( {t - 1} \right)\] Km.............................. (2)

So equate these two equation we have,

\[ \Rightarrow \left( {V + 5} \right)\left( {t - 1} \right) = V.t\]

Now simplify the above equation we have,

\[ \Rightarrow V.t - V + 5t - 5 = V.t\]

\[ \Rightarrow - V + 5t - 5 = 0\]

Now from equation (1) \[t = \dfrac{{360}}{V}\] so substitute this value in above equation we have,

\[ \Rightarrow - V + 5 \times \dfrac{{360}}{V} - 5 = 0\]

Now simplify the above equation we have,

\[ \Rightarrow - {V^2} + 1800 - 5V = 0\]

Divide by -1 throughout we have,

\[ \Rightarrow {V^2} + 5V - 1800 = 0\]

Now factorize this equation we have,

\[ \Rightarrow {V^2} + 45V - 40V - 1800 = 0\]

$ \Rightarrow V\left( {V + 45} \right) - 40\left( {V + 45} \right) = 0$

$ \Rightarrow \left( {V + 45} \right)\left( {V - 40} \right) = 0$

$ \Rightarrow V = 40, - 45$

Negative speed is not possible.

So the speed of the train is 40 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 as well.