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.

Let's solve the problem step by step.

Given:

The train travels a distance of 360 km.
If the speed of the train were 5 km/h more, it would take 48 minutes less to cover the same distance.
Let the original speed of the train be x km/h.

Step 1: Time taken at the original speed
Time taken at the original speed is given by: Time = Distance / Speed = 360 / x hours.

Step 2: Time taken at the increased speed
The new speed of the train would be x + 5 km/h. So, the time taken at this speed would be: Time = Distance / Speed = 360 / (x + 5) hours.

Step 3: Difference in time
The difference in time is given as 48 minutes, which is equal to 48 / 60 = 4/5 hours.

So, we can write the equation for the difference in time as: (360 / x) - (360 / (x + 5)) = 4/5.

Step 4: Solve the equation
To solve this equation, first, let's eliminate the fractions by multiplying both sides by 5x(x + 5): 5x(x + 5) * [(360 / x) - (360 / (x + 5))] = 5x(x + 5) * (4 / 5).

Simplifying both sides: 5(x + 5) * 360 - 5x * 360 = 4x(x + 5).

Now, distribute and simplify: 1800(x + 5) - 1800x = 4x(x + 5).

Expanding both sides: 1800x + 9000 - 1800x = 4x^2 + 20x.

Simplifying: 9000 = 4x^2 + 20x.

Step 5: Solve the quadratic equation
Now, let's solve the quadratic equation: 4x^2 + 20x - 9000 = 0.

Divide through by 4: x^2 + 5x - 2250 = 0.

Now, solve this quadratic equation using the quadratic formula: x = [-b ± √(b² - 4ac)] / 2a.

For the equation x² + 5x - 2250 = 0, a = 1, b = 5, and c = -2250. Substituting these values into the quadratic formula: x = [-5 ± √(5² - 4(1)(-2250))] / 2(1).

x = [-5 ± √(25 + 9000)] / 2.

x = [-5 ± √9025] / 2.

x = [-5 ± 95] / 2.

Now, solving for both possible values of x:

x = (-5 + 95) / 2 = 90 / 2 = 45.
x = (-5 - 95) / 2 = -100 / 2 = -50 (which is not possible, as speed cannot be negative).
Thus, the original speed of the train is 45 km/h.

Final Answer:
The original speed of the train is 45 km/h.