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

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45 km per hour.


Let the original speed of the train = x km per hour.

Total distance = 360 km

\text{Time taken by train to reach at final point =}\left ( \frac{360}{x} \right )\left ( \because speed=\frac{distance}{time} \right )

If train’s speed increased by 5 then speed = (x + 5) km per hour. Time taken by train to cover 360 km at a speed of (x + 5) km per hour=

 \left ( \frac{360}{x+5} \right )\left ( \because speed=\frac{distance}{time} \right )

Now it is given that the difference between the time taken at a speed of x and at a speed of x + 5 is 48 min.
\frac{360}{x} - \frac{360}{x+5} =\frac{48}{60}(hour)\\ \frac{360(x+5)-360x}{x(x+5)}=\frac{48}{60}(hour)\\ 5(360x+1800-360x)=4x(x+5)\\ 9000=4x^{2}+20x\\ 4x^{2}+20x-9000=0


x^{2}+5x-2250=0    (divide by 4)

 x^{2}+5x-2250=0\\ x^{2}+50x-45x-2250=0\\ x(x+50)-45(x+50)=0\\ x=-50,45

We can’t take x = – 50 because speed is always positive.

Hence original speed of train is 45 km per hour.

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