#### 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.

45 km per hour.

Solution

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.
$\frac{360}{x} - \frac{360}{x+5} =\frac{48}{60}(hour)\\ \frac{360(x+5)-360x}{x(x+5)}=\frac{48}{60}(hour)\\ 5(360x+1800-360x)=4x(x+5)\\ 9000=4x^{2}+20x\\ 4x^{2}+20x-9000=0$

Or

$x^{2}+5x-2250=0$    (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.