# A train travels at a certain average speed for a distance of 63 km

Question:

A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?

Solution:

It is given that a train travels a distance of 63 km with a certain average speed and 72 km with a speed which is 6 km/hr more than the original average speed.

Time taken for the whole journey is 3hr.

We have to find out the original average speed of the train.

Let the original average speed of the train = x km/hr

Time taken by the train to cover the distance of $63 \mathrm{~km}=\frac{63 \mathrm{~km}}{x \mathrm{~km} / \mathrm{hr}}$

$=\frac{63}{x} \mathrm{hr}$

While covering the distance of 72 km, the speed of train

Time taken by the train to cover the distance of $63 \mathrm{~km}=\frac{72 \mathrm{~km}}{x \mathrm{~km} / \mathrm{hr}}$

$=\frac{72}{x} \mathrm{hr}$

Therefore time taken by the train to cover the distance of $72 \mathrm{~km}=\frac{72 \mathrm{~km}}{(x+6) \mathrm{km} / \mathrm{hr}}$

$=\frac{72}{x} \mathrm{hr}$

Therefore time taken by the train to cover the distance of $72 \mathrm{~km}=\frac{72 \mathrm{~km}}{(x+6) \mathrm{km} / \mathrm{hr}}$

$=\frac{72}{x+6} \mathrm{hr}$

Total time taken by the train

Therefore

$\frac{63}{x}+\frac{72}{x+6}=3$

$\frac{63(x+6)+72 x}{x(x+6)}=3$

$\frac{63 x+378+72 x}{x^{2}+6 x}=3$

$135 x+378=3 x^{2}+18 b$

Dividing both sides by3

$45 x+126=x^{2}+6 x$

$x^{2}-39 x-126=0$

$x^{2}-42 x+3 x-126=0$

$x(x-42)+3(x-42)=0$

$(x-42)(x+3)=0$

$x=42,-3$

Speed cannot be negative

Hence average speed of the train $=42 \mathrm{~km} / \mathrm{hr}$