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

Given:

A train travels at a certain average speed for a distance 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than the original speed.

It takes 3 hours to complete the total journey.

To do:

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

Solution:

Let the original average speed of the train be $x$ km/hr.

This implies,

The new average speed of the train $=x+6$ km/hr

Time taken by the train to travel 63 km at original average speed$=\frac{63}{x}$ hours

Time taken by the train to travel 72 km at new average speed$=\frac{72}{x+6}$ hours

According to the question,

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

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

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

$135x+378=3(x^2+6x)$   (On cross multiplication)

$3(45x+126)=3(x^2+6x)$

$x^2+6x-45x-126=0$

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

Solving for $x$ by factorization method, we get,

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

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

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

$x+3=0$ or $x-42=0$

$x=-3$ or $x=42$

Speed cannot be negative. Therefore, the value of $x$ is $42$ km/hr.

The original average speed of the train is $42$ km/hr.

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Updated on: 10-Oct-2022

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