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

Given: A train travels at a certain average speed for a distance of $63\ km$ and then travels at a distance of $72\ km$ at an average speed of $6\ km/hr$ more than its original speed. Time taken to complete the journey$=3\ hours$.

To do: To find the original speed of the train.

Solution:

Let x be the original average speed of the train for $63\ km$.

Then, $( x + 6)$ will be the new average speed for remaining $72\ km$.

Total time taken to complete the journey is $3\ hrs$.

According to the given condition,

Time taken to travel for $( 63\ km)=\frac{distance}{speed}$

$=\frac{63}{x}$

Time taken to travel for remaining $( 72\ km)=\frac{72}{x+6}$

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

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

Since speed can not be negative.

Therefore $x = 42 km/hr$.

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