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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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