A train travel at certain average …

Given that first distance = 54 km.
Let x be the first speed of the train.
We know that {tex}\frac { \text { Distance } } { \text { Speed } }{/tex} = Time ={tex}\frac { \text {54} } { \text { x } }{/tex} hours
When  63 km cover at an average speed of 6 km/hr more than the first speed.
then  Time ={tex}\frac { \text {63} } { \text { x+ 6 } }{/tex} hours
As per given condition
{tex}\frac { 54 } { x } + \frac { 63 } { x + 6 }{/tex} = 3 hours
{tex}\Rightarrow \frac { 54 ( x + 6 ) + 63 x } { x ( x + 6 ) }{/tex} = 3
{tex}\Rightarrow{/tex} 54(x + 6) + 63x = 3x(x + 6)
{tex}\Rightarrow{/tex} 54x + 324 + 63x = 3x² + 18x
{tex}\Rightarrow{/tex} 117x + 324 = 3x² + 18x
{tex}\Rightarrow{/tex} 3x² - 117x - 324 + 18x = 0
{tex}\Rightarrow{/tex} 3x² - 99x - 324 = 0
{tex}\Rightarrow{/tex} x² - 33x - 108 = 0
{tex}\Rightarrow{/tex} x² - 36x + 3x - 108 = 0
{tex}\Rightarrow{/tex} x(x - 36) + 3(x - 36) = 0
{tex}\Rightarrow{/tex} x + 3 = 0 or x - 36 = 0
{tex}\Rightarrow{/tex} x = -3 or x = 36
Speed cannot be negative and hence initial speed of the train is 36 km/hour.