A train was moving with uniform …

Suppose the original speed of the train be x km/hr

The length of the journey be y km. Then
Time taken =(y/x)hrs.
CASE I:

When defect in the engine occurs after covering a distance of 30 km.
Speed for first 30km=x km/hr
Speed for the remaining {tex}(y - 30)km{/tex}={tex}= \frac { 4 } { 5 } x \mathrm { km } / \mathrm { hrs }{/tex}
{tex}\therefore{/tex} Time taken to cover {tex}30km{/tex} {tex}= \frac { 30 } { x }hrs{/tex}
Time taken to cover {tex}(y - 30)km{/tex}={tex}\frac { y - 30 } { ( 4 x / 5 ) } \text { hrs. } = \frac { 5 } { 4 x } ( y - 30 )hrs.{/tex}
{tex}\Rightarrow \frac { 30 } { x } + \frac { 5 } { 4 x } ( y - 30 ) = \frac { y } { x } + \frac { 45 } { 60 }{/tex}
{tex}\Rightarrow \quad \frac { 30 } { x } + \frac { 5 y - 150 } { 4 x } = \frac { y } { x } + \frac { 3 } { 4 }{/tex}
{tex}\Rightarrow \quad \frac { 120 + 5 y - 150 } { 4 x } = \frac { 4 y + 3 x } { 4 x }{/tex}
{tex}\Rightarrow{/tex} {tex}5y - 30=4y + 3x{/tex}
{tex}\Rightarrow{/tex}{tex}3x - y +30=0 {/tex}.................................(i)
CASE II:

When defect in the engine occurs after covering a distance of 48 km.
Speed for first {tex}48km =x\ km/hr{/tex}.
Speed for the remaining {tex}(y -48){/tex} {tex}\mathrm { km } = \frac { 4 x } { 5 } \mathrm { km } / \mathrm { hr }{/tex}
{tex}\therefore{/tex}Time taken to cover {tex}48 km{/tex} = {tex}\frac { 48 } { x }hrs.{/tex}
Time taken to cover {tex}(y-48)km{/tex} = {tex}\left( \frac { y - 48 } { 4 x / 5 } \right) \mathrm { hr } = \left\{ \frac { 5 ( y - 48 ) } { 4 x } \right\} \mathrm { hr }{/tex}
According to the given condition, the train now reaches 9 minutes earlier i.e., 36 minutes later.
 {tex}\frac { 48 } { x } + \frac { 5 ( y - 48 ) } { 4 x } = \frac { y } { x } + \frac { 36 } { 60 }{/tex}
{tex}\Rightarrow \quad \frac { 48 } { x } + \frac { 5 y - 240 } { 4 x } = \frac { y } { x } + \frac { 3 } { 5 }{/tex}
{tex}\Rightarrow \quad \frac { 192 + 5 y - 240 } { 4 x } = \frac { 5 y + 3 x } { 5 x }{/tex}
{tex}\Rightarrow \quad \frac { 5 y - 48 } { 4 } = \frac { 5 y + 3 x } { 5 }{/tex}
{tex}\Rightarrow{/tex}{tex}25y - 240 = 20y + 12x{/tex}
{tex}\Rightarrow{/tex}{tex}12x - 5y + 240=0{/tex} ................................(ii)
{tex}3x - y + 30 =0{/tex}
{tex}12x  - 5y + 240=0{/tex}
Using cross-multiplication, we have
{tex}\frac { x } { - 240 + 150 } = \frac { - y } { 720 - 360 } = \frac { 1 } { - 15 + 12 }{/tex}
{tex}\Rightarrow \quad \frac { x } { - 90 } = \frac { - y } { 360 } = \frac { 1 } { - 3 }{/tex}
{tex}\Rightarrow \quad x = \frac { - 90 } { - 3 } = 30 \text { and } y = \frac { - 360 } { - 3 } = 120{/tex}
The original speed of the train is {tex}30km/hr{/tex} and the length of the journey is 120 km.