A train travels 360km/h at uniform …

Let the uniform speed of the train be x km/h.
Then, the time taken by the train to travel 360/km {tex}= \frac { 360 } { \mathrm { x } }{/tex}
.......{tex}\text { Time } = \frac { \text { distance } } { \text { speed } }{/tex}
If the speed had been 5km/h more,
than the time taken by train to travel 360km {tex}= \frac { 360 } { x + 5 }{/tex}
{tex}\ldots \text { Time } = \frac { \text { distance } } { \text { speed } }{/tex}

According to the question, {tex}\frac { 360 } { x + 5 } = \frac { 360 } { x } - 1{/tex}
{tex}\Rightarrow \frac { 360 } { x } - \frac { 360 } { x + 5 } = 1 \Rightarrow 360 \left( \frac { 1 } { x } - \frac { 1 } { x + 5 } \right) = 1{/tex}
{tex}\Rightarrow \frac { 1 } { x } - \frac { 1 } { x + 5 } = \frac { 1 } { 360 } \Rightarrow \frac { x + 5 - x } { x ( x + 5 ) } = \frac { 1 } { 360 }{/tex}
{tex}\Rightarrow \frac { 5 } { x ( x + 5 ) } = \frac { 1 } { 360 } \Rightarrow x ( x + 5 ) = ( 5 ) ( 360 ){/tex}
{tex}\Rightarrow x ( x + 5 ) = 1800 \Rightarrow x ^ { 2 } + 5 x - 1800 = 0{/tex}
Using quadratic formula, {tex}\mathrm { x } = \frac { - \mathrm { b } \pm \sqrt { b ^ { 2 } - 4 \mathrm { ac } } } { 2 \mathrm { a } }{/tex}
we get {tex}= \frac { - 5 \pm \sqrt { ( 5 ) ^ { 2 } - 4 ( 1 ) ( - 1800 ) } } { 2 ( 1 ) }{/tex}
{tex}= \frac { - 5 \pm \sqrt { 25 + 7200 } } { 2 } = \frac { - 5 \pm \sqrt { 7225 } } { 2 } = \frac { - 5 \pm 85 } { 2 }{/tex}
{tex}= \frac { - 5 + 85 } { 2 } , \frac { - 5 - 85 } { 2 } \Rightarrow x = 40 , - 45{/tex}
Since, x is the speed of the train, it cannot be negative.
So, we ignore the root x = -45.
Therefore, x = 40 gives the speed of the train as 40 km/h.