An Engine length $1000 $ m moving at $10$ m/s. A bird is flying from engine to end with $x$ kmph and coming back at $2x$. Take total time of bird traveling as $187.5$ s. Find $x$ and $2x$.
My approach:
$s+10$ and $2s-10$ are the speeds in both directions.
$\dfrac {1000} {s+10} + \dfrac{1000}{2s-10} = 187.5 $ s
I am not getting the lcm of $s+10$ and $2s-10$.
You've got the "process" correct, but we need to account for discrepancies in units:
Namely, the speed of flight of the bird, $x$ and $2x$, are given in kilometers per hour, while the engine length is given in meters, and the speed at which the engine is moving is given in meters per second, with overall time of the flight of the bird given in seconds.
Taking unit conversion into account:
$\dfrac{1\;\text{hour}}{3600\;\text{seconds}}$
$\dfrac{1000\;\text{meters}}{1\;\text{kilometer}}$
we have $x \;\text{kmph} \times \dfrac{1000\;\text{m}}{\text{km}}\times \dfrac {1 \;\text{hour}}{3600\;\text{sec}} = \dfrac 5{18} x \;\text{m/s}$
and we have $2x\;\text{kmph} = \dfrac 59 x \;\text{m/s}$.
This then gives us the equation: $$\begin{align}\dfrac{1000}{\frac 5{18}x + 10} + \dfrac {1000}{\frac 59x - 10} & = 187.5 \\ \\ \iff \frac{3600}{x + 36} + \frac {1800}{x - 18} & = 187.5 \\ \\ \iff \frac {5400x}{(x + 36)(x - 18)} & = 187.5\end{align}$$
With a bit of algebra, we obtain a quadratic equation, with positive root $$x \approx 31.4223 \;\text{meters per second}$$
You can use the conversions bulleted above to obtain $x$ in kilometers per hour.
