A train was delayed by a semaphore for $16$ minutes and made up for the delay on a section of $80$ km travelling with the speed $10$ km per hour higher than that which accorded the schedule. Find the speed of the train which accorded the schedule.
Let the accorded speed of train $= x$ kmph
If the train was travelling with the accorded speed, then the time taken to cover $80$ km will be $= t_1 = \dfrac{80}{x}$ hours
But, distance of $80$ km covered with a speed of $x + 10$ kmph
$\therefore \;$ Time taken to cover $80$ km $= t_2 = \dfrac{80}{x + 10}$ hr
$\therefore \;$ Time for which the train was delayed $= t_1 - t_2 = \dfrac{80}{x} - \dfrac{80}{x + 10}$ hours
Given: Time for which the train was delayed $= 16$ min $= \dfrac{16}{60} = \dfrac{4}{15}$ hours
i.e. $\;$ $\dfrac{80}{x} - \dfrac{80}{x + 10} = \dfrac{4}{15}$
i.e. $\;$ $\dfrac{20}{x} - \dfrac{20}{x + 10} = \dfrac{1}{15}$
i.e. $\;$ $20 \left(x + 10 - x\right) \times 15 = x \left(x + 10\right)$
i.e. $\;$ $x^2 + 10 x - 3000 = 0$
i.e. $\;$ $\left(x + 60\right) \left(x - 50\right) = 0$
i.e. $\;$ $x = -60$ $\;$ or $\;$ $x = 50$
But since the speed of the train cannot be negative
$\implies$ Original speed of train $= x = 50$ kmph