The train left station A for station B. Having traveled $450$ km, which constitutes $75$ percent of the distance between A and B, the train was stopped by a snow drift. Half an hour later the track was cleared and the engine driver, having increased the speed by $15$ kmph, arrived at station $B$ on time. Find the initial speed of the train.
Let distance between stations A and B = $d\left(AB\right) = x$ km
Let the train stop at $A'$
Given: $\;$ $d\left(AA'\right) = 450$ km and $d\left(AA'\right) = 75\% \text{ of } d\left(AB\right)$
i.e. $\;$ $450 = \dfrac{75}{100} \times x$
i.e. $\;$ $x = \dfrac{450 \times 100}{75} = 600$ km
i.e. $\;$ distance between stations A and B $= d\left(AB\right) = 600$ km
$\therefore \;$ Remaining distance $= d\left(A'B\right) = 600 - 450 = 150$ km
Let initial speed of train $= s$ kmph
Then, time taken to cover $d\left(AB\right) = t_{\left(AB\right)} = \dfrac{d\left(AB\right)}{s} = \dfrac{600}{s}$ hr $\;\;\; \cdots \; (1)$
Time taken to cover $d\left(AA'\right) = t_{\left(AA'\right)} = \dfrac{d\left(AA'\right)}{s} = \dfrac{450}{s}$ hr
Train stopped at A' for $\dfrac{1}{2}$ hour
New speed of train $= s_1 = \left(s + 15\right)$ kmph
Time taken to cover distance $d\left(A'B\right) = t_{\left(A'B\right)} = \dfrac{d\left(A'B\right)}{s_1} = \left(\dfrac{150}{s+15} + \dfrac{1}{2}\right)$ hr
Now, time taken to cover $d\left(AB\right) = t_{\left(AB\right)} = t_{\left(AA'\right)} + t_{\left(A'B\right)}$
i.e. $\;$ $t_{\left(AB\right)} = \dfrac{450}{s} + \dfrac{150}{s + 15} + \dfrac{1}{2}$ $\;\;\; \cdots \; (2)$
Since the train arrives at station B on time after stoppage,
we have from equations $\left(1\right)$ and $\left(2\right)$
$\dfrac{450}{s} + \dfrac{150}{s + 15} + \dfrac{1}{2} = \dfrac{600}{s}$
i.e. $\;$ $\dfrac{150}{s + 15} + \dfrac{1}{2} = \dfrac{600}{s} - \dfrac{450}{s} = \dfrac{150}{s}$
i.e. $\;$ $\dfrac{150}{s} - \dfrac{150}{s + 15} = \dfrac{1}{2}$
i.e. $\;$ $150 \times 15 \times 2 = s\left(s + 15\right)$
i.e. $\;$ $s^2 + 15s - 4500 = 0$
i.e. $\;$ $\left(s+75\right)\left(s-60\right) = 0$
i.e. $\;$ $s = -75$ $\;$ or $\;$ $s = 60$
Since the speed of a train cannot be negative $\implies$ $s = -75$ is not a valid solution
$\therefore \;$ The original speed of train $= s = 60$ kmph