A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km/hour, it takes 2 hours less in the journey.
A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km/hour, it takes 2 hours less in the journey. Find the original speed of the train.
Distance covered by the train = 300 km
We know that distance covered $(d)=\operatorname{speed}(s) \times \operatorname{time}(t)$
$\Rightarrow s \times t=300$
$\Rightarrow t=\frac{300}{s} \quad \ldots(\mathrm{i})$
Also, given that if the speed is increased by 5 km/h, time of travel gets reduced by 2 hours.
$\Rightarrow(s+5)(t-2)=300 \quad \ldots \ldots$ (ii)
Put the value of (i) in (ii), we get
$(s+5)\left(\frac{300}{s}-2\right)=300$
$(s+5)\left(\frac{300-2 s}{s}\right)=300$
$(s+5)(300-2 s)=300 s$
$300 s-2 s^{2}+1500-10 s=300 s$
$-2 s^{2}+1500-10 s=0$
$-s^{2}+750-5 s=0$
$s^{2}+5 s-750=0$
$s^{2}+30 s-25 s-750=0$
$s(s+30)-25(s+30)=0$
$(s-25)(s+30)=0$
$(s-25)=0$ or $(s+30)=0$
$s=25$ or $s=-30$
Ignore the negative value
Therefore, the original speed = 25 km/h