A train travels 180 km at a uniform speed. If the speed had been 9 km/hr more, it would have taken 1 hour less for the same journey.

Distance covered by the train = 180 km

We know that distance covered $(d)=\operatorname{speed}(s) \times \operatorname{time}(t)$

$\Rightarrow s \times t=180$

$\Rightarrow s=\frac{180}{t} \quad \ldots(\mathrm{i})$

Also, given that if the speed is increased by 9km/h, time of travel gets reduced by 1 hour. 

$\Rightarrow(s+9)(t-1)=180 \quad \ldots$ (ii)

Put the value of (i) in (ii), we get

$\left(\frac{180}{t}+9\right)(t-1)=180$

$(180+9 t)(t-1)=180 t$

$180 t-180+9 t^{2}-9 t=180 t$

$9 t^{2}-9 t-180=0$

$t^{2}-t-20=0$

$t^{2}-5 t+4 t-20=0$

$t(t-5)+4(t-5)=0$

$(t+4)(t-5)=0$

$(t+4)=0$ or $(t-5)=0$

$t=-4$ or $t=5$

Ignore the negative value
So, time taken = 5 hours
From (i)

$s=\frac{180}{t}=\frac{180}{5}=36$

Hence, the speed is 36 km/h.