A train has $320\,\mathrm{km}$ to run. After going $\tfrac{1}{5}$ of the distance, the engine breaks down and it can only run the remaining part of the journey at $\tfrac{3}{4}$ of the original speed. If it arrives 2 hrs 40 min late, what was its original speed?

Let the original speed be $x\,\mathrm{km/hr}$. Normally, time for $320$ km would be $\frac{320}{x}$ hr. The train covers $\frac{1}{5}\times 320=64$ km at speed $x$, and the remaining $256$ km at speed $\frac{3x}{4}$. So actual time is $\frac{64}{x}+\frac{256}{3x/4}=\frac{64}{x}+\frac{1024}{3x}$. Since the delay is $2$ hr $40$ min $=\frac{8}{3}$ hr, we get $\frac{64}{x}+\frac{1024}{3x}=\frac{320}{x}+\frac{8}{3}$, which gives $x=32$.