A train has to complete a journey of 800 km. If it meets a minor accident, its speed becomes half of the existing speed. If there is a mechanical defect, the speed becomes one-fourth of the existing speed. On its way, the train meets with a minor accident after 200 km; and 400 km thereafter, it develops a mechanical defect. Had the train developed the mechanical defect after 200 km and met the minor accident 400 km thereafter, it would have taken 4 more hours to reach its destination. What was the original speed of the train in km per hour?

Correct Answer: (A) 200

Let the train's original speed be $v$ km/h over the 800 km journey. In the actual run, the train travels 200 km at speed $v$, 400 km at speed $v/2$ (after minor accident), and the remaining 200 km at speed $\frac{1}{4}(v\mathrm{/}2)=v\mathrm{/}8$$\frac{1}{4}$(v/2) = v/8 (after mechanical defect), taking total time $T_1 =$$\frac{200}{v}$ + $\frac{400}{v/2}$ + $\frac{200}{v/8}$ = $\frac{2600}{v}$. In the hypothetical run, it travels 200 km at speed $v$, 400 km at speed $v/4$ (after mechanical defect), and the remaining 200 km at speed $\frac{1}{2}(v\mathrm{/}4)=v\mathrm{/}8$$\frac{1}{2}$(v/4) = v/8 (after minor accident), taking $T_2 =$$\frac{200}{v}$ + $\frac{400}{v/4}$ + $\frac{200}{v/8}$ = $\frac{3400}{v}$. The given time difference is $T_2 - T_1 =$$\frac{800}{v}$ = 4$\text{ hours}$ $\implies$ v = 200$\text{ km/h}$.