A train running at $\frac{7}{11}$ of its own speed reached a place in 22 hours. How much time could be saved if the train would have run at its own speed?

### Concept & Proportionality When distance is constant, speed and time are inversely proportional. $$S_1 \times T_1 = S_2 \times T_2$$ If speed becomes $\frac{a}{b}$ of its original, time becomes $\frac{b}{a}$ of its original. ### Step-by-Step Solution * Let the usual speed be $S$ and usual time be $T$. * The new speed is $\frac{7}{11}S$. * Because speed and time are inversely proportional for a constant distance, the new time taken is $\frac{11}{7}T$. * We are given the new time is 22 hours. * $\frac{11}{7} \times T = 22$ * $T = 22 \times \frac{7}{11} = 14$ hours. * The usual time taken would have been 14 hours. * Time saved = New time - Usual time = $22 - 14 = 8$ hours. ### Exam Strategy & Shortcut Using ratios: Speed ratio is $11:7$ (original to new). Therefore, the time ratio is $7:11$ (original to new). If 11 units $= 22$ hours, 1 unit $= 2$ hours. The time saved is $11 - 7 = 4$ units. $4 \times 2 = 8$ hours. ### Common Pitfall Students often calculate the usual time (14 hours) and incorrectly select it as the answer, forgetting that the question asks for the "time saved". ### Final Answer Therefore, the correct answer is **8 hours**.