A car takes 15 minutes less to cover a distance of 75 km, if it increases its speed by 10 km/hr from its usual speed. How much time would it take to cover a distance of 300 km using this speed?

### Concept & Proportional Time Variation We use the fundamental relation of Time, Speed, and Distance. The difference in time taken at two different speeds can be formulated algebraically to find the original speed. $$\frac{D}{S_{original}} - \frac{D}{S_{new}} = \Delta T$$ ### Step-by-Step Solution 1. Let the usual speed of the car be $v$ km/hr. 2. The normal time to cover $75$ km is $\frac{75}{v}$ hours. 3. If speed increases by $10$ km/hr, the new speed is $(v + 10)$ km/hr, taking $\frac{75}{v + 10}$ hours. 4. The time difference is $15$ minutes, which must be converted to hours: $\frac{15}{60} = \frac{1}{4}$ hour. 5. Set up the equation: $$\frac{75}{v} - \frac{75}{v + 10} = \frac{1}{4}$$ 6. Factor out $75$ and solve: $$75 \left( \frac{v + 10 - v}{v(v + 10)} \right) = \frac{1}{4}$$ $$75 \left( \frac{10}{v^2 + 10v} \right) = \frac{1}{4}$$ $$3000 = v^2 + 10v$$ $$v^2 + 10v - 3000 = 0$$ 7. Factorize the quadratic: $$(v + 60)(v - 50) = 0$$ Since speed is positive, the usual speed $v = 50$ km/hr. 8. The question asks for the time taken to cover 300 km at "this speed" (referring to the usual speed). $$Time = \frac{300}{50} = 6 \text{ hours}$$ ### Exam Strategy & Shortcut To bypass the quadratic, look for factors of $75 \times 40 = 3000$ that differ by $10$. Those are $50$ and $60$. The original slower speed is $50$. The target time is simply $\frac{300}{50} = 6$. ### Common Pitfall A major pitfall is solving for the new speed instead of the usual speed, or failing to convert 15 minutes into $\frac{1}{4}$ of an hour, which disrupts the entire calculation. ### Final Answer Therefore, the correct answer is **6 hours**.