Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph; he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.? (D.M.R.C., 2003)

### Concept & Average Speed Logic When travelling the same distance, the relationship between speeds and times can be established using a common base time. $$Distance = Speed \times Time$$ ### Step-by-Step Solution * **Given:** To reach at 2 P.M., $S_1 = 10$ kmph. To reach at 12 noon, $S_2 = 15$ kmph. We want to find the speed for 1 P.M. * **Establish Times:** Let $t$ be the time in hours to reach at 1 P.M. * Time to reach at 2 P.M. = $(t + 1)$ hours. * Time to reach at 12 noon = $(t - 1)$ hours. * **Equate Distances:** * $10 \times (t + 1) = 15 \times (t - 1)$ * $10t + 10 = 15t - 15$ * $5t = 25 \Rightarrow t = 5$ hours. * **Calculate Distance & Target Speed:** * Distance = $10 \times (5 + 1) = 60$ km. * Speed for 1 P.M. (which takes 5 hours) = $\frac{60}{5} = 12$ kmph. ### Exam Strategy & Shortcut Because 1 P.M. is exactly halfway between 12 noon and 2 P.M. in time, you might be tempted to average the speeds $\frac{10+15}{2} = 12.5$, but average speed doesn't work this way over equal distance. However, you can find distance directly: time gap is 2 hrs. $D = \frac{10 \times 15}{15 - 10} \times 2 = \frac{150}{5} \times 2 = 60$ km. Time at 15kmph is 4 hrs. Start time is 8 A.M. 1 P.M. is 5 hrs from 8 A.M. Speed = $60 / 5 = 12$ kmph. ### Common Pitfall Taking the simple arithmetic mean of the speeds (12.5 kmph) assuming time linearity translates directly to speed linearity. ### Final Answer Therefore, the correct answer is **12 kmph**.