Discussion
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.?
Answer: C
The two speeds produce arrival times two hours apart, so the travel times differ by 2 hours.
\[ \frac{D}{10} - \frac{D}{15} = 2 \quad\Longrightarrow\quad \frac{D}{30} = 2 \quad\Longrightarrow\quad D = 60 \text{ km} \]
At 10 kmph the trip takes 6 hours and ends at 2 p.m., so he starts at 8 a.m. To arrive at 1 p.m. he has 5 hours:
\[ v = \frac{60}{5} = 12 \text{ kmph} \]
Averaging the two given speeds to 12.5 kmph happens to come close, but it is not the reasoning - the times, not the speeds, are what average out here.
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