Discussion
A cab on its way to Bangalore overtakes an auto also going in the same direction at 11 am. The cab reaches Bangalore at 1.30 pm and starts the return journey after taking a break of 1 hour. On its way back it meets the auto at 3 pm. At what time will the auto reach Bangalore?
Answer: D
Let \( D \) be the distance from the overtaking point to Bangalore.
The cab covers it between 11 am and 1:30 pm, so \( v_{\text{cab}} = \tfrac{D}{2.5} \). After a one-hour break it leaves at 2:30 pm and meets the auto at 3 pm, half an hour later:
\[ \text{cab's return distance} = \frac{D}{2.5} \times 0.5 = \frac{D}{5} \]
So they meet \( D - \tfrac{D}{5} = \tfrac{4D}{5} \) from the overtaking point. The auto took from 11 am to 3 pm - four hours - to get there:
\[ v_{\text{auto}} = \frac{4D/5}{4} = \frac{D}{5} \text{ per hour} \]
The auto still has \( \tfrac{D}{5} \) to go, which takes exactly one more hour:
\[ \text{arrival} = 4 \text{ p.m.} \]
The distance \( D \) cancels throughout, so it never needs a value.
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