Question

Arup and Swarup leave point A at 8 AM to point B. To reach B, they have to walk the first 2 km, then travel 4 km by boat and complete the final 20 km by car. Arup and Swarup walk at a constant speed of 4 km/hr and 5 km/hr respectively. Each rows his boat for 30 minutes. Arup drives his car at a constant speed of 50 km/hr while Swarup drives at 40 km/hr. If no time is wasted in transit, when will they meet again?

• A. At 9:15 AM
• B. At 9:18 AM
• C. At 9:21 AM
• D. At 9:24 AM
• E. At 9:30 AM

Hint:

When one traveller has a head start before a faster one begins, convert the time lead into a \emph{distance} lead at the slower person’s speed, then use relative speed to find catch-up time.

Answer 1

The Correct Option is D

Step 1: Walking (2 km).
Arup: \(t_A=\frac{2}{4}=0.5\) hr \(=30\) min \(\Rightarrow\) reaches at 8:30.
Swarup: \(t_S=\frac{2}{5}=0.4\) hr \(=24\) min \(\Rightarrow\) reaches at 8:24.
Swarup is \(6\) minutes ahead.

Step 2: Boat (4 km).
Each rows for \(30\) min \(\Rightarrow\) Swarup boats 8:24–8:54, Arup 8:30–9:00.
Lead after boating remains \(6\) minutes (Swarup at 8:54; Arup at 9:00).

Step 3: Car (20 km).
At 9:00, Swarup already has a \(\,6\)-minute head start and has driven \(40\times 0.1=4\) km. Relative speed \(=50-40=10\) km/hr.
Time for Arup to catch up \(=\frac{4}{10}=0.4\) hr \(=24\) min.
Catch-up time \(=9{:}00+24\text{ min}=9{:}24\) AM.

Check: In these 24 minutes, Arup drives \(50\times 0.4=20\) km and Swarup (from 8:54 to 9:24 is 30 min) drives \(40\times 0.5=20\) km. They meet at point \(B\).

Final Answer: \[ \boxed{\text{(D) 9:24 AM}} \]