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Meeting Time after Multi-modal Travel

XAT 2017 · Quantitative Ability & Data Interpretation · Medium · Speed Time Distance

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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?

  1. A.

    At 9.15 AM

  2. B.

    At 9.18 AM

  3. C.

    At 9.21 AM

  4. D.

    At 9.24 AM

  5. E.

    At 9.30 AM

Answer

D

Explanation

Walking time: Arup: 24=30\frac{2}{4} = 30 mins. Swarup: 25=24\frac{2}{5} = 24 mins.

Boat time: Both take 30 mins.

Thus, Swarup finishes boat leg at 24+30=5424 + 30 = 54 mins after 8 AM and starts driving at 8:548:54 AM. Arup finishes boat leg at 30+30=6030 + 30 = 60 mins after 8 AM and starts driving at 9:009:00 AM. In those 6 mins (8:54 to 9:00 AM), Swarup drives 40×660=440 \times \frac{6}{60} = 4 km ahead.

Relative speed during car driving = 50−40=1050 - 40 = 10 km/hr. Time taken by Arup to catch Swarup = 410\frac{4}{10} hours = 24 minutes. Since Arup started driving at 9:00 AM, they meet at 9:00+24 mins=9:249:00 + 24 \text{ mins} = 9:24 AM.

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