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Walking Speed and Elevation Angles to Clock Tower

XAT 2015 · Quantitative Ability & Data Interpretation · Hard · Quantitative Ability

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A person is standing at a distance of 1800 meters1800\text{ meters} facing a giant clock at the top of a tower. At 5.00 p.m., he can see the tip of the minute hand of the clock at 30∘30^\circ degree elevation from his eye-level. Immediately, the person starts walking towards the tower. At 5.10 p.m., the person noticed that the tip of the minute hand made an angle of 60∘60^\circ degrees with respect to his eye-level. Using three-dimensional vision, find the speed at which the person is walking. The length of the minute hand is 2003 meters200\sqrt{3}\text{ meters} (3=1.732\sqrt{3} = 1.732).

  1. A.

    7.2 km/hour

  2. B.

    7.5 km/hour

  3. C.

    7.8 km/hour

  4. D.

    8.4 km/hour

  5. E.

    None of the above

Answer

D

Explanation

At 5:00 p.m., minute hand points vertically UP (PP). Distance to tower base AM=1800 mAM = 1800\text{ m}. PM=AMtan⁡30∘=18003=6003 mPM = AM \tan 30^\circ = \frac{1800}{\sqrt{3}} = 600\sqrt{3}\text{ m}. Center of clock OO: OM=PM−OP=6003−2003=4003 mOM = PM - OP = 600\sqrt{3} - 200\sqrt{3} = 400\sqrt{3}\text{ m}.

At 5:10 p.m., minute hand is at 60∘60^\circ angle from vertical, but its horizontal distance projection BM=OMtan⁡60∘=40033=400 mBM = \frac{OM}{\tan 60^\circ} = \frac{400\sqrt{3}}{\sqrt{3}} = 400\text{ m}.

Distance walked AB=1800−400=1400 mAB = 1800 - 400 = 1400\text{ m} in 10 minutes (600 s600\text{ s}). Speed =1400600=73 m/s=73×185=8.4 km/hr= \frac{1400}{600} = \frac{7}{3}\text{ m/s} = \frac{7}{3} \times \frac{18}{5} = 8.4\text{ km/hr}.

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