XAT 2015 Question 75

Question 75

A person is standing at a distance of 1800 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 degree elevation from his eye - level. Immediately, the person starts walking towards the tower. At 5.10 pm., the person noticed that the tip of the minute hand made an angle of 60 degrees with respect to his eye - level. Using three - dimensional vision, find the speed at which the person is walking. The length of the minutes hand is 200√3 meters (√3 = 1.732).

Solution

Let O be the centre of the clock. Let the person's eye be at A and the tip of minute hand at 5.00 p.m. is at P and at 5.10 p.m. at Q

AM = 1800 m and OP = OQ = $$200\sqrt{3}$$ m

In $$\triangle$$ APM

=> $$tan 30 = \frac{PM}{AM}$$

=> $$\frac{1}{\sqrt{3}} = \frac{PM}{1800}$$

=> $$PM = \frac{1800}{\sqrt{3}} = 600 \sqrt{3}$$

=> $$OM = PM - OP = 600 \sqrt{3} - 200 \sqrt{3} = 400 \sqrt{3}$$

In $$\triangle$$ OBM

=> $$tan 60 = \frac{OM}{BM}$$

=> $$\sqrt{3} = \frac{400 \sqrt{3}}{BM}$$

=> $$BM = 400$$ m

=> $$AB = AM - BM = 1800 - 400 = 1400$$ m

Time taken to reach B from A = 10 minutes = 600 sec

$$\therefore$$ Speed of the person = $$\frac{1400}{600} = \frac{7}{3}$$ m/s

= $$(\frac{7}{3} \times \frac{18}{5})$$ km/hr = $$8.4$$ km/hr



Create a FREE account and get:

  • All Quant Formulas and shortcuts PDF
  • 40+ previous papers with solutions PDF
  • Top 500 MBA exam Solved Questions for Free

    Comments
    Ashish Kumar

    2 years, 9 months ago

    We should consider BQM not BOM as he is looking at tip not middle part. Kindly revert If I am wrong.

    Register with

    OR
    cracku

    Boost your Prep!

    Download App