XAT 2015 — QA & DI Question 23
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).
Answer & solution
- A
7.2 km/hour
- B
7.5 km/hour
- C
7.8 km/hour
8.4 km/hour
- E
None of the above
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At 5.00 p.m., position of the person be P
PB = 1800 (Given)
m∠DPB = 30°
⇒ DB = 1800 tan 30° = 600√3
At 5.10 p.m. the minute hand of the clock moves by 60°
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DC = CF = 200√3 m (Given)
âFEC is 30° - 60° - 90° triangle.
So, EC = 100√3 m and EF = 300 m
DE = DC – EC = 200√3 – 100√3 = 100√3 m
FG = DB – DE = 600√3 – 100√3 = 500√3 m
In âAFG, m∠FAG = 60° and FG = 500√3 m
By theorem of 30°-60°-90° triangle,
AG = 500 m
BG = EF = 300 m
In âABG, AG = 500 m and BG = 300 m ⇒ AB = 400 m
PA = PB – AB = 1800 – 400 = 1400 m = 1.4 km
Time taken = 10 minutes = (1/6) hours
Speed = 1.4 × 6 = 8.4 km/hr
Hence, option (d).