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Explanation:

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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)

∆FEC 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 ∆AFG, m∠FAG = 60° and FG = 500√3 m

By theorem of 30°-60°-90° triangle,

AG = 500 m

BG = EF = 300 m

In ∆ABG, 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).

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