A drone flies along the perimeter of an equilateral triangular patrol circuit. It flies along the first side at $100\text{ km/h}$, along the second side at $150\text{ km/h}$, and along the third side at $300\text{ km/h}$. The average speed of the drone for the entire closed patrol circuit (in $\text{km/h}$) is:

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### 1. Strategy & Theoretical Principles For equal distances $d$ traversed at different speeds $v_1, v_2, v_3$, the average speed is given by the Harmonic Mean: $$v_{\text{avg}} = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}}$$ ### 2. Analytical Formulation & Calculation Given speeds: - $v_1 = 100\text{ km/h} \implies 1/v_1 = 3/300$ - $v_2 = 150\text{ km/h} \implies 1/v_2 = 2/300$ - $v_3 = 300\text{ km/h} \implies 1/v_3 = 1/300$ Sum of reciprocal speeds: $$\sum \frac{1}{v_i} = \frac{3 + 2 + 1}{300} = \frac{6}{300} = \frac{1}{50}$$ Average speed: $$v_{\text{avg}} = \frac{3}{1/50} = 3 \times 50 = 150.0\text{ km/h}$$ ### 3. Distractor Forensics - Candidate taking arithmetic mean gets $(100 + 150 + 300)/3 = 550/3 = 183.33\text{ km/h}$, which is fundamentally incorrect for equal distances. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Average speed over equal distance segments is always the harmonic mean, never the arithmetic mean. - **Standard Reference:** Aggarwal, R.S., *Quantitative Aptitude*.

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