A river flows at a uniform speed of $3\text{ km/h}$. A motorboat travels $24\text{ km}$ downstream and returns upstream back to the starting point in a total time of $5\text{ hours}$. The speed of the motorboat in still water (in $\text{km/h}$) is:

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### 1. Strategy & Theoretical Principles Let the speed of the motorboat in still water be $v_b\text{ km/h}$ and river current speed be $v_c = 3\text{ km/h}$. - Downstream speed: $v_d = v_b + 3$ - Upstream speed: $v_u = v_b - 3$ Total round-trip travel time for one-way distance $d = 24\text{ km}$ is: $$T = \frac{d}{v_b + 3} + \frac{d}{v_b - 3} = 5$$ ### 2. Analytical Formulation & Calculation $$24 \left( \frac{1}{v_b + 3} + \frac{1}{v_b - 3} \right) = 5$$ $$24 \left( \frac{2 v_b}{v_b^2 - 9} \right) = 5$$ $$48 v_b = 5(v_b^2 - 9)$$ $$5 v_b^2 - 48 v_b - 45 = 0$$ Factorizing the quadratic equation: $$5 v_b^2 - 50 v_b + 2 v_b - 45 = (5v_b + 3)(v_b - 10) = 0$$ Since speed must be positive: $$v_b = 10.0\text{ km/h}$$ ### 3. Distractor Forensics - Candidate neglecting upstream slowdown uses $24/v = 2.5 \implies v = 9.6\text{ km/h}$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Average speed of a round trip in a current is $\frac{v_b^2 - v_c^2}{v_b}$, strictly less than still-water speed $v_b$. - **Standard Reference:** Aggarwal, R.S., *Quantitative Aptitude*.

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