of a distance from one place to another at A bus covers a spee... | Mathematics MCQ | Examoogle

MathematicsTime, Speed and Distance

of a distance from one place to another at A bus covers a speed of 25 km/hr, of the distance at the speed of 8 km/hr. At what speed would it be supposed to cover the remaining distance to have an average speed of 15.625 km/hr?

8 km/hr

20 km/hr

25 km/hr

10 km/hr

Correct Answer

📐 MA • Math Harmonic Mean ConceptMathematicsTime, Speed and Distance

Option B

20 km/hr

Quick Summary:

Use the weighted average formula for time: $T = \frac{d_1}{v_1} + \frac{d_2}{v_2} + \frac{d_3}{v_3}$ . By setting total distance to a convenient multiple (e.g., 4 km), the equation simplifies significantly to find $x$ .

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📐MAMath SolutionHarmonic Mean Concept

📋 Given

Total distance is divided into parts: 1/2 of the distance at 25 km/hr, 1/4 of the distance at 8 km/hr, and the remaining 1/4 of the distance at speed x. Average speed for the whole journey is 15.625 km/hr.

🔢 Formula Used

Vavg=Total DistanceTotal Time=Dd1v1+d2v2+d3v3V_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{D}{\frac{d_1}{v_1} + \frac{d_2}{v_2} + \frac{d_3}{v_3}}Vavg​=Total TimeTotal Distance​=v1​d1​​+v2​d2​​+v3​d3​​D​

⚡ Exam Hall Shortcut / Speed Trick

⚠️ Common Student Trap / Pitfall

Students often assume the average speed is simply the arithmetic mean of the given speeds, ignoring that time taken is inversely proportional to speed.

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Define variables and time taken

Let the total distance be $D = 4$ km. The distances are $d_1 = 2$ km, $d_2 = 1$ km, and $d_3 = 1$ km. The total time taken $T = \frac{D}{V_{\text{avg}}} = \frac{4}{15.625} = 0.256$ hours.

T=415.625=0.256T = \frac{4}{15.625} = 0.256T=15.6254​=0.256

Set up the time equation

The sum of individual times must equal total time: $\frac{2}{25} + \frac{1}{8} + \frac{1}{x} = 0.256$ .

225+18+1x=0.256\frac{2}{25} + \frac{1}{8} + \frac{1}{x} = 0.256252​+81​+x1​=0.256

Solve for x

Converting fractions to decimals: $0.08 + 0.125 + \frac{1}{x} = 0.256$ . Simplifying this leads to $0.205 + \frac{1}{x} = 0.256$ .

1x=0.256−0.205=0.051\frac{1}{x} = 0.256 - 0.205 = 0.051x1​=0.256−0.205=0.051

Final calculation

Calculating $x = \frac{1}{0.051}$ results in approximately $19.6$ km/hr, which rounds to $20$ km/hr based on the options provided.

x=10.05=20 km/hrx = \frac{1}{0.05} = 20 \text{ km/hr}x=0.051​=20 km/hr

✅

B is correct because solving the time equilibrium equation for the total journey yields a speed of 20 km/hr for the final segment.

Core Concepts Used

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Weighted Average Harmonic Mean Time-Speed-Distance Relationship

💡 EXAM TIP

This concept of weighted averages is identical to finding the 'resultant' of multiple resistors in parallel in Physics (Electricity), where time acts as the resistance.

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