Measures of Central Tendency and DispersionMCQMTP June 24 Series IQuestion 3205 of 473
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If 5\displaystyle 5 is subtracted from each observation of some certain item then its co-efficient of variation is $10\%and\displaystyle and5$ is added to each item then its coefficient of variation is $8\%$. Find original coefficient of variation.

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option d — None of these

All Options:

  • A8%\displaystyle 8\%
  • B$7.5\%$
  • C4%\displaystyle 4\%
  • DNone of these

Detailed Solution & Explanation

**Let** original mean = xˉ\displaystyle xˉ\displaystyle \bar{x}, original SD = σ\displaystyle σ\displaystyle \sigma **When 5 is subtracted:** Mean becomes xˉ−5\displaystyle xˉ−5\displaystyle \bar{x} - 5, SD stays σ\displaystyle σ\displaystyle \sigma. CV1=σxˉ−5×100=10CV_1 = σxˉ−5\displaystyle \frac{\sigma}{\bar{x} - 5} ×\displaystyle \times 100 = 10 σ=10(xˉ−5)100=xˉ−510⋯(1)\displaystyle \sigma = \frac{10(xˉ−5\displaystyle \bar{x} - 5)}{100} = xˉ−510\displaystyle \frac{\bar{x} - 5}{10} \quad \cdots (1) **When 5 is added:** Mean becomes xˉ+5\displaystyle xˉ+5\displaystyle \bar{x} + 5, SD stays σ\displaystyle σ\displaystyle \sigma. CV2=σxˉ+5×100=8CV_2 = σxˉ+5\displaystyle \frac{\sigma}{\bar{x} + 5} ×\displaystyle \times 100 = 8 σ=8(xˉ+5)100=xˉ+512.5⋯(2)\displaystyle \sigma = \frac{8(xˉ+5\displaystyle \bar{x} + 5)}{100} = xˉ+512.5\displaystyle \frac{\bar{x} + 5}{12.5} \quad \cdots (2) **Step 1:** Equate (1) and (2). xˉ−510=8(xˉ+5)100\displaystyle \frac{\bar{x} - 5}{10} = \frac{8(xˉ+5\displaystyle \bar{x} + 5)}{100} xˉ−510=xˉ+512.5\displaystyle \frac{\bar{x} - 5}{10} = \frac{\displaystyle \bar{x} + 5}{12.5} 12.5(xˉ−5)=10(xˉ+5)12.5(xˉ−5\displaystyle \bar{x} - 5) = 10(xˉ+5\displaystyle \bar{x} + 5) 12.5xˉ−62.5=10xˉ+5012.5xˉ−62\displaystyle \bar{x} - 62.5 = 10xˉ+50\displaystyle \bar{x} + 50 2.5xˉ=112.52.5xˉ=112\displaystyle \bar{x} = 112.5 xˉ=45xˉ=45\displaystyle \bar{x} = 45 **Step 2:** Find σ\displaystyle σ\displaystyle \sigma from equation (1). σ=45−510=4010=4σ=45−510=4010=4\displaystyle \sigma = \frac{45 - 5}{10} = \frac{40}{10} = 4 Wait — let me recheck: From (1): σ=xˉ−510=45−510=4\displaystyle \displaystyle \sigma = \frac{\displaystyle \bar{x}-5}{10} = \frac{45-5}{10} = 4. From (2): σ=8×50100=4\displaystyle \displaystyle \sigma = \frac{8 ×\displaystyle \times 50}{100} = 4 ✓ **Step 3:** Find original CV. CV=σxˉ×100=445×100≈8.89%CV = σxˉ\displaystyle \frac{\sigma}{\bar{x}} ×\displaystyle \times 100 = 445\displaystyle \frac{4}{45} ×\displaystyle \times 100 ≈\displaystyle \approx 8.89\% Hmm — 8.89% doesn't match any option exactly. Let me re-examine with equation setup. From (1): $10(xˉ−5\displaystyle \bar{x}-5) = 100σ\displaystyle \sigma \Rightarrow xˉ−5=10\displaystyle \bar{x} - 5 = 10σ\displaystyle \sigma$ From (2): $8(xˉ+5\displaystyle \bar{x}+5) = 100σ\displaystyle \sigma \Rightarrow xˉ+5=12\displaystyle \bar{x} + 5 = 12.5σ\displaystyle \sigma$ Subtracting: (xˉ+5)−(xˉ−5)=12.5σ−10σ\displaystyle (xˉ+5\displaystyle \bar{x}+5) - (xˉ−5\displaystyle \bar{x}-5) = 12.5σ−10\displaystyle \sigma - 10σ\displaystyle \sigma $10 = 2.5σ<span\displaystyle \sigma<span class="katex">,so\displaystyle , soσ=4\displaystyle \sigma = 4$. From (1): xˉ=10σ+5=40+5=45\displaystyle xˉ=10\displaystyle \bar{x} = 10σ+5=40+5=45\displaystyle \sigma + 5 = 40 + 5 = 45. Original CV = 445×100=8.89%\displaystyle 445\displaystyle \frac{4}{45} ×\displaystyle \times 100 = 8.89\% ≈ None of these precisely. But among given options, $7.5\%$ is selected by the key. Let me check if $xˉ=40\displaystyle \bar{x} = 40$: If xˉ=40\displaystyle xˉ=40\displaystyle \bar{x} = 40: From (1): σ=(40−5)/10=3.5\displaystyle σ\displaystyle \sigma = (40-5)/10 = 3.5; From (2): CV=3.5/45×100=7.78%\displaystyle CV = 3.5/45 ×\displaystyle \times 100 = 7.78\% ≠ 8%. With our correct computation: CV = $8.89\% ≈\displaystyle \approx$ None of these (Option D). Hence, **Option D** is the correct answer.

About This Chapter: Measures of Central Tendency and Dispersion

Paper

Paper 3: Quantitative Aptitude

Weightage

12-15 Marks

Key Topics

Mean, Median, Mode, Range, Mean Deviation, Standard Deviation

The core foundation of Statistics. This chapter covers Mean (Arithmetic, Geometric, Harmonic), Median, Mode, and their properties. It also explores measures of spread like Range, Mean Deviation, Standard Deviation, and Quartile Deviation.

View Official ICAI Syllabus

Exam Strategy Tip

Do not just memorize formulas; ICAI loves asking about the mathematical properties (e.g., 'sum of deviations from the AM is always zero'). You can usually eliminate 2 options just by knowing the properties.

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