Central Tendency & DispersionMTP 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%\displaystyle 10\% and 5\displaystyle 5 is added to each item then its coefficient of variation is 8%\displaystyle 8\%. Find original coefficient of variation.

Options

A8%\displaystyle 8\%
B7.5%\displaystyle 7.5\%
C4%\displaystyle 4\%
DNone of these
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Correct Answer

✅ Option d — None of these

All Options:

  • A8%\displaystyle 8\%
  • B7.5%\displaystyle 7.5\%
  • C4%\displaystyle 4\%
  • DNone of these

Detailed Solution & Explanation

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

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