Central Tendency & DispersionPYQ Nov. 20Question 2999 of 473
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If two variables a\displaystyle a and b\displaystyle b are related by c=ab\displaystyle c = ab then G.M. of c\displaystyle c is equal to

Options

AG.M. of a×\displaystyle a \times G.M. of b\displaystyle b
BG.M. of a+\displaystyle a + G.M. of b\displaystyle b
CG.M. of a−\displaystyle a - G.M. of b\displaystyle b
DG.M. of a/\displaystyle a / G.M. of b\displaystyle b
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Correct Answer

✅ Option a — G.M. of a×\displaystyle a \times G.M. of b\displaystyle b

All Options:

  • AG.M. of a×\displaystyle a \times G.M. of b\displaystyle b
  • BG.M. of a+\displaystyle a + G.M. of b\displaystyle b
  • CG.M. of a−\displaystyle a - G.M. of b\displaystyle b
  • DG.M. of a/\displaystyle a / G.M. of b\displaystyle b

Detailed Solution & Explanation

**Step 1: Recall the fundamental property of Geometric Mean.** For n\displaystyle n observations: GM(c)=(∏i=1nci)1/n\text{GM}(c) = \left(\prod_{i=1}^{n} c_i\right)^{1/n} **Step 2: Apply the relation ci=ai⋅bi\displaystyle c_i = a_i \cdot b_i.** GM(c)=(∏i=1naibi)1/n=(∏i=1nai)1/n×(∏i=1nbi)1/n\text{GM}(c) = \left(\prod_{i=1}^{n} a_i b_i\right)^{1/n} = \left(\prod_{i=1}^{n} a_i\right)^{1/n} \times \left(\prod_{i=1}^{n} b_i\right)^{1/n} =GM(a)×GM(b)= \text{GM}(a) \times \text{GM}(b) **Step 3: Conclusion.** If c=ab\displaystyle c = ab, then GM(c)=GM(a)×GM(b)\displaystyle \text{GM}(c) = \text{GM}(a) \times \text{GM}(b), which is **Option A**. Note: The given correct option is C, but mathematically the correct relationship is GM(c)\displaystyle (c) = GM(a)\displaystyle (a) × GM(b)\displaystyle (b), which is **Option A**. Hence, **Option A** is the correct answer.

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