If there are two groups with 75 and 65 as harmonic means and containing 15 and 13 observations. Then the combined H.M. is given by:

Option a: 70. Solution: Step 1: Recall the formula for combined HM. Hcombined = n1 + n2/1H1 + n2/H2 Step 2: Substitute values. - n1 = 15 , H1 = 75 → 15/75 = 0.2 - n2 = 13 , H2 = 65 → 13/65 = 0.2 Step 3: Calculate. H = 15 + 13/0.2 + 0.2 = 28/0.4 = 70 The combined HM = 70 = Option A. Note: The exam gives option B ( 80 ) as correct, but the mathematical computation gives 70 (Option A). Hence, Option A is the correct answer.

If there are two groups with $75$ and $65$ as harmonic means and containing $15$ and $13$ observations. Then the combined H.M. is given by:

The correct answer is Option a: 70. **Step 1: Recall the formula for combined HM.** $$H_{\text{combined}} = \frac{n_1 + n_2}{\frac{n_1}{H_1} + \frac{n_2}{H_2}}$$ **Step 2: Substitute values.** - $n_1 = 15$, $H_1 = 75$ → $\frac{15}{75} = 0.2$ - $n_2 = 13$, $H_2 = 65$ → $\frac{13}{65} = 0.2$ **Step 3: Calculate.** $$H = \frac{15 + 13}{0.2 + 0.2} = \frac{28}{0.4} = 70$$ The combined HM = $70$ = **Option A**. Note: The exam gives option B ($80$) as correct, but the mathematical computation gives $70$ (Option A). Hence, **Option A** is the correct answer.

Measures of Central Tendency and DispersionMCQMTP March 21Question 3017 of 473

All Questions

If there are two groups with $\displaystyle 75$ and $\displaystyle 65$ as harmonic means and containing $\displaystyle 15$ and $\displaystyle 13$ observations. Then the combined H.M. is given by:

Options

A70

B80

C70.35

D69.48

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — 70

All Options:

A70
B80
C70.35
D69.48

Detailed Solution & Explanation

**Step 1: Recall the formula for combined HM.**

H_{\text{combined}} = \frac{n_1 + n_2}{\frac{n_1}{H_1} + \frac{n_2}{H_2}}

**Step 2: Substitute values.** -

\displaystyle n_1 = 15

\displaystyle H_1 = 75

→

\displaystyle \frac{15}{75} = 0.2

\displaystyle n_2 = 13

\displaystyle H_2 = 65

→

\displaystyle \frac{13}{65} = 0.2

**Step 3: Calculate.**

H = \frac{15 + 13}{0.2 + 0.2} = \frac{28}{0.4} = 70

The combined HM =

\displaystyle 70

= **Option A**. Note: The exam gives option B (

\displaystyle 80

) as correct, but the mathematical computation gives

\displaystyle 70

(Option A). Hence, **Option A** 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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If there are two groups with 75\displaystyle 7575 and 65\displaystyle 6565 as harmonic means and containing 15\displaystyle 1515 and 13\displaystyle 1313 observations. Then the combined H.M. is given by: