The Harmonic mean H of two numbers is 4 and their arithmetic means A and the geometric mean G satisfy eq. 2A + G2 = 27 , the numbers are

Option c: (6,3). Solution: Step 1: Let the two numbers be a and b . Using the relation AM × HM = GM2 (for two numbers): A × H = G2 ⟹ G2 = 4A Step 2: Substitute into the given equation. 2A + G2 = 27 2A + 4A = 27 6A = 27 A = 4.5 Step 3: Find G2 . G2 = 4A = 4 × 4.5 = 18 G = sqrt(18) = 3sqrt(2) Step 4: Find a + b and ab . a + b = 2A = 9 ab = G2 = 18 Step 5: Solve the quadratic. t2 - 9t + 18 = 0 (t-3)(t-6) = 0 t = 3 or t = 6 So the numbers are 6 and 3. Verification with HM: H = 2ab/a+b = 2 × 18/9 = 36/9 = 4 Hence, Option C is the correct answer.

The Harmonic mean H of two numbers is $4$ and their arithmetic means $A$ and the geometric mean $G$ satisfy eq. $2A + G^2 = 27$, the numbers are

The correct answer is Option c: $(6,3)$. **Step 1: Let the two numbers be $a$ and $b$.** Using the relation $\text{AM} \times \text{HM} = \text{GM}^2$ (for two numbers): $$A \times H = G^2 \Rightarrow G^2 = 4A$$ **Step 2: Substitute into the given equation.** $$2A + G^2 = 27$$ $$2A + 4A = 27$$ $$6A = 27$$ $$A = 4.5$$ **Step 3: Find $G^2$.** $$G^2 = 4A = 4 \times 4.5 = 18$$ $$G = \sqrt{18} = 3\sqrt{2}$$ **Step 4: Find $a + b$ and $ab$.** $$a + b = 2A = 9$$ $$ab = G^2 = 18$$ **Step 5: Solve the quadratic.** $$t^2 - 9t + 18 = 0$$ $$(t-3)(t-6) = 0$$ $$t = 3 \text{ or } t = 6$$ So the numbers are **6 and 3**. **Verification with HM:** $$H = \frac{2ab}{a+b} = \frac{2 \times 18}{9} = \frac{36}{9} = 4 \checkmark$$ Hence, **Option C** is the correct answer.

Measures of Central Tendency and DispersionMCQMTP Dec 23 - Series IQuestion 3022 of 473

All Questions

The Harmonic mean H of two numbers is $\displaystyle 4$ and their arithmetic means $\displaystyle A$ and the geometric mean $\displaystyle G$ satisfy eq. $\displaystyle 2A + G^2 = 27$ , the numbers are

Options

\displaystyle (1,3)

\displaystyle (9,3)

\displaystyle (6,3)

\displaystyle (4.5, 7)

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

Correct Answer

✅ Option c — $\displaystyle (6,3)$

All Options:

A $\displaystyle (1,3)$
B $\displaystyle (9,3)$
C $\displaystyle (6,3)$
D $\displaystyle (4.5, 7)$

Detailed Solution & Explanation

**Step 1: Let the two numbers be

\displaystyle a

and

\displaystyle b

.** Using the relation

\displaystyle \text{AM} \times \text{HM} = \text{GM}^2

(for two numbers):

A \times H = G^2 \Rightarrow G^2 = 4A

**Step 2: Substitute into the given equation.**

2A + G^2 = 27

2A + 4A = 27

6A = 27

A = 4.5

**Step 3: Find

\displaystyle G^2

.**

G^2 = 4A = 4 \times 4.5 = 18

G = \sqrt{18} = 3\sqrt{2}

**Step 4: Find

\displaystyle a + b

and

\displaystyle ab

.**

a + b = 2A = 9

ab = G^2 = 18

**Step 5: Solve the quadratic.**

t^2 - 9t + 18 = 0

(t-3)(t-6) = 0

t = 3 \text{ or } t = 6

So the numbers are **6 and 3**. **Verification with HM:**

H = \frac{2ab}{a+b} = \frac{2 \times 18}{9} = \frac{36}{9} = 4 \checkmark

Hence, **Option C** 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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The Harmonic mean H of two numbers is 4\displaystyle 44 and their arithmetic means A\displaystyle AA and the geometric mean G\displaystyle GG satisfy eq. 2A+G2=27\displaystyle 2A + G^2 = 272A+G2=27, the numbers are