Central Tendency & DispersionMTP Sep 24 Series IQuestion 3073 of 473
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The HM, H\displaystyle H of two numbers is 4\displaystyle 4 and their AM, A\displaystyle A and the GM, G\displaystyle G satisfy the equation 2A+G2=27\displaystyle 2A+G^2=27, the numbers are:

Options

A(1,3)\displaystyle (1,3)
B(9,5)\displaystyle (9,5)
C(6,3)\displaystyle (6,3)
D(12,7)\displaystyle (12,7)
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Correct Answer

✅ Option c — (6,3)\displaystyle (6,3)

All Options:

  • A(1,3)\displaystyle (1,3)
  • B(9,5)\displaystyle (9,5)
  • C(6,3)\displaystyle (6,3)
  • D(12,7)\displaystyle (12,7)

Detailed Solution & Explanation

**Step 1: Use the relationship GM² = AM × HM.** G2=A×H=4AG^2 = A \times H = 4A **Step 2: Substitute into the equation 2A+G2=27\displaystyle 2A + G^2 = 27.** 2A+4A=27⇒6A=27⇒A=4.52A + 4A = 27 \Rightarrow 6A = 27 \Rightarrow A = 4.5 **Step 3: Find G2\displaystyle G^2.** G2=4×4.5=18G^2 = 4 \times 4.5 = 18 **Step 4: Find the sum and product of the two numbers.** a+b=2A=9,ab=G2=18a + b = 2A = 9, \quad ab = G^2 = 18 **Step 5: Solve the quadratic t2−9t+18=0\displaystyle t^2 - 9t + 18 = 0.** (t−3)(t−6)=0⇒t=3 or t=6(t-3)(t-6) = 0 \Rightarrow t = 3 \text{ or } t = 6 **Verification with HM = 4:** H=2aba+b=369=4✓H = \frac{2ab}{a+b} = \frac{36}{9} = 4 \checkmark Hence, **Option C** is the correct answer.

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