Question

The geometric mean of any two positive numbers x and y is \(\sqrt{xy}\)
COLUMN A: The geometric mean of 4 and 8
COLUMN B: The average arithmetic mean of 4 and 8 

• A. The quantity in Column A is greater.
• B. The quantity in Column B is greater.
• C. The two quantities are equal.
• D. The relationship cannot be determined from the information given.

Hint:

For any two distinct positive numbers, the arithmetic mean is always greater than the geometric mean (AM-GM Inequality). You can use this rule to answer the question without any calculation. The means are only equal if the two numbers are identical.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question asks us to calculate and compare the geometric mean and the arithmetic mean of the same two numbers, 4 and 8.
Step 2: Detailed Explanation:
Column A: Geometric Mean
The formula for the geometric mean of \(x\) and \(y\) is given as \( \sqrt{xy} \). For the numbers 4 and 8: \[ \text{Geometric Mean} = \sqrt{4 \times 8} = \sqrt{32} \] We can estimate the value of \( \sqrt{32} \). We know that \( 5^2 = 25 \) and \( 6^2 = 36 \). So, \( \sqrt{32} \) is between 5 and 6, specifically around 5.6 or 5.7. Column B: Arithmetic Mean
The formula for the arithmetic mean (average) of \(x\) and \(y\) is \( \frac{x+y}{2} \). For the numbers 4 and 8: \[ \text{Arithmetic Mean} = \frac{4+8}{2} = \frac{12}{2} = 6 \] Comparison:
We are comparing \( \sqrt{32} \) (Column A) with 6 (Column B). Since \( 6 = \sqrt{36} \), we are comparing \( \sqrt{32} \) with \( \sqrt{36} \). Because \( 32<36 \), it follows that \( \sqrt{32}<\sqrt{36} \). Therefore, the quantity in Column B is greater.
Step 3: Final Answer:
The geometric mean is \( \sqrt{32} \) and the arithmetic mean is 6. Since \( \sqrt{32}<6 \), Column B is greater.