Question

The arithmetic mean (A.M.) of two numbers \( x \) and \( y \) is \( 3 \) and their geometric mean (G.M.) is \( 1 \). Then \( x^2+y^2 \) is equal to:

• A. \( 30 \)
• B. \( 31 \)
• C. \( 32 \)
• D. \( 33 \)
• E. \( 34 \)

Hint:

You don't need to solve for the individual values of \( x \) and \( y \). Working with the sum and product directly is faster and less prone to calculation errors.

Answer 1

Answer: (E)

Concept: For two numbers \( x \) and \( y \), the definitions are: - \( \text{A.M.} = \frac{x+y}{2} \) - \( \text{G.M.} = \sqrt{xy} \) We can use these to find the sum \( (x+y) \) and the product \( (xy) \), then apply the algebraic identity \( x^2+y^2 = (x+y)^2 - 2xy \).

Step 1: Finding the sum and product of \( x \) and \( y \).
From A.M. = 3: \[ \frac{x+y}{2} = 3 \implies x+y = 6 \] From G.M. = 1: \[ \sqrt{xy} = 1 \implies xy = 1^2 = 1 \]

Step 2: Calculating the sum of squares.
Using the identity: \[ x^2 + y^2 = (x+y)^2 - 2xy \] \[ x^2 + y^2 = (6)^2 - 2(1) \] \[ = 36 - 2 = 34 \]