Question

x and y are positive integers and x > y.
COLUMN A: \(\frac{x^2}{y^3}\)
COLUMN B: \(\frac{y^3}{x^2}\)

• 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:

When comparing expressions with variables and exponents, the relative size of the base and the power matters a lot. By picking a small value for \(y\) (like y=1), you can often make Column A large. By picking values for \(x\) and \(y\) that are closer together but larger, the higher power (\(y^3\)) can dominate, making Column B larger.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need to compare two fractional expressions involving powers of positive integers \(x\) and \(y\), where \(x>y\). The relationship might depend on the specific values chosen.
Step 2: Key Formula or Approach:
The best strategy is to test different pairs of integers that satisfy the condition \(x>y\).
Step 3: Detailed Explanation:
We are given that \(x\) and \(y\) are positive integers and \(x>y\). Case 1: Let \(x=2\) and \(y=1\).

Column A: \( \frac{x^2}{y^3} = \frac{2^2}{1^3} = \frac{4}{1} = 4 \)
Column B: \( \frac{y^3}{x^2} = \frac{1^3}{2^2} = \frac{1}{4} \)
In this case, Column A>Column B.
Case 2: Let's try to find a case where the relationship is different. We need to make \(y^3\) large relative to \(x^2\). Let's pick larger numbers. Let \(y=3\) and \(x=4\).

Column A: \( \frac{x^2}{y^3} = \frac{4^2}{3^3} = \frac{16}{27} \). This is a fraction less than 1.
Column B: \( \frac{y^3}{x^2} = \frac{3^3}{4^2} = \frac{27}{16} \). This is a fraction greater than 1.
In this case, Column B>Column A.
Since we found one case where A>B and another where B>A, the relationship cannot be determined from the information given.
Step 4: Final Answer:
The relationship between the two fractions depends on the specific values of \(x\) and \(y\).