Question:

For this question, indicate all of the answer choices that apply.
If \((x^2)(y^3) \gt 0\), and \((x)(y^2)(z) \lt 0\), which of the following must be true?

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In problems involving signs of variables, break down each piece of information. An even power (like \(x^2\)) of a non-zero number is always positive. An odd power (like \(y^3\)) has the same sign as the base. Use these rules to systematically determine the signs of the variables.
Updated On: Oct 3, 2025
  • \(x \gt 0\)
  • \(z \lt 0\)
  • \(xy \gt 0\)
  • \(yz \lt 0\)
  • \(xyz \lt 0\)
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Solution and Explanation

Step 1: Understanding the Concept:
This question tests our ability to deduce the signs (positive or negative) of variables based on inequalities involving their products. Key properties of signs are: (positive) × (positive) = positive, (negative) × (negative) = positive, and (positive) × (negative) = negative. Also, any non-zero number squared is positive.
Step 2: Key Formula or Approach:
We will analyze each inequality separately to determine the signs of \(x\), \(y\), and \(z\). Then we will evaluate each option to see if it "must be true" based on our deductions.
Step 3: Detailed Explanation:
Analyze the first inequality: \((x^2)(y^3) \gt 0\)
The term \(x^2\) must be positive, as \(x\) cannot be zero (otherwise the product would be 0, not greater than 0).
Since \((x^2)\) is positive, for the product \((x^2)(y^3)\) to be positive, \((y^3)\) must also be positive.
If \(y^3 \gt 0\), then \(y\) must be positive. So, our first deduction is: \(\mathbf{y \gt 0}\).
Analyze the second inequality: \((x)(y^2)(z) \lt 0\)
We know \(y \gt 0\), so \(y^2\) must be positive.
The inequality is \((x)(\text{positive})(z) \lt 0\). For this product to be negative, the product of the remaining terms, \((x)(z)\), must be negative.
If \(xz \lt 0\), it means that \(x\) and \(z\) must have opposite signs. One is positive and the other is negative. We don't know which is which.
Summary of Deductions:
\(y\) is positive (\(y \gt 0\)).
\(x\) and \(z\) have opposite signs (\(xz \lt 0\)).
Evaluate the options:
(A) \(x \gt 0\): This is not necessarily true. We could have \(x \lt 0\) and \(z \gt 0\).
(B) \(z \lt 0\): This is not necessarily true. We could have \(z \gt 0\) and \(x \lt 0\).
(C) \(xy \gt 0\): We know \(y \gt 0\). The sign of \(xy\) depends on the sign of \(x\), which is unknown. So, this is not necessarily true.
(D) \(yz \lt 0\): We know \(y \gt 0\). The sign of \(yz\) depends on the sign of \(z\), which is unknown. So, this is not necessarily true.
(E) \(\frac{y^2}{z} \lt 0\): We know \(y^2 \gt 0\). For this fraction to be negative, \(z\) must be negative. But we don't know for sure that \(z \lt 0\). So, this is not necessarily true.
(F) \(xyz \lt 0\): We can group this as \((xz)y\). We deduced that \(xz \lt 0\) (negative) and \(y \gt 0\) (positive). The product of a negative term and a positive term is negative. Thus, \((xz)y \lt 0\). This statement must be true.
Step 4: Final Answer:
Based on the analysis, only the statement in option (F) must be true.
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