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GMAT Club Daily Prep
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17 May 2024, 12:30
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
66% (00:42) correct
33% (00:07) wrong based on 3 sessions
History
Date
Time
Result
Not Attempted Yet
07 Jul 2024, 12:24
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Official Explanation
You are given that \(x^2y>0\), which means that the product of the two numbers \(x^2\) and y is positive. Recall that the product of two numbers is positive only if both numbers are positive or both numbers are negative. The square of a number is always greater than or equal to 0. In this case, \(x^2\) cannot equal 0 because the product \(x^2y\) is not 0. Thus, \(x^2\) is positive and it follows that y is also positive.
You are also given that \(x^2y>0\), which means that the product of the two numbers x and \(y^2\) is negative. The product of two numbers is negative only if one of the numbers is negative and the other number is positive. In this case, \(y^2\) cannot be negative because it is the square of a number, and it cannot be 0 because the product \(x^2y\) is not 0. Thus, \(y^2\) is positive and so x must be negative.
Because x is negative and y is positive, y must be greater than x, and the correct answer is Choice B.
You are given that \(x^2y>0\), which means that the product of the two numbers \(x^2\) and y is positive. Recall that the product of two numbers is positive only if both numbers are positive or both numbers are negative. The square of a number is always greater than or equal to 0. In this case, \(x^2\) cannot equal 0 because the product \(x^2y\) is not 0. Thus, \(x^2\) is positive and it follows that y is also positive.
You are also given that \(x^2y>0\), which means that the product of the two numbers x and \(y^2\) is negative. The product of two numbers is negative only if one of the numbers is negative and the other number is positive. In this case, \(y^2\) cannot be negative because it is the square of a number, and it cannot be 0 because the product \(x^2y\) is not 0. Thus, \(y^2\) is positive and so x must be negative.
Because x is negative and y is positive, y must be greater than x, and the correct answer is Choice B.
