BTGmoderatorDC wrote:If xy > 0 and xz < 0, which of the following must be true?
I. y/z < 0
II. xyz < 0
III. xy - xz > 0
(A) I only
(B) III only
(C) I and II
(D) I and III
(E) II and III
OA D
Source: EMPOWERgmat
Given xy > 0, we know that x and y both are of the same sign: both positive or both negative. Similarly, given xz < 0, we know that x and z are of opposite signs: If x > 0, then z < 0 and if x < 0, then z > 0.
So, we have two cases.
1. Say x > 0 and y > 0. Since x > 0, we have z < 0.
Let's analyze the three statements.
I. y/z < 0: Since y and z are of opposite signs, y/z must be negative; thus, y/z < 0. This is correct for Case 1.
II. xyz < 0: Since xy > 0 and z < 0, we have xyz < 0. This is also correct for Case 1.
III. xy - xz > 0 => x(y - z) > 0 => y - z > 0; dividing the inquality by x. Since x > 0, we need not care about the chnage of sign of the inequality. Thus, y > z. This is also correct for Case 1.
2. Say x < 0 and y < 0. Since x < 0, we have z > 0.
Let's analyze the three statements.
I. y/z < 0: Since y and z are of opposite signs, y/z must be negative; thus, y/z < 0. This is correct for Case 2, too. The correct answer must be among A, C and D.
II. xyz < 0: Since xy > 0 and z > 0, we have xyz > 0. This is not correct for Case 2. The correct answer must be between A and D.
III. xy - xz > 0 => x(y - z) > 0 => y - z < 0; dividing the inquality by x. Since x < 0, we need to reverse the chnage of sign of the inequality. Thus, y < z. This is correct for Case 2, too.
The correct answer:
D
Hope this helps!
-Jay
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