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28 Apr 2017, 03:01
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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:
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Question Stats:
38% (02:55) correct
62%
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wrong
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If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be true?
A. x ≤ 1
B. 0 ≤ y
C. x ≤ 3y + 2
D. |x − 2| ≤ 2 − x
E. |y| ≤ |y + 1|
A. x ≤ 1
B. 0 ≤ y
C. x ≤ 3y + 2
D. |x − 2| ≤ 2 − x
E. |y| ≤ |y + 1|
07 May 2017, 09:43
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Bunuel
Given \(xy > 0\) and \((x − 2)(y + 1) < 0\).
\(xy > 0\) means that x and y must have the same sign.
\((x − 2)(y + 1) < 0\) means that \((x - 2)\) and \((y + 1)\) must have the different signs.
Case 1: if \(x > 0\) and \(y > 0\), then \((y + 1)\) will be positive and thus \((x - 2)\) must be negative.
\(x - 2 < 0\) --> \(x < 2\).
So, for this case we have \(0 < x < 2\) and \(y > 0\).
Case 2: if \(x < 0\) and \(y < 0\), then \((x - 2)\) will be negative and thus \((y + 1)\) must be positive.
\(x - 2 < 0\) --> \(x < 2\).
\(y + 1 > 0\) --> \(y > -1\).
So, for this case we have \(x < 0\) and \(-1 < y < 0\).
Check the options:
A. \(x ≤ 1\). This is not always true. For example, x can be 1.5 (from case 1)
B. \(0 ≤ y\). This is not always true. For example, y can be -0.5 (from case 2)
C. \(x ≤ 3y + 2\). This is not always true. For example, x = -0.1 and y = - 0.9 (from case 2)
D. \(|x − 2| ≤ 2 − x\). This implies that \(x - 2 ≤ 0\) or \(x ≤ 2\). Now, \(x ≤ 2\) covers all values of x possible (\(0 < x < 2\) from case 1 as well as \(x < 0\) from case 2). So, in any case \(x ≤ 2\) must be true.
E. |y| ≤ |y + 1|. This is not always true. For example, consider y = - 0.9 (from case 2).
Answer: D.
07 May 2017, 09:52
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Veritas Prep Official Solution:
This inequality question is best solved by thinking in terms of number properties. A hint to that effect comes from the fact that each inequality discusses whether a product is positive or negative. Let's turn to the positive/negative properties now.
In order to get \(xy<0\), we must either have
\(x>0\) and \(y>0\)
or
\(x<0\) and \(y<0\)
In order to get \((x−2)(y+1)<0\), we must either have
\(x−2>0\) and \((y+1)<0\)
\(x>2\) and \(y<−1\)
or
\(x−2<0\) and \((y+1)>0\)
\(x<2\) and \(y>−1\)
However, note that the case in which \(x>2\) and \(y<−1\) contradicts the other given inequality, since in this case x and y would have different signs and xy would have to be negative (violating the other given inequality).
Conversely, the case in which \(x<2\) and \(y>−1\) is not problematic. If \(0<x<2\) and \(y>0\) then both inequalities will be satisfied. Or if \(x<0\) and \(−1<y<0\) then again both inequalities will work.
We turn now to the answer choices.
Answer A is close, but, since x could be between 1 and 2 with y positive, it is not necessarily true that \(x≤1\). (E.g. \(x=1.1\), \(y=10\))
Answer B is close as well, but, since y could be between −1 and 0 with x negative, it is not necessarily true that \(0≤y\). (E.g. \(x=−5\), \(y=−0.1\))
Answer C is close yet again, but we cannot quite obtain this inequality algebraically (we can get \(x<2y+2\) by solving the system using elimination), and in fact this answer choice can be false if y is negative and close to −1. For instance, \(x=−0.1\), \(y=−0.9\) satisfies both given inequalities but does not satisfy answer choice C since \(3y+2=−0.7<−0.1\).
Answer D is correct. Since we've established that \(x<2\), we also know that \(x−2<0\), and therefore \(|x−2|=−(x−2)=2−x\). It is therefore true that \(|x−2|≤2−x\).
Answer E is again close but wrong. In the case of y close to −1, adding 1 will actually move y closer to 0. For example, if \(x=−0.1\) and \(y=−0.9\), then both given inequalities are satisfied but \(|y|=0.9\) is greater than \(|y+1|=|−0.9+1|=|0.1|=0.1\).
