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If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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28 Apr 2017, 03:01
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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02 May 2017, 05:54
from the stem,, we can conclude that both x and y should be positive and not equal to zero..
only answer D fits in ,, IMO D



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If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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02 May 2017, 06:30
mohshu wrote: from the stem,, we can conclude that both x and y should be positive and not equal to zero..
only answer D fits in ,, IMO D If you say they are both positive then how is D correct? I believe it should be A Posted from my mobile device



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If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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07 May 2017, 07:31
I really don't see a possible solution How is statement possible if the value of X=2 for D value of X can be 2



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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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07 May 2017, 09:31
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IMO ans D
xy>0
So either both positive or both negative
(x2)(y+1)<0
2 posibilities (x2)<0 and (y+1)>0 x <2 and y>1
Or (x2)>0 and (y+1)<o x> 2 and y<1, this doesn't satisfy the first condition
So x<2 x2<=(2x)



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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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07 May 2017, 09:43
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Bunuel wrote: 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 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.
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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07 May 2017, 09:52
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Bunuel wrote: 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 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 \(y+1=−0.9+1=0.1=0.1\).
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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07 May 2017, 10:08
Bunuel wrote: Bunuel wrote: 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 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. Hello Bunuel, But option D says x can be equal to 2 also. If we put the value of x=2 in the premise given the x2 will become 0 and (x2)(y+1)<0 will not hold true. How do I approach such questions? Can you provide more such questions for practice? Regards
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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07 May 2017, 10:12
gmatexam439 wrote: Bunuel wrote: Bunuel wrote: 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 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. Hello Bunuel, But option D says x can be equal to 2 also. If we put the value of x=2 in the premise given the x2 will become 0 and (x2)(y+1)<0 will not hold true. How do I approach such questions? Can you provide more such questions for practice? Regards It is the other way around. We know that \(x < 0\) or \(0 < x < 2\). These are the ranges for all possible values of x. Now, for any possible x (again from \(x < 0\) or from \(0 < x < 2\)) \(x − 2 ≤ 2 − x\) will be true. Hope it's clear.
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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07 May 2017, 10:17
Bunuel wrote: It is the other way around. We know that \(x < 0\) or \(0 < x < 2\). These are the ranges for all possible values of x. Now, for any possible x (again from \(x < 0\) or from \(0 < x < 2\)) \(x − 2 ≤ 2 − x\) will be true.
Hope it's clear.
But the inequality is strict for x. X cannot be 2 else the premise will go for a toss. Regards
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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07 May 2017, 10:23
Bunuel wrote: gmatexam439 wrote: Bunuel wrote: It is the other way around. We know that \(x < 0\) or \(0 < x < 2\). These are the ranges for all possible values of x. Now, for any possible x (again from \(x < 0\) or from \(0 < x < 2\)) \(x − 2 ≤ 2 − x\) will be true.
Hope it's clear.
But the inequality is strict for x. X cannot be 2 else the premise will go for a toss. Regards I'll try to explain it once more. Again, x CANNOT be 2 because we know that \(x < 0\) or \(0 < x < 2\). For any value of x possible \(x − 2 ≤ 2 − x\) will be true. Ok ok, I understand. Sorry for commotion. Can you please share more such questions or point out a few links for practice of such types of question. Thanks a lot Bunuel.
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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07 May 2017, 13:18
1)xy>0 x>0,y>0 or x<0,y<0 1)(x2) (y+1)<0 == (+) 1 () 2 (+) 1<x,y<2 satisfy option D



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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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05 Oct 2017, 20:41
Bunuel wrote: Bunuel wrote: 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 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. How did u get 0<x<2 when x2<0 Because x<2, x can take any value below 2 pls explain Thanks in advance



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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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05 Oct 2017, 21:12
zanaik89 wrote: Bunuel wrote: Bunuel wrote: 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 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. How did u get 0<x<2 when x2<0 Because x<2, x can take any value below 2 pls explain Thanks in advance Case 1 considers the case when: [/b] if \(x > 0\) and \(y > 0\). So, x < 2 and x > 0 > 0 < x < 2.
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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07 Oct 2017, 04:51
xy>0 means x&y have same sign.
(x2)(y+1)<0 means (x2) & (y+1) have opposite sign.
Case 1: If x&y are both positive In such case, it's (x2) that led to the sign change by shifting two units to left on the number line and entering negative territory. This means that x is any value between 0 and 2 when y>0.
Case 2: If x&y are both negative In such case, it's (y+1) that led to the sign change by shifting one units to right on the number line and entering positive territory. This means that y is any value between 1 and 0 when x<0.
Combining Case 1&2, we can deduce that irrespective of the sign, x cannot be greater than +2 and y cannot be smaller than 1.
Now, the only Answer choice that satisfies one of the above two conditions is D.
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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08 Oct 2017, 22:49
No need to calculated because A, B can be eliminated easily. in C, D, E the relation ship for X is given only in D So answer should be D



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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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06 Jan 2018, 05:56
Hi please see my approach as per attached sketch, and let me know if I get to know that D is correct, do I even need to check for other option... Bunuel wrote: Bunuel wrote: 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 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.
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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17 Feb 2018, 06:16
Why do we say D option is correct when at x=2, original question gets void? And question asks which of the following must be true. BunuelPosted from my mobile device



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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be [#permalink]
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17 Feb 2018, 06:23




Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be
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