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If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be
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28 Apr 2017, 02:01
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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
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be
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07 May 2017, 08:43
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
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07 May 2017, 08:31
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
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07 May 2017, 08:52
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
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07 May 2017, 09:12
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
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07 May 2017, 09:20
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.
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be
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07 May 2017, 09:08
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
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07 May 2017, 09:17
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
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05 Oct 2017, 19:41
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
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05 Oct 2017, 20:12
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
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06 Jan 2018, 04:56
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
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17 Feb 2018, 05:16
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
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17 Feb 2018, 05:23
Mudit27021988 wrote: 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 deviceIt's the other way around we know that \(x < 0\) or \(0 < x < 2\). For any x from these possible ranges, \(x − 2 ≤ 2 − x\) will be true.
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be
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26 Feb 2020, 02:33
Hi Bunuel,
I understood solution. But just to be sure, how will we get the range for X and Y respectively in the options D and E. how can we solve algebraically options D and E? Would really appreciate your reply.



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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be
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26 Feb 2020, 03:06
shanks2020 wrote: Hi Bunuel,
I understood solution. But just to be sure, how will we get the range for X and Y respectively in the options D and E. how can we solve algebraically options D and E? Would really appreciate your reply. D. \(x − 2 ≤ 2 − x\); \(x − 2 + (x  2) ≤ 0\). x − 2 is positive or 0. If x  2 is positive then the sum is positive. So, x  2 ≤ 0. E. y ≤ y + 1. Square: y^2 ≤ y^2 + 2y + 1; 1 ≤ 2y 1/2 ≤ y.
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Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be
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