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

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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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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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Collection of Questions:
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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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New post 07 May 2017, 08:43
6
14
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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New post 02 May 2017, 04:54
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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New post 02 May 2017, 05:30
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

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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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New post 07 May 2017, 06:31
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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New post 07 May 2017, 08:31
3
1
IMO ans D

xy>0

So either both positive or both negative

(x-2)(y+1)<0

2 posibilities
(x-2)<0 and (y+1)>0
x <2 and y>-1

Or (x-2)>0 and (y+1)<o
x> 2 and y<-1, this doesn't satisfy the first condition

So
x<2
|x-2|<=(2-x)
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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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New post 07 May 2017, 08:52
1
4
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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Collection of Questions:
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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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New post 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 x-2 will become 0 and (x-2)(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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New post 07 May 2017, 09:12
1
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 x-2 will become 0 and (x-2)(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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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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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New post 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  [#permalink]

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New post 07 May 2017, 09:20
1
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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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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New post 07 May 2017, 09:23
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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New post 07 May 2017, 12:18
1)xy>0
x>0,y>0 or x<0,y<0
1)(x-2) (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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New post 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 x-2<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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New post 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 x-2<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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New post 07 Oct 2017, 03:51
xy>0 means x&y have same sign.

(x-2)(y+1)<0 means (x-2) & (y+1) have opposite sign.

Case 1: If x&y are both positive-
In such case, it's (x-2) 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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New post 08 Oct 2017, 21:49
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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New post 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  [#permalink]

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New post 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. 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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New post 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. Bunuel

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It'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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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