GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

It is currently 14 Jul 2020, 15:46

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 65290
If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

New post 28 Apr 2017, 02:01
7
117
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

36% (02:53) correct 64% (02:58) wrong based on 856 sessions

HideShow timer Statistics

Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 65290
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

New post 07 May 2017, 08:43
9
17
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.
_________________
General Discussion
Current Student
avatar
B
Joined: 23 Apr 2017
Posts: 20
Location: India
GMAT 1: 720 Q50 V36
WE: Marketing (Manufacturing)
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

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)
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 65290
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

New post 07 May 2017, 08:52
2
5
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 \(|y+1|=|−0.9+1|=|0.1|=0.1\).
_________________
Moderator
User avatar
V
Joined: 28 Mar 2017
Posts: 1194
Location: India
GMAT 1: 730 Q49 V41
GPA: 4
CAT Tests
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

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
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 65290
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

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.
_________________
Moderator
User avatar
V
Joined: 28 Mar 2017
Posts: 1194
Location: India
GMAT 1: 730 Q49 V41
GPA: 4
CAT Tests
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

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
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 65290
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

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.
_________________
Manager
Manager
avatar
B
Joined: 19 Aug 2016
Posts: 66
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

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
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 65290
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

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.
_________________
Retired Moderator
User avatar
V
Joined: 27 Oct 2017
Posts: 1859
WE: General Management (Education)
GMAT ToolKit User CAT Tests
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

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.

Attachments

WhatsApp Image 2018-01-06 at 18.07.10.jpeg
WhatsApp Image 2018-01-06 at 18.07.10.jpeg [ 108.6 KiB | Viewed 8920 times ]


_________________
Manager
Manager
avatar
S
Joined: 29 Nov 2016
Posts: 205
Location: India
GMAT 1: 750 Q50 V42
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

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

Posted from my mobile device
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 65290
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

New post 17 Feb 2018, 05:23
Manager
Manager
avatar
B
Joined: 02 Dec 2018
Posts: 53
CAT Tests
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

New post 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.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 65290
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

Show Tags

New post 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.
_________________
GMAT Club Bot
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be   [#permalink] 26 Feb 2020, 03:06

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





cron

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne