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

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

07 May 2017, 09:31

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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]

07 May 2017, 10:08

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

L

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

07 May 2017, 10:12

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gmatexam439 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.

L

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

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]

07 May 2017, 10:20

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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 [#permalink]

05 Oct 2017, 20:41

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

L

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

05 Oct 2017, 21:12

Expert Reply

zanaik89 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]

06 Jan 2018, 05:56

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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:

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]

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. Bunuel

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

17 Feb 2018, 06:23

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

26 Feb 2020, 03:33

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 [#permalink]

26 Feb 2020, 04:06

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

02 Oct 2020, 07:53

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Bunuel VeritasKarishma how did you decide the values for option C. What's the quick way to check this?

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

06 Oct 2020, 04:23

Expert Reply

Suneha123 wrote:

Bunuel VeritasKarishma how did you decide the values for option C. What's the quick way to check this?

Option (C) is put there as a time guzzler - to waylay the test taker. The other 4 are clear and easy to work with since they have a single variable. The test taker should look at (C) and walk away. One needs to come back to (C) only if all other options turn out to be wrong, in which case it would be automatically correct.

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

23 May 2023, 00:55

Timebomb 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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Bunuel KarishmaB

I had the same confusion. Why are we taking the opposite? The premise should always be true, thus the premise should be the super set.
However, if take option D, x=2, the premise is not satisfied.
Since the question is must be true, so the options should have all values staisfying premise?

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

23 May 2023, 01:17

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shanks2020 wrote:

Timebomb 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

Posted from my mobile device

Bunuel KarishmaB

I had the same confusion. Why are we taking the opposite? The premise should always be true, thus the premise should be the super set.
However, if take option D, x=2, the premise is not satisfied.
Since the question is must be true, so the options should have all values staisfying premise?

I've already attempted to explain this several times, now, to avoid redundancy, I'd suggest referring to similar questions from the link:Trickiest Inequality Questions Type: Confusing Ranges. This should help clarify any confusion.

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

23 May 2023, 02:50

Bunuel wrote:

shanks2020 wrote:

Timebomb 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

Posted from my mobile device

Bunuel KarishmaB

I had the same confusion. Why are we taking the opposite? The premise should always be true, thus the premise should be the super set.
However, if take option D, x=2, the premise is not satisfied.
Since the question is must be true, so the options should have all values staisfying premise?

I've already attempted to explain this several times, now, to avoid redundancy, I'd suggest referring to similar questions from the link:Trickiest Inequality Questions Type: Confusing Ranges. This should help clarify any confusion.

Bunuel

Sorry for asking again. Yes, the confusion is in general for those questions.
I didnt see the explanation for this confusion. Why do we take the options as the super set and not the premise?

gmatclubot

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

23 May 2023, 02:50

GMAT Problem Solving (PS) Questions

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