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Math Expert V
Joined: 02 Sep 2009
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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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Question Stats: 36% (02:53) correct 64% (02:58) wrong based on 858 sessions

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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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Math Expert V
Joined: 02 Sep 2009
Posts: 65384
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

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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).

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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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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 V
Joined: 02 Sep 2009
Posts: 65384
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

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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 $$|y+1|=|−0.9+1|=|0.1|=0.1$$.
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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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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).

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 V
Joined: 02 Sep 2009
Posts: 65384
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

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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).

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

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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 V
Joined: 02 Sep 2009
Posts: 65384
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

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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]

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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).

How did u get 0<x<2 when x-2<0

Because x<2, x can take any value below 2

pls explain

Math Expert V
Joined: 02 Sep 2009
Posts: 65384
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

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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).

How did u get 0<x<2 when x-2<0

Because x<2, x can take any value below 2

pls explain

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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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).

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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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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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Math Expert V
Joined: 02 Sep 2009
Posts: 65384
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

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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

Posted from my mobile device

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]

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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?
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Posts: 65384
Re: If xy > 0 and (x − 2)(y + 1) < 0, then which of the following must be  [#permalink]

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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?

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.
_________________ 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  