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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 01 Jan 2018, 13:54
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Tpral
Hello Bunuel,

I arrive to y>4.

When you replace it in (1) it makes x>2: OK

But

When you replace it in (2) it makes x - 2(4) < -6 ; so x <2 and so can be negative or positive.

Could you advise when replacing in (1) X>2 (answer =C) and when replacing in (2) we have answer (E) as x<2.

Thanks in advance,
Show more

We got y > 4 not y = 4, so you cannot substitute the way you did. In the solution, x > 2, is derived by adding y > 4 and x > y - 2 (we can add inequalities when their signs are in the same direction): \(y+x>4+(y-2)\) --> \(x>2\).
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 02 Jun 2018, 10:29
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Bunuel



We got y > 4 not y = 4, so you cannot substitute the way you did. In the solution, x > 2, is derived by adding y > 4 and x > y - 2 (we can add inequalities when their signs are in the same direction): \(y+x>4+(y-2)\) --> \(x>2\).
Show more

Hi Bunuel ,

Kindly clarify -

"In the Solution of X>2" : Instead of adding Y>4 and X > Y-2 (from equation 1) , can we not add Y>4 and -X> 6-2y ? (from equation 2 : x-2y <-6)

If we do this , we will arrive at X < 2
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 02 Jun 2018, 12:31
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Bunuel



We got y > 4 not y = 4, so you cannot substitute the way you did. In the solution, x > 2, is derived by adding y > 4 and x > y - 2 (we can add inequalities when their signs are in the same direction): \(y+x>4+(y-2)\) --> \(x>2\).

Hi Bunuel ,

Kindly clarify -

"In the Solution of X>2" : Instead of adding Y>4 and X > Y-2 (from equation 1) , can we not add Y>4 and -X> 6-2y ? (from equation 2 : x-2y <-6)

If we do this , we will arrive at X < 2
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y > 4
-x > 6 - 2y

Adding those gives:
y - x > 10 - 2y
3y - x > 10
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 02 Jun 2018, 17:42
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Bunuel



We got y > 4 not y = 4, so you cannot substitute the way you did. In the solution, x > 2, is derived by adding y > 4 and x > y - 2 (we can add inequalities when their signs are in the same direction): \(y+x>4+(y-2)\) --> \(x>2\).

Hi Bunuel ,

Kindly clarify -

"In the Solution of X>2" : Instead of adding Y>4 and X > Y-2 (from equation 1) , can we not add Y>4 and -X> 6-2y ? (from equation 2 : x-2y <-6)

If we do this , we will arrive at X < 2

y > 4
-x > 6 - 2y

Adding those gives:
y - x > 10 - 2y
3y - x > 10
Show more

Sorry Bunuel , I still don't follow. Maybe I'm missing something.

Last step =3y - x > 10

Substituting Y >4 : 12 - X > 10 -> 2 > X -> X < 2 : right?
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 03 Jun 2018, 05:10
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Hi Bunuel ,

Kindly clarify -

"In the Solution of X>2" : Instead of adding Y>4 and X > Y-2 (from equation 1) , can we not add Y>4 and -X> 6-2y ? (from equation 2 : x-2y <-6)

If we do this , we will arrive at X < 2

y > 4
-x > 6 - 2y

Adding those gives:
y - x > 10 - 2y
3y - x > 10

Sorry Bunuel , I still don't follow. Maybe I'm missing something.

Last step =3y - x > 10

Substituting Y >4 : 12 - X > 10 -> 2 > X -> X < 2 : right?
Show more

How are you getting 12 - x > 10?

3y - x > 10
3y > 12.

We cannot subtract those two the way you did.

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

For more check Manipulating Inequalities.
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 04 Oct 2018, 18:45
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This question is fantastic for bringing up a strategy not enough people think to do with inequalities: as long as you have the inequality signs pointed in the same direction, you can add the inequalities together to eliminate a variable the same way you would for a system of equations.

Here once you've eliminated A, B, and D by quick conceptual number picking**, you can multiply the Statement 2 inequality by -1 so that you have both inequalities with signs in the same direction:

x - y > -2

and

2y - x > 6

If you then add the two inequalities, you eliminate the x term (one is positive and the other is negative), leaving:

y > 4

So you know y is positive, and then when you plug y > 4 in to the first equation (which can be expressed as x > y - 2), you know that x is greater than "something greater than 4, minus 2" so x must also be positive, so you know that the product xy is greater than 0. Together the statements are sufficient and the answer is (C).

Note that when you do have two inequalities this situation presents itself more often than most people think! If you can get the signs facing in the same direction (generally by multiplying one inequality by -1 if they're not already in the same direction) you can add the inequalities together (don't subtract, since "minus" is basically "plus negative" and can screw up the positive/negative inequality logic) and solve like a system of equations.


** Try x = 4 and y = 6 versus x = -2 and y = 0 for statement 1 and x= 0 and y = 3 and x = 1 and y = 12 for statement 2, for example.
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 04 Oct 2018, 19:32
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Is xy > 0 ?

This basically asks us .."Are x and y of SAME sign?"

(1) x – y > –2
x>y-2..
y=2, x=4...yes
y=-2, x=4...no
Insufficient

(2) x – 2y < –6
x<2y-6..
y=2, x=-4...no
y=-2, x=-15...yes
Insufficient


Combined
x-y>-2
x-2y<-6 or 2y-x>6
Add the two inequalities
x-y+2y-x>-2+6.......y>4
Multiply first by 2 and add the equations to get x
2(x-y)+2y-x>2*(-2)+6.....2x-2y+2y-x>6-4.....x>2

So both x and y are positive
Ans is always yes
Sufficient

C
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 17 Aug 2019, 09:29
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Baten80
Is xy > 0?

(1) x - y > -2
(2) x - 2y < -6
Show more

Is xy>0

1) x-y > -2
x>y-2
xy>y(y-2)=y^2-2y
Value of y is unknown
NOT SUFFICIENT

2) x-2y < -6
x<2y-6
xy<y(2y-6)=2y^2-6y
Value of y is unknown
NOT SUFFICIENT

Combining (1) & (2)
1) x-y > -2
x>y-2. (3)
2) x-2y < -6
x<2y-6
-x>-2y+6 (4)
Adding (3) & (4)
0>-y+4
y>4>0 (5)
From equation (3)
x>y-2>4-2=2>0
x>2>0 (6)
xy>0
SUFFICIENT

IMO C
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 19 Dec 2020, 16:25
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Video solution from Quant Reasoning (starts at 0:00:25):
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 22 May 2021, 21:16
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EgmatQuantExpert Bunuel JeffTargetTestPrep

Hi experts. I tend to get these types of questions right but they take me WAY TOO LONG to do. I normally test cases and it takes a while. Are there any strategies other than testing cases that I can use for these kind of questions to help improve timing?

Thank you in advance!
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 23 May 2021, 03:09
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EgmatQuantExpert Bunuel JeffTargetTestPrep

Hi experts. I tend to get these types of questions right but they take me WAY TOO LONG to do. I normally test cases and it takes a while. Are there any strategies other than testing cases that I can use for these kind of questions to help improve timing?

Thank you in advance!
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UserMaple5 yes, you can use number line reasoning to avoid having to test cases. You can see how I do that in the first part of the video from this post:
https://gmatclub.com/forum/is-xy-0-1-x- ... l#p2686547
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 06 Jul 2021, 08:34
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I think it should be possible to skip all the calculations and graphical solutions and number line drawings and just think about this for a few moments.

So it is clear by just picking a few numbers that neither (1) nor (2) are sufficient by itself.

So now the big question is - are they sufficient together?

We have: x - y > -2 and x - 2y < -6 from statement (1) and (2).

We can rewrite statement (2): x - y - y < -6 and the red-coloured part much be larger than -2, as it is given by statement (1). But statement (2) must be smaller than -6. We can only achieve that if y is positive (notice the negative sign) and must be great than 4 to make statement (2) smaller than -6).

So now we know that y is positive and greater than 4 and we can go back to (1): x - y > -2. If y is greater than 4, we now know that x must be positive so that (1) is great than -2.
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 08 Jul 2021, 12:05
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Taken together:

Subtracting inequalities of different signs -->

(x-y)-(x-2y) > (-2)-(-6)

y > 4

From s1 it is now obvious that x must also be positive.

Posted from my mobile device
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 12 Jun 2023, 03:00
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Bunuel
Is xy>0?

Note that question basically asks whether \(x\) and \(y\) have the same sign.

(1) x-y > -2 --> we can have an YES answer, if for example \(x\) and \(y\) are both positive (\(x=10\) and \(y=1\)) as well as a NO answer, if for example \(x\) is positive and \(y\) is negative (\(x=10\) and \(y=-10\)). Not sufficient.

(2) x-2y <-6 --> again it' easy to get an YES answer, if for example \(x\) and \(y\) are both positive (\(x=1\) and \(y=10\)) as well as a NO answer, if for example \(x\) is negative and \(y\) is positive (\(x=-1\) and \(y=10\)). Not sufficient.

You can get that the the two statement individually are not sufficient in another way too: we have (1) \(y<x+2\) and (2) \(y>\frac{x}{2}+3\). We are asked whether \(x\) and \(y\) have the same sign or whether the points (x,y) are in the I or III quadrant ONLY. But all (x,y) points below the line \(y=x+2\) (for 1) and all (x, y) points above the line \(y=\frac{x}{2}+3\) cannot lie only in I or III quadrant: points above or below some line (not parallel to axis) lie at least in 3 quadrants.

(1)+(2) Now, remember that we can subtract inequalities with the signs in opposite direction --> subtract (2) from (1): \(x-y-(x-2y)>-2-(-6)\) --> \(y>4\). As \(y>4\) and (from 1) \(x>y-2\) then \(x>2\) (because we can add inequalities when their signs are in the same direction, so: \(y+x>4+(y-2)\) --> \(x>2\)) --> we have that \(y>4\) and \(x>2\): both \(x\) and \(y\) are positive. Sufficient.

Answer: C.
Show more

Hey Bunuel!
I am a bit slow when it comes to thinking of numbers to solve DS Questions like these. I am quite strong in visualising the graphs when it comes to equations. So I started by plotting both the lines after quickly ruling out answer choices (A) and (B). After plotting, I selected the common area i.e the intersection of x - y > -2 and x - 2y < -6 and the common area of these two equations lies completely in the first quadrant, therefore, both x and y will always be positive and we can conclude that (C) is the correct answer. Is there any downside to this approach? I know all the graphs of the basic equations that I have seen on 750, or 700 level questions, what other graphs should I know about if I want to solve problems like these using the graph approach?
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 07 Jan 2024, 12:56
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Bunuel


Is xy > 0?

\(xy>0\) means that x and y must have the same sign, so point (x, y) must be either in the first or the third quadrant (green regions).

(1) x-y > -2 --> \(y<x+2\) --> the area below blue line (\(y=x+2\)). (x, y) may or may not be in green region. Not sufficient.

(2) x-2y < -6 --> \(y>\frac{x}{2}+3\) --> the area above red line (\(y>\frac{x}{2}+3\)). (x, y) may or may not be in green region. Not sufficient.

(1)+(2) Below blue line and above red line, is yellow region, which is entirely in I quadrant (where \(y>4\) and \(x>2\)) --> \(xy>0\). Sufficient.

Answer: C.

Hope it helps.

Show Spoiler
Attachment:
xy.png
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Bunuel KarishmaB MartyMurray
Generic question:
I did get it right in my practice exam (1:34) and I plugged in numbers as couple of the solutions show in this discussion. But I typically do not feel confident mostly because I don't want to keep spending time to cross check / revise. What should I do? What is a recommended method to save time and also inadvertently cross check solution for inequality question types? I don't think I have "problem" with the concept and if I have more time, I do these questions well. Thank you.
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 08 Jan 2024, 04:42
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Bunuel


Is xy > 0?

\(xy>0\) means that x and y must have the same sign, so point (x, y) must be either in the first or the third quadrant (green regions).

(1) x-y > -2 --> \(y<x+2\) --> the area below blue line (\(y=x+2\)). (x, y) may or may not be in green region. Not sufficient.

(2) x-2y < -6 --> \(y>\frac{x}{2}+3\) --> the area above red line (\(y>\frac{x}{2}+3\)). (x, y) may or may not be in green region. Not sufficient.

(1)+(2) Below blue line and above red line, is yellow region, which is entirely in I quadrant (where \(y>4\) and \(x>2\)) --> \(xy>0\). Sufficient.

Answer: C.

Hope it helps.

Show Spoiler
Attachment:
xy.png


Bunuel KarishmaB MartyMurray
Generic question:
I did get it right in my practice exam (1:34) and I plugged in numbers as couple of the solutions show in this discussion. But I typically do not feel confident mostly because I don't want to keep spending time to cross check / revise. What should I do? What is a recommended method to save time and also inadvertently cross check solution for inequality question types? I don't think I have "problem" with the concept and if I have more time, I do these questions well. Thank you.
Show more

The graphical solution Bunuel has given above is a 1 min solution and that would be my method of choice.
Plugging in numbers is generally not a good strategy for DS questions. At most, it can help you arrive at the pattern in a question.
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 03 Feb 2025, 19:24
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Hi Bunuel,

Could you please help with this? Are we missing something here?

Tpral
Hello Bunuel,

I arrive to y>4.

When you replace it in (1) it makes x>2: OK

But

When you replace it in (2) it makes x - 2(4) < -6 ; so x <2 and so can be negative or positive.

Could you advise when replacing in (1) X>2 (answer =C) and when replacing in (2) we have answer (E) as x<2.

Thanks in advance,
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 03 Feb 2025, 19:24
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Hi Bunuel,

Could you please help with this? Are we missing something here?

Tpral
Hello Bunuel,

I arrive to y>4.

When you replace it in (1) it makes x>2: OK

But

When you replace it in (2) it makes x - 2(4) < -6 ; so x <2 and so can be negative or positive.

Could you advise when replacing in (1) X>2 (answer =C) and when replacing in (2) we have answer (E) as x<2.

Thanks in advance,
Show more
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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 03 Feb 2025, 23:47
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vibhasharma
Hi Bunuel,

Could you please help with this? Are we missing something here?

Tpral
Hello Bunuel,

I arrive to y>4.

When you replace it in (1) it makes x>2: OK

But

When you replace it in (2) it makes x - 2(4) < -6 ; so x <2 and so can be negative or positive.

Could you advise when replacing in (1) X>2 (answer =C) and when replacing in (2) we have answer (E) as x<2.

Thanks in advance,
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That doubt has already been addressed here: https://gmatclub.com/forum/is-xy-0-1-x- ... l#p1989838
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