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-x + 2y > 6 (to make the signs same for both equations)

and

solve for y to get y>4

Now for solving x

Do you solve the equations again for x

or

how does it work by substituting y > 4 ? Could someone show this method and also tell me if this is the best way to solve these questions.

Thanks Ravi

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.

Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]

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12 Mar 2013, 10:28

Bunuel wrote:

tech3 wrote:

Hello, I have a doubt.

After we have got y >4, do we check both the equations for consistency? With the first equation, m getting x >2 but with the second, I am receiving x<2 upon solving. What am I doing wrong

Can you please show how you get x<2?

_____________________________________________________________________________________________________________________________ Bunuel, I'm struggling to understand how the second statement proves that x is definitely positive. I follow that y>4 and I follow that x>2 if I only test equation (1). But I would think that I should be able to test either and come away with the fact that x is positive. However when I test equation (2) x proves to be positive or negative!

Testing equation (2) x - 2y < -6

since y>4 If y = 4.1, then x - 8.2 < -6 and x < 2.2 (this suggests x can be positive or negative) If y = 6, then x - 12 < -6 and x < 6 (this suggests x can be positive or negative) and so on...

I don't understand how we prove x is definitely positive if I test equation (2)

Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]

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18 Mar 2013, 13:04

adding both equations, you would get y >4. Substituting this to equation 1, you would get x > 2! Hence both x and y are positive, so xy > 0
_________________

Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]

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15 Jul 2013, 17:12

pleaman wrote:

dolumoks wrote:

Bunuel wrote:

Hello, I have a doubt.

After we have got y >4, do we check both the equations for consistency? With the first equation, m getting x >2 but with the second, I am receiving x<2 upon solving. What am I doing wrong

Can you please show how you get x<2?

_____________________________________________________________________________________________________________________________ Bunuel, I'm struggling to understand how the second statement proves that x is definitely positive. I follow that y>4 and I follow that x>2 if I only test equation (1). But I would think that I should be able to test either and come away with the fact that x is positive. However when I test equation (2) x proves to be positive or negative!

Testing equation (2) x - 2y < -6

since y>4 If y = 4.1, then x - 8.2 < -6 and x < 2.2 (this suggests x can be positive or negative) If y = 6, then x - 12 < -6 and x < 6 (this suggests x can be positive or negative) and so on...

I don't understand how we prove x is definitely positive if I test equation (2)

Agree! I cant understand the same )

I have the same doubt, with y>4 , put the value in 2nd equation, and you have -negative values for x.
_________________

And many strokes, though with a little axe, hew down and fell the hardest-timbered oak. - William Shakespeare

_____________________________________________________________________________________________________________________________ Bunuel, I'm struggling to understand how the second statement proves that x is definitely positive. I follow that y>4 and I follow that x>2 if I only test equation (1). But I would think that I should be able to test either and come away with the fact that x is positive. However when I test equation (2) x proves to be positive or negative!

Testing equation (2) x - 2y < -6

since y>4 If y = 4.1, then x - 8.2 < -6 and x < 2.2 (this suggests x can be positive or negative) If y = 6, then x - 12 < -6 and x < 6 (this suggests x can be positive or negative) and so on...

I don't understand how we prove x is definitely positive if I test equation (2)

Agree! I cant understand the same )

I have the same doubt, with y>4 , put the value in 2nd equation, and you have -negative values for x.

But if you use y>4 and x-y > -2 you'll get that x>2. It's more restrictive so it's correct.
_________________

Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]

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31 Jul 2014, 07:19

Bunuel wrote:

Graphic approach:

Attachment:

xy.png

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.

Hi Bunuel,

I have doubts on graphical method -->

why is line y<x+2 passing through x = -2 ? the value of x should be greater than -2 (the line may pass through x = -1, 0, 1, 2, ...) Similarly, why is line y > (x/2) + 3 passing through x = -6 the value of x should be less than -6 (the line may pass through x = -7, -8, -9,...)

With the graphical method, can you illustrate solution of following example ?

\(xy>0\) 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.

Hi Bunuel,

I have doubts on graphical method -->

why is line y<x+2 passing through x = -2 ? the value of x should be greater than -2 (the line may pass through x = -1, 0, 1, 2, ...) Similarly, why is line y > (x/2) + 3 passing through x = -6 the value of x should be less than -6 (the line may pass through x = -7, -8, -9,...)

With the graphical method, can you illustrate solution of following example ?

Is x + y > 0 ?

1) x - y > 0 2) x^2 - y^2 > 0

Regards, Akshay

y < x + 2 is the region which lies BELOW (because of < sign) line y = x + 2.

y > x/2 + 3 is the region which lies ABOVE (because of > sign) line y = x/2 + 3.

Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]

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13 May 2015, 20:37

Hi,

I approached the 1&2 decision the same way as Bunuel but just had a clarifying question. When we subtract the two inequalities from eachother..how do we know that the sign stays as greater than (i.e. how do we know that it's y>4 instead of y<4). I assumed it was y>4 and got the question correct because i was subtracting a less than equation from a greater than equation but I have a feeling that's not the right logic. Thanks!

I approached the 1&2 decision the same way as Bunuel but just had a clarifying question. When we subtract the two inequalities from eachother..how do we know that the sign stays as greater than (i.e. how do we know that it's y>4 instead of y<4). I assumed it was y>4 and got the question correct because i was subtracting a less than equation from a greater than equation but I have a feeling that's not the right logic. Thanks!

Let me help you out with such operations on inequalities.

Assume two inequalities x + y > 10......(1) & x - y < 2........(2)

We see here that both the inequalities have opposite signs.

The basic concept of the inequality says to convert the inequalities into same sign before adding or subtracting them.

We can change the sign of the inequality by multiplying -1 on both sides of the inequality. Let's do it on inequality (2)

x - y < 2, multiplying -1 on both sides we get -x + y > -2.

Since we have inequality (1) & (2) with the same signs we can add them now

x + y -x + y > 10 -2 i.e. y > 4.

Note here that adding inequality (1) to inequality (2) after multiplying inequality (2) by -1 is similar to subtracting inequality (2) from inequality (1). This concept is used when we say that we can subtract two inequalities with opposite signs.

In this question you asked we subtracted inequality in st-II from inequality in st-I which incidentally meant that st-II was multiplied by -1 and then added to st-I. So, the sign of inequality in st-II flipped.

x - 2y < -6, multiplying it with -1 gave the inequality -x + 2y > 6 which was then added to inequality in st-I

Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]

Show Tags

01 Apr 2016, 05:40

Bunuel wrote:

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 \(y=\frac{x}{2}+3\) can not lie only 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.

However, the combined solving of eqns gave me, x<-4 or x>-1, a case in which we cannot comment on sign of x. Can you please help me understand where I have gone wrong?

Attachments

IMG_20160401_163125.jpg [ 2.2 MiB | Viewed 8019 times ]

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 \(y=\frac{x}{2}+3\) can not lie only 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.

However, the combined solving of eqns gave me, x<-4 or x>-1, a case in which we cannot comment on sign of x. Can you please help me understand where I have gone wrong?

Hi, your Equations are further to be simplified..

you got 2y-6 > x> y-2 we can infer 2y-6>y-2... or 2y-y>6-2 that is y>4.. if y>4 and x+2>y, x will be >2.. thus both x and y are +ive and our answer is YES for xy>0..

Do not substitute value in 2y-6 > x> y-2 and find the signs of x and y..
_________________

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