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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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31 Jul 2014, 06:19

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, 19:37

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]

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01 Apr 2016, 04:40

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

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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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chetan2u Looks like how I came with the solution is wrong. Since statement 1 and 2 are not sufficient I tried to combine the two statements.

1. X- Y>-2 2. X-2Y <-6

FROM 1: X>Y-2-------(WE HAVE A VALUE FOR X) substituting this value in statement 2 yields;

Y-2 -2Y<-6 OR, -Y < -4 OR, Y> 4 (MUTLIPLYING BOTH SIDES WITH A NEGATIVE SIGN.)

SINCE Y> 4, FROM STATEMETN ONE

X-Y>-2 OR, X- 4.1> -2, SOLVING FOR THIS WOULD GIVE X>2.1.

Is my process correct?

HI, In this Q it is ok.. But a lot depends on '>' and '<' signs.. you cannot substitute without knowing these.. 1. X- Y>-2 ...... x>y-2 2. X-2Y <-6..... x<2y-6.. so y-2<x<2y-6.... so here clearly y-2<2y-6 .... or y>4... so form equations then substitute...
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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]

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05 Aug 2016, 09:42

I didn't know that you can subtract opposite sign equations and actually I feel that many will still get the sign mixed up as to whether it would be < or >. Instead you can flip one of the equations and then you will end up with one sign and then just add from there. That way, there is less chance for a mistake to occur.

We need to determine whether the product of x and y is positive. We should recall that the product of two numbers is positive only if both the numbers are positive or if both are negative.

Statement One Alone:

x - y > -2

Statement one tells us that the difference between x and y is -2; it does not tell us anything about the signs of x and y. For instance, if x = 2 and y = 1, we have x - y = 1 > -2, and xy is positive. However, if x = 3 and y = -2, 3 - (-2) = 5 > -2, but xy is negative. Statement one alone is not sufficient. We can eliminate answer choices A and D.

Statement Two Alone:

x - 2y < -6

Again, we have a statement that tells us nothing about the signs of x and y. For instance, if x = 3 and y = 5, then x - 2y = 3 - 2(5) = 3 - 10 = -7 < -6, and xy is positive. However, if x = -1 and y = 5, then x - 2y = -1 - 2(5) = -11 < -6, and xy is negative. Statement two alone is not sufficient. We can eliminate answer choice B.

Statements One and Two Together:

Let’s manipulate the first inequality to read: y < x + 2. Similarly, we can manipulate the second inequality to read: y > (1/2)x + 3.

Thus, we can say the following:

(1/2)x + 3 < y < x + 2

(1/2)x + 3 < x + 2

x + 6 < 2x + 4

2 < x

Thus, x is positive.

We also know the following:

y > (1/2)x + 3

Since x is greater than two, let’s see what we can determine about y, if we substitute 2 for x.

y > (1/2)(2) + 3

y > 4

So y is positive as well. Both statements together are sufficient.

Answer: C
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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,

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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