Is xy > 0? (1) x - y > -2 (2) x - 2y < -6

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.
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Please find attached graphical solution. I think its easier and simpler to solve such inequality problem using co-ordinate geometry.

I made this graph in a hurry in paint, so please don't mind the poorly drawn lines.

Consider pressing on Kudos if my post helped you in any way!

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.


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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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(1)

x = 1, y = 0

xy = 0 although x-y = 1 > -2

x = -3, y = -2

xy = 6 > 0 and x-y = -1 > -2

Insufficient

(2)

x - 2y < -6

x = 1, y = 4

x - 2y = 1 - 8 = -7 < -6 and xy > 0

x = 0, y = 4

x - 2y = 0 - 8 = -8 < -6 and xy = 0

Insufficient

(1) + (2)

x - y > -2

2y - x > 6

y > 4

Now x has to be such that x > y - 2

So x is positive, because y is 4.1, 5 etc.
and x is > 2.1, 3 etc.


So xy = positive

Answer - C
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This is a question which is a great example of what the GMAT expects from test takers in terms of combining statements in a DS question on Inequalities.

It is also an excellent question to learn how to prove/disprove the individual statements in a DS question.

Breaking down the question stem, we can rephrase the question as “Are x and y of the same signs?”. The answers to this question could be,

Yes – which means x and y are either both positive or both negative
No – which means one of them could be positive and the other could be negative OR ZERO (many of us ignore ZERO at our own peril).

From statement I alone, x-y > -2. Let us take some values to find out if this information is sufficient to answer the main question.
If x = 10 and y = 5, x-y = 5 which is definitely greater than -2. For these values of x and y, we answer the main question with a Yes.
If x = 10 and y = -5, x-y = 15 which is also greater than -2. For these values of x and y, we answer the main question with a No.

Statement I alone is insufficient to answer the question with a definite YES or NO. Answer options A and D can be eliminated. Possible answer options are B, C or E.

From statement II alone, x-2y < -6. Again, as in the first statement, let us test some values.
If x = 5, y = 10, x – 2y = -15 which is less than -6. These values give a YES to the main question.
If x = -5, y = 10, x – 2y = -25 which is less than -6. These values give a NO to the main question.

Statement II alone is insufficient to answer the question with a definite YES or NO. Answer options B can be eliminated. Possible answer options are C or E.

Now, at this stage, we are required to combine the statements I and II. Here’s where GMAT expects you to combine the statements using an important property of inequalities and that is – Two inequalities can be added if they have the same inequality sign.

The two inequalities are: x – y > -2 and x – 2y < -6.

When we multiply the first inequality with -2, we obtain -2x + 2y < 4. Now, since the inequality sign is the same as the second inequality, we can add these inequalities.
Adding the inequalities, 2y and -2y cancel out each other and we have –x < -2 which can be rewritten as x > 2.

When we multiply the second inequality with -1, we obtain –x + 2y > 6. Since the inequality sign is the same as the first, we can add these inequalities.
Adding them, x and –x cancel out each other and we have y > 4.

This means xy > 0 and we can answer the main question with a Yes. The combination of statements is sufficient, answer option E can be eliminated.

The correct answer option is E.

Remember that inequalities can be added as long as they have the same sign.

Hope that helps!
Aravind B T

Target question: Is xy > 0?

Statement 1: x - y > -2
Let's TEST some values.
There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = 5 and y = 1. In this case, xy = (5)(1) = 5. So, the answer to the target question is YES, xy IS greater than 0
Case b: x = 5 and y = -1. In this case, xy = (5)(-1) = -5. So, the answer to the target question is NO, xy is NOT greater than 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x - 2y < -6
Let's TEST some values again.
There are several values of x and y that satisfy statement 2. Here are two:
Case a: x = 1 and y = 10. In this case, xy = (1)(10) = 10. So, the answer to the target question is YES, xy IS greater than 0
Case b: x = -1 and y = 10. In this case, xy = (-1)(10) = -10. So, the answer to the target question is NO, xy is NOT greater than 0
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
IMPORTANT: There’s a useful inequality property that says If two inequalities have their inequality symbols facing the SAME DIRECTION, we can ADD the inequalities.
We have:
x - y > -2
x - 2y < -6

Take the BOTTOM equation and multiply both sides by -1 to get:
x - y > -2
-x + 2y > 6 [since we multiplied both sides by a NEGATIVE value, we REVERSED the direction of the inequality symbol]

Now ADD the equations to get: y > 4
In other words, y is POSITIVE

Now take:
x - y > -2
-x + 2y > 6

Take the TOP equation and multiply both sides by 2 to get:
2x - 2y > -4
-x + 2y > 6

Now ADD the equations to get: x > 2
In other words, x is POSITIVE

Since x and y are both POSITIVE, we know that xy IS positive
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

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Question; Is xy > 0?

STatement 1: x - y > -2

@x = 1, y may be +1 i.e. xy > 0
@x = 1, y may be -1 i.e. xy < 0

NOT SUFFICIENT

STatement 2:x - 2y < -6

@x = 1, y may be +4 i.e. xy > 0
@x = -1, y may be +4 i.e. xy < 0

NOT SUFFICIENT

Combining the statements


Now we have
x - y > -2 and x - 2y < -6
ie.. x - y > -2 and -x + 2y > 6 (multiplied -1 both sides in second inequation)

Adding the two inequations
x - y -x + 2y > -2+6
i.e. y > 4
Substituting it in inequation 1
x - y > -2 i.e. x > -2+y i.e. x > 2

SInce both x and y are positive, therefore xy > 0

Answer: Option C

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UserMaple5

Do x and y have the same sign?

Statement 1, rephrased: x > y-2
Clearly, x and y can have the same sign or different signs.
INSUFFICIENT

Statement 2, rephrased: x < 2y-6
Clearly, x and y can have the same sign or different signs.
INSUFFICIENT.

Inequalities with multiple variables can be ADDED TOGETHER.
Before adding, we must ensure that the inequality symbol faces the same direction in each inequality.

Statements combined:
Adding together x > y-2 and 2y-6 > x, we get:
x+2y-6 > y-2+x
y > 4

Adding together y > 4 and x > y-2, we get:
y+x > 4+y-2
x > 2

Since x>2 and y>4, x and y have the same sign.
Thus, the answer to the question stem is YES.
SUFFICIENT.


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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.

Check here for more: graphic-approach-to-problems-with-inequalities-68037.html

Hope it helps.
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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!

Hi healthjunkie,

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

Hope its clear

Regards
Harsh
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Dear Bunuel,

Please find my attached process.

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..
_________________

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...
_________________

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.

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,