If x and y are integers, is x > y?

(1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient.

(2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient.

(1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient.

Answer: C.
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Question is x>y ?

(1) x+y>0 we can not determine wether x>y or not, but this statement says that one of the vaiables must be positive to satisfy it --> Not sufficient
(2) Y^x<0 the only way it to be negative is when Y is negative, but X could be also negative (-2^-3 = -1/8) --> Not Sufficient, as we don't know wether x>y

(1)+(2) Statement 1 says that one of the variables must be positive + Statement 2 says that Y is in all cases negative --> if Y is negative X must be positive and is > Y

Answer (C)

Further explaining option c

from 2, we know y is a -ve number and to make x+y > 0 , x has to be positive to make the equation greater than zero .

so c is sufficient

From 2, why we have decuded that y is a -ve number? y^x= 3^-3 is also less than zero and in this case y is positive. Can you please explain?

\(3^{(-3)} = \frac{1}{3^3} = \frac{1}{27} > 0\), not less than 0.
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Would anybody please explain why we ruled out statement 2. I thought the that the only was for y^x<0 is for y to be negative and X to be an odd power. I ruled out the negative values of X thinking that if X is negative then y would be a negative fraction and the question stem says that y is an integer. Why was I wrong here? Please help!

Hi naeln,

From statement-II, we can deduce that \(y\) is a negative integer but we can't say if \(x\) is a negative or a positive odd integer. Let's evaluate both the cases:

Case-I: \(x\) is positive
If \(x\) is a positive odd integer, then \(y^x < 0\). For example, assuming \(y = -2\) and \(x = 3\) would give \(y^x = -2^3 = -8 < 0\)


Case-II: \(x\) is negative
When \(x\) is negative, \(y^x\) would become a negative fraction and not \(y\) itself.

For example: if \(y = -2\) and \(x = -3\), then \(x\) & \(y\) both are integers and \(y^x = -2^{-3} = \frac{-1}{8} < 0\). Here \(y^x = \frac{-1}{8}\) is a fraction and not \(y\) itself.

Hope its clear why statement-II does not give us a unique answer.

Regards
Harsh
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Thanks for the explanation. I got the correct answer using the same approach, however at the beginning of my solution I tried a different way to understand statement 2: Since \(y^0=1\) and \(0<1\), so if \(y^x<0\), it should mean that \(y^x<y^0\) and hence \(x<0\). I know this is not correct but I am unable to understand what am I doing wrong here.
Can you please help confirm that?

Thanks again!

If \(y\neq{0}\), then y^0=1. y^x < 0 does not mean that y^x < (1 = y^0).
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(1) \(x+y>0\) : For this to be true, we will have the following 3 senarios:
a) \(x > 0\) and \(y > 0\)
b) \(x > 0\) and \(y <= 0\), which means \(x > y\)
c) \(y > 0\) and \(x <= 0\), which means \(y > x\)

Therefore, this statement is Insufficient

(2) \(y^x < 0\): For this to be true, \(y\) must be less than 0 and \(x\) must be an odd integer.
Therefore, this statement is insufficient.

(1) and (2) together:
From statement (2), we know \(y < 0\), therefore the only viable scenario for this in statement (1) is option (b): \(x > 0\) and \(y < 0\), which means \(x > y\) => Sufficient.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution

If x and y are integers, is x > y?
(1) x + y > 0
(2) y^x < 0

There are two variables (x and y) in the original condition. In order to match the number of variables and the number of equations, we need 2 equations. Since the condition 1) and 2) each has 1 equation, there is high chance that C is going to be the answer.
Using both the condition 1) and 2), we know that the condition 2) states y^x<0, which means y<0 and x=odd. Also, the condition 1) states x+y>0, which means y<0. So, x>0 and it is always the case that x>y. The answer is ‘yes’ and the conditions are sufficient. Therefore, since the condition 1) and 2) are not sufficient alone, the correct answer is C.

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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How to determine whether the variable approach will it into a particular Q?
There are cases where at first glance ,it seems two options will convert to two equations but they turn out to be a single equation.
Is there any pattern which I can look into ?

Hi,
Since x>y? is a question, we figured out the values of x and y.
Also, in a case where 1)=2) and there is 1 equation, it is 95% that D is an answer and 5% that E is an answer.
Thank you.
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Hi All,

We're told that X and Y are integers. We're asked if X > Y. This is a YES/NO question. We can solve it with a mix of Number Properties and TESTing VALUES.

1) X+Y > 0

With this Fact, we know that at least one of the two variables is positive.
IF.... X = 1, Y = 0 then the answer to the question is YES.
IF.... X = 0, Y = 1 then the answer to the question is NO.
Fact 1 is INSUFFICIENT

2) Y^X < 0

With this Fact, we know that Y CANNOT be positive (since a positive raised to any integer power is a positive). Thus, Y MUST be NEGATIVE and X must be ODD (since raising a negative to an EVEN power will end in a positive result).

IF.... X = 1, Y = -1 then the answer to the question is YES.
IF.... X = -3, Y = -1 then the answer to the question is NO.
Fact 2 is INSUFFICIENT

Combined, we know that there must be at least one positive value and that Y must be negative. By extension, that means that X MUST be positive - and the answer to the question is ALWAYS YES.
Combined, SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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The interesting issue here may be explicitly presented below:

\({y^x} < 0\,\,\,;\,\,\,x + y > 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,x > y\)

To prove that, let´s suppose, on the contrary, that we could have SIMULTANEOUSLY the following:

(1) \({y^x} < 0\)
(2) \(x + y > 0\)
(3) \(x \leqslant y\)

The contradiction follows:
\(\left( 1 \right)\,\,\,\, \Rightarrow \,\,\,\,\,y < 0\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( 3 \right)} \,\,\,x < 0\,\,\,\,\,\,\, \Rightarrow \,\,\,x + y < 0\,\,\,\,\, \Rightarrow \,\,\,\,\left( 2 \right)\,\,{\text{contradicted}}\)
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Dear All

Please can you help me with a general query regarding such questions.

x + y > 0 -

Inequalities is my area of learning
If we subtract y from both sides , will the sign change ?

i.e. x+y<0 when y is subtracted will it become x > -y ?

??

x + y < 0

Subtract y from both sides: (x + y) - y < 0 - y;

x < -y.

9. Inequalities

• Theory
  Inequalities Made Easy!
  Inequalities: Tips and hints
  Arithmetic with Inequalities
  Wavy Line Method Application - Complex Algebraic Inequalities
  Solving Quadratic Inequalities: Graphic Approach
  Inequalities - Quadratic Inequalities
  Graphic approach to problems with inequalities
  Inequalities trick
  INEQUATIONS(Inequalities) Part 1
  INEQUATIONS(Inequalities) Part 2
  
  • Questions
    Inequality and absolute value questions from my collection
    DS Questions
    PS Questions

For more check Ultimate GMAT Quantitative Megathread

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Hi proabhinav,

Adding or Subtracting any number will not affect the sign of the inequality.

Only, when we multiply or divide an inequality by a negative number, we need to change the sign.

So, if x + y < 0, subtracting y from both sides, we get,

• x + y - y < -y
• Thus, x < -y

For more information on Inequalities, go through the following articles

Different methods to solve absolute value equations and inequalities

Solving inequalities- Number Line Method

Wavy Line Method Application - Complex Algebraic Inequalities


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This is a moderately difficult question on Inequalities, with the second statement being slightly tricky to analyse. So you will have to be slightly careful in interpreting it. Option B is one of the trap answers here.

We know that x and y are integers, but we do not know their signs. We are trying to ascertain if x>y. In other words, we are trying to ascertain if (x-y)>0.
From statement I alone, we only know that (x+y) > 0. This is hardly sufficient to uniquely answer whether (x-y)>0.

If x = 3 and y = 2, (x+y) = (3+2) = 5 which is definitely greater than 0; also, for these values of x and y, x>y and the answer to the main question will be a YES.
If x = -1 and y = 2, (x+y) = (-1+2) = 1 which is definitely greater than 0; but, for these values of x and y, x<y and the answer to the main question will be a NO.
Therefore, statement I alone is insufficient. So, answer options A and D can be ruled out. The possible answer options are B, C or E.

From statement II alone, we know that \(y^x\) < 0 which means \(y^x\) is negative. So, we can surely say that y has to be negative i.e. y<0. But we cannot say anything about the sign of x.
If y = -3 and x = 1, \((-3)^1\) <0. Here, x>y; if y = -3 and x = -5, \((-3)^-5\)<0, here x<y.
Therefore, statement II alone is insufficient. So, answer option B can be ruled out. The possible answer options are C or E.

Using the data from statements I and II, from the second statement, we know that y is negative, but x can be positive or negative. But, both x and y cannot be negative since x+y>0. Hence, x has to be positive and y has to be negative.
Therefore, we can surely say that x>y. So, the correct answer option is C.

As mentioned earlier, you have to exercise a fair degree of caution while interpreting statement II, because it can trick you into assuming that x has to be positive. Also, trying out simple values while evaluating the inequalities will help analyse them better..

Hope this helps!
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