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

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Re: If x and y are integers, is x > y?  [#permalink]

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New post 07 Jun 2018, 19:54
I'll try to give the simplest solution I could think of:

---------------------------------------------------------------------------

1. \(x + y > 0\)

Therefore: \(x > -y\)

Example: if x = 3 and y = -2 , this is satisfied. This also satisfied x > y (that is, 3 > -2.)
Counterexample: if x = 1 and y = 1, this is satisfied. However, this violates the x > y equation.
We have found both an example and a counterexample - therefore, 1 by itself is insufficient.

---------------------------------------------------------------------------

2. \(y^x < 0\)

The only way this is true is if x is odd and y is negative.

But there's an infinite number of possibilities where x is odd and y is negative. Therefore, 2 by itself is insufficient.

---------------------------------------------------------------------------

What about equation 1 and equation 2 combined?

If we know y is negative, then the equation in 1:

\(x > -y\)

becomes

\(x > y\) (does that look familiar?)

Therefore, by definition, both 1 and 2 together are necessary to solve the problem.
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Re: If x and y are integers, is x > y?  [#permalink]

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New post 22 Aug 2018, 10:05
Walkabout wrote:
If x and y are integers, is x > y?

(1) x + y > 0
(2) y^x < 0



combining two statements Y must be negative and \(x + y > 0\) means that \(y^x < 0\) could be following values \(-3^5\), \(-3^9\) etc X will always be greater
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Re: If x and y are integers, is x > y?  [#permalink]

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New post 22 Aug 2018, 14:52
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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Re: If x and y are integers, is x > y?  [#permalink]

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New post 16 Sep 2018, 11:12
Hi,
here are my two cents for this questions

Question is IS x>y
or we can rephrase this as IS x-y>0

Lets understand in which conditions is x-y>0

x>0,y>0, and |x|>|y|
x<0,y<0, and |y|>|x|
x>0,y<0, and |x|>|y|
x<0,y>0, this case is not possible

So Statement A:
x+y>0
we can see following cases
x>0, y>0
x<0, y>0 and |y|>|x|
x>0,y<0 and |x|>|y|
x<0,y<0 this case is not possible

We can't conclude anything if x>y
Hence A is insufficient


From Statement B:
\(y^x\)<0
so we have two cases
y is negative and x is positive and odd
y is negative and x is negative and odd

So we can't conclude anything from above if x>y
Hence B is insufficient

Now combining the above two we have

there is only one conditions that satisfies both Statement A & B
which is
x>0 and y<0.
So we can conclude x>y
Hence C is sufficient.
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Re: If x and y are integers, is x > y?  [#permalink]

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New post 08 Oct 2018, 20:38
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 ?

??

Walkabout wrote:
If x and y are integers, is x > y?

(1) x + y > 0
(2) y^x < 0
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Re: If x and y are integers, is x > y?  [#permalink]

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New post 08 Oct 2018, 21:16
1
proabhinav wrote:
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 ?

??

Walkabout wrote:
If x and y are integers, is x > y?

(1) x + y > 0
(2) y^x < 0


x + y < 0

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

x < -y.

9. Inequalities




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If x and y are integers, is x > y?  [#permalink]

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New post 08 Oct 2018, 21:23
1
proabhinav wrote:
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 ?



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

Image

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If x and y are integers, is x > y?  [#permalink]

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New post 16 May 2019, 16:26
I think if you can get used to writing down cases you can get a better understanding than just plugging in numbers to test... my attempt:

(1) x+y > 0
We have 3 cases:
1... x is + & y is + (can be both Yes or No)
2... bigger +x & smaller -y (Yes)
3... smaller -x & bigger +y (No)

Not sufficient.

(2) y^x < 0
We have 3 cases:
1... -y & +Odd number x (Yes)
2... -y & -Odd# x bigger than y(Yes)
3... -y & -Odd# x smaller than/equal to y(No)

Against not sufficient.

(1&2)
Since y must be negative only Case 2 works in (1), concurrently x must be positive so only Case 1 works in (2).
Now there's only 1 way to fulfill both statements so it's sufficient.
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Re: If x and y are integers, is x > y?  [#permalink]

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New post 16 May 2019, 19:16
Walkabout wrote:
If x and y are integers, is x > y?

(1) x + y > 0
(2) y^x < 0



Statement 1.

X Y
2 3
3 2, so not sufficient.

Statement 2.

y has to be a negative Integer and x has to be an odd to satisfy the condition.
Now,
X Y
3 -2 (X>y)
-3 -2 (Y>x)
Not sufficient.

Let's check for condition C (Statement 1+2):-
Since y is a negative integer X has to be a positive (odd) integer >y to satisfy x+y>0
So, C is answer (Combining Statement 1 & 2).
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Re: If x and y are integers, is x > y?  [#permalink]

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New post 19 May 2019, 07:08
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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Re: If x and y are integers, is x > y?   [#permalink] 19 May 2019, 07:08

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