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# Is x^y > 0 ? (1) y = 2 (2) x is an integer

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Re: Is x^y > 0 ? (1) y = 2 (2) x is an integer [#permalink]
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Bunuel wrote:
Is x^y > 0 ?

(1) y = 2
(2) x is an integer

Solution:
Statement1 : x^2 is always greater than or equal to 0. Insufficient.

Statament2 : x is an integer. For x=-1, if y=2,then x^y > 0.If y=3, then x^y < 0. Insufficient.

Combined : Here also we get x^y greater than or equal to 0. Insufficient.

Option E
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Re: Is x^y > 0 ? (1) y = 2 (2) x is an integer [#permalink]
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Bunuel wrote:
Is x^y > 0 ?

(1) y = 2
(2) x is an integer

St 1 :

$$x^2$$ > 0, any positive or negative no. is squared other than zero, it will always be positive, Yes.
When x=0, it will be zero, No
Not sufficient.

St 2 :

If x=-1, y is even, YES,
x=-1, y is odd, NO.
Not sufficient.

Combine 1 & 2
y=2, When x=1 , Yes ...

y=2, when x=0, No ....

Ans E.
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Re: Is x^y > 0 ? (1) y = 2 (2) x is an integer [#permalink]
QZ wrote:
Is $$x^y > 0?$$

1. y = 2
2. x is an integer

Statement 1: Nothing mentioned about $$x$$. If $$x=0$$, then we have $$x^y=0$$ and a NO for our question stem. if $$x$$ is any other number then we have a Yes for our question stem. Insufficient

Statement 2: nothing mentioned about $$y$$. Insufficient

Combining 1 & 2: as explained in statement 1, if $$x=0$$, then we have a No and if $$x$$ is any other number then we have a Yes. Insufficient

Option E
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Re: Is x^y > 0 ? (1) y = 2 (2) x is an integer [#permalink]
Bunuel wrote:
Is x^y > 0 ?

(1) y = 2
(2) x is an integer

both statements alone or together are not sufficient...since in both cases x can simply be zero...and 0 is not > 0
in any other case first statement alone would be sufficient to conclude that x^y is indeed > 0 , except in case where x=0...for statement number 2, that x is an integer we can safely conclude that is insufficient in any possible scenario
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Re: Is x^y > 0 ? (1) y = 2 (2) x is an integer [#permalink]
Bunuel wrote:
Is x^y > 0 ?

(1) y = 2
(2) x is an integer

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have 2 variables ($$x$$ and $$y$$) and 0 equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

If $$x = 1, y = 2$$, then $$x^y = 1^2 = 1 > 0$$ and the answer is 'yes'.
If $$x = 0, y = 2$$, then $$x^y = 0^2 = 0$$ and the answer is 'no'.

Since both conditions together do not yield a unique solution, they are not sufficient.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: Is x^y > 0 ? (1) y = 2 (2) x is an integer [#permalink]
Bunuel wrote:
Is x^y > 0 ?

(1) y = 2
(2) x is an integer

#2. x can be 0 (then x^y = 0 ), x can be 1 (then x^y > 0) Insufficient

#1. 1^2 (then x^y > 0) but 0^2 (then x^y = 0 ), Insufficient

1+2 still doesn't solve.

0^2 is 0 but anything else^2 is >0, so still Insufficient. (0 is an integer) E

[-1,1] Anomaly, my fav. GMAC questions.
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Re: Is x^y > 0 ? (1) y = 2 (2) x is an integer [#permalink]
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Re: Is x^y > 0 ? (1) y = 2 (2) x is an integer [#permalink]
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