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Re: S96-02 [#permalink]
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agarwalneha1 wrote:
I didn't understand the logic of the solution. By knowing |x| = |y|, how can we find out x+y? Also, when it is mentioned that x<y.


Please read the first sentence of the question: If \(x\) and \(y\) are integers and \(x \lt y\),...
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Re: S96-02 [#permalink]
Bunuel wrote:
agarwalneha1 wrote:
I didn't understand the logic of the solution. By knowing |x| = |y|, how can we find out x+y? Also, when it is mentioned that x<y.


Please read the first sentence of the question: If \(x\) and \(y\) are integers and \(x \lt y\),...


Thanks.. I got the point.
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Re: S96-02 [#permalink]
I still don't understand how the second statement in sufficient.
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Re: S96-02 [#permalink]
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A1996J wrote:
I still don't understand how the second statement in sufficient.


If x and y are integers and x<y, what is the value of x+y?

(1) x^y=4 --> as x and y are integers and x<y then only possible solution is (-2)^2=4 (other integer solutions for x^y=4 are: 2^2=4 and 4^1=4) --> x+y=-2+2=0. Sufficient.

(2) |x|=|y| --> as also x<y then they have opposite signs (x<0<y, so |x|=-x and|y|=y) --> -x=y --> x+y=0. Sufficient.

Answer: D.

Hope it helps.
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Re: S96-02 [#permalink]
I love this question, got it wrong because I didnt understand the logic of statement 2. Cheers
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Re: S96-02 [#permalink]
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I think this is a high-quality question and I agree with explanation.
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Re: S96-02 [#permalink]
I think this is a high-quality question and I agree with explanation. interesting ques, though got it wrong(chose option A) but learned a point.
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Re: S96-02 [#permalink]
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i had the question wrong but I really appreciate the solution
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Re: S96-02 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: S96-02 [#permalink]
If x and y are integers and x<y, what is the value of x+y?

(1) \(x^y\)=4
The only number that when squared gives 4 is 2. \(2^2\) = 4, but since x < y, the only option we have is \(-2^2\)

x=-2 and y=2, so x+y = 0
Sufficient

(2) |x|=|y|
With this option we know that x and y are same numbers and from x < y, we know that they are opposite in sign. So x+y =0
Sufficient

Ans D
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Re: S96-02 [#permalink]
I think this is a poor-quality question and I don't agree with the explanation. It is given in the question that x<y, We can't assume either x=2 and y=2, or x=4 and y=1. I believe this breaches the question statement. Let me know if I have wrong understanding.
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Re: S96-02 [#permalink]
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shivammann wrote:
I think this is a poor-quality question and I don't agree with the explanation. It is given in the question that x<y, We can't assume either x=2 and y=2, or x=4 and y=1. I believe this breaches the question statement. Let me know if I have wrong understanding.


The solutions above do take into account that x < y. No?
https://gmatclub.com/forum/s96-184673.html#p1750025
https://gmatclub.com/forum/s96-184673.html#p1416014
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Re: S96-02 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: S96-02 [#permalink]
A very nice question
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Re: S96-02 [#permalink]
S1: x and y can take following assumptions:
Case 1: x = -4; y=1 (x<y)
Case 2: x = -2; y=2 (x<y)

Statement 1 is not sufficient clearly! Please correct me where I'm going wrong. I marked 'C' as the correct choice.
---------------------
Edit 2: My bad, read the question wrong!
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Re S96-02 [#permalink]
I think this the explanation isn't clear enough, please elaborate.
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Re S96-02 [#permalink]
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