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16 Sep 2014, 00:47



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24 Mar 2015, 11:02
Bunuel wrote: Official Solution:
The question asks: is \(\frac{x}{y}+\frac{y}{x} \gt 2\)? Or: is \(\frac{x^2+y^2}{xy} \gt 2\)? Since both unknowns are positive then we can safely multiply by \(xy\): Is \(x^2+y^2 \gt 2xy\)? Is \(x^22xy+y^2\gt 0\)? Is \((xy)^2 \gt 0\)? Now, if \(x\) does not equal \(y\) the answer to this question will be YES, but if \(x=y\), then the answer will be NO, since in this case \((xy)^2=0\). (1) \(x\) does not equal \(y\). Directly answers the question. Sufficient. (2)\(x\) and \(y\) do not share any common divisors except 1. If \(x=y=1\) then the answer is NO, but if \(x=1\) and \(y=2\), then the answer is YES. Not sufficient.
Answer: A Hi Bunuel Doesnt Stmnt 2 mean that x & y are PRIME ? In that case this statement is Sufficient Pls clarify Thanks



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24 Mar 2015, 12:11
buddyisraelgmat wrote: Bunuel wrote: Official Solution:
The question asks: is \(\frac{x}{y}+\frac{y}{x} \gt 2\)? Or: is \(\frac{x^2+y^2}{xy} \gt 2\)? Since both unknowns are positive then we can safely multiply by \(xy\): Is \(x^2+y^2 \gt 2xy\)? Is \(x^22xy+y^2\gt 0\)? Is \((xy)^2 \gt 0\)? Now, if \(x\) does not equal \(y\) the answer to this question will be YES, but if \(x=y\), then the answer will be NO, since in this case \((xy)^2=0\). (1) \(x\) does not equal \(y\). Directly answers the question. Sufficient. (2)\(x\) and \(y\) do not share any common divisors except 1. If \(x=y=1\) then the answer is NO, but if \(x=1\) and \(y=2\), then the answer is YES. Not sufficient.
Answer: A Hi Bunuel Doesnt Stmnt 2 mean that x & y are PRIME ? In that case this statement is Sufficient Pls clarify Thanks There are two examples in the solution which show that x and y are not necessarily primes.
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07 Jan 2016, 14:52
I kinda get lost here.
Is (x−y)2>0? Now, if x does not equal y the answer to this question will be YES, but if x=y, then the answer will be NO, since in this case (x−y)2=0.
(1) x does not equal y. Directly answers the question. Sufficient.
(2)x and y do not share any common divisors except 1. If x=y=1 then the answer is NO, but if x=1 and y=2, then the answer is YES. Not sufficient.
And how does statement not say x and y are not prime?



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08 Jan 2016, 01:34



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08 Jan 2016, 01:56
kkahuja91 wrote: I kinda get lost here.
Is (x−y)2>0? Now, if x does not equal y the answer to this question will be YES, but if x=y, then the answer will be NO, since in this case (x−y)2=0.
(1) x does not equal y. Directly answers the question. Sufficient.
(2)x and y do not share any common divisors except 1. If x=y=1 then the answer is NO, but if x=1 and y=2, then the answer is YES. Not sufficient.
And how does statement not say x and y are not prime? Hi, firstly the statement two does not say that x and y are prime, even if it included the info that x and y are not same.. It would have just meant that the two are prime to each other, say 3,8 or 5,7 etcNow,if you intend to ask the way statement 2 would have been written to say that X and y are not prime... Two scenarios.. X and Y are prime.. 1) X and y are two different positive integers greater than 1, having only two factors( or having only one factor other than 1) X and Y are not prime.. 1) X and Y are two integers having more than One common factor.. I could make this out as your query..
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21 Aug 2016, 04:38
The definition of a prime number is: a number that doesn't have a factor apart from 1 and itself. The "and itself" part is very important as it shows that all prime numbers have EXACTLY 2 factors while 1 just has 1 factor and thus 1 is not a prime number.
Hope it helps.



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25 Dec 2016, 09:12
kindly explain how it could be A from S1 CASE 1:: if X=12 and Y =13 then,x/y+y/x is not >2 CASE 2:: X=1,Y=2 then X/Y+Y/X is >2 NOT SUFFICIENT



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05 Oct 2017, 18:37
Quick question:
Statement 2: would imply that the numbers are consecutive, BUT only if the initial question stem characterized X and Y as two distinct positive integers, correct?
I picked D for this question, upon review I understand why it is not the correct answer. I would like to make sure my understanding of statement two is correct.
Thank you very much.



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05 Oct 2017, 21:10
georgetags10 wrote: Quick question:
Statement 2: would imply that the numbers are consecutive, BUT only if the initial question stem characterized X and Y as two distinct positive integers, correct?
I picked D for this question, upon review I understand why it is not the correct answer. I would like to make sure my understanding of statement two is correct.
Thank you very much. No. Even then, x and y have not to be consecutive integers, just coprime. For example, 1 and any other integer, any two distinct primes (for example 2 and 7), ...
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Re: M1216
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07 May 2018, 05:57
Bunuel wrote: Official Solution:
The question asks: is \(\frac{x}{y}+\frac{y}{x} \gt 2\)? Or: is \(\frac{x^2+y^2}{xy} \gt 2\)? Since both unknowns are positive then we can safely multiply by \(xy\): Is \(x^2+y^2 \gt 2xy\)? Is \(x^22xy+y^2\gt 0\)? Is \((xy)^2 \gt 0\)? Now, if \(x\) does not equal \(y\) the answer to this question will be YES, but if \(x=y\), then the answer will be NO, since in this case \((xy)^2=0\). (1) \(x\) does not equal \(y\). Directly answers the question. Sufficient. (2) \(x\) and \(y\) do not share any common divisors except 1. If \(x=y=1\) then the answer is NO, but if \(x=1\) and \(y=2\), then the answer is YES. Not sufficient.
Answer: A I did the square root of (xy)^2>0 to get x>y. Can we not do that in this case?



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07 May 2018, 06:01
rhnbansal wrote: Bunuel wrote: Official Solution:
The question asks: is \(\frac{x}{y}+\frac{y}{x} \gt 2\)? Or: is \(\frac{x^2+y^2}{xy} \gt 2\)? Since both unknowns are positive then we can safely multiply by \(xy\): Is \(x^2+y^2 \gt 2xy\)? Is \(x^22xy+y^2\gt 0\)? Is \((xy)^2 \gt 0\)? Now, if \(x\) does not equal \(y\) the answer to this question will be YES, but if \(x=y\), then the answer will be NO, since in this case \((xy)^2=0\). (1) \(x\) does not equal \(y\). Directly answers the question. Sufficient. (2) \(x\) and \(y\) do not share any common divisors except 1. If \(x=y=1\) then the answer is NO, but if \(x=1\) and \(y=2\), then the answer is YES. Not sufficient.
Answer: A I did the square root of (xy)^2>0 to get x>y. Can we not do that in this case? The point is that \(\sqrt{x^2}=x\). So, if you take the square root from \((xy)^2 \gt 0\) you'll get \(xy \gt 0\), which is true for all values of x and y but the case when x = y (the same conclusion as we got in the solution).
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Re: M1216
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07 May 2018, 06:15
Bunuel wrote: rhnbansal wrote: Bunuel wrote: Official Solution:
The question asks: is \(\frac{x}{y}+\frac{y}{x} \gt 2\)? Or: is \(\frac{x^2+y^2}{xy} \gt 2\)? Since both unknowns are positive then we can safely multiply by \(xy\): Is \(x^2+y^2 \gt 2xy\)? Is \(x^22xy+y^2\gt 0\)? Is \((xy)^2 \gt 0\)? Now, if \(x\) does not equal \(y\) the answer to this question will be YES, but if \(x=y\), then the answer will be NO, since in this case \((xy)^2=0\). (1) \(x\) does not equal \(y\). Directly answers the question. Sufficient. (2) \(x\) and \(y\) do not share any common divisors except 1. If \(x=y=1\) then the answer is NO, but if \(x=1\) and \(y=2\), then the answer is YES. Not sufficient.
Answer: A I did the square root of (xy)^2>0 to get x>y. Can we not do that in this case? The point is that \(\sqrt{x^2}=x\). So, if you take the square root from \((xy)^2 \gt 0\) you'll get \(xy \gt 0\), which is true for all values of x and y but the case when x = y (the same conclusion as we got in the solution). Understood. I assumed that just because they both are positive integers, we could do the square root and get x>y. Totally missed out on the fact that it could be zero as well. Thank you so much!










