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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 \(x^2-2xy+y^2\gt 0\)?
Is \((x-y)^2 \gt 0\)? 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 \((x-y)^2=0\).
(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.
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 \(x^2-2xy+y^2\gt 0\)?
Is \((x-y)^2 \gt 0\)? 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 \((x-y)^2=0\).
(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.
Answer: A
Hi Bunuel
Doesnt Stmnt 2 mean that x & y are PRIME ? In that case this statement is Sufficient
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 \(x^2-2xy+y^2\gt 0\)?
Is \((x-y)^2 \gt 0\)? 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 \((x-y)^2=0\).
(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.
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.
_________________
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?
Your question is not clear. Only thing we want to know to answer the question is whether x = y. Why do we need to know whether w and y re primes?
_________________
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 etc
Now,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..
_________________
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.
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.
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 co-prime. For example, 1 and any other integer, any two distinct primes (for example 2 and 7), ...
_________________
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 \(x^2-2xy+y^2\gt 0\)?
Is \((x-y)^2 \gt 0\)? 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 \((x-y)^2=0\).
(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.
Answer: A
I did the square root of (x-y)^2>0 to get x>y. Can we not do that in this case?
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 \(x^2-2xy+y^2\gt 0\)?
Is \((x-y)^2 \gt 0\)? 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 \((x-y)^2=0\).
(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.
Answer: A
I did the square root of (x-y)^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 \((x-y)^2 \gt 0\) you'll get \(|x-y| \gt 0\), 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).
_________________
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 \(x^2-2xy+y^2\gt 0\)?
Is \((x-y)^2 \gt 0\)? 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 \((x-y)^2=0\).
(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.
Answer: A
I did the square root of (x-y)^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 \((x-y)^2 \gt 0\) you'll get \(|x-y| \gt 0\), 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.
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42% (01:01) correct 
