Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
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.
close
42% (01:03) correct 
