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M12-16

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Bunuel
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M12-16 [#permalink] New post   16 Sep 2014, 00:47
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If \(x\) and \(y\) are positive integers, is \( \frac{x}{y}+ \frac{y}{x} \gt 2\)?


(1) \(x\) does not equal \(y\)

(2) \(x\) and \(y\) have no common factor other than 1.
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Bunuel
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Re M12-16 [#permalink] New post   16 Sep 2014, 00:47
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Official Solution:


If \(x\) and \(y\) are positive integers, is \( \frac{x}{y}+ \frac{y}{x} \gt 2\)?

Simplify the question:

Is \(\frac{x}{y}+\frac{y}{x} > 2\)?

Is \(\frac{x^2+y^2}{xy} > 2\)?

Since both variables are positive, we can safely multiply by \(xy\):

Is \(x^2+y^2 > 2xy\)?

Is \(x^2-2xy+y^2 > 0\)?

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 \(y\) have no common factor other than 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
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Bunuel
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Re: M12-16 [#permalink] New post   07 May 2018, 06:01
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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 \(x^2-2xy+y^2\gt 0\)?

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).
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georgetags10
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Re: M12-16 [#permalink] New post   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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Bunuel
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Re: M12-16 [#permalink] New post   05 Oct 2017, 21:10
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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 co-prime. For example, 1 and any other integer, any two distinct primes (for example 2 and 7), ...
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rhnbansal
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Re: M12-16 [#permalink] New post   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 \(x^2-2xy+y^2\gt 0\)?

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?
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Re: M12-16 [#permalink] New post   17 Oct 2020, 06:44
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 \(x^2-2xy+y^2\gt 0\)?

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


Hello Bunuel

At this step, Is \((x-y)^2 \gt 0\) ?
I went further and arrived at
Is \((x-y) \gt 0\) ?
--> Is \(x \gt y\) ?

Statement 1 says that x not equal to y
Thus x < y or x > y
Hence not sufficient

Statement 2 says that x and y donot share any common divisors except 1
So let x be 2, y be 3 x<y
OR
let x be 3, y be 2 x>y
Hence not sufficient

Option E

Please tell me what is wrong with my approach.
TIA!
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Re: M12-16 [#permalink] New post   17 Oct 2020, 07:19
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ProfChaos 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 \(x^2-2xy+y^2\gt 0\)?

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


Hello Bunuel

At this step, Is \((x-y)^2 \gt 0\) ?
I went further and arrived at
Is \((x-y) \gt 0\) ?
--> Is \(x \gt y\) ?

Statement 1 says that x not equal to y
Thus x < y or x > y
Hence not sufficient

Statement 2 says that x and y donot share any common divisors except 1
So let x be 2, y be 3 x<y
OR
let x be 3, y be 2 x>y
Hence not sufficient

Option E

Please tell me what is wrong with my approach.
TIA!


The point is "is \((x-y)^2 \gt 0\)?" after taking the square root gives "is \(|x - y| > 0\)?" (recall that \(\sqrt{a^2}=|a|\)). So, \((x-y)^2 \gt 0\) even if x - y < 0, for example, if x - y = -2, then \((x-y)^2=4 \gt 0\). The only case when \((x-y)^2 \gt 0\) is not true is when x = y.

Hope its clear.
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Re: M12-16 [#permalink] New post   17 Sep 2022, 06:07
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 \(x^2-2xy+y^2\gt 0\)?

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).


Hi Buneul ,

Please correct my understanding here , as sqrt (x) = |x| . Here it would be |x-y|, doesnt it mean that x is not equal to y for x>0 and x<0 . Can we derive this from the stem , and hence S2 will also be sufficient ?
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Re: M12-16 [#permalink] New post   17 Sep 2022, 06:27
Expert Reply
Magni03 wrote:
Bunuel wrote:
rhnbansal wrote:
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).


Hi Buneul ,

Please correct my understanding here , as sqrt (x) = |x| . Here it would be |x-y|, doesnt it mean that x is not equal to y for x>0 and x<0 . Can we derive this from the stem , and hence S2 will also be sufficient ?


From the stem we only know that x and y are positive integers. NOTHING more.

(2) says that x and y are co-prime. So, for (2) we know that x and y are positive integers that are co-prime. The solution on page 1 gives an example: if \(x=y=1\) then the answer is NO (\(\frac{x}{y}+ \frac{y}{x}\) is NOT greater than 2), but if \(x=1\) and \(y=2\), then the answer is YES (\(\frac{x}{y}+ \frac{y}{x}\) IS greater than 2). Not sufficient.
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Re: M12-16 [#permalink] New post   20 Apr 2023, 03:48
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M12-16 [#permalink]
20 Apr 2023, 03:48
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