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M12-16

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M12-16 [#permalink]

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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 \(y\) do not share any common divisors except 1
[Reveal] Spoiler: OA

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Re M12-16 [#permalink]

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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
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Re: M12-16 [#permalink]

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New post 24 Mar 2015, 10:02
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 \(x^2-2xy+y^2\gt 0\)?

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
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Re: M12-16 [#permalink]

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New post 24 Mar 2015, 11:11
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 \(x^2-2xy+y^2\gt 0\)?

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.
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Re: M12-16 [#permalink]

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New post 07 Jan 2016, 13:52
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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Re: M12-16 [#permalink]

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New post 08 Jan 2016, 00:34
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?


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?
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Re: M12-16 [#permalink]

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New post 08 Jan 2016, 00:56
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 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..
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Re: M12-16 [#permalink]

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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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Re: M12-16 [#permalink]

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New post 25 Dec 2016, 08:12
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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Re: M12-16 [#permalink]

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New post 25 Dec 2016, 08:16
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Re: M12-16 [#permalink]

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New post 05 Oct 2017, 17:37
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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Re: M12-16 [#permalink]

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New post 05 Oct 2017, 20:10
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 co-prime. For example, 1 and any other integer, any two distinct primes (for example 2 and 7), ...
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M12-16   [#permalink] 05 Oct 2017, 20:10
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