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15 Sep 2014, 23:47



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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 \(x^22xy+y^2\gt 0\)? 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, 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 \(x^22xy+y^2\gt 0\)? 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, 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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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 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, 03:38
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, 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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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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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 coprime. For example, 1 and any other integer, any two distinct primes (for example 2 and 7), ...
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07 May 2018, 04:57
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, 05:01
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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07 May 2018, 05:15
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!



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28 Nov 2018, 10:06
while simplifying we have got is (xy)>0? is x>y?
1 says x does not equal y how this can be correct? whether this statement haven't say x is greater or y is smaller. Please tell me where i fall.



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29 Nov 2018, 06:52
Bunuel wrote: shivam5511 wrote: while simplifying we have got is (xy)>0? is x>y?
1 says x does not equal y how this can be correct? whether this statement haven't say x is greater or y is smaller. Please tell me where i fall. Please read carefully. After simplifying we got "is (x  y) ^2 > 0" Exactly i am asking the same thing. (xy)^2>0 we can also right it as (xy) (x+y) >0 so x and y is positive integer then, we arrived at is (xy)>0? again, is x>y? statement 1  x does not equal y ( This statement tell us x not equal to y. But nothing to say about x is greater than y or smaller than y). Please tell me where i fall. while simplifying is there any probability of x=y?



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shivam5511 wrote: Bunuel wrote: shivam5511 wrote: while simplifying we have got is (xy)>0? is x>y?
1 says x does not equal y how this can be correct? whether this statement haven't say x is greater or y is smaller. Please tell me where i fall. Please read carefully. After simplifying we got "is (x  y) ^2 > 0" Exactly i am asking the same thing. (xy)^2>0 we can also right it as (xy) (x+y) >0 so x and y is positive integer then, we arrived at is (xy)>0? again, is x>y? statement 1  x does not equal y ( This statement tell us x not equal to y. But nothing to say about x is greater than y or smaller than y). Please tell me where i fall. while simplifying is there any probability of x=y? Here is on in a million trick for you  PLUG NUMBERS to check your math and your answers. It can also help you solve the question. Let's say X=10 and Y=3 1. (xy)^2 = 49 2. (xy)(x+y)=91 3. x^22xy+y^2 = 10060+9 = 49 4. (xy)(xy) = 49 You spot your mistake? It is the fourth grade formula. This is a very basic mistake  it shows you have a weak base  you need to go way back to fix it.
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29 Nov 2018, 08:57
shivam5511 wrote: Bunuel wrote: shivam5511 wrote: while simplifying we have got is (xy)>0? is x>y?
1 says x does not equal y how this can be correct? whether this statement haven't say x is greater or y is smaller. Please tell me where i fall. Please read carefully. After simplifying we got "is (x  y) ^2 > 0" Exactly i am asking the same thing. (xy)^2>0 we can also right it as (xy) (x+y) >0so x and y is positive integer then, we arrived at is (xy)>0? again, is x>y? statement 1  x does not equal y ( This statement tell us x not equal to y. But nothing to say about x is greater than y or smaller than y). Please tell me where i fall. while simplifying is there any probability of x=y? Red part is not correct. \((xy)^2 = x^22xy+y^2\) NOT \((xy) (x+y)\), which in turn equals to \(x^2  y^2\).
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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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