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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2
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28 Oct 2012, 09:44
Yes there is something dicey. In the solution he has added (1) and (2) and proved that X^2+y^2>100. Request some expert to reply!
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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2
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29 Oct 2012, 04:41



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Re: Is x^2 + y^2 > 100?
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31 Oct 2012, 15:49
It's B for me.
\((x + y)^2 = x^2 + 2xy + y^2\)
The largest possible value \(2xy\) can reach is \((x + y)^2/2\), that only occurs when \(x = y\).
When \(x = y\), it turns out that \(x^2 + 2xy + y^2 > 200\) is \(x^2 + 2xx + x^2 > 200\), which can be rearranged in \(2x^2 + 2x^2 > 200\), meaning that \(x^2 + y^2\) is at least half of the stated value, while \(2xy\) can be at most the other half.
Any value above 200 will require that \(x^2 + y^2 > 100\), making 2) sufficient.



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Re: Is x^2 + y^2 > 100?
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18 Feb 2013, 19:58
Bunuel wrote: (2) (x + y)^2 > 200 > \(x^2+2xy+y^2>200\). Now, as \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) then \(x^2+y^2\geq{2xy}\) Hi Bunuel, could you please explain why you followed with (xy)^2 instead of (x+y)^2? Shouldn't (xy)^2 be distributed as x^22xy+y^2 in which case it will be different from the original expression? Also, when you transfer 2xy to the other side from x^2+2xy+y^2, why do you keep it positive?



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Re: Is x^2 + y^2 > 100?
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19 Feb 2013, 05:43
LinaNY wrote: Bunuel wrote: (2) (x + y)^2 > 200 > \(x^2+2xy+y^2>200\). Now, as \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) then \(x^2+y^2\geq{2xy}\) Hi Bunuel, could you please explain why you followed with (xy)^2 instead of (x+y)^2? Shouldn't (xy)^2 be distributed as x^22xy+y^2 in which case it will be different from the original expression? Also, when you transfer 2xy to the other side from x^2+2xy+y^2, why do you keep it positive? We have \((x + y)^2 > 200\) which is the same as \(x^2+2xy+y^2>200\). Now, we need to find the relationship between x^2+y^2 and 2xy. Next, we know that \((xy)^2\geq{0}\) > \(x^22xy+y^2\geq{0}\) > \(x^2+y^2\geq{2xy}\). So, we can safely substitute \(2xy\) with \(x^2+y^2\) in \(x^2+2xy+y^2>200\) (as \(x^2+y^2\) is at least as big as \(2xy\) then the inequality will still hold true) > \(x^2+(x^2+y^2)+y^2>200\) > \(2(x^2+y^2)>200\) > \(x^2+y^2>100\). Sufficient. Hope it's clear.
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Re: Is x^2 + y^2 > 100?
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19 Feb 2013, 21:14
Thanks, it's clear now!



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Re: Is x^2 + y^2 > 100?
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30 Mar 2013, 06:32
Bunuel wrote: arvindg wrote: Problem source: Veritas Practice Test Is x^2 + y^2 > 100? (1) 2xy < 100 (2) (x + y)^2 > 200 Is x^2 + y^2 > 100?(1) 2xy < 100 > clearly insufficient: if \(x=y=0\) then the answer will be NO but if \(x=10\) and \(y=10\) then the answer will be YES. (2) (x + y)^2 > 200 > \(x^2+2xy+y^2>200\). Now, as \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) then \(x^2+y^2\geq{2xy}\) so we can safely substitute \(2xy\) with \(x^2+y^2\) (as \(x^2+y^2\) is at least as big as \(2xy\) then the inequality will still hold true) > \(x^2+(x^2+y^2)+y^2>200\) > \(2(x^2+y^2)>200\) > \(x^2+y^2>100\). Sufficient. Answer: B. Are you sure the OA is C? Brunel I have one confusion, As you said that x^2+y^2 is at least as big as 2xy and so 2(x^2 + y^2)>200 but if x^2 +y^2 is = 2xy then wont 2(x^2+y^2)<200 as 4xy<200. (2xy<100, First statement, two statements cannot contradice) So till the time we are not sure how big is the difference between x^2+y^2 and 2xy can we really say if 2(x^2+y^2)>200



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Re: Is x^2 + y^2 > 100?
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31 Mar 2013, 08:46
anujtsingh wrote: Bunuel wrote: arvindg wrote: Problem source: Veritas Practice Test Is x^2 + y^2 > 100? (1) 2xy < 100 (2) (x + y)^2 > 200 Is x^2 + y^2 > 100?(1) 2xy < 100 > clearly insufficient: if \(x=y=0\) then the answer will be NO but if \(x=10\) and \(y=10\) then the answer will be YES. (2) (x + y)^2 > 200 > \(x^2+2xy+y^2>200\). Now, as \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) then \(x^2+y^2\geq{2xy}\) so we can safely substitute \(2xy\) with \(x^2+y^2\) (as \(x^2+y^2\) is at least as big as \(2xy\) then the inequality will still hold true) > \(x^2+(x^2+y^2)+y^2>200\) > \(2(x^2+y^2)>200\) > \(x^2+y^2>100\). Sufficient. Answer: B. Are you sure the OA is C? Brunel I have one confusion, As you said that x^2+y^2 is at least as big as 2xy and so 2(x^2 + y^2)>200 but if x^2 +y^2 is = 2xy then wont 2(x^2+y^2)<200 as 4xy<200. (2xy<100, First statement, two statements cannot contradice) So till the time we are not sure how big is the difference between x^2+y^2 and 2xy can we really say if 2(x^2+y^2)>200 No. If \(x^2 +y^2 = 2xy\), then we can simply substitute 2xy to get the same: \(x^2+(x^2+y^2)+y^2>200\) > \(2(x^2+y^2)>200\). How did you get 2(x^2+y^2) <200?
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Re: Is x^2 + y^2 > 100?
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31 Mar 2013, 12:34
If 2(x^2+y^2)>200 and (x^2+y^2)=2xy then 2(2xy)>200 and 2xy>100 but the statement one says that 2xy<100. So even though we are not considering 1st statement, the answer cannot contradict the first statement.
Pls correct me if I am wrong



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Re: Is x^2 + y^2 > 100?
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Re: Is x^2 + y^2 > 100?
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Re: Is x^2 + y^2 > 100?
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14 Jan 2014, 07:45
Bunuel wrote: arvindg wrote: Problem source: Veritas Practice Test Is x^2 + y^2 > 100? (1) 2xy < 100 (2) (x + y)^2 > 200 Is x^2 + y^2 > 100?(1) 2xy < 100 > clearly insufficient: if \(x=y=0\) then the answer will be NO but if \(x=10\) and \(y=10\) then the answer will be YES. (2) (x + y)^2 > 200 > \(x^2+2xy+y^2>200\). Now, as \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) then \(x^2+y^2\geq{2xy}\) so we can safely substitute \(2xy\) with \(x^2+y^2\) (as \(x^2+y^2\) is at least as big as \(2xy\) then the inequality will still hold true) > \(x^2+(x^2+y^2)+y^2>200\) > \(2(x^2+y^2)>200\) > \(x^2+y^2>100\). Sufficient. Answer: B. Are you sure the OA is C? Hello Bunuel, I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition. So I guess the OA is correct i.e. C ...



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Re: Is x^2 + y^2 > 100?
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14 Jan 2014, 08:03
joe26219 wrote: Bunuel wrote: arvindg wrote: Problem source: Veritas Practice Test Is x^2 + y^2 > 100? (1) 2xy < 100 (2) (x + y)^2 > 200 Is x^2 + y^2 > 100?(1) 2xy < 100 > clearly insufficient: if \(x=y=0\) then the answer will be NO but if \(x=10\) and \(y=10\) then the answer will be YES. (2) (x + y)^2 > 200 > \(x^2+2xy+y^2>200\). Now, as \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) then \(x^2+y^2\geq{2xy}\) so we can safely substitute \(2xy\) with \(x^2+y^2\) (as \(x^2+y^2\) is at least as big as \(2xy\) then the inequality will still hold true) > \(x^2+(x^2+y^2)+y^2>200\) > \(2(x^2+y^2)>200\) > \(x^2+y^2>100\). Sufficient. Answer: B. Are you sure the OA is C? Hello Bunuel, I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition. So I guess the OA is correct i.e. C ... There is no flaw in my reasoning. Statement (2) says: \(x^2+2xy+y^2>200\). Next, we know that \(x^2+y^2\geq{2xy}\) is true for any values of \(x\) and \(y\). So we can manipulate and substitute \(2xy\) with \(x^2+y^2\) in (2) (because \(x^2+y^2\) is at least as large as \(2xy\)): \(x^2+(x^2+y^2)+y^2>200\) > \(x^2+y^2>100\). By the way the OA is B, not C. VeritasPrep corrected it: isx2y108343.html#p860573
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Re: Is x^2 + y^2 > 100?
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14 Jan 2014, 21:30
joe26219 wrote: Bunuel wrote: arvindg wrote: Problem source: Veritas Practice Test Is x^2 + y^2 > 100? (1) 2xy < 100 (2) (x + y)^2 > 200 Is x^2 + y^2 > 100?(1) 2xy < 100 > clearly insufficient: if \(x=y=0\) then the answer will be NO but if \(x=10\) and \(y=10\) then the answer will be YES. (2) (x + y)^2 > 200 > \(x^2+2xy+y^2>200\). Now, as \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) then \(x^2+y^2\geq{2xy}\) so we can safely substitute \(2xy\) with \(x^2+y^2\) (as \(x^2+y^2\) is at least as big as \(2xy\) then the inequality will still hold true) > \(x^2+(x^2+y^2)+y^2>200\) > \(2(x^2+y^2)>200\) > \(x^2+y^2>100\). Sufficient. Answer: B. Are you sure the OA is C? Hello Bunuel, I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition. So I guess the OA is correct i.e. C ... To put it in words, think of it this way: Is x^2 + y^2 > 100? (2) (x + y)^2 > 200 which means: x^2 + y^2 + 2xy > 200 Now you know that 2xy is less than or equal to x^2 + y^2. If 2xy is equal to \((x^2 + y^2)\), \((x^2 + y^2)\) will be greater than 100 since the total sum is greater than 200. If 2xy is less than \((x^2 + y^2)\), then anyway \((x^2 + y^2)\) will be greater than 100 (which is half of 200). So statement 2 is sufficient alone.
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Re: Is x^2 + y^2 > 100?
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16 Jan 2014, 04:04
Bunuel wrote: joe26219 wrote: Bunuel wrote: Is x^2 + y^2 > 100?
(1) 2xy < 100 > clearly insufficient: if \(x=y=0\) then the answer will be NO but if \(x=10\) and \(y=10\) then the answer will be YES.
(2) (x + y)^2 > 200 > \(x^2+2xy+y^2>200\). Now, as \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) then \(x^2+y^2\geq{2xy}\) so we can safely substitute \(2xy\) with \(x^2+y^2\) (as \(x^2+y^2\) is at least as big as \(2xy\) then the inequality will still hold true) > \(x^2+(x^2+y^2)+y^2>200\) > \(2(x^2+y^2)>200\) > \(x^2+y^2>100\). Sufficient.
Answer: B.
Are you sure the OA is C?
Hello Bunuel, I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition. So I guess the OA is correct i.e. C ... There is no flaw in my reasoning. Statement (2) says: \(x^2+2xy+y^2>200\). Next, we know that \(x^2+y^2\geq{2xy}\) is true for any values of \(x\) and \(y\). So we can manipulate and substitute \(2xy\) with \(x^2+y^2\) in (2) (because \(x^2+y^2\) is at least as large as \(2xy\)): \(x^2+(x^2+y^2)+y^2>200\) > \(x^2+y^2>100\). By the way the OA is B, not C. VeritasPrep corrected it: isx2y108343.html#p860573Thank you Bunuel.Got it.Your explanation was really helpful (the highlighted line esp.) I guess any value of 2xy should be assumed to satisfy the given condition, as it is given.



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Re: Is x^2 + y^2 > 100?
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10 May 2014, 08:00
Am still confused. Consider, x^2 + y^2 = 200 for the minimum value. That means, x + y is minimum 200 ^ 0.5 which means for a minimum sum of x + y, x has to be (200^0.5)/2 which gives x = 7.07. Sum up x^2 + y^2 at x = y = 7.07 we get x^2+ y^2 = approx. 98.30. Can someone please explain where I am going wrong.



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Re: Is x^2 + y^2 > 100?
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Re: Is x^2 + y^2 > 100?
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17 Jun 2015, 06:11
Bunuel wrote: Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HEREIs X^2+Y^2>100 .. > is [(x+y)^2 + (xy)^2]/2 >100 So we can change the question stem to.. is (x+y)^2+(xy)^2>200? Statement A can in invalidated easily. Use different values of x and y. statement B says that (X+Y)^2>200 Substitute statement B in question stem  is Something >200 [(X+Y)^2] + something >=0 [(XY)^2] >200 .... Ans is obviously yes. So B is the correct Ans. I hope this makes sense!
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Is x^2 + y^2 > 100?
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15 Jan 2016, 02:06
VeritasPrepKarishma wrote: Bunuel wrote: VeritasPrepKarishma wrote: Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200 Is x^2 + y^2 > 100?(1) 2xy < 100 > clearly insufficient: if \(x=y=0\) then the answer will be NO but if \(x=10\) and \(y=10\) then the answer will be YES. (2) (x + y)^2 > 200 > \(x^2+2xy+y^2>200\). Now, as \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) then \(x^2+y^2\geq{2xy}\) so we can safely substitute \(2xy\) with \(x^2+y^2\) (as \(x^2+y^2\) is at least as big as \(2xy\) then the inequality will still hold true) > \(x^2+(x^2+y^2)+y^2>200\) > \(2(x^2+y^2)>200\) > \(x^2+y^2>100\). Sufficient. Answer: B. Are you sure the OA is C? To put it in words, think of it this way: Is x^2 + y^2 > 100? (2) (x + y)^2 > 200 which means: x^2 + y^2 + 2xy > 200 Now you know that 2xy is less than or equal to x^2 + y^2. If 2xy is equal to \((x^2 + y^2)\), \((x^2 + y^2)\) will be greater than 100 since the total sum is greater than 200. If 2xy is less than \((x^2 + y^2)\), then anyway \((x^2 + y^2)\) will be greater than 100 (which is half of 200). So statement 2 is sufficient alone. Thanks Bunuel and Karishma for detailed explanation... one confusion though... from point 2) if 2xy is less than x^2 + y^2 .... lets say it is  (x^2 +y^2)... than the equation becomes 0 > 200 ? Am I missing some point ( SInce we need to know the minimum value of left side ... we should be able to substitute (x^2 + y^2 ) by 2xy..... but not 2xy by (x^2 + y^2 ) Thanks for any inputs...




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