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Re: Is x^2 + y^2 > 100?
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30 May 2017, 16:21
MacFauz wrote: rajathpanta wrote: Is x^2+y^2>100??
(1) 2xy<100 (2) (x+y)^2>200
To me its only B. because statement 2 boils down to x+y>\(\sqrt{200}\)
Can someone explain the OA Its B for me as well. 2) \((x+y)^2 > 200\), So., \((x+y) > 10\sqrt{2} or (x+y) < 10\sqrt{2}\) I cant find any two numbers that would satisfy statement 2 and give a value of \(x^2 + y^2 < 100\) Confusing. Kudos Please... If my post helped. Bunuel this is the same reasoning I used is this actually valid?



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Re: Is x^2 + y^2 > 100?
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30 May 2017, 21:21
Nunuboy1994 wrote: MacFauz wrote: rajathpanta wrote: Is x^2+y^2>100??
(1) 2xy<100 (2) (x+y)^2>200
To me its only B. because statement 2 boils down to x+y>\(\sqrt{200}\)
Can someone explain the OA Its B for me as well. 2) \((x+y)^2 > 200\), So., \((x+y) > 10\sqrt{2} or (x+y) < 10\sqrt{2}\) I cant find any two numbers that would satisfy statement 2 and give a value of \(x^2 + y^2 < 100\) Confusing. Kudos Please... If my post helped. Bunuel this is the same reasoning I used is this actually valid? Algebraic approach is more precise. The fact that you cannot find some set of values does not necessarily mean that there is no such set of values.
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Re: Is x^2 + y^2 > 100?
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24 Aug 2017, 11:56
VeritasPrepKarishma wrote: Hey arvindg, Thanks for pointing that out. The explanation is indeed incorrect. Sometimes, errors just creep up unwittingly. We will fix it soon. Your strategy of using numbers was spot on and that's exactly what I was thinking while reading the question too... Though the switch from <100 to >100 takes place at decimals so I was a little unhappy about that. The algebraic solution given by Bunuel is neat. VeritasPrepKarishma hi mam for statement 1, we cannot set x to equal to y, as any maximum value can untrue the statement, but we can duly set x equals to y for statement 2, to find out the minimum value that complies to the information provided by statement 2... is that the logic behind this operation, or anything else ....? please say to me, mam... thanks in advance



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Re: Is x^2 + y^2 > 100?
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27 Aug 2017, 11:46
Hi Bunuel / VeritasPrepKarishma, In the post, link provided ( https://gmatclub.com/forum/advancednum ... 94102.html), Quote: "Question: If x and y are positive, is x^2 + y^2 > 100?
Statement 1: 2xy < 100
Statement 2: (x + y)^2 > 200
Solution:
We need to find whether x^2 + y^2 must be greater than 100.
Statement 1: 2xy < 100
Plug in some easy values to see that this is not sufficient alone.
If x = 0 and y = 0, 2xy < 100 and x^2 + y^2 < 100
If x = 40 and y = 1, 2xy < 100 but x^2 + y^2 > 100"
Unquoted: In the question, we are given that x and y are positive then why you considered x=0, y=0 in statement 1 because 0 is neither positive nor negative. Please clarify for the doubt. Thanks.



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Re: Is x^2 + y^2 > 100?
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28 Aug 2017, 02:56
Bunuel wrote: arvindg wrote: Problem source: Veritas Practice Test 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. OMG!! Does this kind of question can be found in the real test? I would have spent 35 min on this and got it wrong.... The quoted part of Brunnel solution is just... ingenious



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Is x^2 + y^2 > 100?
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05 Sep 2017, 11:05
Although there is no substitute for Bunuel's solution, as it is very neat, and requires clever ideas like (x−y)^2≥0 (square of any number is more than or equal to zero) and requires crystal clear knowledge of algebra/equations. Since I could not think of that, I solved it in the following manner and reached the answer B:
1 is insufficient
2 says (x+y)^2>200 > (x+y)^2>(14.something)^2 Let's say either x or y is zero then either of x or y has to be greater than 14, and inequality will hold for both the ve and +ve values both and any combination of values of x and y. Hence, the value of x^2+y^2 will be more than 100



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Re: Is x^2 + y^2 > 100?
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22 Sep 2017, 14:58
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? Hi Bunuel, I understand the algebraic solution now but I had a different reasoning when I did this problem. Let's say (x + y)^2 = 200. In this case x+y would be 10\sqrt{2}. I considered the minimum possibility where x=y which makes each of them 5\sqrt{2}. So x^2+y^2=100. Since it our question says (x + y)^2 > 200, based on above deduction, I assumed that their (x + y) should definitely be more than 10 \sqrt{2} implying that x^2+y^2 should also be greater than 100. Is this a reasonable deduction? I honestly didn't even think of any other numbers to prove my case incorrect after this and I know that doesn't stand well with other questions maybe but I was wondering if it was a good method for this question.



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Re: Is x^2 + y^2 > 100?
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25 Sep 2017, 02:54
rajudantuluri 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? Hi Bunuel, I understand the algebraic solution now but I had a different reasoning when I did this problem. Let's say (x + y)^2 = 200. In this case x+y would be 10\sqrt{2}. I considered the minimum possibility where x=y which makes each of them 5\sqrt{2}. So x^2+y^2=100. Since it our question says (x + y)^2 > 200, based on above deduction, I assumed that their (x + y) should definitely be more than 10 \sqrt{2} implying that x^2+y^2 should also be greater than 100. Is this a reasonable deduction? I honestly didn't even think of any other numbers to prove my case incorrect after this and I know that doesn't stand well with other questions maybe but I was wondering if it was a good method for this question. rajudantuluri hi I have a similar question to Bunuel the great Question stem: Is "x^2 + y^2 > 100" Now taking the minimum values for x and y x^ + x^2 > 100 or, 2x^2 > 100 Now the question becomes: Is 2x^2 > 100 statement 1 says: twice the value, the product of x and y, is less than 100 If maximized, it becomes, 2x^2 < 100 but, we don't know whether the maximum value is less than 100 not sufficient statement 2 says: (x + y)^2 > 200 taking minimum values (x + x)^2 > 200 or (2x)^ 2 > 200 or 4x^2 > 200 reducing by 1/2 2x^2 > 100 sufficient ... B But, rajudantuluri, here I have a question, why it has to be assumed that the minimum value of (x + y)^2 will exceed 200 ...? thanks in advance



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Re: Is x^2 + y^2 > 100?
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11 Jan 2018, 08:14
If we plug in values, we can find that the statement 1 is clearly insufficient. However with the concept Geometric Mean is Max when x=y, to statement 1 we get 2x^2 < 100, As the product of 2 numbers will be maximum when the 2 numbers are set equal to each other.
that is x < √50. and we can say x^2 + y^2 < 100. can be solved.
Am I Applying concept wrong here?



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Is x^2 + y^2 > 100?
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26 Feb 2019, 11:08
I have a less algebraic approach to Statement 2: (x + y) ^ 2 > 200 Let's take x+ y as n, Therefore n^2 > 200: If we ballpark the lowest possible value of n it is somewhere between 14 and 14.25 (because 14^2 is 196, which is almost 200) Now if we take x = 7 and y = 7 we get 49 + 49 < 100 But if we bump up the value of either by even the slightest amount we go over 100, therefore B is sufficient. Bunuel does this make sense?



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Re: Is x^2 + y^2 > 100?
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26 Feb 2019, 11:56
arvindg wrote: Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200 Statement 1: 2xy < 100 Thus, xy < 50. If x=1 and y=1, then x²+y² < 100. If x=2 and y=10, then x²+y² > 100. INSUFFICIENT. Statement 2: (x+y)² > 200 Since the square of a value cannot be negative, (xy)² ≥ 0. Adding together (x+y)² > 200 and (xy)² ≥ 0, we get: (x+y)² + (xy)² > 200+0 (x² + 2xy + y²) + (x²  2xy + y²) > 200 2x² + 2y² > 200 x² + y² > 100 SUFFICIENT.
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Re: Is x^2 + y^2 > 100?
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