Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 24 Jan 2011
Posts: 12

Is x^2 + y^2 > 100? [#permalink]
Show Tags
27 Jan 2011, 08:07
3
This post received KUDOS
12
This post was BOOKMARKED
Question Stats:
33% (02:40) correct
67% (01:23) wrong based on 860 sessions
HideShow timer Statistics
Is x^2 + y^2 > 100? (1) 2xy < 100 (2) (x + y)^2 > 200
Official Answer and Stats are available only to registered users. Register/ Login.
Last edited by Bunuel on 29 Oct 2012, 04:38, edited 1 time in total.
OA edited.



Intern
Joined: 19 Jul 2010
Posts: 11

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
27 Jan 2011, 08:50
1
This post received KUDOS
first option 1:
2xy < 100  1
cant say from option 1.
(x+y)^2 > 200 x^2 + y^2 + 2xy > 200 x^2 + y^2 > 200  2xy
substitute 2xy in above equation so x^2 + y^2 > 200  2xy 2xy is less than 100 from equation 1. implies x^2 + y^2 > 100
so using both options.. Ans C



Intern
Joined: 24 Jan 2011
Posts: 12

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
27 Jan 2011, 08:56
Thanks Tarun. I get that (1) alone doesn't work, but how did you rule out (2)? I chose B by trying different numbers and not finding any numbers such that (2) was true but the stem was not true.



Math Expert
Joined: 02 Sep 2009
Posts: 39723

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
27 Jan 2011, 09:00
20
This post received KUDOS
Expert's post
14
This post was BOOKMARKED
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?
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 24 Jan 2011
Posts: 12

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
27 Jan 2011, 09:15
Thanks Bunuel! Yes, the OA was C, I've attached a screen capture. Hope I don't run into something like this on the actual GMAT Attachment:
veritas_30.JPG [ 41.99 KiB  Viewed 13301 times ]



Math Expert
Joined: 02 Sep 2009
Posts: 39723

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
27 Jan 2011, 09:28



Intern
Joined: 24 Jan 2011
Posts: 12

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
27 Jan 2011, 20:28
Here's the explanation! Attachment:
Veritas_30_expln.jpg [ 63.89 KiB  Viewed 13275 times ]



Intern
Joined: 19 Jul 2010
Posts: 11

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
27 Jan 2011, 22:44
Bunuel's explanation is clear to me.. I think what bunuel said could be correct..
bunuel can you please explain why they say 2xy value cannnot be found hence combining both the equation..
kudos to bunuel, for thinking (xy)^2 >= 0



Math Expert
Joined: 02 Sep 2009
Posts: 39723

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
28 Jan 2011, 03:14



Senior Manager
Joined: 08 Nov 2010
Posts: 408
WE 1: Business Development

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
29 Jan 2011, 10:42
Damn Bunuel  nice trick. very nice!
_________________
GMAT Club Premium Membership  big benefits and savings



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7446
Location: Pune, India

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
29 Jan 2011, 14:01
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.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Intern
Joined: 19 Jul 2010
Posts: 11

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
29 Jan 2011, 18:39
well from (xy)^2 >=0 to x^2 + y^2 >= 2xy case there could be a situation.. wen x^2 + y^2 is almost equal to 4xy or 6xy(since greater than 2xy is also applicable) then in that case we can not decide x^2 + y^2 > 100 can we vouch for this case as x^2 + y^2 equal to 4xy. ??? confused..



Math Expert
Joined: 02 Sep 2009
Posts: 39723

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
29 Jan 2011, 18:59
tarunjagtap wrote: well from (xy)^2 >=0 to x^2 + y^2 >= 2xy case there could be a situation.. wen x^2 + y^2 is almost equal to 4xy or 6xy(since greater than 2xy is also applicable) then in that case we can not decide x^2 + y^2 > 100
can we vouch for this case as x^2 + y^2 equal to 4xy. ??? confused.. Not sure understood what you meant by that but anyway: 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 \(2xy\)): \(x^2+(x^2+y^2)+y^2>200\) > \(x^2+y^2>100\). Hope it's clear.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Director
Joined: 21 Dec 2010
Posts: 627

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
30 Jan 2011, 23:09
x*2+y*2 > 100, Bunuel proved it algebraically . the explanation is clear , so the answer should be B, not C , it is so much easier to fall for the trap under time pressure.
_________________
What is of supreme importance in war is to attack the enemy's strategy.



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2010

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
02 Feb 2011, 05:41
Problem source: Veritas Practice Test I was not able to rule out B as "Not sufficient" at first. After Bunuel's explanation, I tried to solve this with numbers and here's what I got. Question: Is x^2 + y^2 > 100? (1) x=1,y=1,2xy=2<100 but x^2+y^2=1+1=2<100. Answer to Question: NO x=10,y=1,2xy=20<100 but x^2+y^2=100+1=101>100. Answer to Question: YES Not sufficient. (2) \((x+y)^2>200\) \(x+y>200\) \(x+y>\sqrt{200}\) i.e. \((x+y) > \sqrt{200} or (x+y) < \sqrt{200}\)  Restriction Now; if we can prove that at least one value of \(x^2 + y^2\) for the given condition is <=100; this statement becomes NOT SUFFICIENT. Let's find the least value for \(x^2 + y^2\) with above restriction; Considering the above restriction, the least value pair for x and y so that \(x^2 + y^2\) should give us the least value should be greater than the following: \(x=\sqrt{200}/2\) and \(y=\sqrt{200}/2\) or \(x=\sqrt{200}/2\) and \(y=\sqrt{200}/2\) Solve for \(x^2 + y^2\) \((200/4)+(200/4)\) \(50+50=100\) But; this result is for x+y=(\sqrt{200}/2)+(\sqrt{200}/2); we want it for x+y>(\sqrt{200}/2)+(\sqrt{200}/2); even a little fluctuation in x or y will push the result beyond 100. Thus; there is no value pair for x and y that gives us a value for \(x^2 + y^2\) <=100. The second statement is SUFFICIENT. Ans: B However, I will remember the takeaway from Bunuel that \(x^2 + y^2 >= 2xy\) for solving similar types of questions later.
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Manager
Status: Preparing myself to break the sound( 700 )barrier!
Affiliations: IFC  Business Edge, Bangladesh Enterprise Institute
Joined: 15 Feb 2011
Posts: 209
Location: Dhaka, Bangladesh
Schools: Texas A&M  Mays; PennState  Smeal; Purdue  Krannert; JohnsHopkins  Carey; Vanderbilt  Owen; WakeForest  Babcock; UBC  Sauder
WE 1: Financial Analyst/Financial Management Trainer  BRAC Afghanistan  13 months
WE 2: Business Proposal Coordinator  MiDS, Inc. (in Kabul, Afghanistan)  4 Months (did quit the job bcoz of security concerns)
WE 3: Readers For Readers (A Social Impact Generating Organization)  CoFounder

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
30 Apr 2011, 00:37
I am weak (i think the weakest) in DS.
This problem tempted me today to give a try. I solved this and came up with the answer  Option C.
Here is why.
First I simplified the question stem in this was so that I can find both the statements' issues in the stem. x^2 + y^2 > 100, or x^2 + y^2 + 2xy > 100 + 2xy, or (x + y)^2 > 100 + 2xy
Statement 1 is not sufficient. So, we eliminate option A and D. Then, statement 2 is insufficient as well. Thus, we can eliminate the option D. Because, both the statements don't give us any specific values and keep another set of unknown/variable in the inequality.
Now, if we consider Statements 1 and 2 together, we can find different values for (x + y)^2 > 100 + 2xy. Hence, C is the answer.
As I said before, I am really weak in DS; my explanation may be so hilarious and incorrect. I will appreciate if anyone could pinpoint where problem lies with my explanation though the answer is correct.



Senior Manager
Joined: 05 Jul 2010
Posts: 355

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
07 Sep 2011, 15:13
1
This post received KUDOS
1
This post was BOOKMARKED
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? I am feeling proud about myself . I got just the ONE question wrong in my Veritas CAT test today, and this was the one. I marked B as I proved it during the test  took 3.42 mins though. I should have got 100% correct otherwise. Good to know Veritas guys were wrong! I was a bit baffled by their explanation. My proof (lengthy BUT conceptual proof): (x+y)^2 > 200 => x + y > 10*Sqrt(2) => For x^2 + y^2 to MINIMUM x=y : Why? Because squaring a number SPREADS the value exponentially. For given {x,y} such that x+y is known, x^2 + y^2 will ONLY spread MORE as we spread x and y away from their mean which is (x+y)/2  regardless of the signs of x and y; infact opposite signs will spread the sum of squares even further. Hence x,y both MUST be > 5*sqrt(2). Hence MIN(x^2 + y^2) MUST be > 25*2 + 25*2 = 100. Hence, x^2 + y^2 > 100. If there is any Veritas Instructor here, please let me know if I got it wrong in some freakish way.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7446
Location: Pune, India

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
07 Sep 2011, 21:18
abhicoolmax 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? I am feeling proud about myself . I got just the ONE question wrong in my Veritas CAT test today, and this was the one. I marked B as I proved it during the test  took 3.42 mins though. I should have got 100% correct otherwise. Good to know Veritas guys were wrong! I was a bit baffled by their explanation. My proof (lengthy BUT conceptual proof): (x+y)^2 > 200 => x + y > 10*Sqrt(2) => For x^2 + y^2 to MINIMUM x=y : Why? Because squaring a number SPREADS the value exponentially. For given {x,y} such that x+y is known, x^2 + y^2 will ONLY spread MORE as we spread x and y away from their mean which is (x+y)/2  regardless of the signs of x and y; infact opposite signs will spread the sum of squares even further. Hence x,y both MUST be > 5*sqrt(2). Hence MIN(x^2 + y^2) MUST be > 25*2 + 25*2 = 100. Hence, x^2 + y^2 > 100. If there is any Veritas Instructor here, please let me know if I got it wrong in some freakish way. Your reasoning is fine. That's good thinking. Think of it in another way: When 2 numbers are equal, their Arithmetic Mean = Geometric Mean AM is least when it is equal to GM and GM is greatest when it is equal to AM. So sum of the terms is least when the numbers are equal; product is maximum when they are equal. For minimum value of \(x^2 + y^2\), we need \(x^2 = y^2\) or x = y On the same line, if product is given to be constant, sum is minimum when numbers are equal. If the sum is given to be constant, the product is maximum when the numbers are equal.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Senior Manager
Status: Prevent and prepare. Not repent and repair!!
Joined: 13 Feb 2010
Posts: 259
Location: India
Concentration: Technology, General Management
GPA: 3.75
WE: Sales (Telecommunications)

Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
Show Tags
28 Oct 2012, 02:54
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
_________________
I've failed over and over and over again in my life and that is why I succeedMichael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+



Moderator
Joined: 02 Jul 2012
Posts: 1223
Location: India
Concentration: Strategy
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
Show Tags
28 Oct 2012, 06:54
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.
_________________
Did you find this post helpful?... Please let me know through the Kudos button.
Thanks To The Almighty  My GMAT Debrief
GMAT Reading Comprehension: 7 Most Common Passage Types




Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2
[#permalink]
28 Oct 2012, 06:54



Go to page
1 2 3
Next
[ 44 posts ]




