Author 
Message 
TAGS:

Hide Tags

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



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)

Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
Show Tags
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!
_________________
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, 21:56
rajathpanta wrote: 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! I found this solution from Bunuel. Statement 2. \(x^2 + y^2 + 2xy > 200\) Since \((xy)^2\) is cannot be lesser than 0, (square of a number is always positive or 0), the minimum value that 2xy can take is equal to \(x^2 + y^2\). So statement 2 can be changed to be, \(x^2 + y^2 + x^2 + y^2 > 200\). So, \(x^2 + y^2 > 100\). That is a convincing method to show that B is infact the right answer. 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



Intern
Joined: 31 Oct 2012
Posts: 2

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
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.



Intern
Joined: 06 Feb 2013
Posts: 23

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
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?



Intern
Joined: 06 Feb 2013
Posts: 23

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
19 Feb 2013, 21:14
Thanks, it's clear now!



Intern
Joined: 19 Feb 2013
Posts: 10

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
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



Math Expert
Joined: 02 Sep 2009
Posts: 39759

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
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?
_________________
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: 19 Feb 2013
Posts: 10

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
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



Math Expert
Joined: 02 Sep 2009
Posts: 39759

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
31 Mar 2013, 12:44



Math Expert
Joined: 02 Sep 2009
Posts: 39759

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
05 Jul 2013, 02:44



Current Student
Joined: 10 Oct 2013
Posts: 33
Concentration: Marketing, Entrepreneurship

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
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 ...



Math Expert
Joined: 02 Sep 2009
Posts: 39759

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
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
_________________
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



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7453
Location: Pune, India

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



Current Student
Joined: 10 Oct 2013
Posts: 33
Concentration: Marketing, Entrepreneurship

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
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.



Intern
Joined: 11 Apr 2014
Posts: 12
Location: India
Concentration: Strategy, Finance
GPA: 4
WE: Management Consulting (Consulting)

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
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.



Math Expert
Joined: 02 Sep 2009
Posts: 39759

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
10 May 2014, 08:29



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16031

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
27 May 2015, 12:36
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Current Student
Joined: 29 Mar 2015
Posts: 77
Concentration: Strategy, Operations
WE: Research (Other)

Re: Is x^2 + y^2 > 100? [#permalink]
Show Tags
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!
_________________
If you like my post, Pl. do not hesitate to press kudos!!!!
Q51 on GMAT  PM me if you need any help with GMAT QUANTS!!!




Re: Is x^2 + y^2 > 100?
[#permalink]
17 Jun 2015, 06:11



Go to page
Previous
1 2 3
Next
[ 44 posts ]




