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Is x^2 + y^2 > x^2  y^2?
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Updated on: 05 Jun 2013, 00:12
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Is x^2 + y^2 > x^2  y^2? (1) x > y (2) x > 0
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Originally posted by makhija1 on 04 Jun 2013, 20:45.
Last edited by Bunuel on 05 Jun 2013, 00:12, edited 1 time in total.
Renamed the topic and added the OA.



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Re: Answer without doing any algebra
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04 Jun 2013, 20:49
IMO, the answer should be E. 1) For (2,1), the answer to target ques is Yes; but for (1,0), the answer is No. Thus the statement is insuff. 2) The same values from above can be used to get two contradicting answers as there is no indication of what y should be. So this statement is also insuff. Any thoughts? FYI, I got this question from an article posted in Beat the GMAT forum; it didn't say the what the OA was... Sorry!



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Re: Answer without doing any algebra
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04 Jun 2013, 21:41
I am also going with E. When Y = 0, it's never true.



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Re: Answer without doing any algebra
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04 Jun 2013, 22:01
Yahtzeefish wrote: I am also going with E. When Y = 0, it's never true. I like this succinct answer from Yahtzeefish! The question is baiting you into thinking about negatives, but answer will never be negative because all the terms are squared. Since the terms are squared, they must necessarily be positive (or zero). The absolute values thus change nothing to the equation and can be ignored. However, all values of Y, positive or negative, will end up being positive under the square, and therefore smaller in the second half of the equation than in the first. The number that will mess everything up is zero. Had statement 2 indicated that y>0, we'd have a different answer, but since the statements only care about the value of x and offer no real limitations on the value of y, the answer will be that both statements combined are still not sufficient (answer E) Hope this helps! Ron
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Re: Is x^2 + y^2 > x^2  y^2?
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Updated on: 06 Jun 2013, 02:42
WholeLottaLove wrote: Could someone post a step by step explanation? This is an interesting problem but I am having some trouble figuring it out!
Thanks! x is always a nonnegative entity. Thus, the given expression can be safely squared on both sides > Is \((x^2+y^2)^2>(x^2y^2)^2\) or \((x^2+y^2 + x^2y^2)[x^2+y^2 ( x^2y^2)] >0 > IS 2x^2*2y^2>0.\) Thus, we have to answer the questions : Is \(4x^2y^2>0\). Now we know that this expression WILL always be greater than zero, irrespective of any nonzero real values of x and y. However, if x=0 OR y=0 OR both x = y = 0, then we wouldn't be able to say that the given expression is greater than 0. The two fact statements, provide some idea about both x and y, both they don't guide us to the conclusion as to whether both x and y are nonzero values or not. Hence, we can't conclude even after using the 2 fact statements. For the same problem,imagine 2 fact statements like this : 1. x>0 2. xy<0. From F.S 1, all we know is that x is a nonzero(positive) number. However, y could still be zero. From F.S 2, we know that both x AND y are nonzero numbers, thus, Sufficient. You could also visualize the given problem as this: \(x^2\) and \(y^2\) are both positive quantities(assuming x and y are nonzero numbers). Thus, ADDING two positive numbers will always be greater than the DIFFERENCE of the same two positive numbers. Thus, what the given expression asks, is invariably ALWAYS true, except for the one case where x=0 OR y=0 OR x = y = 0.
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Originally posted by mau5 on 06 Jun 2013, 00:25.
Last edited by mau5 on 06 Jun 2013, 02:42, edited 1 time in total.



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Re: Is x^2 + y^2 > x^2  y^2?
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06 Jun 2013, 01:33
WholeLottaLove wrote: Could someone post a step by step explanation? This is an interesting problem but I am having some trouble figuring it out!
Thanks! The easiest way to solve this is by picking numbers Is \(x^2 + y^2 > x^2  y^2\)? The left term will always be \(\geq{0}\) because is the sum of two \(\geq{0}\) numbers, so we can safely rewrite it as: \(x^2 + y^2 > x^2  y^2\), here x^2, and y^2 can be seen as \(number\geq{0}\), so is \((num_1\geq{0})+(num_2\geq{0})>(num_1\geq{0})(num_2\geq{0})\) Even though this seems to be true, it does not hold true if \(y=0\), \(num_2=0\) for example So in this case we have \(x^2=x^2\), and since both statement do not exclude this possibility (1) x > y (2) x > 0 the correct answer is E
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Re: Is x^2 + y^2 > x^2  y^2?
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06 Jun 2013, 02:14
Ans is E This is how I solved... \(x^2 + y^2\)> \(x^2  y^2\) given simply say squaring on both sides and solving gives us Is XY > 0 ? i.e do X and Y have same signs Stmt 1 Not Sufficient As X > Y Does not provide sufficient the information on the signs, they may be of the may sign or may be not. Stmt 2 Not sufficient As X > 0 , Y can be + ve or ve Combined Stmt 1 & 2 , still Y can be + ve or ve. Therefore Ans E.
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Re: Is x^2 + y^2 > x^2  y^2?
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Re: Is x^2 + y^2 > x^2  y^2?
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06 Jun 2013, 02:48
Thanks Bunuel for correcting me, I learned the trick of squaring of mods from one of your posts I took Square root of the \(x^2y^2>0\) Can't we take square root ??? please explain if we cannot.
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Re: Is x^2 + y^2 > x^2  y^2?
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Re: Is x^2 + y^2 > x^2  y^2?
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23 Jul 2013, 15:11
I thought we can only square both inequalities only if we know that both sides are positive (for example, what if x^2  y^2 =2^2  5^2 = 19)? Thanks! Bunuel wrote: Manhnip wrote: Is x^2 + y^2 > x^2  y^2?
(1) x > y (2) x > 0
Ans is E
This is how I solved...
\(x^2 + y^2\)> \(x^2  y^2\) given simply say squaring on both sides and solving gives us Is XY > 0 ? i.e do X and Y have same signs
Stmt 1 Not Sufficient As X > Y Does not provide sufficient the information on the signs, they may be of the may sign or may be not.
Stmt 2 Not sufficient As X > 0 , Y can be + ve or ve
Combined Stmt 1 & 2 , still Y can be + ve or ve.
Therefore Ans E. That's not correct. If you square, the questions becomes: is \(x^2y^2>0\)? Now, this holds true if \(xy\neq{0}\). But even when we consider the statements together we cannot say whether y=0 or not. Therefore the answer is E. Similar questions to practice: isxyxy123108.htmlifayzbisyayb82673.htmlisxy146991.htmlisxyxy1xy2xy132654.htmlisxyxy137050.htmlisxyxz1yz2x86132.htmlHope it helps.



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Re: Is x^2 + y^2 > x^2  y^2?
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23 Jul 2013, 15:18
WholeLottaLove wrote: I thought we can only square both inequalities only if we know that both sides are positive (for example, what if x^2  y^2 =2^2  5^2 = 19)? Thanks! Bunuel wrote: Manhnip wrote: Is x^2 + y^2 > x^2  y^2?
(1) x > y (2) x > 0
Ans is E
This is how I solved...
\(x^2 + y^2\)> \(x^2  y^2\) given simply say squaring on both sides and solving gives us Is XY > 0 ? i.e do X and Y have same signs
Stmt 1 Not Sufficient As X > Y Does not provide sufficient the information on the signs, they may be of the may sign or may be not.
Stmt 2 Not sufficient As X > 0 , Y can be + ve or ve
Combined Stmt 1 & 2 , still Y can be + ve or ve.
Therefore Ans E. That's not correct. If you square, the questions becomes: is \(x^2y^2>0\)? Now, this holds true if \(xy\neq{0}\). But even when we consider the statements together we cannot say whether y=0 or not. Therefore the answer is E. Similar questions to practice: isxyxy123108.htmlifayzbisyayb82673.htmlisxy146991.htmlisxyxy1xy2xy132654.htmlisxyxy137050.htmlisxyxz1yz2x86132.htmlHope it helps. We can square an inequality if both sides are nonnegative. ADDING/SUBTRACTING INEQUALITIES: You can only add inequalities when their signs are in the same direction:If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) > \(a+c>b+d\). Example: \(3<4\) and \(2<5\) > \(3+2<4+5\). You can only apply subtraction when their signs are in the opposite directions:If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). RAISING INEQUALITIES TO EVEN/ODD POWER: A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are nonnegative (the same for taking an even root of both sides of an inequality).For example: \(2<4\) > we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) > we can square both sides and write: \(x^2<y^2\); But if either of side is negative then raising to even power doesn't always work. For example: \(1>2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are nonnegative. B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).For example: \(2<1\) > we can raise both sides to third power and write: \(2^3=8<1=1^3\) or \(5<1\) > \(5^2=125<1=1^3\); \(x<y\) > we can raise both sides to third power and write: \(x^3<y^3\). For multiplication check here: helpwithaddsubtractmultdividmultipleinequalities155290.html#p1242652THEORY ON INEQUALITIES: x24x94661.html#p731476inequalitiestrick91482.htmldatasuffinequalities109078.htmlrangeforvariablexinagiveninequality109468.htmleverythingislessthanzero108884.htmlgraphicapproachtoproblemswithinequalities68037.htmlinequationsinequalitiespart154664.htmlinequationsinequalitiespart154738.htmlQUESTIONS: All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189700+ Inequalities problems: inequalityandabsolutevaluequestionsfrommycollection86939.htmlHope it helps.
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Re: Is x^2 + y^2 > x^2  y^2?
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24 Jul 2013, 14:08
Odd man Out.. My answer is A. The question boils down to is XY>0? 1) x>y => xy>0 It does mean both are positive and hence XY>0.  Sufficient
2) X> 0 => we dont know any thing about Y and hence IN Sufficient
please let me know what I am missing here..



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Re: Is x^2 + y^2 > x^2  y^2?
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24 Jul 2013, 14:24



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Re: Is x^2+y^2 > x^2y^2?
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24 Sep 2015, 01:53
Bunuel wrote: Is x^2+y^2 > x^2y^2?
(1) x > y (2) x > 0
Kudos for a correct solution. Solution : x^2+y^2 > x^2y^2 is always true except for x=y=0. Statement1 : x != y. Sufficient Statement2 : x != 0. Sufficient Option D



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