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Is x^2>y^2? [#permalink]
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03 Jul 2016, 05:54
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Is x^2>y^2? 1) x<y 2) –y>x *An answer will be posted in 2 days.
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Re: Is x^2>y^2? [#permalink]
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03 Jul 2016, 06:17
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Is x^2>y^2? 1) x<y 2) –y>x Statement 1:Insufficient  e.g. x = 2 , y = 3 Ans = N. But, If x = 10 , Y = 9 Ans = Y. Statement 2: Insufficient  e.g. x = 10 , y = 2 Ans = Y. But, if x = 10 , y = 20 Ans = N. Combined > x < y and x < y form these 2: x^2 will always be greater than y^2. Hence sufficient. please give kudos if this all makes sense !



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Re: Is x^2>y^2? [#permalink]
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04 Jul 2016, 13:02
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Q. x^2> y^2? this can be rewritten as : x^2y^2>0 => (x+y) (xy)>0?? i.e do (x+y) and (xy) have same signs. Going to the stmts: Stmt1: x<y => (xy)<0; this does not tells us anything about the sign of x+y. Therefore insufficient. stmt2: y>x => (x+y)<0; this does not tells us anything about the sign of xy. Therefore insufficient. Combining, we know signs of both. Hence sufficient. Answer =C.
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Re: Is x^2>y^2? [#permalink]
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04 Jul 2016, 23:36
From the original condition, we can see that there are 2 variables. Hence, the correct answer is C.  Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
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Re: Is x^2>y^2? [#permalink]
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09 Jul 2016, 05:19
Can someone elaborate solution for this question Qstn: x^2y^2>0 ? (x+y)(xy)>0 ? x<y or x>y? statement1: x<y ( means x>y is not true so x<y) statement 2: x<y I got D..pls advise where am i going wrong
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Re: Is x^2>y^2? [#permalink]
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25 Sep 2016, 05:23
sairam595 wrote: Can someone elaborate solution for this question Qstn: x^2y^2>0 ? (x+y)(xy)>0 ? x<y or x>y? statement1: x<y ( means x>y is not true so x<y) statement 2: x<y I got D..pls advise where am i going wrong Qstn: x^2y^2>0 ? (x+y)(xy)>0 ? Now here (x+y)(xy)>0 There are two cases possible that Both (x+y) and (xy) have to positive in order to be greater than 0. Both (x+y) and (xy) have to negative in order to be greater than 0. Now stat 1: x< y i.e. we can write xy < 0.. i.e. ve value...we are not sure whether (x+y) is also negative or positive to comment that product is greater than 0...Insufficient. Stat 2: y > x => y x > 0 => (x+y) > 0 => (x+y) < 0 ( Note..when we multiply by ve value on both the sign changes ) we have (x+y) < 0 ...but we are not sure whether (xy) is also negative or positive to comment that product is greater than 0...Insufficient. Both (x+y) < 0 and xy < 0 i.e. ve * ve > 0...Hence sufficient... Hope this clears..



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Re: Is x^2>y^2? [#permalink]
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25 Sep 2016, 06:53
MathRevolution wrote: From the original condition, we can see that there are 2 variables. Hence, the correct answer is C.
 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Hi, I am sorry Mathrevolution BUT this must be the one of the worst ways to do these type of Qs.. Two variables and two equations would be OK if these are linear equations and THEN too, check for similarities in the two equations... In INEQUALITIES, keep this method at an arm's length..
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Re: Is x^2>y^2? [#permalink]
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25 Sep 2016, 07:07
MathRevolution wrote: Is x^2>y^2? 1) x<y 2) –y>x
*An answer will be posted in 2 days. Hi, A way to do it .... Main point is ... Is NUMERIC value of x more than y, irrespective of SIGN 1) x<y.... X can be 4, y = 6 ans YES X can be 7, y=6...ans NO Insuff 2) y>x... x can be 4, y can be 5...ans NO.. x can be 6, y can be 3, ans YES... hi Insuff Combined (1) is x<y and (2) is y>x or y<x.... From above TWO, \(x<y<x\) SO y lies between a x and x and thus it's NUMERIC value is less than x Ans is YES always Suff C
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Re: Is x^2>y^2? [#permalink]
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25 Sep 2016, 07:14
To reword the question: is (xy)(x+y)>0? To determine this, we need more information about whether the above equation satisfies either of these conditions: 1) x>y and x>y 2) x<y and x<y Since we do not know whether x & y is negative is positive, we cannot sum up the above equations as "x>y" or "x<y", hence we need both the first & second conditions to answer the question (for example, if y=2 and x=1, x<y but x>y) > C is correct



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Re: Is x^2>y^2? [#permalink]
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25 Sep 2016, 11:44
chetan2u wrote: MathRevolution wrote: Is x^2>y^2? 1) x<y 2) –y>x
*An answer will be posted in 2 days. Hi, A way to do it .... Main point is ... Is NUMERIC value of x more than y, irrespective of SIGN 1) x<y.... X can be 4, y = 6 ans YES X can be 7, y=6...ans NO Insuff 2) y>x... x can be 4, y can be 5...ans NO.. x can be 6, y can be 3, ans YES... hi Insuff Combined (1) is x<y and (2) is y>x or y<x.... From above TWO, \(x<y<x\) SO y lies between a x and x and thus it's NUMERIC value is less than x Ans is YES always Suff C Hi chetan can we multiply both equations and get X^2Y^2<0.........X^2<Y^2 But then answer will be No in such case. thanks



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Re: Is x^2>y^2? [#permalink]
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08 Oct 2016, 08:14
Is x^2>y^2? 1) x<y 2) –y>x clearly ST1 and ST2 are not sufficient alone but when u combine them  x < y and x <  y When x = 5 and y = 3 both eqn statmnts satisfied  hence x^2 > y^2 now when x =  0.5 and y = 0.3 both eqn statmnts satisfied x^2 = 0.25 y^2 = 0.09 Hence x^2 > y^2 So answer is clearly C ; Simple advice i would say , it does help sometimes is to consider values bw 0 and 1 and bw 0 and 1 . These can sometimes change the answer too . also you could ADD stmnt 1 and 2 2x<0 or, x<0 for all values x^2 > y^2
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Re: Is x^2>y^2? [#permalink]
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06 Jan 2017, 11:17
hi this method can also result in "E"? how to avoid that? thanks MathRevolution wrote: From the original condition, we can see that there are 2 variables. Hence, the correct answer is C.
 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.



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Is x^2>y^2? [#permalink]
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23 Feb 2017, 04:47
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PROMPT ANALYSIS x and y are real numbers.
SUPERSET The answer will either come as ‘YES’ or ‘NO’
TRANSLATION We need to know if x >y. In order to know that, we need: The exact value of x and y. Any relation between x and y Any characteristics of x and y that can give the range of x and y.
STATEMENT TRANSLATION ST 1: x<y. We will 2 test value of (x,y). Case#1: (x,y) is (1,2). For this result the answer is no. Case#2: (x,y) is (2,1). For this result, the answer is yes.
Therefore cannot be determined. INSUFFICIENT. Hence option a and d eliminated. St 2: x <y Again we take 2 values Case#1: (x,y) = (2,3). For this the result is NO. Case#2: (x,y) = (4, 3). For this the answer is YES.
Therefore cannot be determined. INSUFFICIENT. Hence option b is also eliminated.
St1 & St2: We will solve with the assumption of data. Consider the value of y to be 4. For this the value of x could be in the range of (infinity. 4). From statement 2, the final domain cones as (infinity. 4). All the values in this range have greater modulus value as compared to y. Hence SUFFICIENT. Option e eliminated
Answer option C.



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Is x^2>y^2? [#permalink]
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02 Nov 2017, 20:04
MathRevolution wrote: From the original condition, we can see that there are 2 variables. Hence, the correct answer is C.
 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. How did you determine that? There are many questions having 2 variables so can the answer always be C? I tried to solve this question by trying different values like everybody else but it takes more time than it actually should. What is Variable approach?



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Is x^2>y^2? [#permalink]
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02 Nov 2017, 23:15
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simplifying the given question, X^2>y^2 X^2y^2>0 (x+y)(xy)>0 for this to be true,either both term should be negative or both term should be positive. 1) tells only about (xy) <0 insuff. 2) tells only one cond. X+Y <0 insuff. combining both satisfying both cond. hence ans c.
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