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Is x  x^2 > y  y^2?
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05 Jul 2017, 00:31
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53% (02:29) correct 47% (02:10) wrong based on 92 sessions
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Is x  x^2 > y  y^2? (1) x > y (2) x^2 > y^2
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Re: Is x  x^2 > y  y^2?
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05 Jul 2017, 02:11
anurag16 wrote: Is x  x^2 > y  y^2?
(1) x > y (2) x^2 > y^2 \(xx^2>yy^2....... xy+y^2x^2>0......\).. \((xy)(xy)(x+y)>0...... (xy)(1(x+y))>0\).. So two case.. A) both POSITIVE.. xy>0 or x>y AND 1>x+y.. B) both NEGATIVE X<y and x+y>1.. Let's see if any statement fulfills any of the above cases.. 1) x>y.. see A, NOTHING about if 1>x+yInsufficient 2) \(x^2>y^2\).. It just tells us that NUMERIC value of (x> y)... Nothing about values with sign.. Insuff Combined.. We can surely say by adding two statements that.. x+x^2>y+y^2 but nothing about given equation.. You can find various combinations in FRACTION or INTEGERS, in + or  that will give us answer as YES and NO Insufficient
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Current Student
Joined: 04 Feb 2014
Posts: 223
Location: India
Concentration: General Management, Entrepreneurship
Schools: ISB '19 (D), MBS '20 (A), Sauder '20 (S), GMBA '19, Copenhagen (D), XLRI (II), Alberta"20, Trinity College (A$), Smurfit (D), UConn (WD), HHL Leipzig"20 (A), Ryerson"19 (S)
GPA: 3
WE: Project Management (Manufacturing)

Re: Is x  x^2 > y  y^2?
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06 Jul 2017, 06:02
Hi chetan2u, Thanks for the reply. Can you please explain why statement B is insufficient by taking some values? x^2>y^2 implies mod x> mod y So in x  x^2 > y  y^2 will always give a no x=2 y=1 NO x=2 y=1 NO x=2 y=1 NO x=2 y=1 NO Thanks in advance.
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Re: Is x  x^2 > y  y^2?
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06 Jul 2017, 06:13
anurag16 wrote: Hi chetan2u, Thanks for the reply. Can you please explain why statement B is insufficient by taking some values? x^2>y^2 implies mod x> mod y So in x  x^2 > y  y^2 will always give a no x=2 y=1 NO x=2 y=1 NO x=2 y=1 NO x=2 y=1 NO Thanks in advance. Hi, What are you missing on is that they can be fractions too.. X=1/2, Y=1/4 YES Whenever variables are given, check for ,+,0, fractions. We know this but tend to miss while doing a Q
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Re: Is x  x^2 > y  y^2?
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16 Jul 2017, 04:22
anurag16 wrote: Is x  x^2 > y  y^2?
(1) x > y (2) x^2 > y^2 It can be easily derived: 1)x > y : x= 2 and y = 3 ==> answer Yes x = 3 and y = 2 ===> answer no insufficient 2) x^2>y^2==> x^2y^2 >0 therefore from original condition xy> x^2y^2>0==>xy>0===> x>y...same as statement 1st which proved insufficient Since statement 1 and 2 are same and insufficient answer will always be E
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Re: Is x  x^2 > y  y^2?
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11 Dec 2018, 12:07
Supermaverick wrote: anurag16 wrote: Is x  x^2 > y  y^2?
(1) x > y (2) x^2 > y^2 It can be easily derived: 1)x > y : x= 2 and y = 3 ==> answer Yes x = 3 and y = 2 ===> answer no insufficient 2) x^2>y^2==> x^2y^2 >0 therefore from original condition xy> x^2y^2>0==>xy>0===> x>y...same as statement 1st which proved insufficient Since statement 1 and 2 are same and insufficient answer will always be E statement 2 is not equal to stat1, consider x and y as negative integers...



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Re: Is x  x^2 > y  y^2?
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25 Apr 2019, 20:04
chetan2u wrote: anurag16 wrote: Is x  x^2 > y  y^2?
(1) x > y (2) x^2 > y^2 \(xx^2>yy^2....... xy+y^2x^2>0......\).. \((xy)(xy)(x+y)>0...... (xy)(1(x+y))>0\).. So two case.. A) both POSITIVE.. xy>0 or x>y AND 1>x+y.. B) both NEGATIVE X<y and x+y>1.. Let's see if any statement fulfills any of the above cases.. 1) x>y.. see A, NOTHING about if 1>x+yInsufficient 2) \(x^2>y^2\).. It just tells us that NUMERIC value of (x> y)... Nothing about values with sign.. Insuff Combined.. We can surely say by adding two statements that.. x+x^2>y+y^2 but nothing about given equation.. You can find various combinations in FRACTION or INTEGERS, in + or  that will give us answer as YES and NO Insufficient Hi, But if we add both the inequalities we get x+ x^2 > y + y^2 Now after rearranging 1*(xy)> (xy)(x+y) As from statement 1 we know that x>y so xy is +ve, thus 1>x+y now putting this in the main equation i.e. (xy)(1(x+y))>0 we get that it will be always greater than 0. So, answer should be C instead. Please correct me if I am wrong. Regards, Avi



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Re: Is x  x^2 > y  y^2?
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25 Apr 2019, 23:02
aviaggarwal.ism wrote: chetan2u wrote: anurag16 wrote: Is x  x^2 > y  y^2?
(1) x > y (2) x^2 > y^2 \(xx^2>yy^2....... xy+y^2x^2>0......\).. \((xy)(xy)(x+y)>0...... (xy)(1(x+y))>0\).. So two case.. A) both POSITIVE.. xy>0 or x>y AND 1>x+y.. B) both NEGATIVE X<y and x+y>1.. Let's see if any statement fulfills any of the above cases.. 1) x>y.. see A, NOTHING about if 1>x+yInsufficient 2) \(x^2>y^2\).. It just tells us that NUMERIC value of (x> y)... Nothing about values with sign.. Insuff Combined.. We can surely say by adding two statements that.. x+x^2>y+y^2 but nothing about given equation.. You can find various combinations in FRACTION or INTEGERS, in + or  that will give us answer as YES and NO Insufficient Hi, But if we add both the inequalities we get x+ x^2 > y + y^2 Now after rearranging 1*(xy)> (xy)(x+y) As from statement 1 we know that x>y so xy is +ve, thus 1>x+y now putting this in the main equation i.e. (xy)(1(x+y))>0 we get that it will be always greater than 0. So, answer should be C instead. Please correct me if I am wrong. Regards, Avi Hi Avi, you made a mistake here x+ x^2 > y + y^2 Now after rearranging 1*(xy)> (xy)(x+y) As from statement 1 we know that x>y so xy is +ve, thus 1>x+yx+x^2>y+Y^2 => xy>(x^2Y^2) => (xy) > (x+y)(xy) As x>y, this is reduced to 1>(x+y) or (x+y)>1Now putting this in expression (xy)(1(x+y)) we cant say if the second term (1(x+y) is positive or negative. For x+y between 1 and <1, it will be positive, For x+y>1, it will be negative. Therefore Insufficient.Please hit +1 Kudos if the answer helps.




Re: Is x  x^2 > y  y^2?
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25 Apr 2019, 23:02






