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Is x + y > 1? [#permalink]
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09 Apr 2016, 14:31
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32% (01:35) correct 68% (01:17) wrong based on 73 sessions
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Is x + y > 1? (1) x^2  y^2 > 1 (2) x/y + 1 > 0, where y is positive.
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Last edited by chetan2u on 26 Oct 2017, 03:07, edited 3 times in total.
Corrected the OA



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Re: Is x + y > 1? [#permalink]
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09 Apr 2016, 19:54
iMyself wrote: is x+y>1?
1) x^2y^2>1 2) x/y+1>0, where y is positive. Hi, lets see the statements 1) \(x^2y^2>1\)OR (xy)(x+y)>1.. Insuff 2) \(\frac{x}{y}+1>0\), where y is positive(x+y)/y>0 since y is positive, x+y is also positive.. but we cannot say if x+y>1 insuff combined..x+y is positive , so xy is also +ive.. which means x>y..Now \(x^2y^2>1....x^2>1+y^2\) so so \(x^2>1\) and as x is POSITIVE, x>1 so x+y will be >1 sufficient C
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Re: Is x + y > 1? [#permalink]
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09 Apr 2016, 22:38
Thanks chetan2u. So, what will be the difficulty level of this question?
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Re: Is x + y > 1? [#permalink]
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09 Apr 2016, 22:41
iMyself wrote: Thanks chetan2u. So, what will be the difficulty level of this question? hi, It should be close to 600650 since it contains two variables and inequality too..
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Re: Is x + y > 1? [#permalink]
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09 Apr 2016, 22:47
chetan2u wrote: iMyself wrote: Thanks chetan2u. So, what will be the difficulty level of this question? hi, It should be close to 600650 since it contains two variables and inequality too.. how can we be sure that it is close to 600650 by seeing ''since it contains two variables and inequality too''. I want to know just for curiosity.
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Re: Is x + y > 1? [#permalink]
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09 Apr 2016, 23:59
iMyself wrote: chetan2u wrote: iMyself wrote: Thanks chetan2u. So, what will be the difficulty level of this question? hi, It should be close to 600650 since it contains two variables and inequality too.. how can we be sure that it is close to 600650 by seeing ''since it contains two variables and inequality too''. I want to know just for curiosity. Personal Opinion => This is a 650  700 level.. Again => This is a personal opinion ..
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Re: Is x + y > 1? [#permalink]
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10 Apr 2016, 00:26
iMyself wrote: chetan2u wrote: iMyself wrote: Thanks chetan2u. So, what will be the difficulty level of this question? hi, It should be close to 600650 since it contains two variables and inequality too.. how can we be sure that it is close to 600650 by seeing ''since it contains two variables and inequality too''. I want to know just for curiosity. Hi, generally inequalities and two variables Q are considered to be difficult by any average person, BUT since the Q has some straight forward equations, it can be 600650 level.. However difficulty level is each ones perception. A difficult topic such as prob may be easy for you but very difficult for others
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Re: Is x + y > 1? [#permalink]
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10 Apr 2016, 00:42
Thanks @chetan2u. I got it
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Last edited by iMyself on 10 Apr 2016, 06:56, edited 1 time in total.



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Re: Is x + y > 1? [#permalink]
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10 Apr 2016, 11:05
GMAT is nothing but all about technique. So could you please give a technique of this problem? Can you share unconventional method/shortcut way for THIS problem? Thanks...
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Is x + y > 1? [#permalink]
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26 Oct 2017, 02:44
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This post received KUDOS
chetan2u wrote: Hi, I think you are subconsciously taking x and y as integers.. but say if x and y are fractions.. x= 1/2 and y = 1/4 .. x+ y <1.. But x= 1/2 and y = 1/4 > x^2  y^2 < 1 > statement 1 is not satisfied. Hi Bunuel! From (1) and (2), we have y > 0, x + y > 0 and x  y > 0 (Or x > y). Assume x + y <= 1 > x  y <= x + y <= 1 > x^2  y^2 = (xy)(x+y) <= 1 > contradict the statement (1) > x + y> 1 > C I think the OA must be C, please confirm! Many thanks ^^
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Last edited by leanhdung on 26 Oct 2017, 04:33, edited 1 time in total.



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Re: Is x + y > 1? [#permalink]
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26 Oct 2017, 03:05
leanhdung wrote: chetan2u wrote: Hi, I think you are subconsciously taking x and y as integers.. but say if x and y are fractions.. x= 1/2 and y = 1/4 .. x+ y <1.. But x= 1/2 and y = 1/4 > x^2  y^2 < 1 > statement 1 is not satisfied. Hi Bunuel! From (1) and (2), we have y > 0, x + y > 0 and x  y > 0 (Or x > y). Assume x + y <= 1 > x  y <= x + y <= 1 > x^2  y^2 = (xy)(x+y) <= 1 > contradict the statement (1) > x^2  y^2> 1 > C I think the OA must be C, please confirm! Many thanks ^^ Hi.. You are absolutely correct OA has to be C.. editing the OA +1 kudos
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