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Is 1/(xy) < (y  x) ? [#permalink]
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15 Jan 2012, 06:26
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Is 1/(xy) < (y  x) ? (1) y is positive. (2) x is negative.
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Re: Inequality...involving reciprocals [#permalink]
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15 Jan 2012, 06:35



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Re: Is 1/xy<yx? (1) y is positive (2) x is negative [#permalink]
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15 Jan 2012, 10:07
Is 1/(xy) < y  x ?
1 ______  (yx) < 0 (xy)
(1) y is positive. (2) x is negative.
From 1 and 2 
(xy) is always negative. (yx) will be positive (yx) will be negative
Adding 2 negatives = ve
If it were 1/(xy) < x  y then answer would have been E



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Re: Is 1/xy<yx? (1) y is positive (2) x is negative [#permalink]
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09 May 2012, 17:18
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Slightly different approach, I hope everyone can see the attached images with solutions. Dabral Apoorva81 wrote: Is 1/xy < y  x ?
(1) y is positive. (2) x is negative.
can you please provide detailed explanation..??
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Re: Inequalities [#permalink]
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02 Apr 2013, 20:52
In order to separate your x's and y's from each other, you need to know which term is larger. This is because you need to know whether xy or yx is negative (one of them will be, unless they are equal). Taking both statements together, you know that the two variables are not equal, and you can manipulate the inequality, keeping in mind that you need to flip the sign when multiplying or dividing by a negative number.
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Re: Inequalities [#permalink]
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02 Apr 2013, 20:53
I explained it in kind of a backwards way. The first thing you need to know in that problem is whether x and y are equal. No single statement tells you that. Sorry for the double post, but it seems I can't edit my previous post while viewing the forum with Tapatalk. Sent from my HTC Glacier using Tapatalk 2
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Re: Is 1/(xy) < (y  x) ? [#permalink]
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22 Feb 2014, 14:15



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Re: Inequality...involving reciprocals [#permalink]
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09 May 2014, 08:38
Bunuel wrote: Apoorva81 wrote: Is 1/xy < y  x ?
(1) y is positive. (2) x is negative.
can you please provide detailed explanation..?? Is \(\frac{1}{xy}<y  x\)? (1) \(y\) is positive, clearly insufficient, as no info about \(x\); (2) \(x\) is negative, also insufficient, as no info about \(y\); (1)+(2) We have \(y=positive\) and \(x=negative\), thus \(y>x\) (this can be rewritten as \(yx>0\) or \(0>xy\)). Now: \(LHS=\frac{1}{xy}=\frac{1}{negative}=negative\), and \(RHS=yx=positive\) thus \(\frac{1}{xy}=negative<yx=positive\). Sufficient. Answer: C. Why can't we do the following rephrase? 1/(xy) < (yx) 1/(yx) < (yx) 1 < (yx)^2 since the RHS is a square, irrespective of x and y values the equation is satisfied.



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Re: Inequality...involving reciprocals [#permalink]
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09 May 2014, 09:25
rishiroadster wrote: Bunuel wrote: Apoorva81 wrote: Is 1/xy < y  x ?
(1) y is positive. (2) x is negative.
can you please provide detailed explanation..?? Is \(\frac{1}{xy}<y  x\)? (1) \(y\) is positive, clearly insufficient, as no info about \(x\); (2) \(x\) is negative, also insufficient, as no info about \(y\); (1)+(2) We have \(y=positive\) and \(x=negative\), thus \(y>x\) (this can be rewritten as \(yx>0\) or \(0>xy\)). Now: \(LHS=\frac{1}{xy}=\frac{1}{negative}=negative\), and \(RHS=yx=positive\) thus \(\frac{1}{xy}=negative<yx=positive\). Sufficient. Answer: C. Why can't we do the following rephrase? 1/(xy) < (yx) 1/(yx) < (yx) 1 < (yx)^2 since the RHS is a square, irrespective of x and y values the equation is satisfied. The point is that we don't know whether yx is positive or negative. If it's positive, then yes we'd have 1 < (yx)^2 but if it's negative, then we'd have 1 > (yx)^2 (flip the sign when multiplying by negative value). Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.
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Is 1/(xy) < (yx) ? [#permalink]
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09 Aug 2014, 01:53
Is 1/(xy) < (yx) ? 1) x is +ve 2) y is ve I took a very long time to solve this Does anyone know a shortcut? or a simpler method? Source: 4gmat



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Re: Is 1/(xy) < (yx) ? [#permalink]
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09 Aug 2014, 02:15
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alphonsa wrote: Is 1/(xy) < (yx) ? 1) x is +ve 2) y is ve I took a very long time to solve this Does anyone know a shortcut? or a simpler method? Source: 4gmat The question can be re written as \(\frac{1}{(xy)} +( xy )<0\) Or \(\frac{1+ (xy)^2}{(xy)} <0\) For the expression to be less than zero,numerator and denominator should be of opp sign Now\((xy)^2\geq{0}\) So numerator is positive ,therefore if denominator is negative then answer to the question is yes or else no st 1 x is positive let x=1 and y=10 then xy=9answer is no But if x=1 and y=3 then answer to the question is yes St 1 not sufficient .. Option A and D ruled out St 2 y is negative If y= 3 and x =8 then ans to the question is yes If y=3 and x =8 then answer to the q. Is no St 2 alone not sufficient Combining we see that xy=+ve So ans is C
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Re: Is 1/(xy) < (yx) ? [#permalink]
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09 Aug 2014, 02:54
With the way you put it, this question seemed like a below 500 question :\
I don't know why I took so long to decipher it was option C .



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Re: Is 1/(xy) < (yx) ? [#permalink]
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09 Aug 2014, 03:05
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WoundedTiger wrote: The question can be re written as
\(\frac{1}{(xy)} +( xy )<0\)
Or \(\frac{1+ (xy)^2}{(xy)} <0\) For the expression to be less than zero,numerator and denominator should be of opp sign Now\((xy)^2\geq{0}\)
So numerator is positive ,therefore if denominator is negative then answer to the question is yes or else no
st 1 x is positive let x=1 and y=10 then xy=9answer is no But if x=1 and y=3 then answer to the question is yes
St 1 not sufficient .. Option A and D ruled out
St 2 y is negative If y= 3 and x =8 then ans to the question is yes If y=3 and x =8 then answer to the q. Is no
St 2 alone not sufficient
Combining we see that xy=+ve
So ans is C
WoundedTiger, it's better to do one step more. In data sufficiency if you have very simple statements you need to understand what you are looking for. So, a little bit different solution: The question can be re written as \(\frac{1}{(xy)} +( xy )<0\) Or \(\frac{1+ (xy)^2}{(xy)} <0\) Since \(1+(xy)^2>0\), denominator must be negative: \(xy<0\) or \(x<y\). Hence, we need to find: is \(x\) less than \(y\)? (1) Don't know anything about \(y\). Insufficient(2) Don't know anything about \(x\). Insufficient(1)+(2) Since \(x>0\) and \(y<0\), we have \(x>y\). SufficientThe correct answer is C.
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Re: Is 1/(xy) < (yx) ? [#permalink]
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09 Aug 2014, 03:31
alphonsa wrote: With the way you put it, this question seemed like a below 500 question :\
I don't know why I took so long to decipher it was option C . It happens to the best of people especially if the question is done during exam time.... One more thing..I think you need to look at these links for books for your preparation bestgmatmathprepbooksreviewsrecommendations77291180.htmlgmatprepsoftwareanalysisandwhatifscenarios146146.html#p1171760
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Re: Is 1/(xy) < (yx) ? [#permalink]
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09 Aug 2014, 04:43
smyarga wrote: WoundedTiger wrote: The question can be re written as
\(\frac{1}{(xy)} +( xy )<0\)
Or \(\frac{1+ (xy)^2}{(xy)} <0\) For the expression to be less than zero,numerator and denominator should be of opp sign Now\((xy)^2\geq{0}\)
So numerator is positive ,therefore if denominator is negative then answer to the question is yes or else no
st 1 x is positive let x=1 and y=10 then xy=9answer is no But if x=1 and y=3 then answer to the question is yes
St 1 not sufficient .. Option A and D ruled out
St 2 y is negative If y= 3 and x =8 then ans to the question is yes If y=3 and x =8 then answer to the q. Is no
St 2 alone not sufficient
Combining we see that xy=+ve
So ans is C
WoundedTiger, it's better to do one step more. In data sufficiency if you have very simple statements you need to understand what you are looking for. So, a little bit different solution: The question can be re written as \(\frac{1}{(xy)} +( xy )<0\) Or \(\frac{1+ (xy)^2}{(xy)} <0\) Since \(1+(xy)^2>0\), denominator must be negative: \(xy<0\) or \(x<y\). Hence, we need to find: is \(x\) less than \(y\)? (1) Don't know anything about \(y\). Insufficient(2) Don't know anything about \(x\). Insufficient(1)+(2) Since \(x>0\) and \(y<0\), we have \(x>y\). SufficientThe correct answer is C. I tried this way. we have 1/(xy) < yx 1) says x is positive, we have no value of Y, not sufficient 2) says y is negative, no value of x defined, not sufficient. 1+2 , x is +ve and y is ve => 1/+ve < ve => +ve < ve, not possible. Answer is C



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Re: Is 1/(xy) < (y  x) ? [#permalink]
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13 Mar 2016, 07:16
Excellent Explanation dabral Also i did it a bit differently.. I made two cases => one for Y>x and the other for Y<x hence i marked C
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Re: Is 1/(xy) < (y  x) ? [#permalink]
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02 Dec 2016, 02:27
In the question it does not say it has to be an integer. The solution does not work for rational numbers. Could you explain? Posted from GMAT ToolKit



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