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Is 1/(x  y) < (y  x) ?
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21 Sep 2010, 04:22
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Is 1/(x  y) < (y  x) ? (1) y is positive (2) x is negative
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Is 1/(xy) < (y  x) ?
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Re: Algebra DS
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21 Sep 2010, 04:33
rxs0005 wrote: Is 1 / ( x  y) < ( y  x )
y is positive
x is negative
Can someone explain the quickest way to solve this Let \(A=(xy)\), then we have : \(\frac{1}{A} < A\) if A>0, then A^2<1. So no solution if A<0, then A^2>1. So all A<0 is a solution Therefore, if we can impose conditions on x,y such that (xy) is >0, then we know the answer is always "No" conversely if (xy) < 0 then we know the answer is always "Yes" (1) not sufficient as it only talks of y (2) not sufficient as it only talks of x (1)+(2) y positive & x negative means (xy)<0. So this inequality is always true. SUFFICIENT Ans is (c)
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Re: Algebra DS
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21 Sep 2010, 04:36
(1) y greater than 0 : example 1 : x = 0.5 and y = 0.2 => 1/0.7>0.3 example 2 : x = 0.2 and y = 0.5 => 1/0.3<0.7 INSUFFICIENT
(2) x less than 0 is the same problem as before ! INSUFFICIENT
(Both) x negative, y positive but whose greater than the other (in absolute value of course)? example 1: x = 2 and y = 3 => 1/(5) < 1 example 2: x = 3 and y = 2 => 1/(5) = 0.2 > 1
ANS: E.
Hope it is clear



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Re: Algebra DS
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Updated on: 21 Sep 2010, 04:47
alexn49 wrote: (Both) x negative, y positive but whose greater than the other (in absolute value of course)? example 1: x = 2 and y = 3 => 1/(5) < 1 example 2: x = 3 and y = 2 => 1/(5) = 0.2 > 1
ANS: E.
Hope it is clear Example 2 is incorrect. RHS is yx which will be +5 not 1. The answer should be c not e
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Originally posted by shrouded1 on 21 Sep 2010, 04:40.
Last edited by shrouded1 on 21 Sep 2010, 04:47, edited 2 times in total.



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Re: Algebra DS
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21 Sep 2010, 04:42
rxs0005 wrote: Is 1 / ( x  y) < ( y  x )
y is positive
x is negative
Can someone explain the quickest way to solve this 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) Since y is positive and x is negative, then \(y>x\). We can rewrite this as \(xy<0\), as well as \(yx>0\). Evaluate LHS and RHS from the question: \((LHS=\frac{1}{xy})<0\), and \((RHS=yx)>0\), therefore \((LHS=negative)<(RHS=positive)\). Sufficient. Answer: C. OR: Is \(\frac{1}{xy}<y  x\) > is \(\frac{1}{xy}+xy<0\) > is \(\frac{1+(xy)^2}{xy}<0\)? (1) \(y\) is positive, clearly insufficient, as no info about \(x\); (2) \(x\) is negative, also insufficient, as no info about \(y\); (1)+(2) Is \(\frac{1+(xy)^2}{xy}<0\)? Now, nominator in this fraction is always positive (1 plus some nonnegative number), but denominator is always negative as \(xy=negativepositive=negative\) (for example: 32=5).So we would have is \(\frac{positive}{negative}<0\)? Which is true. Sufficient. Answer: C. Hope it helps.
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Re: Algebra DS
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23 Dec 2010, 09:01
1 / ( x  y) < ( y  x )  > 1< (yx)*(xy)  > 1<x^2+2xyy^2  > the question is 1> x^22xy+y^2 ? With regard to the stmt (1) it follows from 1> x^22xy+y^2 that x^2+y^2 is always greater than 2xy for any term of x and y. Hence, x^22xy+y^2>1. Even if x were equal to zero, the result will not change. SUFF.
The same solution, i.e. x^22xy+y^2>1, is true also for the case when x is negative according to stmt (2). Here, again it does not matter whether x and y are positive or negative because x^2+y^2 is always greater than 2xy. Here we do not need to know the term of y, anyway the result will not change. Hence stmt (2)  SUFF. Hence, the ans. is D.



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Re: Algebra DS
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23 Dec 2010, 09:18
feruz77 wrote: 1 / ( x  y) < ( y  x )  > 1< (yx)*(xy)  > 1<x^2+2xyy^2  > the question is 1> x^22xy+y^2 ? With regard to the stmt (1) it follows from 1> x^22xy+y^2 that x^2+y^2 is always greater than 2xy for any term of x and y. Hence, x^22xy+y^2>1. Even if x were equal to zero, the result will not change. SUFF.
The same solution, i.e. x^22xy+y^2>1, is true also for the case when x is negative according to stmt (2). Here, again it does not matter whether x and y are positive or negative because x^2+y^2 is always greater than 2xy. Here we do not need to know the term of y, anyway the result will not change. Hence stmt (2)  SUFF. Hence, the ans. is D. Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it.So you can not multiply 1/(xy)<(yx ) by xy and write 1<(yx)*(xy) because you don't know whether xy is positive or negative: if it's positive then you should write 1 <(yx)*(xy) but if its negative then you should flip the sign and write 1 >(yx)*(xy) OA for this question is C not D, refer to the posts above for a solution.
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Re: Is 1/xy<yx? (1) y is positive (2) x is negative
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09 May 2012, 17:18
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
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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
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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: Inequality...involving reciprocals
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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
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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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Re: Is 1/(xy) < (yx) ?
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09 Aug 2014, 02:15
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) ?
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09 Aug 2014, 03:05
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) ?
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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) ?
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Is 1/(xy) < (y  x) ?
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09 Feb 2017, 11:04
good memory, Bunuel you did. And I always forget where the stuff is thanks!




Is 1/(xy) < (y  x) ?
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