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If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) >

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If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > [#permalink] New post 04 Feb 2011, 16:29
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If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > (x^(-1)*y^(-1))^(-1) ?

(1) x = 2y

(2) x + y > 0


I need to clarify a doubt in these types of questions. Firstly apologies guys for giving away too much info but it is important to clarify a doubt. If you solve the above equations you'll get 1/x+y > 1 ...so my question is why cant you solve it one step further and say is x+y < 1. Now if i follow the initial approach i get 1/3y > 1 for 1st statement so why cant we say 3y< 1. I got my ans wrong because of this . Please can some clarify if i am doing some thing conceptually wrong.

Apologies again for giving out too much.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 04 Dec 2013, 07:51, edited 3 times in total.
Edited the question.
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Re: If x and y are nonzero integers [#permalink] New post 04 Feb 2011, 17:22
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ajit257 wrote:
If x and y are nonzero integers, is (x-1 + y-1)-1 > [(x-1)(y-1)]-1 ?

(1) x = 2y

(2) x + y > 0

I need to clarify a doubt in these types of questions. Firstly apologies guys for giving away too much info but it is important to clarify a doubt. If you solve the above equations you'll get 1/x+y > 1 ...so my question is why cant you solve it one step further and say is x+y < 1. Now if i follow the initial approach i get 1/3y > 1 for 1st statement so why cant we say 3y< 1. I got my ans wrong because of this . Please can some clarify if i am doing some thing conceptually wrong.

Apologies again for giving out too much.


First of all the question should be:

If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > (x^(-1)*y^(-1))^(-1) ?

Is (x^{-1}+y^{-1})^{-1}> (x^{-1}*y^{-1})^{-1}? --> is (\frac{1}{x}+\frac{1}{y})^{-1}>(\frac{1}{xy})^{-1}? --> is (\frac{x+y}{xy})^{-1}>xy --> is \frac{xy}{x+y}>xy?

Now, from this point you can not divide both parts of the inequality by xy and write \frac{1}{x+y}>1 (as you did), because you don't know whether xy is positive or negative: if xy>0 then you should write \frac{1}{x+y}>1 BUT if xy<0 then you should flip the sign and write \frac{1}{x+y}<1. But even if you knew that xy>0 then the next step of writing x+y<1 from \frac{1}{x+y}>1 would still be incorrect for the same exact reason: you don't k now whether x+y is positive or negative, hence you can not muliply both sides of the inequality by x+y.

Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

Thus the question is boiled down to: is \frac{xy}{x+y}>xy? Actually we can manipulate further but there is no need.

(1) x = 2y --> question becomes: is \frac{2y^2}{3y}>2y^2? Now, as we know that y is nonzero then 2y^2>0 and we can divide both parts by it --> is \frac{1}{3y}>1? As y is an integer (no matter positive or negative) then the answer to this question is always NO (if it's a positive integer then \frac{1}{3y}<1 and if it's a negative integer then again: \frac{1}{3y}<0<1). Sufficient.

(2) x + y > 0 --> if x=y=1 then the answer will be NO but if x=3 and y=-1 then the answer will be YES. Not sufficient.

Answer: A.
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Re: If x and y are nonzero integers [#permalink] New post 04 Feb 2011, 17:43
Damn :x
I thought it was x-1 :P

Silly me! Good one man :)
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Re: If x and y are nonzero integers [#permalink] New post 05 Feb 2011, 06:37
apologies AmrithS...i forgot to structure the question after posting it...

Thanks a ton ! Bunuel ....a major concept cleared.
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Re: If x and y are nonzero integers, is (x-1 + y-1)-1 > -1 ? [#permalink] New post 25 Jun 2013, 03:46
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Re: is (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ? [#permalink] New post 28 Jul 2013, 20:25
abilash10 wrote:
If x and y are nonzero integers, is (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ?
(1) x = 2y
(2) x + y > 0
I'm not quite satisfied with the official answer for this question from MGMAT


The original question can be reduced to
Is \frac{xy}{(x+y)} > xy ?

Statement 1
2y^2/3y >2y^2
2y^2 (1-3y)/3y >0 --------eq(2)

If Y is positive, the answer will be + -/+ --->Negative (-)
If Y is negative, the answer will be + +/- --->Negative (-)
Sufficient

Statement 2
x+y>0
If x = + & y = + , then \frac{xy}{(x+y)} > xy will be false
If x = - & y = + , then \frac{xy}{(x+y)} > xy will be True
Thus Insufficient
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Re: is (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ? [#permalink] New post 28 Jul 2013, 20:29
The original question can be reduced to
Is \frac{xy}{(x+y)} > xy ?

why can i not cancel out xy on both sides ? x & y are both nonzero.
if i do that, i'm left with : is (x+y)^-1 >1.

in which case, statement 2 gives me the answer, since it has a positive denominator
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Re: is (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ? [#permalink] New post 28 Jul 2013, 20:31
damn.. i jus realised that they may be non-zero, but xy cud still be negative.. my bad.. thanks... :)
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Re: is (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ? [#permalink] New post 28 Jul 2013, 21:35
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abilash10 wrote:
If x and y are nonzero integers, is (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ?

(1) x = 2y

(2) x + y > 0


I'm not quite satisfied with the official answer for this question from MGMAT


Merging similar topics.
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Re: If x and y are nonzero integers, is (x-1 + y-1)-1 > -1 ? [#permalink] New post 04 Dec 2013, 06:30
ajit257 wrote:
If x and y are nonzero integers, is (x-1 + y-1)-1 > [(x-1)(y-1)]-1 ?

(1) x = 2y

(2) x + y > 0


I need to clarify a doubt in these types of questions. Firstly apologies guys for giving away too much info but it is important to clarify a doubt. If you solve the above equations you'll get 1/x+y > 1 ...so my question is why cant you solve it one step further and say is x+y < 1. Now if i follow the initial approach i get 1/3y > 1 for 1st statement so why cant we say 3y< 1. I got my ans wrong because of this . Please can some clarify if i am doing some thing conceptually wrong.

Apologies again for giving out too much.


This is a very good question. Its such a shame that it is so poorly formatted. Would it be possible to format it correctly just as Bunuel did in his post below? It keeps confusing me every time I attempt to do it with the clock.

Cheers!
J :)
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Re: If x and y are nonzero integers, is (x-1 + y-1)-1 > -1 ? [#permalink] New post 04 Dec 2013, 07:52
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jlgdr wrote:
ajit257 wrote:
If x and y are nonzero integers, is (x-1 + y-1)-1 > [(x-1)(y-1)]-1 ?

(1) x = 2y

(2) x + y > 0


I need to clarify a doubt in these types of questions. Firstly apologies guys for giving away too much info but it is important to clarify a doubt. If you solve the above equations you'll get 1/x+y > 1 ...so my question is why cant you solve it one step further and say is x+y < 1. Now if i follow the initial approach i get 1/3y > 1 for 1st statement so why cant we say 3y< 1. I got my ans wrong because of this . Please can some clarify if i am doing some thing conceptually wrong.

Apologies again for giving out too much.


This is a very good question. Its such a shame that it is so poorly formatted. Would it be possible to format it correctly just as Bunuel did in his post below? It keeps confusing me every time I attempt to do it with the clock.

Cheers!
J :)

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Edited. Thank you.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If x and y are nonzero integers, is (x-1 + y-1)-1 > -1 ?   [#permalink] 04 Dec 2013, 07:52
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