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If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > [#permalink]
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.

Re: If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > [#permalink]
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.

Re: If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > [#permalink]
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 ?

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: If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > [#permalink]
21 Oct 2014, 06:12

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