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

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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.

Originally posted by ajit257 on 04 Feb 2011, 16:29.
Last edited by Bunuel on 16 Feb 2019, 02:24, edited 5 times in total.
Edited the question.
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Re: If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) >  [#permalink]

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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 cannot 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.

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

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Damn I thought it was x-1 Silly me! Good one man _________________
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Re: If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) >  [#permalink]

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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) >  [#permalink]

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

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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

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

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

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damn.. i jus realised that they may be non-zero, but xy cud still be negative.. my bad.. thanks... Intern  Joined: 03 Aug 2014
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Re: If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) >  [#permalink]

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Hi Bunuel - in your explanation above, how do you get from (1/x + 1/y)^(-1) to ((x+y)/xy)^(-1)?
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Re: If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) >  [#permalink]

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cg0588 wrote:
Hi Bunuel - in your explanation above, how do you get from (1/x + 1/y)^(-1) to ((x+y)/xy)^(-1)?

Does not $$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$$ ?
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Re: If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) >  [#permalink]

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I solved it this way:

from the original inequality --> is xy/y+x > xy --> is xy((1/x+y) -1) > 0? to satisfy the inequality either xy > 0 and (1/x+y)-1 > 0 or xy < 0 and (1/x+y)-1 < 0.

Notice that 1/x+y is a proper fraction. So unless x+y = 1 this expression (1/x+y) is going to be negative. If that expression is negative we want xy to be negative as well.

statement 1. x=2y --> x/y=2/1 x and y have the same sign. If x and y are negative ((1/x+y) -1) is surely negative and xy is positive. Thus the overall expression is not > 0. If x and y are positive ((1/x+y) -1) is also negative and xy is positive. Thus the overall expression is not > 0.

sufficient

statement 2. x+y >0 we don't know the exact values of x and y. Assume that x is negative and y is positive and x<y then x+y > 0 still ((1/x+y) -1) is negative and xy is negative too, making the overall expression > 0. Assume that both x and y are positive and you end up with the same scenario as statement 1. Assume that x+y=1 and the overall expression becomes 0.

not sufficient.

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

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Bunuel 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.

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 cannot 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.

Dear Bunuel, Could you elaborate on the highlighted sentence as I am confused with the sign  _________________
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Re: If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) >  [#permalink]

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ziyuenlau wrote:
Bunuel 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.

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 cannot 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.

Dear Bunuel, Could you elaborate on the highlighted sentence as I am confused with the sign  We know that y is an integer. Now, for integer y, the answer to the question whether $$\frac{1}{3y}>1$$ will always be NO (no matter whether y is negative or positive). If y > 0, then $$\frac{1}{3y}<1$$ and if y < 0, then again $$\frac{1}{3y}<1$$.

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

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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

Target question: Is $$(x^{-1}+y^{-1})^{-1}> (x^{-1}*y^{-1})^{-1}$$
This is a good candidate for rephrasing the target question.

Take: $$(x^{-1}+y^{-1})^{-1}> (x^{-1}*y^{-1})^{-1}$$

Rewrite as: $$(\frac{1}{x}+\frac{1}{y})^{-1}> (\frac{1}{x}*\frac{1}{y})^{-1}$$

Rewrite as: $$(\frac{y}{xy}+\frac{x}{xy})^{-1}> (\frac{1}{xy})^{-1}$$

Simplify to get: $$(\frac{x+y}{xy})^{-1}> (\frac{1}{xy})^{-1}$$

Apply exponents to get: $$\frac{xy}{x+y}> xy$$

REPHRASED target question: Is $$\frac{xy}{x+y}> xy$$?

Aside: the video below has tips on rephrasing the target question

Statement 1: $$x = 2y$$

Replace $$x$$ with $$2y$$ to get: Is $$\frac{(2y)y}{(2y)+y}> (2y)y$$?

Simplify to get: Is $$\frac{2y^2}{3y}> 2y^2$$?

Since $$2y^2$$ must be POSITIVE, we can safely divide both sides of the inequality by $$2y^2$$ to get: Is $$\frac{1}{3y}> 1$$?

IMPORTANT: Since y is an INTEGER, 1/3y will ALWAYS be less than 1.
For example, if y = 1, we get: 1/3 < 1
If y = 2, we get: 1/6 < 1
If y = 3, we get: 1/9 < 1
etc.
Likewise, if y is NEGATIVE, then 1/3y will always be NEGATIVE, and a negative value is always less than 1.

So, the answer to the REPHRASED target question is NO, $$\frac{xy}{x+y}$$ is NOT greater than $$xy$$
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x + y > 0
Let's TEST some values.
There are several values of x and y that satisfy statement 2. Here are two:

Case a: x = 2 and y = 1. Plug into REPHRASED target question to get: $$\frac{(2)(1)}{2+1}> (2)(1)$$.
Simplify: $$\frac{2}{3}> 2$$ In this case, the answer to the REPHRASED target question is NO, $$\frac{xy}{x+y}$$ is NOT greater than $$xy$$

Case b: x = 3 and y = -1. Plug into REPHRASED target question to get: $$\frac{(3)(-1)}{3+(-1)}> (3)(-1)$$.
Simplify: $$\frac{-3}{2}> -3$$ In this case, the answer to the REPHRASED target question is YES, $$\frac{xy}{x+y}$$ IS greater than $$xy$$

Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Cheers,
Brent

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

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Bunuel 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.

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 cannot 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.

Hi

Liked how easy your solution is I have a silly doubt/clarification that I need to seek, while solving the question I arrived at 2y^2/3y but I further simplified it to 2y/3>2y^2, I wanted to ask the following:
Is this step conceptually correct? I assumed values for Y (positive and negative) and therefore A wasn't sufficient.
Would request experts to highlight the gap in my understanding

EDIT- After posting my query I realized the operation has to be the same on both sides of the inequality so even if I consider the case 2y/3>2y I will still arrive that A is not sufficient. Please address where I am going wrong. Thanks If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) >   [#permalink] 27 Dec 2019, 20:42
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