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Is xy > x/y?

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28 May 2014, 09:00
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Is xy > x/y?

(1) 0 < y < 1

(2) xy > 1

[Reveal] Spoiler:
Bunnel could you please look into this.

official ans is C. but i am getting A

Last edited by Bunuel on 28 May 2014, 09:51, edited 1 time in total.
Edited the question.
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28 May 2014, 09:50
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PathFinder007 wrote:
Is xy>xy?
(1) 0<y<1

(2) xy>1

Bunnel could you please look into this.

official ans is C. but i am getting A

I guess you are talking about question m18 q30.

Is $$xy>\frac{x}{y}$$?

Is $$xy>\frac{x}{y}$$? --> is $$x(y-\frac{1}{y})>0$$?

(1) $$0<y<1$$. From this statement it follows that $$(y-\frac{1}{y})<0$$. Now, if $$x>0$$ then $$x(y-\frac{1}{y})<0$$ but if $$x<0$$ then $$x(y-\frac{1}{y})>0$$. Not sufficient.

(2) $$xy>1$$. If $$x=y=2$$ then the answer is YES but if $$x=4$$ and $$y=\frac{1}{2}$$ then the answer is NO. Not sufficient.

(1)+(2) Since from (1) $$y>0$$ and from (2) $$xy>1$$ then $$x>0$$. So, we have that $$x>0$$ and $$(y-\frac{1}{y})<0$$ (from 1), which means that their product $$x(y-\frac{1}{y})<0$$. Sufficient.

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29 May 2014, 11:03
Bunuel wrote:
PathFinder007 wrote:
Is xy>xy?
(1) 0<y<1

(2) xy>1

Bunnel could you please look into this.

official ans is C. but i am getting A

I guess you are talking about question m18 q30.

Is $$xy>\frac{x}{y}$$?

Is $$xy>\frac{x}{y}$$? --> is $$x(y-\frac{1}{y})>0$$?

(1) $$0<y<1$$. From this statement it follows that $$(y-\frac{1}{y})<0$$. Now, if $$x>0$$ then $$x(y-\frac{1}{y})<0$$ but if $$x<0$$ then $$x(y-\frac{1}{y})>0$$. Not sufficient.

(2) $$xy>1$$. If $$x=y=2$$ then the answer is YES but if $$x=4$$ and $$y=\frac{1}{2}$$ then the answer is NO. Not sufficient.

(1)+(2) Since from (1) $$y>0$$ and from (2) $$xy>1$$ then $$x>0$$. So, we have that $$x>0$$ and $$(y-\frac{1}{y})<0$$ (from 1), which means that their product $$x(y-\frac{1}{y})<0$$. Sufficient.

Bunuel,

How come 0<y<1 leads to (y-1/y)<0 ? Isn't it (1-1/y)<0 ?

Regards,
Ammu
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29 May 2014, 11:19
ammuseeru wrote:
Bunuel wrote:
PathFinder007 wrote:
Is xy>xy?
(1) 0<y<1

(2) xy>1

Bunnel could you please look into this.

official ans is C. but i am getting A

I guess you are talking about question m18 q30.

Is $$xy>\frac{x}{y}$$?

Is $$xy>\frac{x}{y}$$? --> is $$x(y-\frac{1}{y})>0$$?

(1) $$0<y<1$$. From this statement it follows that $$(y-\frac{1}{y})<0$$. Now, if $$x>0$$ then $$x(y-\frac{1}{y})<0$$ but if $$x<0$$ then $$x(y-\frac{1}{y})>0$$. Not sufficient.

(2) $$xy>1$$. If $$x=y=2$$ then the answer is YES but if $$x=4$$ and $$y=\frac{1}{2}$$ then the answer is NO. Not sufficient.

(1)+(2) Since from (1) $$y>0$$ and from (2) $$xy>1$$ then $$x>0$$. So, we have that $$x>0$$ and $$(y-\frac{1}{y})<0$$ (from 1), which means that their product $$x(y-\frac{1}{y})<0$$. Sufficient.

Bunuel,

How come 0<y<1 leads to (y-1/y)<0 ? Isn't it (1-1/y)<0 ?

Regards,
Ammu

0<y<1, so y is some fraction between 0 and 1. In this case y<1/y, isn't it? For example, say y=1/2, then 1/y=2 --> y<1/y --> (y-1/y)<0.

Does this make sense?
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Re: Is xy > x/y? [#permalink]

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11 Aug 2014, 17:42
Bonuel,

What's wrong in my reasoning?

Q) xy > x/y ? ---- reasoning: first thing I notice is the trap related to the Y sign. If Y is positive, I may pass Y to the left side.. but if Y is negative, I cannot do this).

(1) 0 < y < 1 ---- reasoning: so Y is positive, the question becomes is XY.Y > X, or X(Y^2)>X? Then I divide both sides by X, then the question becomes: is Y^2 > 1? Since Y is a fraction (because is greater than 0 but less than 1), I can assume that the square of Y will yield a number below 1, so I can answer the question with a "NO". Stat. A is sufficient

What I did wrong?

Thank you so much!
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12 Aug 2014, 02:14
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rodrigocani wrote:
Bonuel,

What's wrong in my reasoning?

Q) xy > x/y ? ---- reasoning: first thing I notice is the trap related to the Y sign. If Y is positive, I may pass Y to the left side.. but if Y is negative, I cannot do this).

(1) 0 < y < 1 ---- reasoning: so Y is positive, the question becomes is XY.Y > X, or X(Y^2)>X? Then I divide both sides by X, then the question becomes: is Y^2 > 1? Since Y is a fraction (because is greater than 0 but less than 1), I can assume that the square of Y will yield a number below 1, so I can answer the question with a "NO". Stat. A is sufficient

What I did wrong?

Thank you so much!

You cannot divide xy^2 > x by x because you don't know the sign of x. If x is positive then yes, y^2 > 1 but if x is negative, then when dividing by it you should flip the sign and you get y^2 < 1.

Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know its sign.

Other tips on Inequalities are here: inequalities-tips-and-hints-175001.html
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13 Aug 2014, 08:16
In this case, we can see:
(1) doesn't mean anything of x --> (1) is insufficent
(2) xy>1.
_ this case is too risky ( when x = 1 and y = 2 --> xy = 2 but x/y = 1/2, contrastly, x = 3 and y =2 --> xy=6 and x/y = 3/2 = 1,5 )
In combination of 2 cases, xy>1 and 0<y<1, we easily see that x must be superior than 1. When x>1 and 0<y<1, x/y is always bigger than xy. Example: 3*0,5= 4,5 but 3/0,5 = 6
So the answer is C !
Is xy > x/y?   [#permalink] 13 Aug 2014, 08:16
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