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Is xy > x/y?

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Is xy > x/y? [#permalink]

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New post 25 Feb 2011, 08:15
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Is xy > x/y?

(1) xy > 0
(2) y < 0
[Reveal] Spoiler: OA
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Re: how to crack this kind of problems [#permalink]

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New post 25 Feb 2011, 08:27
naaga wrote:
Is xy > x/y?
(1) xy > 0
(2) y < 0



(1) => Both should either be positive or negative .
\(5\)*\(2\) >\(\frac{5}{2}\) (yes)
-5*-2 > -5/-2 (Yes)
\(0.5 *0.5\) > \(0.5/0.5\) (No) Insufficient

(2) => y<0 ; x can be positive or negative Insufficient

(1) + (2) =>
\(-5*-2\) > \(-5/-2\)(Yes)
\(-0.5 * -0.5\)> \(-0.5/-0.5\) (No)

Hence , Clearly E .
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Re: how to crack this kind of problems [#permalink]

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New post 25 Feb 2011, 08:52
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naaga wrote:
Is xy > x/y?
(1) xy > 0
(2) y < 0


Is xy > x/y?

Is \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? is \(\frac{x(y^2-1)}{y}>0\)? --> is \(\frac{x(y+1)(y-1)}{y}>0\)? This inequality holds true when:

A. \(x>0\) and \(y>1\) or \(-1<y<0\);
B. \(x<0\) and \(0<y<1\) or \(y<-1\).

(1) xy > 0 --> \(x\) and \(y\) have the same sign. Not sufficient.
(2) y < 0. Not sufficient.

(1)+(2) Both \(x\) and \(y\) are negative, so we are in scenario B, though we still need more precise range for \(y\) (if \(y<-1\) then the answer will be YES but if \(-1<y<0\) then the answer will be NO). Not sufficient.

Or: we can reduce our expression by x/y (which is positive since \(x\) and \(y\) have the same sign) and the question becomes: is \((y+1)(y-1)>0\)? and as \(y<0\) then the question reduces whether \(y<-1\). But we don't know that and thus even taken together statements are not sufficient.

Answer: E.
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Re: how to crack this kind of problems [#permalink]

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New post 25 Feb 2011, 09:26
Bunuel wrote:
naaga wrote:
Is xy > x/y?
(1) xy > 0
(2) y < 0


Is xy > x/y?

Is \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? is \(\frac{x(y^2-1)}{y}>0\)? --> is \(\frac{x(y+1)(y-1)}{y}>0\)? This inequality holds true when:

A. \(x>0\) and \(y>1\) or \(-1<y<0\);
B. \(x<0\) and \(0<y<1\) or \(y<-1\).


(1) xy > 0 --> \(x\) and \(y\) have the same sign. Not sufficient.
(2) y < 0. Not sufficient.

(1)+(2) Both \(x\) and \(y\) are negative, so we are in scenario B, though we still need more precise range for \(y\) (if \(y<-1\) then the answer will be YES but if \(-1<y<0\) then the answer will be NO). Not sufficient.

Or: we can reduce our expression by x/y (which is positive since \(x\) and \(y\) have the same sign) and the question becomes: is \((y+1)(y-1)>0\)? and as \(y<0\) then the question reduces whether \(y<-1\). But we don't know that and thus even taken together statements are not sufficient.

Answer: E.


Bunuel, how did you deduce A and B above(the range for x and y) from given inequality and how did you appropriately put "and" and "or" to the respective places? I always get confused at this. Could you please show me your way of thinking for that?
thanks
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Re: how to crack this kind of problems [#permalink]

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New post 25 Feb 2011, 10:00
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Expert's post
fluke wrote:
Bunuel wrote:
naaga wrote:
Is xy > x/y?
(1) xy > 0
(2) y < 0


Is xy > x/y?

Is \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? is \(\frac{x(y^2-1)}{y}>0\)? --> is \(\frac{x(y+1)(y-1)}{y}>0\)? This inequality holds true when:

A. \(x>0\) and \(y>1\) or \(-1<y<0\);
B. \(x<0\) and \(0<y<1\) or \(y<-1\).


(1) xy > 0 --> \(x\) and \(y\) have the same sign. Not sufficient.
(2) y < 0. Not sufficient.

(1)+(2) Both \(x\) and \(y\) are negative, so we are in scenario B, though we still need more precise range for \(y\) (if \(y<-1\) then the answer will be YES but if \(-1<y<0\) then the answer will be NO). Not sufficient.

Or: we can reduce our expression by x/y (which is positive since \(x\) and \(y\) have the same sign) and the question becomes: is \((y+1)(y-1)>0\)? and as \(y<0\) then the question reduces whether \(y<-1\). But we don't know that and thus even taken together statements are not sufficient.

Answer: E.


Bunuel, how did you deduce A and B above(the range for x and y) from given inequality and how did you appropriately put "and" and "or" to the respective places? I always get confused at this. Could you please show me your way of thinking for that?
thanks


To check when \(\frac{x(y+1)(y-1)}{y}>0\) holds true consider expressions with x and y separately: the product of two multiples (x and (y+1)(y-1)/y) to be positive they must have the same sign.

So either: \(x>0\) AND \(\frac{(y+1)(y-1)}{y}>0\), which is true when \(y>1\) OR \(-1<y<0\) (for this check: everything-is-less-than-zero-108884.html, hilit=extreme#p868863, here: inequalities-trick-91482.html, xy-plane-71492.html?hilit=solving%20quadratic#p841486, data-suff-inequalities-109078.html);

Or: \(x<0\) AND \(\frac{(y+1)(y-1)}{y}<0\), which is true when \(0<y<1\) OR \(y<-1\).

Hope it's clear.
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Re: how to crack this kind of problems [#permalink]

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New post 25 Feb 2011, 10:57
Well the first statement told that its not gonna help. The second does not add any information. How can you expect to see the answer? mark E and move on !

1) statement just tells x and y are same sign. Insuff
2) statement tells me y is -ve. What about x? Is x = 0? Insuff
Combine 1) + 2)
No new information received.
x/y : numerator can be -1<x<0 or denominator can be -1<x<0. Inequality can face both ways. E
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Re: Is xy > x/y? [#permalink]

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Re: Is xy > x/y? [#permalink]

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New post 29 Mar 2016, 16:29
Bunuel wrote:
naaga wrote:
Is xy > x/y?
(1) xy > 0
(2) y < 0


Is xy > x/y?

Is \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? is \(\frac{x(y^2-1)}{y}>0\)? --> is \(\frac{x(y+1)(y-1)}{y}>0\)? This inequality holds true when:

A. \(x>0\) and \(y>1\) or \(-1<y<0\);
B. \(x<0\) and \(0<y<1\) or \(y<-1\).

(1) xy > 0 --> \(x\) and \(y\) have the same sign. Not sufficient.
(2) y < 0. Not sufficient.

(1)+(2) Both \(x\) and \(y\) are negative, so we are in scenario B, though we still need more precise range for \(y\) (if \(y<-1\) then the answer will be YES but if \(-1<y<0\) then the answer will be NO). Not sufficient.

Or: we can reduce our expression by x/y (which is positive since \(x\) and \(y\) have the same sign) and the question becomes: is \((y+1)(y-1)>0\)? and as \(y<0\) then the question reduces whether \(y<-1\). But we don't know that and thus even taken together statements are not sufficient.

Answer: E.




Couple of questions:

1. How can we multiply both sides by y when we do not know if y is +ve or -ve.
2. I did not understand \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? why is there a y in the denominator?
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Re: Is xy > x/y? [#permalink]

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New post 10 Apr 2016, 10:37
Avinashs87 wrote:
Bunuel wrote:
naaga wrote:
Is xy > x/y?
(1) xy > 0
(2) y < 0


Is xy > x/y?

Is \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? is \(\frac{x(y^2-1)}{y}>0\)? --> is \(\frac{x(y+1)(y-1)}{y}>0\)? This inequality holds true when:

A. \(x>0\) and \(y>1\) or \(-1<y<0\);
B. \(x<0\) and \(0<y<1\) or \(y<-1\).

(1) xy > 0 --> \(x\) and \(y\) have the same sign. Not sufficient.
(2) y < 0. Not sufficient.

(1)+(2) Both \(x\) and \(y\) are negative, so we are in scenario B, though we still need more precise range for \(y\) (if \(y<-1\) then the answer will be YES but if \(-1<y<0\) then the answer will be NO). Not sufficient.

Or: we can reduce our expression by x/y (which is positive since \(x\) and \(y\) have the same sign) and the question becomes: is \((y+1)(y-1)>0\)? and as \(y<0\) then the question reduces whether \(y<-1\). But we don't know that and thus even taken together statements are not sufficient.

Answer: E.




Couple of questions:

1. How can we multiply both sides by y when we do not know if y is +ve or -ve.
2. I did not understand \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? why is there a y in the denominator?


We do not multiply by y. We re-arrange xy > x/y as xy - x/y > 0 by subtracting x/y from both sides.
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Resources:
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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: Is xy > x/y?   [#permalink] 10 Apr 2016, 10:37
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