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Is xy > x/y?

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Is xy > x/y? [#permalink] New post 25 Feb 2011, 07:15
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A
B
C
D
E

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Question Stats:

80% (02:15) correct 20% (01:09) wrong based on 49 sessions
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] New post 25 Feb 2011, 07:27
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] New post 25 Feb 2011, 07:52
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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] New post 25 Feb 2011, 08:26
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] New post 25 Feb 2011, 09:00
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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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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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: how to crack this kind of problems [#permalink] New post 25 Feb 2011, 09:57
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
Re: how to crack this kind of problems   [#permalink] 25 Feb 2011, 09:57
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