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Is 1/(x-y) < (y - x) ?

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Is 1/(x-y) < (y - x) ? [#permalink] New post 15 Jan 2012, 05:26
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Is 1/(x-y) < (y - x) ?

(1) y is positive.
(2) x is negative.
[Reveal] Spoiler: OA
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Re: Inequality...involving reciprocals [#permalink] New post 15 Jan 2012, 05:35
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Apoorva81 wrote:
Is 1/x-y < y - x ?

(1) y is positive.
(2) x is negative.

can you please provide detailed explanation..??


Is \frac{1}{x-y}<y - x?

(1) y is positive, clearly insufficient, as no info about x;
(2) x is negative, also insufficient, as no info about y;

(1)+(2) We have y=positive and x=negative, thus y>x (this can be rewritten as y-x>0 or 0>x-y). Now: LHS=\frac{1}{x-y}=\frac{1}{negative}=negative, and RHS=y-x=positive thus \frac{1}{x-y}=negative<y-x=positive. Sufficient.

Answer: C.
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Re: Is 1/x-y<y-x? (1) y is positive (2) x is negative [#permalink] New post 15 Jan 2012, 09:07
Is 1/(x-y) < y - x ?

1
______ - (y-x) < 0
(x-y)


(1) y is positive.
(2) x is negative.

From 1 and 2 -

(x-y) is always negative.
(y-x) will be positive
-(y-x) will be negative

Adding 2 negatives = -ve

If it were 1/(x-y) < x - y then answer would have been E
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Re: Is 1/x-y<y-x? (1) y is positive (2) x is negative [#permalink] New post 09 May 2012, 16:18
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Slightly different approach, I hope everyone can see the attached images with solutions.

Dabral


Apoorva81 wrote:
Is 1/x-y < y - x ?

(1) y is positive.
(2) x is negative.

can you please provide detailed explanation..??

Attachments

image1.png
image1.png [ 57.59 KiB | Viewed 1470 times ]

image2.png
image2.png [ 44.46 KiB | Viewed 1470 times ]


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Re: Inequalities [#permalink] New post 02 Apr 2013, 19:52
In order to separate your x's and y's from each other, you need to know which term is larger. This is because you need to know whether x-y or y-x is negative (one of them will be, unless they are equal). Taking both statements together, you know that the two variables are not equal, and you can manipulate the inequality, keeping in mind that you need to flip the sign when multiplying or dividing by a negative number.
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Re: Inequalities [#permalink] New post 02 Apr 2013, 19:53
I explained it in kind of a backwards way. The first thing you need to know in that problem is whether x and y are equal. No single statement tells you that.

Sorry for the double post, but it seems I can't edit my previous post while viewing the forum with Tapatalk.

Sent from my HTC Glacier using Tapatalk 2
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Re: Is 1/(x-y) < (y - x) ? [#permalink] New post 22 Feb 2014, 13:15
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Re: Inequality...involving reciprocals [#permalink] New post 09 May 2014, 07:38
Bunuel wrote:
Apoorva81 wrote:
Is 1/x-y < y - x ?

(1) y is positive.
(2) x is negative.

can you please provide detailed explanation..??


Is \frac{1}{x-y}<y - x?

(1) y is positive, clearly insufficient, as no info about x;
(2) x is negative, also insufficient, as no info about y;

(1)+(2) We have y=positive and x=negative, thus y>x (this can be rewritten as y-x>0 or 0>x-y). Now: LHS=\frac{1}{x-y}=\frac{1}{negative}=negative, and RHS=y-x=positive thus \frac{1}{x-y}=negative<y-x=positive. Sufficient.

Answer: C.



Why can't we do the following rephrase?

1/(x-y) < (y-x)

-1/(y-x) < (y-x)

-1 < (y-x)^2

since the RHS is a square, irrespective of x and y values the equation is satisfied.
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Re: Inequality...involving reciprocals [#permalink] New post 09 May 2014, 08:25
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rishiroadster wrote:
Bunuel wrote:
Apoorva81 wrote:
Is 1/x-y < y - x ?

(1) y is positive.
(2) x is negative.

can you please provide detailed explanation..??


Is \frac{1}{x-y}<y - x?

(1) y is positive, clearly insufficient, as no info about x;
(2) x is negative, also insufficient, as no info about y;

(1)+(2) We have y=positive and x=negative, thus y>x (this can be rewritten as y-x>0 or 0>x-y). Now: LHS=\frac{1}{x-y}=\frac{1}{negative}=negative, and RHS=y-x=positive thus \frac{1}{x-y}=negative<y-x=positive. Sufficient.

Answer: C.



Why can't we do the following rephrase?

1/(x-y) < (y-x)

-1/(y-x) < (y-x)

-1 < (y-x)^2

since the RHS is a square, irrespective of x and y values the equation is satisfied.


The point is that we don't know whether y-x is positive or negative. If it's positive, then yes we'd have -1 < (y-x)^2 but if it's negative, then we'd have -1 > (y-x)^2 (flip the sign when multiplying by negative value).

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.
_________________

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: Inequality...involving reciprocals   [#permalink] 09 May 2014, 08:25
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