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Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b

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Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 30 Jun 2006, 05:49
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Is 1/(a-b) < b-a ?

(1) a<b
(2) 1<|a-b|
[Reveal] Spoiler: OA
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Re: DS - 1/(a-b) < b-a (from GMATPrep) [#permalink] New post 30 Jun 2006, 09:41
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likar wrote:
Is 1/(a-b) < b-a ?

(1) a<b
(2) 1<|a-b|


To start with a,b belong to a set of real numbers, since nothing else is known.

Let b=a+k (where k is a +ve or -ve real)
=>k = b-a

To prove

1/(a-b) < b-a
=>1/(a-a-k)<a+k-a

=> -1/k < k ....(1)

From 1:
a<b.
=>0<b-a
=>0<k
Therefore k is +ve. Therefore 1 is sufficient.

From 2:
1<|a-b|
=> (a-b)>1 OR -(a-b)>1
=> -k>1 OR k>1
=> k<-1 OR k>1
If k<-1 say -2, then (1) becomes:-
1/2 < -1 which means the inequality does not hold. So 2 is insufficient.

I'd say A.
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 [#permalink] New post 30 Jun 2006, 10:18
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A

1/(a-b) < b-a ?

St1: a-b <0. This means left hand side term will always be -ve and right hand side term will always be +ve.: SUFF

St2: 1 < |a-b| we can't say anything about the sign of (a-b) so either side can be either positive or negative. : INSUFF
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Re: DS - 1/(a-b) < b-a (from GMATPrep) [#permalink] New post 02 Jul 2006, 07:31
likar wrote:
Is 1/(a-b) < b-a ?

(1) a<b
(2) 1<|a-b|


AD-BCE

rephrasing the Q stem,

1/(a-b) < b-a => 1 < (a-b)(b-a) => 1 < - (a-b)^2,

(a-b)^2 will always be +ve regardless of whether a < b or not, so INSUFF

therefore, BCE are choices.

taking second condition,
1 < |a-b| => -(a-b)^2 is definitely < 1 therefore B is SUFF

(B)

but i'm not too sure of my answer...now let me see others solutions :)
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Re: DS - 1/(a-b) < b-a (from GMATPrep) [#permalink] New post 10 Feb 2011, 11:36
guys im concerned with (2)

im getting:

1/a-b< b-a?
1/a-b< - (a-b) ?

1st scenario if a-b >
1< - (a-b)^2

2nd scenario:
of a-b<0
a
1< - (a-b)^2

why im geting the same answer? could somebody advice whats wrong?
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Re: DS - 1/(a-b) < b-a (from GMATPrep) [#permalink] New post 10 Feb 2011, 12:14
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tinki wrote:
guys im concerned with (2)

im getting:

1/a-b< b-a?
1/a-b< - (a-b) ?

1st scenario if a-b >
1< - (a-b)^2

2nd scenario:
of a-b<0
a
1< - (a-b)^2


why im geting the same answer? could somebody advice whats wrong?


How did you even find this thread?

Anyway: when you consider 2nd scenario (a-b<0) and multiply the inequality by this expression then you must flip the sign as you multiply be the negative value --> 1>- (a-b)^2.

But you don't actually need to manipulate this way.

Is 1/(a-b) < b-a ?

(1) a<b --> we can rewrite this as: a-b<0 so LHS is negative, also we can rewrite it as: b-a>0 so RHS is positive --> negative<positive. Sufficient.

(2) 1<|a-b| --> if a-b=2 (or which is the same b-a=-2) then LHS>0 and RHS<0 and in this case the answer will be NO if a-b=-2 (or which is the same b-a=2) then LHS<0 and RHS>0 and in this case the answer will be YES. Not sufficient.

Answer: A.
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Re: DS - 1/(a-b) < b-a (from GMATPrep) [#permalink] New post 10 Feb 2011, 12:33
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1/(a-b) < b-a ?

1/(a-b) - (b-a) < 0
1/(a-b) + (a-b) < 0
(1+(a-b)^2)/(a-b) < 0

(1+(a-b)^2)= +ve

a-b < 0

Q: Is a<b?

(1) a<b. Sufficient.

(2) 1<|a-b|

|a-b| > 1

(a-b) > 1
a > 1+b
Answer: No.

or

a-b < -1
a < b-1
Answer: Yes
Not sufficient.

Ans: "A"
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Re: DS - 1/(a-b) < b-a (from GMATPrep) [#permalink] New post 11 Feb 2011, 02:29
Bunuel wrote:
tinki wrote:
guys im concerned with (2)

im getting:

1/a-b< b-a?
1/a-b< - (a-b) ?

1st scenario if a-b >
1< - (a-b)^2

2nd scenario:
of a-b<0
a
1< - (a-b)^2


why im geting the same answer? could somebody advice whats wrong?


How did you even find this thread?

Anyway: when you consider 2nd scenario (a-b<0) and multiply the inequality by this expression then you must flip the sign as you multiply be the negative value --> 1>- (a-b)^2.

But you don't actually need to manipulate this way.

Is 1/(a-b) < b-a ?

(1) a<b --> we can rewrite this as: a-b<0 so LHS is negative, also we can rewrite it as: b-a>0 so RHS is positive --> negative<positive. Sufficient.

(2) 1<|a-b| --> if a-b=2 (or which is the same b-a=-2) then LHS>0 and RHS<0 and in this case the answer will be NO if a-b=-2 (or which is the same b-a=2) then LHS<0 and RHS>0 and in this case the answer will be YES. Not sufficient.

Answer: A.


Thanks Bunuel, u re my Math angels, always there when help is needed.
1 more request : could you specify definition for abbreviations for LHS and RHS ?
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Re: DS - 1/(a-b) < b-a (from GMATPrep) [#permalink] New post 11 Feb 2011, 02:37
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tinki wrote:
Thanks Bunuel, u re my Math angels, always there when help is needed.
1 more request : could you specify definition for abbreviations for LHS and RHS ?


LHS = Left Hand Side;
RHS = Right Hand Side.
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Re: DS - 1/(a-b) < b-a (from GMATPrep) [#permalink] New post 11 Feb 2011, 02:40
thanks bunuel , thanks fluke
+kudos to both
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Re: DS - 1/(a-b) < b-a (from GMATPrep) [#permalink] New post 23 Feb 2011, 20:19
Agree with A.

More on part B, if :

1< |a-b|, this gives two inequalities

1< a-b or

1< -(a-b) = 1< b-a = -1> a-b

This gives two inequalities, which allow (a-b) to be anywhere on interval (-Infinite, -1) or (1, infininty)

If a-b is positive, then b-a is negative....proof a-b>0.....a>b......0>b-a

Therefore 1/positive < negative is false

If a-b is negative, then b-a is positive

Therefore 1/negative< positive is true

Cannot determine based on B

Answer is A
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 17 Jun 2013, 07:06
I want to know where is the error in following method for statement 2

1 < la-bl
la-bl > 1

then

a-b > 1 or a-b< -1

a-b > 1 or b-a> 1

since both sides are positive for first possibility

1/(a-b) < 1 or 1 < b-a ( i think problem is here) :x

combining

1/(a-b) < 1 < b-a

1/(a-b)< b-a

therefore yes... i know i am wrong somewhere.. but where i dontknow
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 17 Jun 2013, 08:34
Expert's post
GMAT98 wrote:
I want to know where is the error in following method for statement 2

1 < la-bl
la-bl > 1

then

a-b > 1 or a-b< -1

a-b > 1 or b-a> 1

since both sides are positive for first possibility

1/(a-b) < 1 or 1 < b-a ( i think problem is here) :x

combining

1/(a-b) < 1 < b-a

1/(a-b)< b-a

therefore yes... i know i am wrong somewhere.. but where i dontknow


Please elaborate your question.
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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

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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 18 Jun 2013, 08:35
Bunuel wrote:
GMAT98 wrote:
I want to know where is the error in following method for statement 2

1 < la-bl
la-bl > 1

then

a-b > 1 or a-b< -1

a-b > 1 or b-a> 1

since both sides are positive for first possibility

1/(a-b) < 1 or 1 < b-a ( i think problem is here) :x

combining

1/(a-b) < 1 < b-a

1/(a-b)< b-a

therefore yes... i know i am wrong somewhere.. but where i dontknow


Please elaborate your question.


Hi Bunuel

Pl help me with following problem

Is 1/ a-b < b-a
(1) a<b
(2) 1<|a-b|

My thought process is as follows
(1) a <b
a-b < 0 or 0 < b-a
1/(a-b) < 0 or 0 < b-a
negative will always be less than positive, therefore, statement one is sufficient


(2) 1 < la-bl
la-bl > 1

then opening absolute bracket

a-b > 1 or a-b< -1

a-b > 1 or b-a> 1

since a-b > 1 it implies a-b is positive and 1 is also positive therefore taking reciprocal

1/(a-b) < 1 or b-a > 1

combining

1/(a-b) < 1 < b-a

1/(a-b)< b-a

therefore statement two is also sufficient Therefore answer D but answer is A
... i know i am wrong somewhere.. but where i dont know
can you please go through my solution for second statement and pin-point the error
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 21 Jun 2013, 10:18
Hi Bunuel

Pl help me with following problem

Is 1/ a-b < b-a
(1) a<b
(2) 1<|a-b|

My thought process is as follows
(1) a <b
a-b < 0 or 0 < b-a
1/(a-b) < 0 or 0 < b-a
negative will always be less than positive, therefore, statement one is sufficient


(2) 1 < la-bl
la-bl > 1

then opening absolute bracket

a-b > 1 or a-b< -1

a-b > 1 or b-a> 1

since a-b > 1 it implies a-b is positive and 1 is also positive therefore taking reciprocal

1/(a-b) < 1 or b-a > 1

combining

1/(a-b) < 1 < b-a

1/(a-b)< b-a

therefore statement two is also sufficient Therefore answer D but answer is A
... i know i am wrong somewhere.. but where i dont know
can you please go through my solution for second statement and pin-point the error
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 27 Jun 2013, 12:19
Is 1/(a-b) < b-a ?

(1) a<b
(2) 1<|a-b|

(1) a<b

If a<b then (b-a) will always be a positive number. Therefore, a-b will always be negative. Even if b-a turned out to be a fraction, a-b will negative.
SUFFICIENT

(2) 1<|a-b|

a-b could be 4-2 = 2 or it could be b-a = 2-4 = -2. In other words, we don't know if it's positive or negative meaning we cannot solve for (a-b) or (b-a)
INSUFFICIENT

(A)
Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b   [#permalink] 27 Jun 2013, 12:19
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