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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.

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.

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 ?

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.
_________________

Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink]
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)

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

Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink]
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

Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink]
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)

gmatclubot

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