Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink]
27 Aug 2014, 02:29

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________