Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 05 May 2016, 20:00

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 09 Feb 2006
Posts: 90
Location: Prague, CZ
Followers: 1

Kudos [?]: 36 [2] , given: 0

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

### Show Tags

30 Jun 2006, 06:49
2
KUDOS
8
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

62% (02:17) correct 38% (01:26) wrong based on 497 sessions

### HideShow timer Statictics

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

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-1-a-b-b-a-168100.html
[Reveal] Spoiler: OA
Director
Joined: 28 Dec 2005
Posts: 755
Followers: 1

Kudos [?]: 9 [1] , given: 0

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

### Show Tags

30 Jun 2006, 10:41
1
KUDOS
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.
CEO
Joined: 20 Nov 2005
Posts: 2911
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 20

Kudos [?]: 187 [5] , given: 0

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

### Show Tags

30 Jun 2006, 11:18
5
KUDOS
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
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Director
Joined: 07 Jun 2006
Posts: 513
Followers: 11

Kudos [?]: 106 [0], given: 0

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

### Show Tags

02 Jul 2006, 08:31
1
This post was
BOOKMARKED
likar wrote:
Is 1/(a-b) < b-a ?

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

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 :)
Manager
Joined: 18 Aug 2010
Posts: 90
Followers: 1

Kudos [?]: 6 [0], given: 22

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

### Show Tags

10 Feb 2011, 12:36
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

Math Expert
Joined: 02 Sep 2009
Posts: 32657
Followers: 5660

Kudos [?]: 68759 [9] , given: 9818

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

### Show Tags

10 Feb 2011, 13:14
9
KUDOS
Expert's post
3
This post was
BOOKMARKED
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

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.

_________________
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2022
Followers: 154

Kudos [?]: 1421 [4] , given: 376

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

### Show Tags

10 Feb 2011, 13:33
4
KUDOS
1
This post was
BOOKMARKED
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

or

a-b < -1
a < b-1
Not sufficient.

Ans: "A"
_________________
Manager
Joined: 18 Aug 2010
Posts: 90
Followers: 1

Kudos [?]: 6 [0], given: 22

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

### Show Tags

11 Feb 2011, 03:29
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

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.

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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 32657
Followers: 5660

Kudos [?]: 68759 [1] , given: 9818

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

### Show Tags

11 Feb 2011, 03:37
1
KUDOS
Expert's post
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.
_________________
Manager
Joined: 18 Aug 2010
Posts: 90
Followers: 1

Kudos [?]: 6 [0], given: 22

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

### Show Tags

11 Feb 2011, 03:40
thanks bunuel , thanks fluke
+kudos to both
Manager
Joined: 07 Jun 2010
Posts: 86
Followers: 1

Kudos [?]: 28 [0], given: 0

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

### Show Tags

23 Feb 2011, 21:19
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

Intern
Joined: 14 Jan 2012
Posts: 14
GMAT 1: 600 Q45 V28
GMAT 2: 700 Q48 V38
Followers: 0

Kudos [?]: 1 [0], given: 13

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

### Show Tags

17 Jun 2013, 08:06
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)

combining

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

1/(a-b)< b-a

therefore yes... i know i am wrong somewhere.. but where i dontknow
Math Expert
Joined: 02 Sep 2009
Posts: 32657
Followers: 5660

Kudos [?]: 68759 [0], given: 9818

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

### Show Tags

17 Jun 2013, 09:34
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)

combining

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

1/(a-b)< b-a

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

_________________
Intern
Joined: 14 Jan 2012
Posts: 14
GMAT 1: 600 Q45 V28
GMAT 2: 700 Q48 V38
Followers: 0

Kudos [?]: 1 [0], given: 13

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

### Show Tags

18 Jun 2013, 09:35
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

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
Intern
Joined: 14 Jan 2012
Posts: 14
GMAT 1: 600 Q45 V28
GMAT 2: 700 Q48 V38
Followers: 0

Kudos [?]: 1 [0], given: 13

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

### Show Tags

21 Jun 2013, 11:18
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
Senior Manager
Joined: 13 May 2013
Posts: 472
Followers: 2

Kudos [?]: 126 [0], given: 134

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

### Show Tags

27 Jun 2013, 13:19
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)
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 9298
Followers: 456

Kudos [?]: 115 [0], given: 0

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

### Show Tags

27 Aug 2014, 03:29
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.
_________________
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 338
Followers: 87

Kudos [?]: 699 [1] , given: 84

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

### Show Tags

08 May 2015, 01:40
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
Going through the solutions posted above, I realized that this question is a good illustration of the importance of first analyzing the question statement before moving to St. 1 and 2 in a DS question.

Most students dived straight into Statement 1 and then tried to draw inferences from St. 1 to determine whether the inequality given in the question statement was true or not.

Imagine if you had done this instead before going to St. 1:

$$\frac{1}{a-b} < b-a$$

Case 1: a - b is positive (that is, a > b)

Multiplying both sides of an inequality with with the positive number (a-b) will not change the sign of inequality.

So, the question simplifies to: Is 1 < (b-a)(a-b)

Now, b - a will be negative.

So, the question simplifies to: Is $$1 < -(a-b)^2$$ . . . (1)

$$(a-b)^2$$, being a square term, will be >0 (Note: a - b cannot be equal to zero because then the fraction given in the question: 1/a-b becomes undefined)

So, $$-(a-b)^2$$ will be < 0

Therefore, the question simplifies to: Is 1 < (a negative number?)

Case 2: a - b is negative (that is, a < b)

Multiplying both sides of an inequality with the negative number (a-b) will change the sign of inequality.

So, the question simplifies to: Is 1 > (b-a)(a-b)

Now, b - a will be positive.

So, the question simplifies to: Is $$1 > -(a-b)^2$$ . . . (2)

Again, by the same logic as above, we see that the question simplifies to: Is 1 > (a negative number?)

Thus, from the question statement itself, we've inferred that:

If a > b, the answer to the question asked is NO
If a < b, the answer to the question asked is YES

Thus, the only thing we need to find now is whether a > b or a < b.

Please note how we are going to Statement 1 now with a much simpler 'To Find' task now.

One look at St. 1 and we know that it will be sufficient.

One look at St. 2 and we know that it doesn't give us a clear idea of which is greater between a and b, and so, is not sufficient.

To sum up this discussion, spending time on analyzing the question statement before going to the two statements usually simplifies DS questions a good deal.

Hope this helped!

Japinder
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 32657
Followers: 5660

Kudos [?]: 68759 [0], given: 9818

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

### Show Tags

08 May 2015, 03:49
Expert's post
OPEN DISCUSSION OF THIS QUESTION IS HERE: is-1-a-b-b-a-168100.html
_________________
Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b   [#permalink] 08 May 2015, 03:49
Similar topics Replies Last post
Similar
Topics:
3 If c ≠ 1, is ab = a+b? 2 26 Sep 2015, 06:11
Is (A/B)^3<(AB)^3 1) A>0 2) AB>0 6 05 Sep 2011, 02:18
3 Is 1/(a-b) < b-a ? 8 24 Jul 2011, 13:03
2 What is the perimeter of right triangle ABC? (1) AB = 5 (2) 9 01 Jun 2011, 07:07
(a^3)b+(a^2)(b^2)+a(b^3)>0? 1. ab>0 2. b<0 4 05 Nov 2010, 09:43
Display posts from previous: Sort by