Last visit was: 19 Nov 2025, 09:50 It is currently 19 Nov 2025, 09:50
Close
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
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.
Close
Request Expert Reply
Confirm Cancel
655-705 Level|   Absolute Values|   Inequalities|               
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,288
 [388]
19
Kudos
Add Kudos
366
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,288
 [78]
25
Kudos
Add Kudos
53
Bookmarks
Bookmark this Post
User avatar
anairamitch1804
Joined: 26 Oct 2016
Last visit: 20 Apr 2019
Posts: 506
Own Kudos:
3,564
 [23]
Given Kudos: 877
Location: United States
Concentration: Marketing, International Business
Schools: HBS '19
GMAT 1: 770 Q51 V44
GPA: 4
WE:Education (Education)
Schools: HBS '19
GMAT 1: 770 Q51 V44
Posts: 506
Kudos: 3,564
 [23]
16
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
General Discussion
avatar
lool
Joined: 30 Jan 2014
Last visit: 05 Aug 2014
Posts: 12
Own Kudos:
3
 [1]
Given Kudos: 15
Posts: 12
Kudos: 3
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
SOLUTION

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.


for the second statement;

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

AND

a-b < -1 --> 1< b-a therefore LHS will be negative and RHS will be positive -->negative<positive. Sufficient.


what is wrong with my approach ?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,288
 [7]
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,288
 [7]
4
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
lool
Bunuel
SOLUTION

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.


for the second statement;

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

AND

a-b < -1 --> 1< b-a therefore LHS will be negative and RHS will be positive -->negative<positive. Sufficient.


what is wrong with my approach ?

From \(1 < |a - b|\) we can have two cases:

A. \(1 < a-b\) --> \((\frac{1}{a - b}=positive) > (b - a=negative)\) --> answer YES.

B. \(a-b<-1\) --> \((\frac{1}{a - b}=negative) < (b - a=positive)\) --> answer NO.

Two different answers, hence insufficient.

Hope it's clear.
User avatar
coolredwine
Joined: 18 Nov 2013
Last visit: 27 Apr 2015
Posts: 57
Own Kudos:
46
 [2]
Given Kudos: 7
Location: India
GMAT Date: 12-26-2014
WE:Information Technology (Computer Software)
Posts: 57
Kudos: 46
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I did it like this:

We can reduce the given statement as:

1/(a-b)>(b-a) -> Multiplying both sides by (a-b), we get: 1>(a-b)(b-a)

Taking negative out from RHS: 1<(a-b)(a-b), which is what we have to prove.

Now, Statement 1: a<b -> a-b<0. Squaring both sides: (a-b)(a-b)<0. Hence, the answer would be No. Thus, sufficient.

Statement 2: 1<|a-b| -> 1<(a-b) or 1>(a-b). Thus insufficient.

Hence, answer is A.

Please do let me know if this is a good way to proceed with such questions.

Thanks.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,288
 [13]
7
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
coolredwine
I did it like this:

We can reduce the given statement as:

1/(a-b)>(b-a) -> Multiplying both sides by (a-b), we get: 1>(a-b)(b-a)

Taking negative out from RHS: 1<(a-b)(a-b), which is what we have to prove.

Now, Statement 1: a<b -> a-b<0. Squaring both sides: (a-b)(a-b)<0. Hence, the answer would be No. Thus, sufficient.

Statement 2: 1<|a-b| -> 1<(a-b) or 1>(a-b). Thus insufficient.

Hence, answer is A.

Please do let me know if this is a good way to proceed with such questions.

Thanks.

Unfortunately, most of it is wrong.

We cannot multiply 1/(a-b)>(b-a) by a-b, because we don't know its sign.
If a-b is positive, then we would have 1 > (b-a)(a-b);
If a-b is negative, then we would have 1 < (b-a)(a-b): 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 its sign.

Next, you cannot square a-b<0 and write (a-b)^2<0. This is obviously wrong: the square of a number cannot be less than zero.

We can only raise both parts of an inequality to an even power if we know that both parts of the inequality are non-negative (the same for taking an even root of both sides of an inequality).

Adding/subtracting/multiplying/dividing inequalities: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html

Hope this helps.
User avatar
Lucky2783
Joined: 07 Aug 2011
Last visit: 08 May 2020
Posts: 418
Own Kudos:
2,056
 [7]
Given Kudos: 75
Concentration: International Business, Technology
GMAT 1: 630 Q49 V27
GMAT 1: 630 Q49 V27
Posts: 418
Kudos: 2,056
 [7]
6
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

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

Data Sufficiency
Question: 120
Category: Arithmetic; Algebra Arithmetic operations; Inequalities
Page: 161
Difficulty: 650

GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!

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

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

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

\(\frac{1+(a-b)^2}{(a-b)}\) > 0 ? we can see that numerator is always +ive . all we need to know is a>b?

(1) a < b ; then \(\frac{1+(a-b)^2}{(a-b)}\) < 0 . Sufficient.

(2) 1 < |a - b|
case 1 : a-b < -1 or a<b-1 --> a<b
case 2: a-b>1 or a> b+1------> a>b
not sufficient.


Ans : A
User avatar
EgmatQuantExpert
User avatar
e-GMAT Representative
Joined: 04 Jan 2015
Last visit: 02 Apr 2024
Posts: 3,663
Own Kudos:
20,165
 [10]
Given Kudos: 165
Expert
Expert reply
Posts: 3,663
Kudos: 20,165
 [10]
8
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
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:

The question is asking if:
\(\frac{1}{a-b} < b-a\)

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?)

And the answer is NO.

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?)

And the answer is YES

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. :-D

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
User avatar
AbdurRakib
Joined: 11 May 2014
Last visit: 08 Nov 2025
Posts: 465
Own Kudos:
Given Kudos: 220
Status:I don't stop when I'm Tired,I stop when I'm done
Location: Bangladesh
Concentration: Finance, Leadership
GPA: 2.81
WE:Business Development (Real Estate)
Posts: 465
Kudos: 42,848
Kudos
Add Kudos
Bookmarks
Bookmark this Post
[quote="Bunuel"]SOLUTION

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 YES 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. Not sufficient.


Dear,

Please Clarify me Rewrite like this one.I thought rewite variable without considering its Sign is Wrong
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,288
 [1]
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,288
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
AbdurRakib
Bunuel
SOLUTION

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 YES 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. Not sufficient.


Dear,

Please Clarify me Rewrite like this one.I thought rewite variable without considering its Sign is Wrong

We cannot multiply an inequality by a variable if don't know its sign but we can add/subtract a value to both sides. To get a-b<0 from a < b we are subtracting b from both sides and to get b - a > 0 we are subtracting a from both sides.
User avatar
CrackverbalGMAT
User avatar
Major Poster
Joined: 03 Oct 2013
Last visit: 19 Nov 2025
Posts: 4,844
Own Kudos:
8,945
 [5]
Given Kudos: 225
Affiliations: CrackVerbal
Location: India
Expert
Expert reply
Posts: 4,844
Kudos: 8,945
 [5]
2
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
Let us look at this question differently here:

While solving DS inequality questions, the best approach is to always breakdown the question stem if possible. To breakdown the question stem, there are three hygiene factors, that if followed will simplify your analysis of the question.

1. Always keep the RHS of the inequality as 0
2. Simplify the LHS to a product or division of values (product and division of terms are easier to analyze)
3. Always try and maintain even powered terms (as the sign of them will always be 0 or positive)


Let us breakdown the question stem here:

1/(a - b) > b - a
(1/(a - b)) - (b - a) > 0

Since we have an option to get a squared term, before we take the LCM let us change b - a to a - b.

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

Now since we have deconstructed the question stem to a division we just need to worry about the signs of the numerator and denominator.

The RHS here is > 0, so we the numerator and denominator have to have the same sign. The numerator however will always be positive, since 1 + (a - b)^2 will always be positive. So we just need need the denominator to also be positive.

The entire question stem can now be rephrased as 'Is a - b > 0'?

Statement 1 : a - b < 0

This gives us a definite NO. So sufficient.

Statement 2 : 1 < |a - b|

|a - b| > 1. This only tells us that a - b > 1 or a - b < -1. So a - b can be both positive or negative. Insufficient.

Answer: A

Hope this helps!

Aditya
CrackVerbal Academic Team
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 19 Nov 2025
Posts: 5,794
Own Kudos:
5,510
 [2]
Given Kudos: 161
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
GMAT 1: 690 Q50 V34
Posts: 5,794
Kudos: 5,510
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
Is 1/(a - b) > b - a ?

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


The Official Guide For GMAT® Quantitative Review, 2ND Edition

Asked: Is 1/(a - b) > b - a ?
1/(a-b) + (a-b) >0
(a-b)^2 + 1 / (a-b) >0
Since numerator >0
Q. a -b >0

(1) a < b
a-b<0
a-b is NOT >0
SUFFICIENT

(2) 1 < |a - b|
a-b > 1 or a-b < -1
NOT SUFFICIENT

IMO A
User avatar
siddharth287
Joined: 20 Feb 2018
Last visit: 19 Feb 2025
Posts: 5
Given Kudos: 14
Location: India
Posts: 5
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
SOLUTION

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 YES 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. Not sufficient.

Answer: A.


statement 1 says a<b so if we substitute values for a and b.
example a = 2 and b = 3
then from original equation we get
1/(2-3) > 3-2 ie 1/-1 > 1 ie -1>1 how is that possible? And why is statment 1 sufficient then
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,288
 [1]
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,288
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
siddharth287
Bunuel
SOLUTION

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 YES 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. Not sufficient.

Answer: A.


statement 1 says a<b so if we substitute values for a and b.
example a = 2 and b = 3
then from original equation we get
1/(2-3) > 3-2 ie 1/-1 > 1 ie -1>1 how is that possible? And why is statment 1 sufficient then

The question asks whether 1/(a - b) > b - a. If we can give a definite YES answer to the question (YES, 1/(a - b) IS greater than b - a) OR a definite NO answer to the question (No, 1/(a - b) is NOT greater than b - a), then the statement is sufficient (recall that a definite NO answer to a DS question is also sufficient).

Now, since a < b, then a - b < 0 and b - a > 0, thus 1/(a - b) = 1/negative = negative, while b - a = positive, so 1/(a - b) < b - a, which means that the answer to the question is NO, 1/(a - b) is NOT greater than b - a. That's why the first statement is sufficient.

Check the links below for DS Strategies and Tactics




For other subject check Ultimate GMAT Quantitative Megathread.

Hope it helps.
User avatar
HoudaSR
Joined: 08 Aug 2022
Last visit: 24 Nov 2023
Posts: 72
Own Kudos:
Given Kudos: 182
Location: Morocco
WE:Advertising (Non-Profit and Government)
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Is 1/(a - b) > b - a ?

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



The question asks is 1/(a - b) > b - a ?

We can tell that a-b and b-a have opposite signs and if we suppose that 1/(a-b)> b-a then a-b must be positive.
The question becomes is a>b?

Statement 1 tells us that a<b, so the answer to our question is a definte NO.
Statement 1 alone is sufficient.

Statement 2 tells us that |a - b|>1 which means that the distance between a and b on the number line is greater than 1 but we don't know which is to the right of which. a can be to the right of b, the same way b can be to the right of a (both the scenarios are possible).
Therefore, statement 2 is not sufficient.

Correct answer is A
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,587
Own Kudos:
Posts: 38,587
Kudos: 1,079
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
Moderators:
Math Expert
105389 posts
496 posts