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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

Re: False inequality [#permalink]
15 Nov 2009, 20:00

6

This post received KUDOS

3

This post was BOOKMARKED

Essentially, we need to find the statement that could be false if all the other statements are true.

The first thing that jumps out to me is statements A and C:

(A) a < b (C) a+c < b+c

Regardless of the value of c, these statement are either both false or both true. Since only statement CAN be false, we can eliminate these two.

(E) a < b+c+d

Since c and d are both positive integers, and we have determined that statement A is true, statement E must be true as well. This leaves us with:

(B) c < d (D) a+c < b+d

Since we know statement A (a < b) to be true, if statement B is true, statement D MUST be true. However, if statement D is true, as long as c-d < b-a, statement B does NOT have to be true.

Therefore, the answer is B: c < d . Good question, +1.

Re: a, b, c, d are positive integers such that exactly one of [#permalink]
17 Jun 2013, 09:48

2

This post received KUDOS

Bunuel wrote:

a, b, c, d are positive integers such that exactly one of the following inequalities is false. Which inequality is false?

(A) a<b

(B) c<d

(C) a+c<b+c

(D) a+c<b+d

(E) a<b+c+d

I'm not sure whether we can see such problem on GMAT, but I found this one quite interesting so posting for discussion.

An algebrical approach:

first of all (A) a<b and (C) a+c<b+c are the same: both tell us that \(a<b\). Since exactly one inequality is false, both A and C must be true. With \(a<b\) enstablished we can focus on

(B) c<d

(D) a+c<b+d

(E) a<b+c+d

E must be true since \(a<b\) also \(a<b+(+veNumber)+(+veNumber)\) is true as well.

(B) \(0<d-c\) (D) \(a-b<d-c\) but since a-b is negative => \(-ve<d-c\)

(B) \(d-c>0(>-ve)\) number (D) \(d-c>-ve\) number

if B is true, also D is true for sure (\(d-c=7\) \(7>-ve\) and \(7>0\)) this goes against the text of the question if D is true, B could be false (\(d-c=-1\) \(-1>-ve(-2)\)(for example) and \(-1>0\) FALSE) Hence B must be the false inequality

Simpler approach: given \(a<b\) if B is true \(c<d\) also D is true \(a+c<b+d\), once more this is against the question _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: a, b, c, d are positive integers such that exactly one of [#permalink]
09 Sep 2013, 05:32

C, D & E depend on the true/false of the equation A (a < b) C ( a+c<b+c ) will be true for any a < b D ( a+c<b+d ) can be true for any a < b . It can be true when c > d, if c-d < b-a E is true always if a < b. Since C D E depend on A for the major part, the one inequality that can be wrong is B

Re: a, b, c, d are positive integers such that exactly one of [#permalink]
02 Nov 2014, 09:15

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. _________________

Re: a, b, c, d are positive integers such that exactly one of [#permalink]
14 Jun 2015, 04:14

Bunuel wrote:

a, b, c, d are positive integers such that exactly one of the following inequalities is false. Which inequality is false?

(A) a<b

(B) c<d

(C) a+c<b+c

(D) a+c<b+d

(E) a<b+c+d

I'm not sure whether we can see such problem on GMAT, but I found this one quite interesting so posting for discussion.

If (A) is true then (C) and (E) also true. If (A) is wrong, then (C) is also wrong. That means either (B) or (D) is wrong. For the value 2, 5, 3, and 1, (D) is true but (B) is wrong.

a, b, c, d are positive integers such that exactly one of [#permalink]
27 Jul 2015, 00:24

No plugging in is necessary.

(A) a < b (C) a+c < b+c

These two statements are equal. Subtract c from both sides in (C). Since we must find one of the answer choices that does not belong, then we know that (A) and (C) must be true.

(E) a<b+c+d

This must be true because adding two positive integers to b, which is greater than a, will always be greater than a.

This does not mean that D and E are necessarily contradictory.

Just E: a-b-d<c; -c<-a+b+d (Must be true) Just D: a-b-d<-c; c<-a+b+d

Combining D and E we get: \(a-b-d<c<-a+b+d a-b-d<-c<-a+b+d\)

Possible scenarios (think of it as numbers added/subtracted in order of smallest to biggest): \(a-b-d<-c<c<-a+b+d\) OR \(-c<a-b-d<c<-a+b+d\) OR \(-c<a-b-d<-a+b+d<c\)

The values in question, c and d, are never contradictory. They're always positive on the right side and negative on the left side. For instance, you don't see a positive d value on the left with a negative d value on the right. All of the choices are consistent/possible. c can be larger than d, d can be larger than c; all we know for sure is that b is larger than a.

Combining (E) and (B) we get: Just E: \(a-b-d<c\); \(-c<-a+b+d\) (Must be true) Just B: \(d>c\); \(-c>-d\)

Possible scenarios: \(-c<a-b-d<-a+b+d<c<d\) (Not possible to have \(d> -a +b +d\)) OR \(-d<-c<a-b-d<-a+b+d<c\) (Not possible to have \(-d< a -b -d\)) OR \(-c<a-b-d<c<d<-a+b+d\)

(B) does not work with the inequality.

gmatclubot

a, b, c, d are positive integers such that exactly one of
[#permalink]
27 Jul 2015, 00:24

Interested in applying for an MBA? In the fourth and final part of our live QA series with guest expert Chioma Isiadinso, co-founder of consultancy Expartus and former admissions...