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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]

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17 Jun 2013, 10:48

3

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

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09 Sep 2013, 06:32

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]

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02 Nov 2014, 10:15

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).

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

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14 Jun 2015, 05:14

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]

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27 Jul 2015, 01:24

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 \(-c<a-b-d<c<d<-a+b+d\)

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

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13 Oct 2016, 20:41

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.
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a, b, c, d are positive integers such that exactly one of [#permalink]

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30 Jan 2017, 04:12

Zarrolou wrote:

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

Hi, it could be a little more faster though

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.

I agree with all of above but finally we conclude that either B or D must be incorrect. if B is going to be correct we then can add it with A and it will be a+c<b+d which is D so D must be correct also and we don't have a false inequality. therefore we can conclude that B must be incorrect.