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# a, b, c, d are positive integers such that exactly one of

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

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15 Nov 2009, 18:15
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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.
[Reveal] Spoiler: OA

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Re: False inequality [#permalink]

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15 Nov 2009, 18:27
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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.

I'm going to guess B

A). a < b to prove this false let's say a = 4 and b = 3 this would make C) false as well and only one can be false

we are left with B, D, E

c<d

let's say c = 4 and d = 3 to prove this false

and let's say a = 1 and b = 5

d) 1 + 4 < 5 + 3 is still true
e)1 < 5 + 3 + 4 still true

so my guess is B if i'm reading the question correctly

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Re: False inequality [#permalink]

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15 Nov 2009, 21:00
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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.

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Re: False inequality [#permalink]

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16 Nov 2009, 12:38
Yes, OA is B.

Not hard but good question.
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Re: False inequality [#permalink]

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16 Nov 2009, 21:10
Definitely an interesting question... I have to work on my inequalities

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Re: False inequality [#permalink]

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16 Nov 2009, 21:13
do you have a set of ds geometry?

thanks for the inequality set

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

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17 Jun 2013, 05:51
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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

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17 Jun 2013, 10:48
3
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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

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

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10 Sep 2013, 01:24
I substituted a,b,c,d as 2,1,3,4 and 1,2,4,3 (as only A or B is true ).. plugged in the options.. got the right anser in 2 mins..

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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!

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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.

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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.

(D) a−b<d−c

Rearranging (E):

(E) $$a<b+c+d$$
$$a-b-d<c$$
$$c>a-b-d$$

Rearranging (D):
(D) $$a−b<d−c$$
$$a-b-d<-c$$
$$-c>a-b-d$$
$$c<-a+b+d$$

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.

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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!

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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.

anyway thanks for your astute approach.

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a, b, c, d are positive integers such that exactly one of   [#permalink] 30 Jan 2017, 04:12
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# a, b, c, d are positive integers such that exactly one of

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