Jul 16 03:00 PM PDT  04:00 PM PDT Join a free live webinar and find out which skills will get you to the top, and what you can do to develop them. Save your spot today! Tuesday, July 16th at 3 pm PST Jul 16 08:00 PM EDT  09:00 PM EDT Strategies and techniques for approaching featured GMAT topics. Tuesday, July 16th at 8 pm EDT Jul 19 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 20 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 56244

a, b, c, d are positive integers such that exactly one of
[#permalink]
Show Tags
15 Nov 2009, 18:15
Question Stats:
26% (02:26) correct 74% (01:50) wrong based on 693 sessions
HideShow timer Statistics
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.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________




Manager
Joined: 11 Sep 2009
Posts: 127

Re: False inequality
[#permalink]
Show Tags
15 Nov 2009, 21:00
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 cd < ba, statement B does NOT have to be true.
Therefore, the answer is B: c < d . Good question, +1.




VP
Joined: 05 Mar 2008
Posts: 1390

Re: False inequality
[#permalink]
Show Tags
15 Nov 2009, 18:27
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



Math Expert
Joined: 02 Sep 2009
Posts: 56244

Re: False inequality
[#permalink]
Show Tags
16 Nov 2009, 12:38
Yes, OA is B. Not hard but good question.
_________________



Manager
Joined: 22 Sep 2009
Posts: 179
Location: Tokyo, Japan

Re: False inequality
[#permalink]
Show Tags
16 Nov 2009, 21:10
Definitely an interesting question... I have to work on my inequalities



VP
Joined: 05 Mar 2008
Posts: 1390

Re: False inequality
[#permalink]
Show Tags
16 Nov 2009, 21:13
do you have a set of ds geometry?
thanks for the inequality set



Math Expert
Joined: 02 Sep 2009
Posts: 56244

Re: a, b, c, d are positive integers such that exactly one of
[#permalink]
Show Tags
17 Jun 2013, 05:51
Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE
_________________



VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1044
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8

Re: a, b, c, d are positive integers such that exactly one of
[#permalink]
Show Tags
17 Jun 2013, 10:48
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) \(0<dc\) (D) \(ab<dc\) but since ab is negative => \(ve<dc\) (B) \(dc>0(>ve)\) number (D) \(dc>ve\) number if B is true, also D is true for sure (\(dc=7\) \(7>ve\) and \(7>0\)) this goes against the text of the question if D is true, B could be false (\(dc=1\) \(1>ve(2)\)(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.
Kant , Critique of Pure Reason Tips and tricks: Inequalities , Mixture  Review: MGMAT workshop Strategy: SmartGMAT v1.0  Questions: Verbal challenge SC III CR New SC set out !! , My QuantRules for Posting in the Verbal Forum  Rules for Posting in the Quant Forum[/size][/color][/b]



Intern
Joined: 01 Feb 2013
Posts: 36
Location: India
Concentration: Technology, Leadership
GPA: 3.49
WE: Engineering (Computer Software)

Re: a, b, c, d are positive integers such that exactly one of
[#permalink]
Show Tags
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 cd < ba 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



Manager
Joined: 25 Sep 2012
Posts: 237
Location: India
Concentration: Strategy, Marketing
GMAT 1: 660 Q49 V31 GMAT 2: 680 Q48 V34

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



Manager
Joined: 10 Jun 2015
Posts: 117

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



Manager
Joined: 08 Jun 2015
Posts: 102

a, b, c, d are positive integers such that exactly one of
[#permalink]
Show Tags
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) \(a<b+c+d\) \(abd<c\) \(c>abd\)
Rearranging (D): (D) \(a−b<d−c\) \(abd<c\) \(c>abd\) \(c<a+b+d\)
This does not mean that D and E are necessarily contradictory.
Just E: abd<c; c<a+b+d (Must be true) Just D: abd<c; c<a+b+d
Combining D and E we get: \(abd<c<a+b+d abd<c<a+b+d\)
Possible scenarios (think of it as numbers added/subtracted in order of smallest to biggest): \(abd<c<c<a+b+d\) OR \(c<abd<c<a+b+d\) OR \(c<abd<a+b+d<c\)
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: \(abd<c\); \(c<a+b+d\) (Must be true) Just B: \(d>c\); \(c>d\)
Possible scenarios: \(c<abd<a+b+d<c<d\) (Not possible to have \(d> a +b +d\)) OR \(d<c<abd<a+b+d<c\) (Not possible to have \(d< a b d\)) OR \(c<abd<c<d<a+b+d\)
(B) does not work with the inequality.



Intern
Joined: 08 Jan 2017
Posts: 26

a, b, c, d are positive integers such that exactly one of
[#permalink]
Show Tags
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) \(0<dc\) (D) \(ab<dc\) but since ab is negative => \(ve<dc\) (B) \(dc>0(>ve)\) number (D) \(dc>ve\) number if B is true, also D is true for sure (\(dc=7\) \(7>ve\) and \(7>0\)) this goes against the text of the question if D is true, B could be false (\(dc=1\) \(1>ve(2)\)(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.



Intern
Joined: 19 Apr 2018
Posts: 8

Re: a, b, c, d are positive integers such that exactly one of
[#permalink]
Show Tags
10 Jul 2019, 15:29
IMO the question would have to specify 1 of the following is false when each of the others are true. This question is way to ambiguous with out correct parameters or specifications there is no way this question could be allowed.




Re: a, b, c, d are positive integers such that exactly one of
[#permalink]
10 Jul 2019, 15:29






