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.