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

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

Yes, OA is B.

Not hard but good question.

Definitely an interesting question... I have to work on my inequalities

do you have a set of ds geometry?

thanks for the inequality set

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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

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..

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.

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.

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.

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.

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