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.