If a, b and c are integers such that b > a, is b+c > a

If a, b and c are integers such that b > a, is b+c > a ?

Question: is \(b+c > a\) ? --> or: is \(b+c-a > 0\)?

(1) c > a. If \(a=1\), \(b=2\) and \(c=3\), then the answer is clearly YES but if \(a=-3\), \(b=-2\) and \(c=-1\), then the answer is NO. Not sufficient.

(2) abc > 0. Either all three unknowns are positive (answer YES) or two unknowns are negative and the third one is positive. Notice that in the second case one of the unknowns that is negative must be \(a\) (because if \(a\) is not negative, then \(b\) is also not negative so we won't have two negative unknowns). To get a NO answer for the second case consider \(a=-3\), \(b=1\) and \(c=-4\) and . Not sufficient.

(1)+(2) We have that \(b > a\), \(c > a\) (\(c-a>0\)) and that either all three unknowns are positive or \(a\) and any from \(b\) and \(c\) is negative. Again for the first case the answer is obviously YES. As for the second case: say \(a\) and \(c\) are negative and \(b\) is positive, then \(b+(c-a)=positive+positive>0\) (you can apply the same reasoning if \(a\) and \(b\) are negative and \(c\) is positive). So, we have that in both cases we have an YES answer. Sufficient.

Answer: C.