If p q and r are integers, is p+q+r 0 ?

Question

If p, q, and r are integers, is p+q+r> 0 ??

(1) p + r = q

(2) r^2 < -q

The first one is insufficient : 3 + 2 = 5 > 0 YES or 0+0=0 NO

The second one we do not know about p

Combined: from 2 q is negative (this is the meaning), so 1 is also as consequence = to a negative value .......... negative + negative < 0 always. answer is NO Suff.

here is the tricky thing: if i look at the problem as if I were during the exam and under time pressure I could say: p + r = q so we have 2q > 0 .....NO sufficient and the answer would be A not C.

BUT if I use the two statement combined with the substitution method I have

q is postive ( we know from second statement that q per se is negative but time minus we have q: positive in the end) so p + r = q --------> r = q - p

So we have p + q + q - p > 0 -------- > 2q >0 Suff, answer would be C, again.

Please clarify this weird situation. Thanks

Bunuel · Accepted Answer

If p, q, and r are integers, is p+q+r> 0 ? (1) p + r = q --> original question becomes is q+q>0 ? --> is 2q>0 ? --> is q>0 ? We don't know that, hence this statement is not sufficient. (2) r^2 < -q --> q<0 (because if q is positive (or zero) then we would have that r^20 ?" while (2) say that q<0 , hence the answer to the question is NO. Sufficient. Answer: C.

Carcass · Answer

mmmmmmmm................study, study, study.......practice, practice, practice.

Thanks Mod.

yezz · Answer

If p, q, and r are integers, is p+q+r> 0 ?? (1) p + r = q (2) r^2 < -q FROM 1 it is possible that p+r = q = 0 or p+r = q = 3 .... insuff from 2 r^2 < -q .... r is 0 or any integer and q is -ve integer .... insufficient both p+r = -ve = q , thus question becomes is is 2q > 0 , answer is definitely not .......C