Inequality

[chandak_anand]

Is P<Q

1) p+q/2 < 0
2) P^2 = Q^2

[tublai]

Is P<Q

1) p+q/2 < 0
2) P^2 = Q^2

From(1):

(p+q) / 2 < 0 => (p+q) < 0

p = -1 and q = -2 ; (p+q) < 0 but p>q , answer is NO
p = -2 and q = -1 ; (p+q) < 0 but p<q , answer is YES

INSUFFICIENT

From(2):

p^2 = q^2 i.e |p| = |q|

p = 0 , q = 0 , but, p=q answer NO
p = -2 , q = 2 , |p| = |q| , but p<q answer is YES

INSUFFICIENT

From both:

p^2 = q^2
=> (p+q)(p-q) = 0 but (p+q) < 0 i.e (p+q) != 0

=> (p-q) = 0

i.e p = q

answer to question: is p<q , definitive NO

SUFFICIENT from both the statement

Choice (C )

[Crazygmat]

Statement one to me: reads as p + q/2 < 0 and not (p+q) / 2 < 0 . Please clarify,chandak_anand
Answer will be E..

[chandak_anand]

crazygmat it is (p+q)/2
Tublai C is the Official Answer

[cognos]

If it is (p+q)/2 the answer is C.

[CrackXam]

Agreed C.

Stmnt I : (p+q)/2 < 0
=> either both are < 0 or numerically smaller of the two is < 0
So, Insufficient.

Stmnt II : P^2 = Q^2
=> lpl = lql

Combine I & II :
Both are numerically equal. So if (p+q) < 0, then p = q < 0.
So sufficient.