Is a < 0, if a and b are integers?

(1) \(2a^2-8b^2>0\)
(2) \(a<2b\)
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GMAT Expert

A) a^2-4b^2>0
(a+2b)(a-2b)>0
There are two conditions now
1.
a+2b>0 & a-2b>0 or a>0
or 2.
a+2b<0 & a-2b<0 or a<0. Hence insufficient.

B)
a<2b, a=3 and b=2 satisfies...a>0
a=-10 b=2 still satisfies a<0. Hence insufficient.

Combining A & B,
Since a<2b, condition 2 will satisfy.
a+2b<0 & a-2b<0 or a<0. Hence C.

(1) 2a^2 - 8b^2 > 0 . Dividing both sides by 2, we get a^2 - 4b^2 > 0 OR (a)^2 - (2b)^2 > 0 OR (a+2b)(a-2b) > 0
This doesnt tell us anything about whether a is positive or negative, so insufficient.

(2) a < 2b Or a-2b < 0. Insufficient on its own.

Combining the two statements, a-2b < 0 But (a+2b)(a-2b) > 0. This is possible only when a+2b is also < 0.
So we have both a-2b<0 and a+2b<0 OR we have a < 2b as well as a < -2b. Whether b is positive or negative, 2b and -2b will have opposite signs. a is less than both so a must be negative. And if b=0, both 2b=-2b=0, thus a<0. So in any case a will be negative only. Sufficient.

Hence C answer