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I think it's E

From Statement 1, we find that ab=-30. While there area a range of possible values for either, we know that a and b have different signs, i.e. one is positive and one is negative.

From statement 2, multiplying the left side gives us a^2=b^2-11. This means that, while we still don't know the actual values, it must be that |b|>|a|, as b's square is larger.

Combining the two statements, we know that one the two, a and b, is negative and one is positive (neither can be zero per statement 1) and that |b|>|a|. However, we still don't know which is positive and which is negative, and therefore we don't know which is larger than the other.
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If a and b are integers, is b>a?

Stat1: −30/b=a or, ab= -30; now (a,b) = (2,-15) or (-15,2)....So, Not sufficient to know, if b>a

Stat2: (a–b)(a+b)=–11
Or, a^2- b^2 = -11 or, a= +-5 and b= +-6, So, Not sufficient to know, if b>a

Combining both stats
if a = +5, then b= -6 and b<a but, a=-5, then b=6 and b>a, again Not sufficient

So, I think E. :)
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Is b > a?

(1) −30/b = a

-30 = ab
Either a or b can be greater.
Not sufficient

(2) (a–b)(a+b)= –11

a^2 - b^2 = -11.
Either a or b can be greater.
Not sufficient

(1) + (2)

a^2 - b^2 = -11
b^2 - a^2 = 11
And ab = -30

Either a or b can be greater
Not sufficient

Option E

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Bunuel
If \(a\) and \(b\) are integers, is \(b > a\)?

(1) \(\frac{−30}{b}= a\)

(2) \((a – b)(a + b) = –11\)



(1) \(\frac{−30}{b}= a\)
\(-30=ab=-1*30=-30*1=-6*5\) and so on
insuff

(2) \((a – b)(a + b) = –11\)
\((a – b)(a + b) = –11=-1*11=-11*1=11*-1=1*-11\)
Insuff

Combined
a=-5 and b=6......b>a
a=5 and b=-6....a>b

Insuff
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