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Since A=BC, then the question can be re-written as "is BC>B?" --> is B(C-1)>0? Basically the question asks whether B and C-1 have the same sign.

(1) 0<C<1 --> C-1<0. We don't know the sign of B. Not sufficient.

For example, if A=1/4 and B=C=1/2, then A<B but if A=-1/2, B=-1 and C=1/2, then A>B. Not sufficient.

(2) A>0. If A=B=C=1, then A=B but if A=1 and B=C=-1, then A>B. Not sufficient.

(1)+(2) From above we have that both A and C are positive, hence B must also be positive (from A=BC). So, we have that B>0 and C-1<0: B and C-1 have opposite signs, therefore B(C-1)<0 (we have a definite NO answer to the question). Sufficient.

Answer: C.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to the rules #1 (Choose the Right Forum) and #3 (the name of a topic (subject field) MUST be the first 40 characters (~the first two sentences) of the question).
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Starting with the easy statement: Statement 2 says that A>0, but no information has been given about B. So it is insufficient. Statement 1 says that C is a gractional value, lying between 0 and 1. No info about A and B. Clearly Insufficient. The answer has to be either C or E. On combining the statements we get: A>0 and 0<C<1. Do notice that A and B can be fractional values also. Now consider, A=0.1 and C=0.9, since A=BC therefore B=1/9. Here B>A. Now consider, A=0.9 and C=0.1, since A=BC, therefore B=9. Here also B>A. If one takes integer values of A, then also B>A. Hence sufficient. +1C
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No, B is not sufficient. it just tells A>0 , But you dont know anything about B/C individually.

Each statement alone is insufficient, clearly as none has more than one variable when we need relation between 2.

On combining, we get A>0 and 0<C<1. Since A, C>0 thus B>0

Notice, multiplication by a fraction will alsways result in a lower number for any postive number. Therefore if A =BC where 0<C<1 then A <B Sufficient.
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Case 1: A,B are -ve => A=-3 and B=-4 => A>B Case 2: A,B are +ve => A=3 and B=4 => A<B

=>Insufficient

(2). A>0

Case 1: B is -ve => A=3 , B=-1 => A>B Case 2 : B is +ve => A=3 , B=4 => A <B =>Insufficient

Combined A>0 => Case 2 only of Statement 1 Hence (C).
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