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If A=BC, is A>B ?

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If A=BC, is A>B ? [#permalink]

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New post 28 Jan 2013, 02:30
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If A=BC, is A>B ?

(1) 0<C<1
(2) A>0

[Reveal] Spoiler:
Isn't B alone sufficient to answer the question ?

Appreciate if some one could explain ..
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Jan 2013, 02:40, edited 1 time in total.
Renamed the topic, edited the question and moved to SS forum.
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Re: If A=BC, is A>B ? [#permalink]

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New post 28 Jan 2013, 03:04
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If A=BC, is A>B ?

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.

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Re: If A=BC, is A>B ? [#permalink]

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New post 28 Jan 2013, 03:14
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guerrero25 wrote:
If A=BC, is A>B ?

(1) 0<C<1
(2) A>0

[Reveal] Spoiler:
Isn't B alone sufficient to answer the question ?

Appreciate if some one could explain ..


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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Re: If A=BC, is A>B ? [#permalink]

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New post 28 Jan 2013, 03:32
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guerrero25 wrote:
If A=BC, is A>B ?

(1) 0<C<1
(2) A>0

[Reveal] Spoiler:
Isn't B alone sufficient to answer the question ?

Appreciate if some one could explain ..

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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Re: If A=BC, is A>B ? [#permalink]

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New post 29 Jan 2013, 22:43
My Say

(1). C belongs to (0,1)

=> C is +ve => A and B should have same signs

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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Re: If A=BC, is A>B ? [#permalink]

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New post 30 Jan 2013, 06:52
Expert's post
A=BC;
1. 0<C<1 = 0<(A/B)<1

or 0<AB<B^2 = B(A-B)<0.

Now either B>0;A-B<0 or the other way around. Hence, not sufficient.

2.A>0. Not sufficient.

Taken together, C>0,A>0 hence B has to be positive. Thus, A-B<0.

C.
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Re: If A=BC, is A>B ?   [#permalink] 30 Jan 2013, 06:52
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