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M11 #26 : Retired Discussions [Locked]

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* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient * Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient * BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient * EACH statement ALONE is sufficient * Statements (1) and (2) TOGETHER are NOT sufficient

both will be reqd. just need to determine if both are adequate. B^2*C^2>0 AC>0 So AB^2C^3>0 Now, A is also nonzero from 2. So, A^6>0 Thus, A^6*A*B^2*C^3>0 Hence, sufficient. C. _________________

I simplified this equation to A*B^2*C > 0 to make it slightly less daunting. B will always be positive unless it is zero.

1) BC < 0 means either B or C is Negative, but they can't both be negative. Also, neither can be zero. No other clues are given about A. Insufficient. 2) AC > 0 means A and C are either both negative or both positive. Can't be zero. Insufficient.

(2) states that the product of AC is always positive, since B^2 is also always positive and B isn't zero, then the equation must be positive. (C) is the answer. _________________

"What we obtain too cheap, we esteem too lightly." -Thomas Paine

Here 3 unknown in equations i.e. A, B and C. so to solve this equations require details of all three. (1) statement 1 provides information about B and C (2) Statement 2 provides information about A and C

Over all by multiplication using statement 1 and 2 we can get the answers.

Inequality a^7*b^2*c^3>0 to be true a and c must be either both positive or both negative AND b must not be zero (in order a^7*b^2*c^3 not to equal to zero).

(1) bc<0 --> b\neq{0}. Don't know about a and c. Not sufficient.

(2) ac>0 --> a and c are either both positive or both negative. Don't know about b. Not sufficient.

1. BC < 0 implies either B or C is negative, but not both 2. AC > 0 implies neither A nor C are negative, but both A and C may be negative

Using both 1 and 2, we have two options:

I - A is negative; B is positive; C is negative => A^7 B^2 C^3 > 0 is positive II - A is positive; B is negative; C is positive => A^7 B^2 C^3 > 0 is positive

Hence Option C.! :D _________________

KUDOS-ing does'nt cost you anything, but might just make someone's day!!!

I will split the question as [(A)^6]*[(BC)^2]*(AC)

So (BC)^2>0 or =0 A^6>0 or =0 AC can be either >0 <0 or 0

so {[(A)^6]*[(BC)^2]*(AC)} will look like {(+/0) * (+/0) *(+/-/0)}

1:)BC<0 =>[(BC)^2]>0 but we are unsure about AC wheter its >0 or <0 so reject

strike off options A,D

Remaining options B,C,E

2.)AC>0 now this implies [(A)^6]>0 (as A,C!=0) [(BC)^2]>=0(we are unsure of B, whether its 0 or not) (AC)>0 so result will be >=0, but its insufficient, as its not >0 but we have dual case of >=0

Strike off option B

options E,C are left, lets go for C ,combining BC<0 and AC>0 implies that [(BC)^2]>0 [(A)^6]>0 [AC]>0 so end result will be >0

So C is the ANS... _________________

Whatever one does in life is a repetition of what one has done several times in one's life! If my post was worth it, then i deserve kudos

From the given question I found that IF A>0 than C>0 IF A<0 than C<0

(I answered this question in less than a minute and choose B, because I didn't consider B to be equal to zero)

Good question for GMAT starter, easy to get into trap...

I did this sometime back...bt I dont seem to have considered that B may be equal to 0, and still chose C.

CAn anyone show how B=0 case make a difference?

Sure. For example if we put b\neq{0} in the stem then the answer would become B, instead of C.

IF THE QUESTION WERE:

If b\neq{0} is a^7*b^2*c^3>0 ?

Since b\neq{0} then b^2>0, so we can reduce by it (b^2 does not affect given inequality at all) and the question becomes: is a^7*c^3>0? So, the question basically asks whether a and c have the same sign. (Notice that if b\neq{0} were not given then b^2\geq{0}, and we could not reduce by it)

(1) bc<0. Not sufficient.

(2) ac>0 --> a and c have the same sign. Sufficient.

Answer: B.

Hope it's clear.

P.S. Solution to he original question is given here: m11-75085.html#p1073868 _________________