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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.

Another kudos to you.... BTW, with so many of these already on your name, you must be going crazy with the amount of auto-notification emails coming your way??
_________________

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

Another kudos to you.... BTW, with so many of these already on your name, you must be going crazy with the amount of auto-notification emails coming your way??

That's not a problem.

Also one should consider kudos not only as a "thank you" to user whose post was helpful, but also as a tool to distinguish a valuable post. Notice that posts with more than 1 kudos have different color, also notice that total # of kudos is shown just beside the topic name, so by giving a kudos to a post you are drawing an attention of other users to the helpful material and thus are contributing to the community.
_________________

Another kudos to you.... BTW, with so many of these already on your name, you must be going crazy with the amount of auto-notification emails coming your way??

That's not a problem.

Also one should consider kudos not only as a "thank you" to user whose post was helpful, but also as a tool to distinguish a valuable post. Notice that posts with more than 2 kudos have different color, also notice that total # of kudos is shown just beside the topic name, so by giving a kudos to a post you are drawing an attention of other users to the helpful material and thus are contributing to the community.

Gee...didnt know that...thanks for the info bunuel!

Quote:

lso notice that total # of kudos is shown just beside the topic name,

Didn't get this part though.
_________________

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

Here is how: Question: Is (A^7) * (B^2) * (C^3) > 0?

Statement 1: BC < 0 Evaluate the Y/N question: Is (A^7) * (B*C)^2 * (C) > 0? (B*C)^2 > 0 But, if A > 0 and C > 0 then expression > 0 => Yes However, if A > 0 and C < 0 then expression < 0 => No So, S1 is not sufficient. Eliminate A and D.

Statement 2: AC > 0 Evaluate the Y/N question: Is expression (A^7) * (B^2) * (C^3) > 0?

Combine A and C.

Is (A*C)^3 * A^4 * B^2 > 0

S2 says A*C > 0 ; so (A*C)^3 > 0 A^4 is always positive even if A were negative; so A^4 > 0 B^2 is always positive even if B were negative; so B^2 > 0 So, the expression is always positive even if A and/or B were negative. expression > 0

Only exception to this is: if B = 0, then expression = 0 S2 is NOT sufficient.

Take both S1 and S2. This eliminate the possibility that B may be 0. So, C is correct.

Last edited by vshrivastava on 23 Apr 2013, 06:59, edited 1 time in total.

I'm sorry if I'm missing something, but I think the right answer is B.

Here is how: Question: Is (A^7) * (B^2) * (C^3) > 0?

Statement 1: BC < 0 Evaluate the Y/N question: Is (A^7) * (B*C)^2 * (C) > 0? (B*C)^2 > 0 But, if A > 0 and C > 0 then expression > 0 => Yes However, if A > 0 and C < 0 then expression < 0 => No So, S1 is not sufficient. Eliminate A and D.

Statement 2: AC > 0 Evaluate the Y/N question: Is (A^7) * (B^2) * (C^3) > 0?

Combine A and C.

Is (A*C)^3 * A^4 * B^2 > 0

S2 says A*C > 0 ; so (A*C)^3 > 0 A^4 is always positive even if A were negative; so A^4 > 0 B^2 is always positive even if B were negative; so B^2 > 0

So, the expression is always positive even if A and/or B were negative. S2 is sufficient.

B is correct.

What do you guys think?

Notice that when we consider the second statement we know nothing about the value of b, so if b=0, then \(a^7*b^2*c^3=0\) (\(a^7*b^2*c^3\) is NOT greater than 0). That's why (2) is not sufficient.

I did the question too quick to fall in the trap. Very good question!. C is correct.

A^7*B^2*C^3 = (A^6*B^2*C^2)*(C*A)

If B = 0, (A^6*B^2*C^2)*(C*A) = 0. If B #0, the sign of (A^6*B^2*C^2)*(C*A) depends on (C*A)

We must combine statement 1 and 2 to conclude: B #0 and (C*A) > 0 ==> (A^6*B^2*C^2)*(C*A) > 0.
_________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

for AC>0 either both are negative or positive we dont have depend on value of B,bcause square of anything is positive in this case if A and c ARE NEGATIVE THEN 2 NEGATIVE SIGNS LEFT CANCEL OUT EACH OTHER AND SQUARE OF B IS POSITIVE,SO ANSWER IS POSITIVE AND IF A AND C ARE POSITIVE AGAIN SQUARE OF B IS POSITIVE so answer is B

for AC>0 either both are negative or positive we dont have depend on value of B,bcause square of anything is positive in this case if A and c ARE NEGATIVE THEN 2 NEGATIVE SIGNS LEFT CANCEL OUT EACH OTHER AND SQUARE OF B IS POSITIVE,SO ANSWER IS POSITIVE AND IF A AND C ARE POSITIVE AGAIN SQUARE OF B IS POSITIVE so answer is B

S1 cannot account for A at all, and A^7 could be either positive or negative so S1 is out.

Looking at S2, B^2 is automatically positive, so we have only A^7 and C^3 to worry about. Since AC>0, then (AC)^3>0 so we have only A^4 to worry about, but A^4 is positive by definition.

So S2 alone is sufficient. S1 doesn't work at all.

S1 cannot account for A at all, and A^7 could be either positive or negative so S1 is out.

Looking at S2, B^2 is automatically positive, so we have only A^7 and C^3 to worry about. Since AC>0, then (AC)^3>0 so we have only A^4 to worry about, but A^4 is positive by definition.

So S2 alone is sufficient. S1 doesn't work at all.