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If abc is not equal to 0, what is the value of (a^3 + b^3 + [#permalink]
16 Jul 2003, 01:16

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

0% (00:00) correct
100% (00:49) wrong based on 2 sessions

If abc is not equal to 0, what is the value of (a^3 + b^3 + c^3)/(abc)?

(1). |a|=1, |b|=2, |c|=3
(2). a + b + c = 0

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
_________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

The answer is C.
1 alone is not enough. 2 alone is not enough, either.
1 and 2 taken together are possible only when
a) a=1, b=2, c=-3 or
b) a=-1, b=-2, c=3

but the results in both cases are the same
a) (1+8-27)/-6
b) (-1-8+27)/6 = (1+8-27)/-6

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

Can't challenge a counterexample. Okay, we eliminate A and D. What now?
_________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

I guess if such type of problem appaear on GMAT, all what is required is nice clean trick striking at the correct moment othwerwise, one would land up wasting 4 to 5 minutes solving it for actual.

(1) can be shown to be insufficient with a quick example as most of you have shown.

Many of you also said the (2) does not tell you anything. Did you even try any numbers? The answer will always be the same. Although that does not PROVE it, it would certainly give you a clue that there is a way to simplify the sum of cubes. (Frankly, on the real test, if I tried 4 different sets of simply numbers and they all worked, I would assume it is true). You can prove it as follows:

In (2), we are given a + b + c = 0. This means c = -(a + b). Make this substitution in the sum of cubes, expand what used to be the c^3 term, then simplify. You will get -(a+b)*3*a*b. Now substitute c for the -(a+b) term and you will get 3*a*b*c.

If a + b + c = 0, then (a^3 + b^3 + c^3)/(abc) = 3abc/abc = 3 for ANY a, b, or c not equal to 0, hence B is the correct answer.
_________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993