The numbers a, b, and c are all non-zero integers. Is a 0?

Question

The numbers a, b, and c are all non-zero integers. Is a > 0?

(1) a = b^3
(2)  a=\sqrt {c}

viktorija · Answer

(1) Insufficient. If b<0, then the answer will be no, because it will give an a<0. If b>0, then a will be positive as well. So two different answers.
(2) Insufficient. If c <0, the root, hence, a will be negative, but if c>0, then a will be more than zero too.

(1)+(2) Insufficient. Statements do not provide any information about the relationship between c and b other than if b,c>0 then a>0 and if b,c<0 then a<0. Again, there two possible answers.

Answer E

DudeYou · Answer

(1) a = b^3
If b = -1 a = -1
b = 1 , a = 1
Not Suff

(2) a=\sqrt {c}
Roots are always positive, So a is always poitive

ANS: B

EMPOWERgmatRichC · Answer

Hi Randude,

While square-roots are ALWAYS positive (or 0), cube-roots CAN be negative. Knowing this, what would your answer to this question be?

GMAT assassins aren't born, they're made,
Rich

KarishmaB · Answer

Using both statements, consider two values of a.

a = 8 = 2^3 = \sqrt {512} 
 a = -8 = (-2)^3 = \sqrt {-512}

So a may be positive or negative.

Answer (E)

Bingo123 · Answer

Hi Karishma,

I got the answer as B on the premise that a, b and c are give as integers. Hence, c has to be positive. Because if c is negative, it will result in complex roots and a being equal to cube root of c has to be an integer.

When I read your answer as E, I tried to find out the reason. Please confirm my understanding mentioned below:

"Square root of -1 is "iota"(i.e. complex) however cube root of -1 is -1. Though it has complex conjugate roots as well. but since a, is given has integer it has to the real cube root of -1."

KarishmaB · Answer

Square root of a negative number is complex but cube root of a negative number has a real value. 
Say, if a = -2 and c = -8, can we say  a = \sqrt {c} ? We can! 
So a and c can take negative values.

Bunuel · Answer

MAGOOSH OFFICIAL SOLUTION:

Again, all that is given in the prompt is that a, b, and c are non-zero integers.

Statement #1: now, if we cube a positive, we get a positive, but if we cube a negative, we get a negative. The numbers a & b are either both positive or both negative, but since we don’t know the sign of b, we cannot determine the sign of a. This statement, by itself, is insufficient.

Statement #2: if we take the cube root of a positive, we will get a positive, but if we take the cube-root of a negative, we get a negative. The numbers a & c are either both positive or both negative, but since we don’t know the sign of c, we cannot determine the sign of a. This statement, by itself, is insufficient.

Combined Statements: If we put both statements together, we get that all three numbers, a, b, and c, have to have the same sign: either all three are positive, or all three are negative. We have no further information that would allow us to determine which of those two is the case. Thus, even with combined statements, we still do not have enough information to give a definitive answer to the prompt question. Combined, the statements are still insufficient.

Answer = E

saurabhsavant · Answer

i am not able to comprehend it properly. For statement No.2, aren't we supposed to "CLOSE" our minds to knowledge of complex numbers, like they didn't exist. So to get a cube root of an integer, the integer has to be +ve. No? So, c can only take positive values coz a cube rot of a negative number "IS NOT DEFINED"...so is the correct way to approach it is- substitute c=-8, cube both sides and get a=-ve 2???
Am I right?

EMPOWERgmatRichC · Answer

Hi saurabhsavant,

While square-roots are ALWAYS positive (or 0), cube-roots CAN be negative. Knowing this, what would your answer to this question be?

GMAT assassins aren't born, they're made,
Rich

saurabhsavant · Answer

Ahhh!!elementary Watson....obviously, cube root can be negative....sorry and thnx a ton!!

RaghavKhanna · Answer

Statement 1:
a = b^3
case 1: if b > 0
=> a > 0
case 2: if b < 0
=> a < 0

Statement 1 is insufficient

Statement 2:
a = (c)^1/3
case1 : if c > 0
=> (c)^1/3 > 0
case 2: if c < 0
=> (c)^1/3 < 0 [eg: (-8)^1/3 = -2]

Statement 2 is insufficient

Statement 1 and Statement 2:
Together also, they do not give a definite result about 'a'
=> Statement 1 and Statement 2 together are insufficient

Answer: E

karanbishti · Answer

isnt it is given that abc are integers and if we apply a to be -ve then c will not be integer hencce I belive that the answer will be B

Bunuel · Answer

I believe you haven't reviewed the previous discussion.

If a = b = c = 1, the answer is YES.
If a = b = c = -1, the answer is NO.

bumpbot · Answer

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The numbers a, b, and c are all non-zero integers. Is a > 0?

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The numbers a, b, and c are all non-zero integers. Is a > 0?

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