Set K ={ a, b, c, -a, -b, -c, ab, bc, ac, aab, abc, aac, abbc, abac, ........}
--> no relation between numbers..

To find --> 12 number in "K"

Statement 1:
a = 2
No, info of others..
INSUFFICIENT

Statement 2:
b = 3
No, info of others..
INSUFFiCIENT

Combining,,.
a = 2
b = 3
ab = 6
aab = 12

SUFFICIENT

Answer [spoiler]{C}[/spoiler]
What is the OA????

rakeshd347 wrote:K is a set of numbers such that
(i) If x is in K, then -x is in K, and
(ii) if each of x and y is in K, then xy is in K
Is 12 in K?

(1) 2 is in K
(2) 3 is in K

If "a" is present "-a" is also present
If "a" is present and "b" is present then "ab" is present

Statement 1

"2" and "-2" are present, also we don't know "b" here. So insufficient

Statement 2

"3" and "-3" are present, also no info about "b". Insufficient

Combining 1 and 2

We know that 2, -2, 3, -3 are present in the set. But I think even this is not sufficient.

vinay1983 wrote: We know that 2, -2, 3, -3 are present in the set. But I think even this is not sufficient.

Vinay, it should be sufficient

In this question, i have a doubt.. whether the question means
==>SERIES 1: a , b , -a, -b, ab, aab, bab, .....
Or,
==>SERIES 2: a , b, -a, -b, ab

But, for both the cases.. taking your case 2, -2, 3, -3

SERIES 1: aab = (2)(2)(3) = 12 YES
SERIES 2: ab = (2)(3) = 6 NO.


So, whether you select any of the series.. you will have DEFINITE Answer.. I think Question Stem means SERIES 2 only.

So, Answer {C}

theCodeToGMAT wrote:

vinay1983 wrote: We know that 2, -2, 3, -3 are present in the set. But I think even this is not sufficient.

Vinay, it should be sufficient

In this question, i have a doubt.. whether the question means
==>SERIES 1: a , b , -a, -b, ab, aab, bab, .....
Or,
==>SERIES 2: a , b, -a, -b, ab

But, for both the cases.. taking your case 2, -2, 3, -3

SERIES 1: aab = (2)(2)(3) = 12 YES
SERIES 2: ab = (2)(3) = 6 NO.


So, whether you select any of the series.. you will have DEFINITE Answer.. I think Question Stem means SERIES 2 only.

So, Answer {C}

Answer is Indeed C. Rahul's reasoning is spot on.

K is a set of numbers such that
a) if x is in K, then -x is in K and
b) if each of x and y is in K, then xy is in K.

Is 12 in K?

1)2 is in K.
2)3 is in K.

If x is in K, then -x is in K.
In other words:
If a particular value is in K, then -(THAT VALUE) also is in K.
If x and y are in K, then xy is in K.
In other words:
If any two particular values are in K, then THEIR PRODUCT also is in K.

The conditions above apply to EVERY value in K.
Thus, each condition will yield an INFINITE number of values in K, as we will see when we evaluate the two statements.

Statement 1: 2 is in K
Thus, -2 is in K.
Thus, 2 * -2 = -4 is in K.
Thus, -(-4) = 4 is in K.
Thus, 2*4 = 8 is in K.
Thus, -2*4 = -8 is in K.
Thus, 4 * - 4 = -16 is in K.
Thus, -(-16) = 16 is in K.
Thus, K = {...-16, -8, -4, -2, 2, 4, 8, 16...}.
But we don't know what other values might be in K, so 12 might be in K or 12 might not be in K.
Insufficient.

Statement 2: 3 is in K
Thus, -3 is in K.
Thus, 3 * -3 = -9 is in K.
Thus, -(-9) = 9 is in K.
Thus, 3*9 = 27 is in K.
Thus, -3*9 = -27 is in K.
Thus, K = {...-27, -9, -3, 3, 9, 27...}.
But we don't know what other values might be in K, so 12 might be in K or 12 might not be in K.
Insufficient.

Statements 1 and 2 combined:
Since both 4 and 3 are in K, 4*3 = 12 is in K.
Sufficient.

The correct answer is C.