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by sachindia » Mon Jul 23, 2012 4:54 am
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.
Regards,
Sach
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Source: — Data Sufficiency |

by GMATGuruNY » Mon Jul 23, 2012 5:51 am
sachindia wrote: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.
Last edited by GMATGuruNY on Wed Feb 27, 2013 7:04 am, edited 1 time in total.
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by Anurag@Gurome » Mon Jul 23, 2012 6:45 am
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