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OG Quant Guide 2nd Ed. # 70 (DS)

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by neelgandham » Tue Aug 28, 2012 9:33 pm
I will solve the question first and you can then leave your questions if you are still not happy.

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

Is 12 in K?
(1) 2 is in K
If 2 is in K, -2 is in K
If 2 and -2 are in K then -4(xy) is in K.
If -4 is in K, 4 is in K
If -4 and 2 are in K then -8(xy) is in K and so on.
For the integer 12 to be in the set, there should be a 3(3*4 = xy, since you already know that there is an integer 4) in the set or a -12(if X(-12)is present -X(12) is also present) in the set or the integer 12 itself or few other cases.
Since we don't have a definitive answer to the question, statement I is insufficient to answer the question.
(2) 3 is in K
If 3 is in K, -3 is in K
If 3 and -3 are in K then -9(xy) is in K.
If -9 is in K, 9 is in K
If -9 and 3 are in K then -8(xy) is in K and so on.
For the integer 12 to be in the set, there should be a 4(3*4 = xy, since you already know that there is an integer 3) in the set or a -12(if X(-12)is present -X(12) is also present) in the set or the integer 12 itself or few other cases.
Since we don't have a definitive answer to the question, statement II is insufficient to answer the question.
From I + II
2 is in K and 3 is in K
If 2 is in K, then -2(-X) is in K
If 2 and -2 are in K, then -4(XY) is in K
If -4 is in K, then 4(-X) is in K
If 3 and 4 are in K, then 12(XY) is in K.
So, we are sure that the integer 12 is present in the set K.

Answer C
I don't understand why statement (1) and (2) aren't sufficient since they both show 12 not being in the set.
Answer) 2 is in K, doesn't mean that 12 isn't in K. Similarly, it also doesn't mean that 12 is in K.
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by nisagl750 » Thu Aug 30, 2012 2:28 am
oxfordbound wrote:Hello all,

I've seen several solutions to this problem, but I am unsure of something.

The question states that K is a set with two properties:
i) if x is in K, -x is in K
ii) If x and y are in K, then xy is in K

stmt (1) and (2) are similar, but i'll use stmt (1) as an example.

1) 2 is in K.

K = 2, -2, -4, 4, -8, 8, 16, -16...

Now, the OG and most posters say that this statement is insufficient because K could contain a 12.

This doesn't make sense to me. If we know that both statements and the stem are undoubtedly true, the series illustrated above clearly shows that 12 IS NOT in the set.

I don't understand why statement (1) and (2) aren't sufficient since they both show 12 not being in the set.

I know I'm missing something subtle here, I am hoping someone can clarify this for me.

OxfordBound.

For 12 to be in K, we need, 2 & 3 (with any sign, positive or negative) to be in K
Statement1 states that 2 is in K, but it doesn't tell that 3 is NOT in K. If we are definite that 3 is THERE OR NOT in K, then Statement 1 is sufficient,

similarly, Statement 2 says that 3 is in K, but the existence of 2, -2 is unknown. Hence Insufficient

Statement 1 & 2 together says both 2(x) and 3(y) are in K so 6(xy) will be there and since 2(x) and 6(y) are there in K, 12 will also be there

Hence, C is the answer
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by GMATGuruNY » Fri Aug 31, 2012 5:47 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.

If x is in K, then -x is in K means that if any value is in K, then -(that value) also is in K.
If x and y are in K, then xy is in K means that if any two values are in K, then their product also is in K.

Since the conditions above apply to any values that are in K, each condition will generate an infinite number of values that are 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 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 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.
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