Hmmm. IMO - [spoiler]E (neither)[/spoiler]

I could see if statement one was 4, because you would now that 3 and 4 are in the set and thus 12 (see statement two). The fact that it's negative has no relevance to me.

The only way I could see C being the correct answer is if that set can grow. So you would say if I know 2 and 3 is in the set, I know 6 is in the set. If I know 2, 3, and 6 are in the set, I know that 12 (2*6), and 18 (3*6) are also in the set.... But then that chain would continue to grow with no end. Maybe that's the answer. If someone else has a better idea, please post. It never said the set was bounded....hmm???

umasastry wrote:Can anybody explain the below problem in terms of data sufficiency?

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.

I thought the answer for this would be D which is "Each statement alone is sufficient to answer the question asked".

But the answer is C whcih is "Both statements together are sufficient to answer the question but not alone".

I would appreciate if anybody can give a sensible explanation for this problem.

Thanks

yes the ans wud BE C


individually bth are insuff

s1) for it to be suff we must have 4 or 6 in the set but we don't have that info

s2) for it to be true we must have a $ but we are not sure ...insuff

s1+s2---->set K=2,-2,3,6,-3,-6,12,-12,-.......so on
12 is in K suff

Thread is moved to "Data Sufficiency" folder!

Please post the thread in an appropriate folder. Thanks!

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.

i. if 2 is in K, then -2 is in K and
ii. if each f 2 and y is in K, then 2*y is in K.
--> not sufficient by itself --> A and D are eliminated

2. 3 is in K.

i. if 3 is in K, then -3 is in K and
ii. if each f 3 and y is in K, then 3*y is in K.
--> not sufficient by itself --> B is eliminated

we are left with C & E .. combining both statements -> 2 and 3 are in K
--> ii. if each 2 and 3 is in K, then 2*3 is in K --> 6 is also in K
since 6 is in K and we know what 2 is also in K --> using (ii) --> 6*2 is in K --> 12 is in K

Thus, we have got the answer from combining both statements --> C is the answer

hope this helps...

umasastry wrote:Can anybody explain the below problem in terms of data sufficiency?

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.

12 = 2*2*3

1. 2 is in K; by using i. -2 is in K; using ii. 2*-2 is in K...

We can prove that all the powers of 2 and -2 and combination of these are in K (-2,2,-4,4,8,16 ...) but we cannot prove that 12 is in K

2. Similarly using 2. we can prove that all the powers of 3 and -3 and the combination of these are in K (3, -3, 9, -9 ...) but not 12.

Combining;

2 is in K
3 is in K

using ii 2*3 = 6 is in K

using ii again, 2*6 = 12 is in K

hence C

Hence