Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

(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

for (1) know that 2, -2 is in K for (2) know that 3, -3 is in K Together have [-3, -2, 2, 3, 6]

So I would say neither is sufficient??

Hi, and welcome to the club. Below is the solution for your problem.

(1) 2 is in K --> according to (i) -2 is n K --> according to (ii) -2*2=-4 is in K --> according to (i) -(-4)=4 is in K and so on. Thus we know that 2, -2, -4, 4, 8, -8, 16, -16, ... are in K, so basically powers of 2 and their negative pairs. Is 12 in K? We don't know. Not sufficient.

(2) 3 is in K --> according to (i) -3 is n K --> according to (ii) -3*3=-9 is in K --> according to (i) -(-9)=9 is in K and so on. Thus we know that 3, -3, -9, 9, 27, -27, 81, -81, ... are in K, so basically powers of 3 and their negative pairs. Is 12 in K? We don't know. Not sufficient.

(1)+(2) From (1) 4 is in K and from (2) 3 is in K, hence according to (ii) 4*3=12 must also be in K. Sufficient.

Ahh I guess so but I'm having difficulty understanding why x * -x is in the set because it states xy is in the set and not x* -x everything else below made sense.

I'm just starting out so hopefully I get a better sense of these conditions as I progress

Ahh I guess so but I'm having difficulty understanding why x * -x is in the set because it states xy is in the set and not x* -x everything else below made sense.

I'm just starting out so hopefully I get a better sense of these conditions as I progress

Posted from my mobile device

\(x\) and \(y\) just represent some different numbers in the set K. (ii) says that if two different numbers (\(x\) and \(y\)) are in the set, then their product (\(xy\)) is also in the set.

For example: as we know that 2 and -2 (two different numbers) are in the set, then their product (-2*2=-4) must also be in the set.

Although I'm still missing something fundamental, if (2, -2 and -4) is in the set, why is (2 * -4 = -8 etc) in the set? As you mentioned below we know 2 numbers are in the set and the multiple of the two numbers. Is there wording in there that implies every number in the set can be multiplied by any other number in the set apart from just x or y. To me x, y and xy implies 3 values.

I. 2 is in k; -2,-4,4,8,-8,16,-16. 12 is not there in this series. But, it may be there. II. 3 is in k; 3,-3,-9,9,27,-27. 12 is not there in this series. But, it may be there.

Using both; 2,3,-2,-3,-4,-9,6,-6,12. 12 is definitely there.

I. 2 is in k; -2,-4,4,8,-8,16,-16. 12 is not there in this series. But, it may be there. II. 3 is in k; 3,-3,-9,9,27,-27. 12 is not there in this series. But, it may be there.

Using both; 2,3,-2,-3,-4,-9,6,-6,12. 12 is definitely there.

Ans: "C".

Hi Fluke,

The question asks us if '12 is in K'. After writing out the sequence (as Bunnel has done), I was NOT able to get a 12. Therefore, my answer was suffcient, because I could answer definitively and say that 12 is NOT in K. I reached the same conclusion with statement (2). Therefore, my answer to this question was D.

In your quote above you mention that "[12] may be there". This is what I do not understand. If the statement given to us is a fact, then how can we assume that the sequence COULD have a 12 ? (even after we draw out the set and no 12 is present)

Bunnel, if you can, can you jump in on this one too please!

I. 2 is in k; -2,-4,4,8,-8,16,-16. 12 is not there in this series. But, it may be there. II. 3 is in k; 3,-3,-9,9,27,-27. 12 is not there in this series. But, it may be there.

Using both; 2,3,-2,-3,-4,-9,6,-6,12. 12 is definitely there.

Ans: "C".

Hi Fluke,

The question asks us if '12 is in K'. After writing out the sequence (as Bunnel has done), I was NOT able to get a 12. Therefore, my answer was suffcient, because I could answer definitively and say that 12 is NOT in K. I reached the same conclusion with statement (2). Therefore, my answer to this question was D.

In your quote above you mention that "[12] may be there". This is what I do not understand. If the statement given to us is a fact, then how can we assume that the sequence COULD have a 12 ? (even after we draw out the set and no 12 is present)

Bunnel, if you can, can you jump in on this one too please!

For (i) the data that is given is just enough to say that powers of 2 are present in the set. Data is "INSUFFICIENT" to "DEFINITELY" say that 12 isn't there in the set. It could be so that 12 is present but it hasn't been mentioned. So we can't categorically rule out the presence of 12 in the set, and say NO to the question "Is 12 in K? "

Same with (ii). The data given is "INSUFFICIENT" to say "DEFINITELY" that 12 isn't there.

Taking i and ii together, we can DEFINITELY say with this data that 12 is present. ie the statements together are SUFFICIENT

Re: K is a set of numbers such that (i) If x is in K, then -x [#permalink]

Show Tags

05 Sep 2012, 03:37

1

This post received KUDOS

Just in case someone made my same mistake: always write down the numbers (ex. 2; -2; -4 and so on) in order to realize that you have to deal with a new number everytime you have one: if 2 is there, -2 is there; so -4 is there, so 4 is there and so on, just like Bunuel showed.

I did not write down any of that and ended up with E.

Re: K is a set of numbers such that (i) If x is in K, then -x [#permalink]

Show Tags

16 Dec 2012, 08:14

1

This post received KUDOS

Expert's post

Sachin9 wrote:

bunuel,

This is how I solved..

For 12 to be in the set, 2 and 3 must be there.. as 12's prime factors are 2 and 3.. hence C. Is this correct?

No, that's not correct. 12 can be in the set even if 2 and 3 are not.

For example, (i) says that "if x is in K, then -x is in K", then if we were told that -12 is in the set then -(-12)=12 would be in the set. Or, (ii) say that "if each of x and y is in K, then xy is in K", then if we were told that both 2 and 6 are in the set, then 2*6=12 would be in the set.

Re: K is a set of numbers such that (i) If x is in K, then -x [#permalink]

Show Tags

16 Dec 2012, 20:58

Bunuel wrote:

Sachin9 wrote:

bunuel,

This is how I solved..

For 12 to be in the set, 2 and 3 must be there.. as 12's prime factors are 2 and 3.. hence C. Is this correct?

No, that's not correct. 12 can be in the set even if 2 and 3 are not.

For example, (i) says that "if x is in K, then -x is in K", then if we were told that -12 is in the set then -(-12)=12 would be in the set. Or, (ii) say that "if each of x and y is in K, then xy is in K", then if we were told that both 2 and 6 are in the set, then 2*6=12 would be in the set.

Of course there are many other possibilities.

Hope it's clear.

Yeah so, for 12 to be in the set, according to the given conditions, presence of 2 of the factors (other than 1) of 12 is required .. Since 2 and 3 are present, it follows that 12 will be there.. Is this correct? _________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

Re: K is a set of numbers such that (i) If x is in K, then -x [#permalink]

Show Tags

16 Dec 2012, 23:22

Expert's post

Sachin9 wrote:

Bunuel wrote:

Sachin9 wrote:

bunuel,

This is how I solved..

For 12 to be in the set, 2 and 3 must be there.. as 12's prime factors are 2 and 3.. hence C. Is this correct?

No, that's not correct. 12 can be in the set even if 2 and 3 are not.

For example, (i) says that "if x is in K, then -x is in K", then if we were told that -12 is in the set then -(-12)=12 would be in the set. Or, (ii) say that "if each of x and y is in K, then xy is in K", then if we were told that both 2 and 6 are in the set, then 2*6=12 would be in the set.

Of course there are many other possibilities.

Hope it's clear.

Yeah so, for 12 to be in the set, according to the given conditions, presence of 2 of the factors (other than 1) of 12 is required .. Since 2 and 3 are present, it follows that 12 will be there.. Is this correct?

The red part it not correct. In the post you are quoting you can see that 12 can be there if -12 is in the set. _________________

Re: K is a set of numbers such that (i) If x is in K, then -x [#permalink]

Show Tags

17 Dec 2012, 23:44

1

This post received KUDOS

Expert's post

Sachin9 wrote:

bunuel,

This is how I solved..

For 12 to be in the set, 2 and 3 must be there.. as 12's prime factors are 2 and 3.. hence C. Is this correct?

Responding to a pm:

Given in the question:

(ii) if each of x and y is in K, then xy is in K

What does this mean? It means that if x and y are in K, then xy must also be there e.g. x = 6, y = 8 If 6 and 8 are in K, 6*8 = 48 must also be in K.

Does it also mean that 2 and 3 (i.e. factors of 6) must also be in K? No. It is not necessary. I am building the set K. I could have put 6 in on my own. I don't need to start from 2 and 3 necessarily. If 6 and 8 are in the set K, I necessarily need to put their product in too. But I needn't put in their prime factors. We do not know whether their prime factors were put in and hence 6 and 8 were obtained or whether they were put in by set maker's choice.

So, the statement 'if each of x and y is in K, then xy is in K' only implies that product of x and y must be in K too. It doesn't imply that factors of x and y must be in K. x and y could have been put in by choice. Who says that only prime factors can be added to the set? You can pick any number and add it to the set. The only thing is that once you put in that number, you must put in its product with every number already there in the set and so on...

As pointed out by Bunuel, I hope you see that your logic is not correct. _________________

Re: K is a set of numbers such that (i) If x is in K, then -x [#permalink]

Show Tags

13 Apr 2013, 08:31

Thank you Karishma. Your statement that we don't have to start with 2 or 3 to make the set was the KEY for me. It really told me that may be 12 is possible for statement 1 and 2 because earlier I thought it was ALWAYS NO and was wondering how one could get 12 in the set K. Of course, if we don't assume that we always have to start building the set with 2 or 3, then 12 is entirely possible. Since this is ALWAYS no or ALWAYS yes question, C makes much more sense now.

VeritasPrepKarishma wrote:

Sachin9 wrote:

bunuel,

This is how I solved..

For 12 to be in the set, 2 and 3 must be there.. as 12's prime factors are 2 and 3.. hence C. Is this correct?

Responding to a pm:

Given in the question:

(ii) if each of x and y is in K, then xy is in K

What does this mean? It means that if x and y are in K, then xy must also be there e.g. x = 6, y = 8 If 6 and 8 are in K, 6*8 = 48 must also be in K.

Does it also mean that 2 and 3 (i.e. factors of 6) must also be in K? No. It is not necessary. I am building the set K. I could have put 6 in on my own. I don't need to start from 2 and 3 necessarily. If 6 and 8 are in the set K, I necessarily need to put their product in too. But I needn't put in their prime factors. We do not know whether their prime factors were put in and hence 6 and 8 were obtained or whether they were put in by set maker's choice.

So, the statement 'if each of x and y is in K, then xy is in K' only implies that product of x and y must be in K too. It doesn't imply that factors of x and y must be in K. x and y could have been put in by choice. Who says that only prime factors can be added to the set? You can pick any number and add it to the set. The only thing is that once you put in that number, you must put in its product with every number already there in the set and so on...

As pointed out by Bunuel, I hope you see that your logic is not correct.

Re: K is a set of numbers such that (i) If x is in K, then -x [#permalink]

Show Tags

29 May 2014, 21:29

can any one tell me why (1) and (2) are insufficient ? (1) shows that the set is …. -16,-8,-4,-,2,4,8,16….( there is no 12 here) so it is sufficient. (2) shows that the set is …. -27,-9,-3,3,9,27,…… ( there is no 12 here) so it is sufficient. so the answer is D … each alone is sufficient.

Re: K is a set of numbers such that (i) If x is in K, then -x [#permalink]

Show Tags

29 May 2014, 22:31

Expert's post

shagalo wrote:

can any one tell me why (1) and (2) are insufficient ? (1) shows that the set is …. -16,-8,-4,-,2,4,8,16….( there is no 12 here) so it is sufficient. (2) shows that the set is …. -27,-9,-3,3,9,27,…… ( there is no 12 here) so it is sufficient. so the answer is D … each alone is sufficient.

any explanation please? thanks

How does statement 1 show that 12 is not in the set? All statement 1 tells you is that 2 is there and hence -2 is there. We don't know anything about other elements. How did you get 4... We are not given that if 2 is there, only powers of 2 will be there. _________________

Last year when I attended a session of Chicago’s Booth Live , I felt pretty out of place. I was surrounded by professionals from all over the world from major...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...