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(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 --> 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.

Let's analyze statements with what's given in the question.

Statement (1) 2 is in K (i) -2 is also in K (ii) 2 and -2 are both in K, so -4 is also in K .... then +4... and then -8 and +8 are in K ... looks like 2^n and -2^n are included where n is integer which does not include 12 definitely... So Statement (1) is sufficient to answer the question "Is 12 in K?", answer being no.

Statement (2) Similarly results in a set 3^n and -3^n , which again answers our question, that 12 is definitely not part of the set K. So Statement (2 is sufficient to answer the question "Is 12 in K?", answer being no.

Answer D; Both statement 1 & statement 2 are ALONE sufficient.

[oops It turns out my answer was wrong ... Just leaving the post as it is, so you know what not to do ]
_________________

Kudos (+1) if you find this post helpful.

Last edited by code19 on 31 Jan 2014, 13:59, edited 4 times in total.

Ans. C From S1:if 2 is in the series,then -2 will also be there. And if 2 & -2 are there -4 will be there.If -4 is in the series, 4 will also be there...and so on The series becomes:2,-2,4,-4...powers of 2 But the stimulus remains silent about what is not there in this series.So insufficient.(12 might or might not be there.)

Same explanation for S2:The series will have numbers with powers of 3.

Together for S1 & S2,at some point we'll have multiple of 3 and 4 because if 3 and 4 are there in the series,their multiple will definitely be there as implied by the second statement in stimulus.Sufficient.

(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 --> 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.

Bunuel, one quick query -> When we say that (from stmt 1) 2 is there in the set and hence -2 is also there -> Here we take 2 and -2 as x and -x, but then we also apply the logic x*y = -4 (here we consider 2 as x and -2 as y (and not as -x)). Could there be a flaw in the problem statement?

Bunuel wrote:

pradeepss wrote:

Bunel can you update the oa? It shows as d on gmat timer.

Bunuel, one quick query -> When we say that (from stmt 1) 2 is there in the set and hence -2 is also there -> Here we take 2 and -2 as x and -x, but then we also apply the logic x*y = -4 (here we consider 2 as x and -2 as y (and not as -x)). Could there be a flaw in the problem statement?

Bunuel wrote:

pradeepss wrote:

Bunel can you update the oa? It shows as d on gmat timer.

________________ Done. Thank you.

(i) and (ii) are general rules for the set, meaning that they apply to any numbers in the set:

(i) if a number is in K, then - that number is also in K (ii) for any two numbers in the set, their product is also in the set.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Let us say that x and y are part of K Hence K ={x,y,-x,-y,-xy,-y*y, -x*x….}

SUPERSET The answer will be either YES or NO.

TRANSLATION To know the answer 1# exact value of the element of K 2# some values to calculate the rest of the value

STATEMENT ANALYSIS

St 1: if 2 is in K, the -2 is in K, -4 is in K and so on. Hence, all the calculated elements will be in the form of 2n.we cannot say about rest of the element. Option a and d eliminated

St 2: if 3 is in K, the -3 is in K, -9 is in K and so on. Hence, all the calculated elements will be in the form of 3n.we cannot say about rest of the element.option b eliminated

St 1 & St 2: If 2 and 3 is in K, -2 and -3 is in K, -4 is also in K, 12 is in K.ANSWER

(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 --> 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.

How are we getting 8, -8 in the first set and 27, -27 in the 2nd? Shouldn't it be 2,-2,-4,4,-16,16... and 3,-3,-9,9,-81,81...

The reason is that if we are taking xy as x and -x in statement (i), how do we get (for example) 8,-8 if we get -4 and 4 for x and -x? Is it that we are keeping y constant at 2 throughout? Please clarify.

(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 --> 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.

How are we getting 8, -8 in the first set and 27, -27 in the 2nd? Shouldn't it be 2,-2,-4,4,-16,16... and 3,-3,-9,9,-81,81...

The reason is that if we are taking xy as x and -x in statement (i), how do we get (for example) 8,-8 if we get -4 and 4 for x and -x? Is it that we are keeping y constant at 2 throughout? Please clarify.

Next: (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 --> according to (ii) 2*4 = 8 is in K --> according to (i) -8 is in K...

(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 --> 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.

How are we getting 8, -8 in the first set and 27, -27 in the 2nd? Shouldn't it be 2,-2,-4,4,-16,16... and 3,-3,-9,9,-81,81...

The reason is that if we are taking xy as x and -x in statement (i), how do we get (for example) 8,-8 if we get -4 and 4 for x and -x? Is it that we are keeping y constant at 2 throughout? Please clarify.

Thanks.

Graeme

First of all please read this:

Next: (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 --> according to (ii) 2*4 = 8 is in K --> according to (i) -8 is in K...

Hope it's clear.

Thanks for responding. Think I got it.

Therefore, we keep the pattern, substituting xy found in (ii) into "x" found in (i) but then, when coming back to (ii), leaving the original "x" = 2 in xy. Correct?

gmatclubot

Re: K is a set of numbers such that
[#permalink]
06 May 2017, 10:36

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