2nd Edition - OG - Question 70 - The Beat The GMAT Forum - Expert GMAT Help & MBA Admissions Advice

2nd Edition - OG - Question 70

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shanice: Master | Next Rank: 500 Posts; Posts: 110; Joined: Thu Apr 05, 2012 8:48 amFollowed by: 1 member; Thanked: 3 times

by shanice » Sun Aug 05, 2012 7:56 pm

How do I approach the below question?

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.

Answer is C.

Thank you in advance.

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Source: Beat The GMAT — Data Sufficiency | Sun Aug 05, 2012 7:56 pm

alex.gellatly: Master | Next Rank: 500 Posts; Posts: 435; Joined: Wed Nov 16, 2011 7:27 amFollowed by: 16 members; Thanked: 48 times

by alex.gellatly » Sun Aug 05, 2012 9:04 pm

shanice wrote:How do I approach the below question?

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.

This question is just dealing with prime factorization. We want to know if 12 is in K. When you factor 12 down to its prime numbers you get 2*2*3. So... we need to know if 2*2*3 is in K. Now this question makes it a little more complicated with the two conditions which I'll talk about later.

clearly statements 1 and 2 are not sufficient.

Now we're left with either C or E. On first glance we want to say E because we only know that 2*3=6 is in K, and we are looking for 2*2*3=12. However by condition ii we know that if x and y is in k, than xy is in K. That means if 2 and 3 are in K, then 6 is also in K. If 6 is in K and we know that 2 is in K (statement 1), then 2*6=12 must also be in K.

Hope this helps.

A useful website I found that has every quant OG video explanation:

https://www.beatthegmat.com/useful-websi ... tml#475231

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Aug 06, 2012 3:43 am

I posted a solution here:

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