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If \(L \not= 0\) , is \(\frac{18K}{L}\) an integer? 1. \(\frac{K^2}{L^2}\) is an integer. 2. \(K  L = L\) Source: GMAT Club Tests  hardest GMAT questions Can you please explain to me how K^2/L^2 could equal 3? Obviously my brain is not working today because I have plugged many different possibilities in and can't come up with one that equals three and also how that figure does not ensure k/L integer. Thank you. Very frustrating.



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30 Sep 2008, 04:33
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30 Sep 2008, 06:54
Hello dzyubam.
Thank you, but an example can only prove to insufficient if it is explained where that example was derived from. My question still is? What K^2/L^2 = 3? Or please give me an example that can be proven that also shows the insufficiency.
Thank you!!



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The number 3 was picked because it's not a perfect square. If you pick a perfect square instead of 3 then you see that S1 works fine. Therefore only the numbers that are not perfect squares can be used as examples to prove that S1 is insufficient. We can be sure that some statement is insufficient if we can think of two examples one of which works and another one doesn't. Let's see: \(\frac{K^2}{L^2} = 5\) > this one doesn't work because \(\sqrt{\frac{K^2}{L^2}} = \sqrt{5}\) > not an integer \(\frac{K^2}{L^2} = 25\) > works fine because \(\sqrt{\frac{K^2}{L^2}} = \sqrt{25} = 5\) > an integer Do you agree?
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01 Oct 2008, 16:53
Thank you dzyubam.
I was not considering that K or Y were anything but integers. I need to watch for that. In your first instance, K could be the square root of 5 and L could be 1, resulting in 5 after the fraction is squared, but resulting in the square root of 5 if the fraction is not squared. Thank you.



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I think the answer to this question should be 'E'. (1) is insufficient as the OE correctly indicates. (2) does not work when K=L=0. Hence, E. What do people think?
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11 Jan 2010, 03:02
Congratulations on your first post and welcome to GMAT Club! Yep, you're right. I think we'll have to change the question a bit to include a statement that \(L \not= 0\). +1. Thanks for pointing this out! wiredo wrote: I think the answer to this question should be 'E'. (1) is insufficient as the OE correctly indicates. (2) does not work when K=L=0.
Hence, E.
What do people think?
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Re: M07 #2 [#permalink]
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26 Oct 2010, 06:15
dczuchta wrote: If \(L \not= 0\) , is \(\frac{18K}{L}\) an integer? 1. \(\frac{K^2}{L^2}\) is an integer. 2. \(K  L = L\) Source: GMAT Club Tests  hardest GMAT questions Can you please explain to me how K^2/L^2 could equal 3? Obviously my brain is not working today because I have plugged many different possibilities in and can't come up with one that equals three and also how that figure does not ensure k/L integer. Thank you. Very frustrating. to answer your question: if we take L = 2 and K = 2(3^0.5) [i.e. 2 times sq.rt. of 3], K^2/L^2 will equal 3. 1. If we take above mentioned values, original expression IS NOT int. If we take K=1 & L=1, original expression IS int. NOT SUFFICIENT. 2. Substituting K=2L in original exprerssion gives us 36, an int. SUFFICIENT. Answer: B. 2 alone is sufficient.
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26 Oct 2010, 09:05
Sorry, but can someone tell me where I asm supposed to see the multiple choice answers? I can't see them in the email or when I click on the link am I doing something wrong?



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26 Oct 2010, 09:13
scardin wrote: Sorry, but can someone tell me where I asm supposed to see the multiple choice answers? I can't see them in the email or when I click on the link am I doing something wrong? In Data Sufficiency questions, answer choices are always same. A: 1 alone is sufficient. B: 2 alone is sufficient. C: Both together are sufficient: D: Each alone is sufficient. E: Data insufficient. Hope this is what you were asking about.
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27 Oct 2010, 09:40
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B because.. 1. If the integer is not a perfect square the answer will be insufficient (ex: 3, 5, 7). If the integer is a perfect square the problem works. 2. K  L = L so that means K = 2L. If K is 8 then L is 4 etc... K is divisible by L making the statement YES at all times so it is sufficient.
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12 Dec 2010, 20:19
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stmt 1 will stand true only in case of integers, but nothing is given about K and L Stms 2 : sufficient An eye opener question for not considring variables as Integers by default...
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Re: M07 #2 [#permalink]
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28 Oct 2011, 08:34
Hi Friends,
I feel the answer should be C. If we square the question term it will become 18^2k^2/l^2. since k^2/l^2 is integer ,18k/l should also be an integer please correct me if I m missing something thanks
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Re: M07 #2 [#permalink]
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18 Nov 2011, 00:32
statement 2 is sufficient as kl = l then k = 2l hence 18k/l is integer statement 1 is not sufficient for example k^2/ L^2 = 7 then sqrt of K^2/l^2 = not an integer
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Re: M07 #2 [#permalink]
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30 Oct 2012, 05:04
dczuchta wrote: If \(L \not= 0\) , is \(\frac{18K}{L}\) an integer? 1. \(\frac{K^2}{L^2}\) is an integer. 2. \(K  L = L\) Source: GMAT Club Tests  hardest GMAT questions Can you please explain to me how K^2/L^2 could equal 3? Obviously my brain is not working today because I have plugged many different possibilities in and can't come up with one that equals three and also how that figure does not ensure k/L integer. Thank you. Very frustrating. I have checked the OA, which is B. This implies that statement 1 is not sufficient to answer the question. The root or primary question was 18k/L is an integer. So the answer to this should be a yes or a no. If we are choosing B as the answer so we must have a yes or a no situation. So all those who have chosen B as the answer, request you to please show both the cases.



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davidfrank wrote: dczuchta wrote: If \(L \not= 0\) , is \(\frac{18K}{L}\) an integer? 1. \(\frac{K^2}{L^2}\) is an integer. 2. \(K  L = L\) Source: GMAT Club Tests  hardest GMAT questions Can you please explain to me how K^2/L^2 could equal 3? Obviously my brain is not working today because I have plugged many different possibilities in and can't come up with one that equals three and also how that figure does not ensure k/L integer. Thank you. Very frustrating. I have checked the OA, which is B. This implies that statement 1 is not sufficient to answer the question. The root or primary question was 18k/L is an integer. So the answer to this should be a yes or a no. If we are choosing B as the answer so we must have a yes or a no situation. So all those who have chosen B as the answer, request you to please show both the cases. If \(L \not= 0\) , is \(\frac{18K}{L}\) an integer?(1) \(\frac{K^2}{L^2}\) is an integer. If \(K=L=1\), then \(\frac{18K}{L}=1\) and the answer to the question is YES but if \(K=\sqrt{2}\) and \(L=1\), then \(\frac{18K}{L}=18\sqrt{2}\neq{integer}\) and the answer to the question is NO. Not sufficient. (2) \(K  L = L\) > \(K=2L\). In this case \(\frac{18K}{L}=\frac{18*2L}{L}=36=integer\), so the asnwer to the question is YES. Sufficient. Answer: B. Hope it's clear.
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