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msand
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If L 0, is 18K/L an integer? 09 May 2010, 03:28
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55% (hard)

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52% (01:25) correct 48% (01:28) wrong based on 179 sessions

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If L ≠ 0, is 18K/L an integer?


(1) K^2/L^2 is an integer.

(2) K - L = L

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Bunuel
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If L 0, is 18K/L an integer? 30 Oct 2012, 05:32
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davidfrank
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If \(L \not= 0\) , is \(\frac{18K}{L}\) an integer?

1. \(\frac{K^2}{L^2}\) is an integer.
2. \(K - L = L\)

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B
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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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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If L 0, is 18K/L an integer? 09 May 2010, 04:10
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msand
If L does not equal to 0, is (18*K)/L an integer?

(1) K^2/L^2 is an integer.
(2) K-L=L
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If \(l\neq{0}\) is \(\frac{18k}{l}=integer\)?

(1) \(\frac{k^2}{l^2}=integer\). \(k=4\) and \(l=2\) - answer YES but \(k=\sqrt{6}\) and \(l=\sqrt{2}\) - answer NO. Not sufficient.

(2) \(k=2l\) --> \(\frac{18k}{l}=\frac{18*2l}{l}=36=integer\). Sufficient.

Answer: B.
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If L 0, is 18K/L an integer? 16 May 2010, 12:32
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according to statement A K^2/L^2 is integer but that doesn't mean that K & L are integer; square of 1.732 is 3 which is an integer but 1.732 itself is not a integer. hence statement A - i nsufficient

statment B - k=2l sufficeint we get a integer from it
so B must be the answer.
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If L 0, is 18K/L an integer? 26 Apr 2023, 22:58
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msand
If L ≠ 0, is 18K/L an integer?

(1) K^2/L^2 is an integer.

(2) K - L = L
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Solution:
Pre Analysis:
  • Given \(L ≠ 0\)
  • We are asked if \(\frac{18K}{L}\) is an integer or not

Statement 1: K^2/L^2 is an integer
    According to this statement,
  • \(\frac{k^2}{L^2}=integer\)
    or \(L^2=\frac{K^2}{integer}\)
    or \(L=\frac{K}{\sqrt{integer}}\)
  • Plugging \(L=\frac{K}{\sqrt{integer}}\) in \(\frac{18K}{L}\), we get \(\frac{18\times K\times \sqrt{integer}}{K}\)
    or \(18\times \sqrt{integer}\) which we cannot be sure if it is integer or not
  • Thus, statement 1 alone is not sufficient and we can eliminate options A and D

Statement 2: K - L = L
  • According to this statement, \(2L=K\) or \(L=\frac{K}{2}\)
  • Plugging \(L=\frac{K}{2}\) in \(\frac{18K}{L}\), we get \(\frac{18\times K\times 2}{K}=36\)
  • It is definitely an integer

Hence the right answer is Option B
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If L 0, is 18K/L an integer? 26 Apr 2023, 23:43
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Official Solution:


If \(L \not= 0\), is \(\frac{18K}{L}\) an integer?

It is crucial to remember that we are not given any information about whether K and L are integers. Keep this consideration in mind when evaluating the statements.

(1) \(\frac{K^2}{L^2}\) is an integer.

If \(K=L=1\), then \(\frac{18K}{L}=18\) and the answer to the question is YES. However, if \(K=\sqrt{2}\) and \(L=1\), then \(\frac{18K}{L}=18\sqrt{2}\), which is not an integer, and the answer to the question is NO. This statement is not sufficient.

Note that if we were told that K and L are integers, the first statement would have been sufficient since \(\frac{K^2}{L^2}\) being an integer would imply that \(\frac{K}{L}\) is an integer. This is because the ratio of two integers (\(\frac{K}{L}\)) can only be an integer or a non-integer rational number, but it cannot be an irrational number. However, \(\frac{K}{L}\) cannot be a non-integer rational number, because its square would not be an integer. Therefore, \(\frac{K}{L}\) must be an integer, making \(\frac{18K}{L}\) an integer.

(2) \(K - L = L\). \(K=2L\).

In this case, \(\frac{18K}{L}=\frac{18(2L)}{L}=36\), which is an integer. Thus, the answer to the question is YES. This statement is sufficient.


Answer: B
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