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M07-02

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M07-02  [#permalink]

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New post 15 Sep 2014, 23:34
1
5
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

53% (00:49) correct 47% (00:59) wrong based on 158 sessions

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Re M07-02  [#permalink]

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New post 15 Sep 2014, 23:34
Official Solution:


Statement (1) by itself is insufficient. S1 does not guarantee that \(\frac{K}{L}\) is an integer. For example, let's examine 3:

\(\frac{K^2}{L^2} = 3 -\) integer

\(\frac{K}{L} = \sqrt{3} -\) not an integer

Statement (2) by itself is sufficient. S2 tells us that \(K=2L\). Replace \(K\) with \(2L\) and solve as follows:
\(\frac{18*2L}{L} = 18*2 = 36\)


Answer: B
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M07-02  [#permalink]

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New post 22 Nov 2014, 13:18
Curious to see what other people's first reaction to seeing this is.

My first thought was: for (18k)/L to be an integer, then L must be a factor of either 18, or K, or both. I thought stmt 1 was sufficient to determine that L was a factor of K but I guess this is incorrect.

I'm assuming it is always the case that (x^2)/(y^2) will never be sufficient to determine that y is a factor of x.

Stmt 2 tells us that L is a factor of K (because K=2L) therefore we know the L in the denominator will cancel out one factor of L in the numerator. This leaves a 1 in the denominator as well.
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Re: M07-02  [#permalink]

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New post 17 Aug 2015, 03:27
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JackSparr0w wrote:
Curious to see what other people's first reaction to seeing this is.

My first thought was: for (18k)/L to be an integer, then L must be a factor of either 18, or K, or both. I thought stmt 1 was sufficient to determine that L was a factor of K but I guess this is incorrect.

I'm assuming it is always the case that (x^2)/(y^2) will never be sufficient to determine that y is a factor of x.

Stmt 2 tells us that L is a factor of K (because K=2L) therefore we know the L in the denominator will cancel out one factor of L in the numerator. This leaves a 1 in the denominator as well.



The take-away here is that assuming variables always carry rational numbers is not very becoming of a 700-level tester. Question caught me on that assumption too. That's why I love gmatclub problems; they keep you honest!
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Re: M07-02  [#permalink]

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New post 04 Dec 2015, 22:10
whafif, both k and l were integer?
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Re: M07-02  [#permalink]

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New post 24 Apr 2016, 00:02
Statement (1) by itself is insufficient. S1 does not guarantee that \(\frac{K}{L}\) is an integer. For example, let's examine 3:

\(\frac{K^2}{L^2} = 3 -\) integer


When \(\frac{K^2}{L^2}\) will give a square number. Can you give a example how can we get a value such as 3, which is not a square number.
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Re: M07-02  [#permalink]

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New post 24 Apr 2016, 03:14
atturhari wrote:
Statement (1) by itself is insufficient. S1 does not guarantee that \(\frac{K}{L}\) is an integer. For example, let's examine 3:

\(\frac{K^2}{L^2} = 3 -\) integer


When \(\frac{K^2}{L^2}\) will give a square number. Can you give a example how can we get a value such as 3, which is not a square number.


Consider K^2 = 3 and L^2 = 1.
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Re: M07-02  [#permalink]

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New post 24 Apr 2016, 23:12
Bunuel wrote:
atturhari wrote:
Statement (1) by itself is insufficient. S1 does not guarantee that \(\frac{K}{L}\) is an integer. For example, let's examine 3:

\(\frac{K^2}{L^2} = 3 -\) integer


When \(\frac{K^2}{L^2}\) will give a square number. Can you give a example how can we get a value such as 3, which is not a square number.


Consider K^2 = 3 and L^2 = 1.


Is it possible to get 3 as the squared value? If yes, how do you determine that?
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Re: M07-02  [#permalink]

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New post 24 Apr 2016, 23:42
atturhari wrote:
Bunuel wrote:
atturhari wrote:
Statement (1) by itself is insufficient. S1 does not guarantee that \(\frac{K}{L}\) is an integer. For example, let's examine 3:

\(\frac{K^2}{L^2} = 3 -\) integer


When \(\frac{K^2}{L^2}\) will give a square number. Can you give a example how can we get a value such as 3, which is not a square number.


Consider K^2 = 3 and L^2 = 1.


Is it possible to get 3 as the squared value? If yes, how do you determine that?


Yes, if \(K=\sqrt{3}\), then \(K^2=(\sqrt{3})^2=3\).
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M07-02  [#permalink]

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New post 08 Jan 2017, 03:56
what if L=1.4
18*1.4 =/ integer
pls suggest
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Re: M07-02  [#permalink]

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New post 08 Jan 2017, 04:47
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Re: M07-02  [#permalink]

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New post 10 Nov 2017, 15:57
Is this a 700 level question?
I did it on a Quiz and it was saying so.. but hardly looks like it.
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Re: M07-02  [#permalink]

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New post 16 Nov 2018, 12:57
Bunuel wrote:
gupta87 wrote:
what if L=1.4
18*1.4 =/ integer
pls suggest


If L = 1.4, then from (2) K = 2.8 and \(\frac{18K}{L}=36=integer\).


Hi Bunnel -

K^2/L^2 = Integer

K^2 = L^2 x Integer

K = L x (Integer)^1/2.

As it dependes of the value of the integer, the statment one is insufficient. Is this logic right?
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Re: M07-02 &nbs [#permalink] 16 Nov 2018, 12:57
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