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# M07-02

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Math Expert
Joined: 02 Sep 2009
Posts: 42280

Kudos [?]: 132894 [0], given: 12391

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16 Sep 2014, 00:34
Expert's post
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Difficulty:

65% (hard)

Question Stats:

51% (00:49) correct 49% (00:54) wrong based on 143 sessions

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

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

(2) $$K - L = L$$
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 42280

Kudos [?]: 132894 [0], given: 12391

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16 Sep 2014, 00:34
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:
$$\frac{18*2L}{L} = 18*2 = 36$$

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Joined: 08 Feb 2014
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Location: United States
Concentration: Finance
GMAT 1: 650 Q39 V41
WE: Analyst (Commercial Banking)

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22 Nov 2014, 14:18
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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Senior Manager
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17 Aug 2015, 04:27
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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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04 Dec 2015, 23:10
whafif, both k and l were integer?

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Manager
Joined: 19 Jan 2016
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24 Apr 2016, 01:02
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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Math Expert
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24 Apr 2016, 04:14
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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25 Apr 2016, 00:12
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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Math Expert
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25 Apr 2016, 00:42
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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08 Jan 2017, 04:56
what if L=1.4
18*1.4 =/ integer
pls suggest

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Math Expert
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08 Jan 2017, 05:47
gupta87 wrote:
what if L=1.4
18*1.4 =/ integer
pls suggest

If L = 1.4, then from (2) K = 2.8 and $$\frac{18K}{L}=36=integer$$.
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10 Nov 2017, 16:57
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] 10 Nov 2017, 16:57
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# M07-02

Moderators: Bunuel, chetan2u

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