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# In the fraction x/y, where a and y are positive integers,

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Director
Joined: 15 Aug 2005
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In the fraction x/y, where a and y are positive integers, [#permalink]  14 Oct 2005, 01:59
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In the fraction x/y, where a and y are positive integers, what is the value of y?

(1) The least common dinominator of x/y and 1/3 is 6
(2) x = 1

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Cheers, Rahul.

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Joined: 07 Jul 2004
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In the fraction x/y, where a and y are positive integers, what is the value of y?

(1) The least common dinominator of x/y and 1/3 is 6
(2) x = 1

Using (2), we have 2/y and we're asked for the value of y. Clearly, it's not sufficient.

Using (1), we're told the LCD of y and 3 is 6. So y could be 2,3,6. So statement 1 is not sufficient.

Using both, we still can't solve the question. 1/2, 1/3 and 1/6 all have 6 as the LCD when paired with 1/3.

I'll take E.
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Joined: 24 Sep 2005
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Re: DS - Integers [#permalink]  14 Oct 2005, 03:03
rahulraao wrote:
In the fraction x/y, where a and y are positive integers, what is the value of y?

(1) The least common dinominator of x/y and 1/3 is 6
(2) x = 1

(1) we can point out many examples : 1/2, 1/6, 5/2 , 7/2, etc ..... Thus , (1) is insuff
(2) x=1, say nothing
Combine (1) and (2) , we can still get several values of y : 2,6 ( x/y= 1/2, 1/6) ----> insuff

E it is.
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ywilfred>

I agree that both statements are insuff, either together or independently, but how is it that y can be 3? If the LCD of both x/y and 1/3 is 6 and y were 3, then wouldnt the LCD be 3??

Ex: 1/3 and x/3---->For any value X, shouldnt the LCD be 3?
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Joined: 07 Jul 2004
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GMATT73 wrote:
ywilfred>

I agree that both statements are insuff, either together or independently, but how is it that y can be 3? If the LCD of both x/y and 1/3 is 6 and y were 3, then wouldnt the LCD be 3??

Ex: 1/3 and x/3---->For any value X, shouldnt the LCD be 3?

ah yes, a hindsight on my part!! but the method will still be the same
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# In the fraction x/y, where a and y are positive integers,

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