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

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In the fraction x/y, where a and y are positive integers, [#permalink] New post 14 Oct 2005, 01:59
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E

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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

Please explain your answer!
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Cheers, Rahul.

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 [#permalink] New post 14 Oct 2005, 02:57
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

Start with the second statement as it's the most straightforward.

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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Re: DS - Integers [#permalink] New post 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

Please explain your answer!


(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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 [#permalink] New post 14 Oct 2005, 05:50
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 wouldn`t the LCD be 3??

Ex: 1/3 and x/3---->For any value X, shouldn`t the LCD be 3?
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 [#permalink] New post 14 Oct 2005, 07:53
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 wouldn`t the LCD be 3??

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


ah yes, a hindsight on my part!! :oops: but the method will still be the same
  [#permalink] 14 Oct 2005, 07:53
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In the fraction x/y, where a and y are positive integers,

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