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In the fraction x/y, where x and y are +ve integers, what is

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Manager
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In the fraction x/y, where x and y are +ve integers, what is [#permalink] New post 02 Mar 2005, 03:45
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A
B
C
D
E

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In the fraction x/y, where x and y are +ve integers, what is the value of y?


I) The least common denominator of x/y and 1/3 is 6.
II) x=1.
Manager
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 [#permalink] New post 02 Mar 2005, 04:37
Why not C?

when x=1, and we know least common (call it LCD)is 6, it comes down to,
LCD of 1/y and 1/3 is 6.
Bet y & 3, LCD is 6. Assuming LCD as LCmultiple, value of y is 2.

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 [#permalink] New post 02 Mar 2005, 06:37
"Bet y & 3, LCD is 6. Assuming LCD as LCmultiple, value of y is 2"
But y can be 6 as well. So the answer is E
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 [#permalink] New post 02 Mar 2005, 06:51
E
from 1 it can be 2 or 6
by x=1 we dont get anything conclusive either
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Re: DS [#permalink] New post 04 Mar 2005, 15:58
I) The least common denominator of x/y and 1/3 is 6.
(LCD i believe is LCM. LCM of 1/3 and 6 is 18, since 1/3 * 18 = 6. Therefore x/y = 18 => x = 18y, meaning that y cannot be determined) insufficient

II) x=1.[/quote]
(definitely out) insufficient

Combining, we have from (I) that x=18y and from (II) that x=1, therefore y=1/18. but y is given to be an integer, therefore (C) is out.

Ans. is E. My answer assumes that LCD = LCM
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 [#permalink] New post 04 Mar 2005, 16:01
Y can be 2 or 6
X=1 takes us nowhere

E it is
  [#permalink] 04 Mar 2005, 16:01
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In the fraction x/y, where x and y are +ve integers, what is

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