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

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In the fraction x/y, where x and y are positive integers, [#permalink] New post 12 Jul 2007, 18:00
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In the fraction x/y, where x and y are positive integers, what is the value of y?

1) The least common denominator of x/y and 1/3 is 6.

2) x = 1
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Re: DS - number properties [#permalink] New post 12 Jul 2007, 19:15
asaf wrote:
In the fraction x/y, where x and y are positive integers, what is the value of y?

1) The least common denominator of x/y and 1/3 is 6.

2) x = 1
E?? 2 is obviously insufficient.
from 1), we know that y is a multiple of 6.
1/3 = 2/6 = 4/12 (not sure if this reasoning is sound) whose denominators are all multiples of 6.

1 and 2 lets us know that x/y = 1/6(N) and that 1/3 = 2/6, 4/12 etc.
still nothing
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Re: DS - number properties [#permalink] New post 12 Jul 2007, 19:40
asaf wrote:
In the fraction x/y, where x and y are positive integers, what is the value of y?

1) The least common denominator of x/y and 1/3 is 6.
2) x = 1


E:

1: in x/y, y could be 2 or 6. so nsf.
2: x = 1 doesnot tell much about y.
1 & 2: nothing
so insufficient and E.
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Re: DS - number properties [#permalink] New post 13 Jul 2007, 07:26
Guys,
What is missing in my approach?

From 1:
Since we know LCM is 6 then x/y = 2

Combinig 1 and 2: we can find y, Cant we?
So C.
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answer [#permalink] New post 13 Jul 2007, 07:59
I think the answer is C.with bothstatements you get 2 as the denominator for X/Y.
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Re: DS - number properties [#permalink] New post 13 Jul 2007, 08:24
Himalayan wrote:
asaf wrote:
In the fraction x/y, where x and y are positive integers, what is the value of y?

1) The least common denominator of x/y and 1/3 is 6.
2) x = 1


E:

1: in x/y, y could be 2 or 6. so nsf.
2: x = 1 doesnot tell much about y.
1 & 2: nothing
so insufficient and E.


exact agree with E for same reasons
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Re: DS - number properties [#permalink] New post 13 Jul 2007, 08:52
asaf wrote:
In the fraction x/y, where x and y are positive integers, what is the value of y?

1) The least common denominator of x/y and 1/3 is 6.

2) x = 1


Statement 1 - Not sufficient.

x/y * 1/3 = 6
x/3y = 6
x = 18y

Statement 2 - Not sufficient.
1/y = ?

Combined - Sufficient

1 = 18y => y=1/18

My answer is C
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Re: DS - number properties [#permalink] New post 13 Jul 2007, 17:12
asaf wrote:
Guys,
What is missing in my approach?

From 1:
Since we know LCM is 6 then x/y = 2

Combinig 1 and 2: we can find y, Cant we?
So C.


OA is E on this one but I still dont get it whats am I doing wrong here.
Re: DS - number properties   [#permalink] 13 Jul 2007, 17:12
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