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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 27 Oct 2012, 10:16
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In the fraction x/y, where x and y are positive integers, what is the value of y ?

(1) x is an even multiple of y.
(2) x - y = 2

An alternative method rather than picking numbers?
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Oct 2012, 03:43, edited 1 time in total.
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Re: In the fraction [m][fraction]x/y[/fraction][/m] , where x an [#permalink] New post 27 Oct 2012, 22:06
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danzig wrote:
In the fraction \frac{x}{y} , where x and y are positive integers, what is the value of y ?
(1) x is an even multiple of y.
(2) x - y = 2

An alternative method rather than picking numbers?


1) Insufficient. We only get \frac{x}{y}=2a. a can be any positive integer
2) Insufficient. x and y can be any two positive integers with a difference of 2 between them

1 & 2 together

x = 2ay.,
2ay - y = 2
y = \frac{2}{2a - 1}

Since y is an integer, 2a - 1 should be lesser than 2.

So "a" can only be 1. So we can get values of x and y.

Sufficient.

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Re: In the fraction [m][fraction]x/y[/fraction][/m] , where x an [#permalink] New post 28 Oct 2012, 11:44
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statement 1 : clearly insuff as it says x is an even multiple of y but no info on y

statement 2 : again insuff as it only says y is 2 less than x

taking both together :

we know from statement 1, x is even

from statement 2 : x-y = 2 ie x-y = even, thus from both we know y has to be even ( since,even - even = even )

now notice, if y is 4, 6 ........ min value of x will be 8, 12 and thus x - y will be 4 , 6 ie greater than 2

thus y is an even no and less than 4 , thus the only possibility of y is 2 ( since x n y are integers ) : SUFF

leading to C , my take
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Re: In the fraction [m][fraction]x/y[/fraction][/m] , where x an [#permalink] New post 28 Oct 2012, 12:10
I don't understand why you assume that \frac{x}{y} = 2a

The statement indicates that x is an even multiple of y. So, there is the possibility that y is even. In that sense:
\frac{x}{y} = a In other words, "the even part" of x is provided by y. So x = ay, just that.

This fact could change the answer.

Please, your comments.
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Re: In the fraction [m][fraction]x/y[/fraction][/m] , where x an [#permalink] New post 28 Oct 2012, 14:01
my understanding :

let the fraction be x/y

if x is a multiple of y : we can re-write the fraction as x = k * y ( where k is any integer )

now from stat1 : x is an even multiple, thus k is even

for any even no we can express it in the form = 2a ( where a is any integer )

thus k can be rewritten as k = 2a

hence, x = k * y OR x = 2a * y
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Re: In the fraction [m][fraction]x/y[/fraction][/m] , where x an [#permalink] New post 29 Oct 2012, 03:59
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danzig wrote:
I don't understand why you assume that \frac{x}{y} = 2a

The statement indicates that x is an even multiple of y. So, there is the possibility that y is even. In that sense:
\frac{x}{y} = a In other words, "the even part" of x is provided by y. So x = ay, just that.

This fact could change the answer.

Please, your comments.


Correct: x=(2a)y if y is odd. But if y itself is even, then this won't necessarily be true. Consider x=y=2.

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

(1) x is an even multiple of y --> x=even=my, for some positive integer m. Clearly insufficient: consider x=y=2 and x=2 and y=1. Not sufficient.

(2) x - y = 2 --> x=y+2. Not sufficient.

(1)+(2) Since from (1) x=my, then from (2) y+2=my --> y=\frac{2}{m-1} --> m-1 must be a factor of 2, thus it can be 1 (for m=2) or 2 (for m=3). But if m=3, then y=1 and x=3, which is not even. Therefore, m=2, y=2 and x=4=even. Sufficient.

Answer: C.

Hope it's clear.
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Re: In the fraction [m][fraction]x/y[/fraction][/m] , where x an [#permalink] New post 29 Oct 2012, 05:32
Bunuel wrote:
danzig wrote:
I don't understand why you assume that \frac{x}{y} = 2a

The statement indicates that x is an even multiple of y. So, there is the possibility that y is even. In that sense:
\frac{x}{y} = a In other words, "the even part" of x is provided by y. So x = ay, just that.

This fact could change the answer.

Please, your comments.


Correct: x=(2a)y if y is odd. But if y itself is even, then this won't necessarily be true. Consider x=y=2.

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

(1) x is an even multiple of y --> x=even=my, for some positive integer m. Clearly insufficient: consider x=y=2 and x=2 and y=1. Not sufficient.

(2) x - y = 2 --> x=y+2. Not sufficient.

(1)+(2) Since from (1) x=my, then from (2) y+2=my --> y=\frac{2}{m-1} --> m-1 must be a factor of 2, thus it can be 1 (for m=2) or 2 (for m=3). But if m=3, then y=1 and x=3, which is not even. Therefore, m=2, y=2 and x=4=even. Sufficient.

Answer: C.

Hope it's clear.


Since x was given as an even multiple,(i.e y times an even number would be x) I had taken \frac{x}{y} to be equal to 2a. If x were 2 and y were 2. Then x would not be an even multiple of y. Am I correct in my understanding of the term even multiple??
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Re: In the fraction [m][fraction]x/y[/fraction][/m] , where x an [#permalink] New post 29 Oct 2012, 05:35
Expert's post
MacFauz wrote:
Bunuel wrote:
danzig wrote:
I don't understand why you assume that \frac{x}{y} = 2a

The statement indicates that x is an even multiple of y. So, there is the possibility that y is even. In that sense:
\frac{x}{y} = a In other words, "the even part" of x is provided by y. So x = ay, just that.

This fact could change the answer.

Please, your comments.


Correct: x=(2a)y if y is odd. But if y itself is even, then this won't necessarily be true. Consider x=y=2.

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

(1) x is an even multiple of y --> x=even=my, for some positive integer m. Clearly insufficient: consider x=y=2 and x=2 and y=1. Not sufficient.

(2) x - y = 2 --> x=y+2. Not sufficient.

(1)+(2) Since from (1) x=my, then from (2) y+2=my --> y=\frac{2}{m-1} --> m-1 must be a factor of 2, thus it can be 1 (for m=2) or 2 (for m=3). But if m=3, then y=1 and x=3, which is not even. Therefore, m=2, y=2 and x=4=even. Sufficient.

Answer: C.

Hope it's clear.


Since x was given as an even multiple,(i.e y times an even number would be x) I had taken \frac{x}{y} to be equal to 2a. If x were 2 and y were 2. Then x would not be an even multiple of y. Am I correct in my understanding of the term even multiple??


x is an even multiple of y means that x is even AND a multiple of y.
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Re: In the fraction x/y, where x and y are positive integers [#permalink] New post 29 Oct 2012, 05:54
Thanks for clearing that up... Guess I over thought it a bit.
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Re: In the fraction x/y, where x and y are positive integers [#permalink] New post 29 Aug 2013, 08:30
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In the fraction x/y, where x and y are positive integers, what is the value of y ?

(1) x is an even multiple of y.
(2) x - y = 2

Stmt 1: it says that x=y*even integer. So, if Y=2, X can be 0, 4, 8, 12, ... so on. But if Y=3, X can be 0, 6, 12, so on. So Y can be anything basically. Insufficient.

Stmt 2: Again, 5-3=2. Also 6-4=2. So Y can again be anything as long as X is 2 more than Y. Insufficient.

Together: We see that If Y=0 and X=2, Both statement 1 and 2 are satisfied but divisibility by 0 is not defined. So Y cannot be 0. If Y=1 and X=3, then 3-1=2 but 3 is not a even multiple of 1. If Y=2 and x=4, both conditions met. If Y=3, X will have to be 5, but again is not an even multiple of 3. Thinking of the patter here, if Y>2, then X will never produce x-y=2 when X is an even multiple of Y. Hence, y can only be 2. Satisfied.
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Re: In the fraction x/y, where x and y are positive integers [#permalink] New post 14 Oct 2014, 20:38
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Re: In the fraction x/y, where x and y are positive integers   [#permalink] 14 Oct 2014, 20:38
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