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In the fraction x/y, where x and y are positive integers [#permalink]
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27 Oct 2012, 11:16
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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?
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Last edited by Bunuel on 29 Oct 2012, 04:43, edited 1 time in total.
Edited the question.



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Re: In the fraction [m][fraction]x/y[/fraction][/m] , where x an [#permalink]
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27 Oct 2012, 23:06
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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. Kudos Please... If my post helped.
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Re: In the fraction [m][fraction]x/y[/fraction][/m] , where x an [#permalink]
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28 Oct 2012, 12:44
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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 : xy = 2 ie xy = 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]
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28 Oct 2012, 13:10
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]
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28 Oct 2012, 15:01
my understanding :
let the fraction be x/y
if x is a multiple of y : we can rewrite 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]
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29 Oct 2012, 04:59
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) \(y+2=my\) > \(y=\frac{2}{m1}\) > \(m1\) 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]
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29 Oct 2012, 06:32
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) \(y+2=my\) > \(y=\frac{2}{m1}\) > \(m1\) 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]
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29 Oct 2012, 06:35
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) \(y+2=my\) > \(y=\frac{2}{m1}\) > \(m1\) 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]
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29 Oct 2012, 06:54
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]
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29 Aug 2013, 09:30
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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, 53=2. Also 64=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 31=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 xy=2 when X is an even multiple of Y. Hence, y can only be 2. Satisfied.



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