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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

5

This post received KUDOS

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

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.

Re: In the fraction x/y, where x and y are positive integers [#permalink]

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29 Aug 2013, 09:30

1

This post received KUDOS

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.

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.

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) \(y+2=my\) --> \(y=\frac{2}{m-1}\) --> \(m-1\) 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??
_________________

Did you find this post helpful?... Please let me know through the Kudos button.

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.
_________________

Re: In the fraction x/y, where x and y are positive integers [#permalink]

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14 Oct 2014, 21:38

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Re: In the fraction x/y, where x and y are positive integers [#permalink]

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12 Jun 2016, 05:03

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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