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Re: In the fraction [m][fraction]x/y[/fraction][/m] , where x an [#permalink]
27 Oct 2012, 22:06

4

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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]
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.

Re: In the fraction [m][fraction]x/y[/fraction][/m] , where x an [#permalink]
29 Oct 2012, 03:59

1

This post received KUDOS

Expert's post

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.

Re: In the fraction [m][fraction]x/y[/fraction][/m] , where x an [#permalink]
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?? _________________

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

Re: In the fraction [m][fraction]x/y[/fraction][/m] , where x an [#permalink]
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. _________________

Re: In the fraction x/y, where x and y are positive integers [#permalink]
29 Aug 2013, 08:30

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.

gmatclubot

Re: In the fraction x/y, where x and y are positive integers
[#permalink]
29 Aug 2013, 08:30