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Manager  Joined: 11 Aug 2012
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Schools: HBS '16, Stanford '16
In the fraction x/y, where x and y are positive integers  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 70% (01:39) correct 30% (01:28) wrong based on 305 sessions

### HideShow timer Statistics 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?

Originally posted by danzig on 27 Oct 2012, 11:16.
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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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]

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1
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]

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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.

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

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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]

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1
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.

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.

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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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.

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.

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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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.

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.

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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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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1
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]

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_________________ Re: In the fraction x/y, where x and y are positive integers   [#permalink] 05 Jan 2019, 09:28
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