In the fraction x/y, where x and y are positive integers

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.

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

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.

Thanks for clearing that up... Guess I over thought it a bit.

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.

Statement 1) X is an even multiple of Y

I am just keeping it aside as I am not able to infer anything

Statement 2) X-Y=2

So possible pairs (3,1) ; (4,2) ; (5,3)............. infinity

so each statement not sufficient so let me check if both can help

from the possible pairs only (4,2) satisfy the statement 1 so I can eliminate all

so both statement required

I imagine most people will solve here by testing numbers, since it's easy enough to get a feeling for what's happening by doing that. There is a theoretical justification for the answer here that does not involve number picking:

The answer is clearly C or E. Using Statement 2, we know x and y are 2 apart. Many test takers learn that when two integers are 1 apart (i.e. when integers are consecutive), then their GCD is 1. We can extend that fact. Say two numbers are 2 apart, so our numbers are y and y + 2. Imagine y is a multiple of 3. If y is a multiple of 3, the next larger multiple of 3 will be y + 3, not y + 2. Similarly if y is a multiple of k, where k > 2, then the next multiple of k will be y + k, not y + 2. So y and y + 2 can't possibly share any divisor larger than 2, and their GCD is either 1 (if they are both odd) or 2 (if they are both even). In general, the GCD of two numbers y and y + m, i.e. two numbers that are m apart, must always be a factor of m, so for example, the GCD of c and c + 21 can only ever be 1, 3, 7 or 21.

So here, when we use both Statements, we know both our numbers are even (since x is, and x and y are 2 apart). We know their GCD can only be 1 or 2 from Statement 2, and if both numbers are even, their GCD must be 2. But we also know x is a multiple of y. If that's true, the GCD of x and y is automatically equal to y (since y is clearly then a divisor of both x and y, but no larger number is a divisor of y). If the GCD of x and y is equal both to 2 and to y, then y = 2. So the answer is C.

That said, I find it mystifying why someone would word a question in this way. What does "the fraction x/y" have to do with anything? Why doesn't the question just ask "If x and y are positive integers, what is the value of y"? The wording is copied from an old official question, but in the official question, the fraction was crucially important (the question went on to talk about common denominators), and the fraction is irrelevant here.

Statement 1 is also potentially ambiguous, and if this were an official problem, I'm almost certain they would have chosen a non-ambiguous wording. I expect Statement 1 is meant to be interpreted to mean "x is even and x is a multiple of y". But I don't think it's completely unreasonable to interpret Statement 1, as written, to mean "x is equal to an even number times y", which is not the same thing. And colloquially, sometimes people say one number is "evenly divisible" by another, and when people say that, the meaning of "even" has nothing to do with multiples of 2 -- it means there is no remainder. So here, I also think it's not completely unreasonable to interpret the phrase "even multiple" to mean "a multiple with no remainder". The Statement really should be worded differently so the test taker isn't left to guess what it intends.

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