You do not need an official explanation 9even though I can provide). YOU need to use your logic. During the test you cannot have an OE or a break

\(\frac{x}{4}<y<\frac{x}{2}\)

Now the stem says that x and y must be integers and positive: 1,2,3......

LCD

you have

\(x<4y<2x\)

Suppose y=1 then 4y=4

So x CANNOT be 2 because 2*2=4 but the inequality tells you that 2x must be > of 4y. So the LEAST value of x must be 3

\(x=3\)

\(3 <4 < 6\)

A is the only viable option

regards

Silly me, don't know for what reason I keep multiplying 4 with x to get middle term and wondering why no equation is satisfying. I would have just selected y = 1 and tested different scenarios.

Thanks a lot Carcass

\(x\) and \(y\) are positive integers.

\(\frac{x}{4} < y < \frac{x}{2}\)

We need the smallest value of \(x\), and it is given that \(x\) is a +ve integer.
So the smallest value of \(x\) can be 1.

For \(x = 1\), \(y\) will not be an integer as \(0.25 < y < 0.5\)
Same,
For \(x = 2\), \(y\) will not be an integer as \(0.50 < y < 1\)

And

for \(x = 3\), \(y\) can be an integer as \(0.75 < y < 1.5\)

So for \(x = 3\), \(y\) can be \(1\)

Answer \(A\)