$$Statement\ 1\ ;\ Greastest\ common\ factor\ of\ \frac{x}{2}\ and\ \frac{y}{2}=\ is\ 5$$
$$Therefore,\ \frac{x}{2}\ =\ 5a\ $$ $$\frac{y}{2}\ =\ 5b\ where\ a\ and\ b\ are\ co\ -\ primes.\ $$
x = 10a and y = 10b.
Making their highest common factor = 10 hence Statement 1 is SUFFICIENT.

Statement 2 ; x and y are even.
If x = 4 and y = 8 highest common factor = 4
If x = 2 and y = 4 highest common factor = 2 .

However, Statement 1 alone is SUFFICIENT.
Option A is correct.

we know that GCF of fraction = (GCF of numerator) / (LCM of denominator)
therefore GCF(x/2,y/2) = GFC(x,y) / LCM(2,2)

1) from statement 1: we know that GCF(x/2,y/2) = 5
2) LCM of 2,2 = 2

therefore
5 = GCF(x,y) / 2
hence, GCF (x,y) = 10

A is sufficient $$$$

AAPL wrote:Source: Princeton Review

What is the greatest common factor of positive integers x and y?

(1) The greatest common factor of x/2 and y/2 is 5.

Has anyone another strategic approach to solving this DS question?

\[\left. \begin{gathered}
A,B \geqslant 1\,\,{\text{ints}} \hfill \\
GCD\left( {A,B} \right) = M\,\,\, \hfill \\
\end{gathered} \right\}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,GCD\left( {A \cdot k\,\,,\,\,B \cdot k} \right) = k \cdot M\,\,\,\,\,,\,\,\,\,\,\,{\text{for}}\,\,{\text{all}}\,\,k \geqslant 1\,\,\,\operatorname{int} \,\,\,\,\,\,\left( * \right)\]
\[GCF\left( {\frac{x}{2};\frac{y}{2}} \right) = 5\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,\,GCF\left( {x;y} \right)\,\,\, = \,\,\,GCF\left( {2\left( {\frac{x}{2}} \right);2\left( {\frac{y}{2}} \right)} \right)\,\,\, = \,\,\,2 \cdot 5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,\frac{x}{2}\,\,;\,\,\frac{y}{2}\,\,\,\, \geqslant 1\,\,\,{\text{ints}}\,} \right]\]

This argument follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.