You can create the equation $\frac{x+1}{y+1}=\frac{11 x}{10 y}$.  Cross multiplying and combining like terms gives $x y+11 x-10 y=0$.  This can be factored into $(x-10)(y+11)=-110$.  $x$ and $y$ must be positive, so $x\textgreater0$ and $y\textgreater0$, so $x-10\textgreater-10$ and $y+11\textgreater11$.  Using the factors of 110 , we can get the factor pairs: $(-1,110),(-2,55)$, and $(-5,22)$.  But we can\textquotesingle t stop here because $x$ and $y$ must be relatively prime.  $(-1,110)$ gives $x=9$ and $y=99.9$ and 99 are not relatively prime, so this doesn\textquotesingle t work.  $(-2,55)$ gives $x=8$ and $y=44$. This doesn\textquotesingle t work.  $(-5,22)$ gives $x=5$ and $y=11$. This does work. We found one valid solution so the answer is (B)$1$. 
Consider the set of all fractions $\frac{x}{y}$, where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by $1$, the value of the fraction is increased by $10 \textbackslash\%$? 
Options:
A: $$0$$ 
B: $$1$$ 
C: $$2$$ 
D: $$3$$ 
E: infinitely many 