Given that $\frac{x}{y},$ is a fraction where $x$ and $y$ are relatively prime positive integers. We have to find out the numbers of fraction if both numerator and denominator are increased by 1.

According to the question we have $\frac{x+1}{y+1}=\frac{11 x}{10 y}$

Can you now finish the problem ..........

Now from the equation we can say that $x+1>\frac{11}{10} \cdot x$ so $x$ is at most 9
By multiplying by $\frac{y+1}{x}$ and simplifying we can rewrite the condition as $y=\frac{11 x}{10-x}$. since $x$ and $y$ are integer, this only has solutions for $x \in{5,8,9} .$ However, only the first yields a $y$ that is relative prime to $x$

can you finish the problem........

Therefore the Possible answer will be \(1\)