2015 AMC 10A Problem 15

The condition is $$\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}.$$ Cross-multiplying gives $10y(x+1)=11x(y+1)$, or $xy+11x-10y=0$. Factoring by grouping after subtracting $110$ gives $$(x-10)(y+11)=-110.$$ Since $x,y$ are positive, the useful negative factor pairs are $(-1,110)$, $(-2,55)$, and $(-5,22)$, producing $(x,y)=(9,99),(8,44),(5,11)$. Only $\frac{5}{11}$ has relatively prime numerator and denominator, so exactly one fraction works. Thus, B is the correct answer.