2015 AMC 10A Problem 15

The condition is
$$\frac{x + 1}{y + 1} = \frac{11}{10} \cdot \frac{x}{y}.$$
Cross-multiplying: $10y(x + 1) = 11x(y + 1)$, so $10xy + 10y = 11xy + 11x$, which simplifies to
$$10y = xy + 11x \quad\Longrightarrow\quad y(10 - x) = 11x.$$
Since $x, y > 0$, the right side is positive, so $10 - x > 0$: $x$ is one of $1, 2, \ldots, 9$, and $y = \dfrac{11x}{10 - x}$.

Now test each $x$. For $x = 1, 2, 3, 4, 6, 7$ the denominators $9, 8, 7, 6, 4, 3$ do not divide $11x$ (they share no factor with $11$ and do not divide $x$). The survivors:
- $x = 5$: $y = 55/5 = 11$, and $\gcd(5, 11) = 1$. Valid.
- $x = 8$: $y = 88/2 = 44$, but $\gcd(8, 44) = 4$. Not in lowest terms.
- $x = 9$: $y = 99/1 = 99$, but $\gcd(9, 99) = 9$. Not in lowest terms.

Only $\frac{5}{11}$ works: indeed $\frac{6}{12} = \frac{1}{2}$ and $\frac{11}{10} \cdot \frac{5}{11} = \frac{1}{2}$.

The answer is $\boxed{\textbf{(B)}\ 1}$.

A "fraction changes by a fixed percent" condition is a Diophantine equation after cross-multiplying. Isolating one variable, $y = \frac{11x}{10 - x}$, both bounds $x$ (the denominator must be positive) and makes the integrality test mechanical. The relatively-prime clause is not decoration: it is exactly what eliminates the extra solutions.

• Forgetting the relatively prime condition and also counting $8/44$ and $9/99$ (which reduce to $2/11$ and $1/11$), giving choice (D).
• Setting up "increased by $10\%$" as adding $0.1$ to the fraction instead of multiplying by $1.1$.