Two positive numbers $x$ and $y$ are in the ratio $a: b$ where $0 < a < b$. If $x+y = c$, then the smaller of $x$ and $y$ is
$\text{(A)} \ \frac{ac}{b} \qquad  \text{(B)} \ \frac{bc-ac}{b} \qquad  \text{(C)} \ \frac{ac}{a+b} \qquad  \text{(D)}\ \frac{bc}{a+b}\qquad \text{(E)}\ \frac{ac}{b-a}$
We can write 2 equations.
$\frac{x}{y}=\frac{a}{b}$
and
$x+y=c$
Solving for $x$ and $y$ in terms of $a, b, c$ we get :
$x=\frac{ac}{a+b}$ and $y=\frac{bc}{a+b}$
Since we know $a$ is less than $b$ and $\frac{x}{y}=\frac{bc}{a+b}$, the smaller of $x$ and $y$ must be $x$. Therefore the answer is $\boxed{\textbf{(C) }\frac{ac}{a+b}}$.

~superagh