1982 AHSME Problems/Problem 5


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.




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 $\boxed{\textbf{(C) }\frac{ac}{a+b}}$.


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