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Difference between revisions of "1982 AHSME Problems/Problem 5"

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Since we know <math>a</math> is less than <math>b</math> and <math>\frac{x}{y}=\frac{bc}{a+b}</math>, the smaller of <math>x</math> and <math>y</math> must be <math>x</math>. Therefore the answer is <math>\boxed{\textbf{(C) }\frac{ac}{a+b}}</math>.
 
Since we know <math>a</math> is less than <math>b</math> and <math>\frac{x}{y}=\frac{bc}{a+b}</math>, the smaller of <math>x</math> and <math>y</math> must be <math>x</math>. Therefore the answer is <math>\boxed{\textbf{(C) }\frac{ac}{a+b}}</math>.
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~superagh

Latest revision as of 19:06, 25 March 2020

Problem

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

$\text{(A)} \ \frac{ac}{b} \qquad \text{(B)} \ \frac{bc-ac}{b} \qquad \text{(C)} \ \frac{ac}{a+b} \qquad \text{(D)}\ \frac{bc}{a+b}\qquad \text{(E)}\ \frac{ac}{b-a}$


Solution

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

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