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

(Problem)
 
(Solution)
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<math>\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}</math>
 
<math>\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}</math>
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==Solution==
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We can write 2 equations.
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<math>\frac{x}{y}=\frac{a}{b}</math>
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and
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<math>x+y=c</math>
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Solving for <math>x</math> and <math>y</math> in terms of <math>a, b, c</math> we get :
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<math>x=\frac{ac}{a+b}</math> and <math>y=\frac{bc}{a+b}</math>
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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>.

Revision as of 20:05, 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 $\boxed{\textbf{(C) }\frac{ac}{a+b}}$.