Difference between revisions of "1982 AHSME Problems/Problem 5"
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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 19:05, 25 March 2020
Problem
Two positive numbers 
and 
are in the ratio 
where 
. If 
, then the smaller of and is
Solution
We can write 2 equations.
and
Solving for and in terms of 
we get :
and 
Since we know 
is less than 
and 
, the smaller of and must be . Therefore the answer is 
.
