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

(Problem)
 
(added navbox)
 
(2 intermediate revisions by one other user not shown)
Line 4: Line 4:
  
 
<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>
 +
 +
 +
==Solution==
 +
 +
We can write 2 equations.
 +
 +
<math>\frac{x}{y}=\frac{a}{b}</math>
 +
 +
and
 +
 +
<math>x+y=c</math>
 +
 +
Solving for <math>x</math> and <math>y</math> in terms of <math>a, b, c</math> we get :
 +
 +
<math>x=\frac{ac}{a+b}</math> and <math>y=\frac{bc}{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>.
 +
 +
 +
 +
~superagh
 +
 +
==See Also==
 +
{{AHSME box|year=1982|num-b=4|num-a=6}}
 +
 +
{{MAA Notice}}

Latest revision as of 22:08, 29 June 2025

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

See Also

1982 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions


These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png