Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(added navbox)
 
Line 25: Line 25:
  
 
~superagh
 
~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 (Problems • Answer Key • Resources)
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems/Problem_5&oldid=252866"