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 "1955 AHSME Problems/Problem 4"

(→‎Solution)
(→‎See Also)
 
Line 15: Line 15:
  
 
== See Also ==
 
== See Also ==
−
{{AHSME 4p box|year=1955|num-b=3|after=Last Question}}
+
{{AHSME box|year=1955|num-b=3|num-a=5}}
 +
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:14, 9 July 2018

Problem

The equality $\frac{1}{x-1}=\frac{2}{x-2}$ is satisfied by:

$\textbf{(A)}\ \text{no real values of }x\qquad\textbf{(B)}\ \text{either }x=1\text{ or }x=2\qquad\textbf{(C)}\ \text{only }x=1\\ \textbf{(D)}\ \text{only }x=2\qquad\textbf{(E)}\ \text{only }x=0$

Solution

From the equality, $\frac{1}{x-1}=\frac{2}{x-2}$, we get ${(x-1)}\times2={(x-2)}\times1$.

Solving this, we get, ${2x-2}={x-2}$.

Thus, the answer is $\fbox{{\bf(E)} \text{only} x = 0}$.

Solution by awesomechoco

See Also

1955 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_4&oldid=96076"