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 "1950 AHSME Problems/Problem 4"

(→‎Solution)
m (→‎Solution)
 
(6 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
  
−
Reduced to lowest terms, <math> \frac{a^{2}-b^{2}}{ab} </math> - <math> \frac{ab-b^{2}}{ab-a^{2}} </math> is equal to:
+
Reduced to lowest terms, <math> \frac{a^{2}-b^{2}}{ab} - \frac{ab-b^{2}}{ab-a^{2}} </math> is equal to:
  
 
<math> \textbf{(A)}\ \frac{a}{b}\qquad\textbf{(B)}\ \frac{a^{2}-2b^{2}}{ab}\qquad\textbf{(C)}\ a^{2}\qquad\textbf{(D)}\ a-2b\qquad\textbf{(E)}\ \text{None of these} </math>
 
<math> \textbf{(A)}\ \frac{a}{b}\qquad\textbf{(B)}\ \frac{a^{2}-2b^{2}}{ab}\qquad\textbf{(C)}\ a^{2}\qquad\textbf{(D)}\ a-2b\qquad\textbf{(E)}\ \text{None of these} </math>
  
−
==Solution==
+
== Solution ==
−
We can use difference of two squares to expand <math>a^2-b^2=(a-b)(a+b)</math> and factor to get <math>ab-b^{2}=b(a-b)</math> and <math>a^2-ab=a(a-b)</math>
 
−
<math> \dfrac{a^{2}-b^{2}}{ab}  -  \dfrac{ab-b^{2}}{ab-a^{2}} =\dfrac{(a-b)(a+b)}{ab}-\dfrac{b(a-b)}{a(a-b)}</math>
 
  
−
We can further factor to get <math>\dfrac{(a-b)(a+b)}{ab}-\dfrac{b(a-b)}{a(a-b)}</math> If we assume b is not equal to a, this is equal to <math>\dfrac{(a-b)(a+b)}{ab}-\dfrac{b}{a}=\dfrac{(a-b)(a+b)}{ab}-\dfrac{b^2}{ab}=\dfrac{(a-b)(a+b)-b^2}{ab}=\dfrac{a^2-2b^2}{ab}</math>
+
We start off by factoring the second fraction.
  
−
The answer is \boxed{\textbf{(B)}\ \frac{a^{2}-2b^{2}}{ab}}
+
<cmath>-\frac{ab-b^2}{ab-a^2} = -\frac{b(a-b)}{-a(a-b)} = \frac{b}{a}.</cmath>
 +
 
 +
Now create a common denominator and simplify.
 +
 
 +
<cmath>\frac{a^2-b^2}{ab}+\frac{b}{a}=\frac{a^2-b^2}{ab}+\frac{b^2}{ab} = \frac{a^2}{ab} = \boxed{\mathrm{(A) }\frac{a}{b}}</cmath>
 +
 
 +
obs: Assume that <math>a \not = 0, b\not = 0, a\not = b</math>.
  
 
==See Also==
 
==See Also==
  
−
{{AHSME box|year=1950|num-b=3|num-a=5}}
+
{{AHSME 50p box|year=1950|num-b=3|num-a=5}}
 +
 
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 21:21, 10 January 2023

Problem

Reduced to lowest terms, $\frac{a^{2}-b^{2}}{ab} - \frac{ab-b^{2}}{ab-a^{2}}$ is equal to:

$\textbf{(A)}\ \frac{a}{b}\qquad\textbf{(B)}\ \frac{a^{2}-2b^{2}}{ab}\qquad\textbf{(C)}\ a^{2}\qquad\textbf{(D)}\ a-2b\qquad\textbf{(E)}\ \text{None of these}$

Solution

We start off by factoring the second fraction.

\[-\frac{ab-b^2}{ab-a^2} = -\frac{b(a-b)}{-a(a-b)} = \frac{b}{a}.\]

Now create a common denominator and simplify.

\[\frac{a^2-b^2}{ab}+\frac{b}{a}=\frac{a^2-b^2}{ab}+\frac{b^2}{ab} = \frac{a^2}{ab} = \boxed{\mathrm{(A) }\frac{a}{b}}\]

obs: Assume that $a \not = 0, b\not = 0, a\not = b$.

See Also

1950 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
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=1950_AHSME_Problems/Problem_4&oldid=186292"