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 "1969 AHSME Problems/Problem 1"

(Created page with "== Problem == When <math>x</math> is added to both the numerator and denominatorof the fraction <math>\frac{a}{b},a\neb,b\ne0</math>, the value of the fraction is changed to <ma...")
 
(→‎Solution)
 
(2 intermediate revisions by one other user not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
  
−
When <math>x</math> is added to both the numerator and denominatorof the fraction <math>\frac{a}{b},a\neb,b\ne0</math>, the value of the fraction is changed to <math>\frac{c}{d}</math>. Then <math>x</math> equals:
+
When <math>x</math> is added to both the numerator and denominator of the fraction  
 +
<math>\frac{a}{b},a \ne b,b \ne 0</math>, the value of the fraction is changed to <math>\frac{c}{d}</math>.  
 +
Then <math>x</math> equals:
  
 
<math>\text{(A) } \frac{1}{c-d}\quad
 
<math>\text{(A) } \frac{1}{c-d}\quad
Line 10: Line 12:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{B}</math>
+
<math>\frac{a+x}{b+x}=\frac{c}{d}</math>,
 +
 
 +
<math>bc+cx=ad+dx</math>,
 +
 
 +
<math>(c-d)x=ad-bc</math>,
 +
 
 +
<math>x=\frac{ad-bc}{c-d}</math>. The answer is <math>\fbox{B}</math>.
  
 
== See also ==
 
== See also ==
−
{{AHSME box|year=1969|num-b=1|num-a=2}}   
+
{{AHSME 35p box|year=1969|num-b=1|num-a=2}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:43, 10 July 2015

Problem

When $x$ is added to both the numerator and denominator of the fraction $\frac{a}{b},a \ne b,b \ne 0$, the value of the fraction is changed to $\frac{c}{d}$. Then $x$ equals:

$\text{(A) } \frac{1}{c-d}\quad \text{(B) } \frac{ad-bc}{c-d}\quad \text{(C) } \frac{ad-bc}{c+d}\quad \text{(D) }\frac{bc-ad}{c-d} \quad \text{(E) } \frac{bc+ad}{c-d}$

Solution

$\frac{a+x}{b+x}=\frac{c}{d}$,

$bc+cx=ad+dx$,

$(c-d)x=ad-bc$,

$x=\frac{ad-bc}{c-d}$. The answer is $\fbox{B}$.

See also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 2
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
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=1969_AHSME_Problems/Problem_1&oldid=71098"