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...")
 
m (Problem)
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

Revision as of 16:50, 30 September 2014

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

$\fbox{B}$

See also

1969 AHSME (ProblemsAnswer KeyResources)
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
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