Difference between revisions of "1969 AHSME Problems/Problem 1"

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== 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>.  
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Then <math>x</math> equals:
  
 
<math>\text{(A) } \frac{1}{c-d}\quad
 
<math>\text{(A) } \frac{1}{c-d}\quad

Revision as of 15:50, 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

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