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

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== Solution ==
 
== Solution ==
<math>\fbox{B}</math>
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<math>\frac{a+x}{b+x}=\frac{c}{d}</math>,
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<math>bc+cx=ad+dx</math>,
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<math>(c-d)x=ad-bc</math>,
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<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}}   
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{{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 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 2
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All AHSME Problems and Solutions

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