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

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 $\fbox{B}$.

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