Aptitude Question ID : 94721
$latex \frac{13}{48}&s=1$ is equal to :
[A]
$latex \frac{1}{3+\frac{1}{1+\frac{1}{16}}}&s=2$
[B]
$latex \frac{1}{2+\frac{1}{1+\frac{1}{8}}}&s=2$
[C]
$latex \frac{1}{3+\frac{1}{1+\frac{1}{1+\frac{1}{8}}}}&s=2$
[D]
$latex \frac{1}{3+\frac{1}{1+\frac{1}{2+\frac{1}{4}}}}&s=2$
$latex \mathbf{\frac{1}{3+\frac{1}{1+\frac{1}{2+\frac{1}{4}}}}}&s=2$
Check through options, by option [D]
$latex \frac{1}{3+\frac{1}{1+\frac{1}{2+\frac{1}{4}}}}&s=2$
$latex = \frac{1}{3+\frac{1}{1+\frac{1}{\frac{8+1}{4}}}} = \frac{1}{3+\frac{1}{1+\frac{4}{9}}}&s=2$
$latex = \frac{1}{3+\frac{1}{\frac{9+4}{9}}} = \frac{1}{3+\frac{9}{13}}&s=2$
$latex = \frac{1}{\frac{39+9}{13}} = \frac{13}{48}&s=2$
Hence option [D] is right answer.