Aptitude Question ID: 106413
If $latex \frac{3}{4}&s=1$ of the difference of $latex 2\frac{1}{4}&s=1$ and $latex 1\frac{2}{3}&s=1$ is subtracted from $latex \frac{2}{3}&s=1$ of $latex 3\frac{1}{4}&s=1$ the result is :
[A]$latex \frac{-48}{83}&s=1$
[B]$latex \frac{48}{83}&s=1$
[C]$latex \frac{-83}{48}&s=1$
[D]$latex \frac{83}{48}&s=1$
$latex \mathbf{\frac{83}{48}}&s=1$
$latex \frac{13}{4}\times \frac{2}{3}-\left ( \frac{9}{4}-\frac{5}{3} \right )\times \frac{3}{4}&s=1$
$latex = \frac{13}{6}-\left ( \frac{27-20}{12} \right )\times \frac{3}{4}&s=1$
$latex = \frac{13}{6}-\frac{7}{12}\times \frac{3}{4}&s=1$
$latex = \frac{13}{6}-\frac{7}{16}&s=1$
$latex = \frac{104-21}{48} = \frac{83}{48}&s=1$
Hence option [D] is right answer.