NIMCET 2015 Question Paper with solutions

$\mathit{P}={\left(\mathit{F}87\mathit{B}\right)}_{16}=1111100001111011$
2's complement$=1111100001111011$
              $+1$
        ––––––––––––––––
        $0000011110000101$
        ––––––––––––––––
${\left(00011110000101\right)}_2$
   $\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow$
  $10245122561286432168421$
Value in decimal is $\left(1925\right)10$, Hence, $\mathit{P}=-1925$
So, now doing $8\prime \mathit{P}={\left(-15400\right)}_{10}$
Since, a negative number we have to find the 2's complement of $-15400$
Binary of 15400 = 0011 1100 00101000$={\left(\mathit{C}3\mathit{D}8\right)}_{16}$