What will be the difference in the mass number of the neutrons is halved and the number of electrons is doubled in ${}_8\mathrm{O}^{16}$?

${}_8\mathrm{O}^{16}$                       ${}_{16}{}^{20}\mathrm{X}$

$\mathrm{n}=\mathrm{A}-\mathrm{Z}=8\stackrel{\mathrm{neutrons} \mathrm{is} \mathrm{halved}}{\to }$$\mathrm{n}=\mathrm{A}-\mathrm{Z}=4$

$\mathrm{no}.\mathrm{of} {\mathrm{e}}^{-}=8\stackrel{\mathrm{neutron} \mathrm{is} \mathrm{double}}{\to }\mathrm{no}. \mathrm{of} {\mathrm{e}}^{-}=16$

Mass number difference is 4

25% increase in mass number$\left[6\times \frac{25}{100}=4\right]$