V litre decinormal solution of NaCl is prepared. Half of the solution is converted into centinormal and added to the left decinormal solution. Then

Given: Initial volume = V ; Normality = $\frac{\mathrm{N}}{10}$
Equivalents $=\frac{\mathrm{V}\times 1}{10}=\frac{\mathrm{V}}{10}$
50% solution is taken
$\frac{\mathrm{V}}{\mathrm{\alpha }}\times \frac{1}{10}=\frac{{\mathrm{V}}^{\mathrm{\prime }}\times 1}{100}\Rightarrow {\mathrm{V}}^{\mathrm{\prime }}=\frac{100\times \mathrm{V}}{20}=5\mathrm{V}$
Final volume = (5V)
Now there are two solutions
I solution (Left solution) $\frac{\mathrm{V}}{2};\frac{\mathrm{N}}{10}$
II-Prepared solution      $5\mathrm{V};\frac{\mathrm{N}}{100}$
$\Rightarrow$ On mixing
$\begin{array}{l}{\mathrm{N}}^{\mathrm{\prime }}=\frac{{\mathrm{N}}_1{\mathrm{V}}_1+{\mathrm{N}}_2{\mathrm{V}}_2}{\left({\mathrm{V}}_1+{\mathrm{V}}_2\right)}=\left(\frac{\mathrm{V}}{20}+\frac{\mathrm{V}}{20}\right)/\left(\frac{\mathrm{V}}{2}+5\mathrm{V}\right)\\ \therefore {\mathrm{N}}^{\mathrm{\prime }}=\frac{\mathrm{V}}{10}\times \frac{2}{11\mathrm{V}}=\frac{2}{110}=0.018\mathrm{N}=0.018\mathrm{M}\end{array}$
And no. of equivalent does not change on dilution.