Let the mean of 6 observations  $1, 2, 4, 5, x \text{and} y$ be 5 and their variance be 10. Then their mean deviation about the mean is equal to

Mean A = 5 $\Rightarrow 12+\mathrm{x}+\mathrm{y}=6\times 5\Rightarrow \mathrm{x}+\mathrm{y}=18$
Variance ${\mathrm{\sigma }}^2=\frac{{\displaystyle \sum {\mathrm{x}}_{\mathrm{i}}^2}}{\mathrm{n}}-{\mathrm{A}}^2=\frac{1+4+16+25+{\mathrm{x}}^2+{\mathrm{y}}^2}{6}-5^2=10$
$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^2=164 \\ \Rightarrow {\mathrm{x}}^2+{(18-\mathrm{x})}^2=164 \\ \Rightarrow {\mathrm{x}}^2-18\mathrm{x}+80=0 \boxed{\therefore \mathrm{x}=10,\mathrm{y}=8} \end{gathered}$$
${(\mathrm{M}.\mathrm{D})}_{\mathrm{A}}=\frac{{\displaystyle \sum |{\mathrm{x}}_{\mathrm{i}}-\mathrm{A}|}}{\mathrm{n}}=\frac{4+3+1+0+5+3}{6}=\frac{8}{3}$