Let set A and B have 5 elements each.  Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively.  A new set C of 10 elements is formed by subtracting 3 from each element of A and adding 2 to each element of B.  Then the sum of the mean and variance of the elements of C is _____

$$\begin{gathered} \mathrm{A} {\mathrm{m}}_{\mathrm{A}}=5 \mathrm{B}{\mathrm{m}}_{\mathrm{B}}=8 \\ {\mathrm{V}}_{\mathrm{A}}=12 {\mathrm{V}}_{\mathrm{B}}=70 \\ \sum {\mathrm{A}}_{\mathrm{i}}=5\times 5=25 \sum {\mathrm{B}}_{\mathrm{i}}=40 \end{gathered}$$
Then  new mean
$$\begin{gathered} \sum {\mathrm{C}}_{\mathrm{i}}=25-3\left(5\right)+40+2\left(5\right) \\ \sum {\mathrm{C}}_{\mathrm{i}}=65-5=60 \end{gathered}$$
Mean  C = 6

$V_A=\frac{{\displaystyle \sum _{ }}{A_i}^2}{5}-{\left(m_A\right)}^2$
$$\begin{gathered} \sum {{\mathrm{A}}_{\mathrm{i}}}^2=5(12+25)=\left(37\right)5=185, \\ \sum {{\mathrm{B}}_{\mathrm{i}}}^2=5(20+64)=\left(84\right)5=420 \end{gathered}$$
$$\begin{gathered} {\mathrm{V}}_{\mathrm{A}}=\frac{\sum {({\mathrm{A}}_{\mathrm{i}}-3)}^2}{5}-{\left(\frac{\sum ({\mathrm{A}}_{\mathrm{i}}-3)}{5}\right)}^2 \\ 12=\frac{\sum {({\mathrm{A}}_{\mathrm{i}}-3)}^2}{5}-4 \mathrm{then} \sum {({\mathrm{A}}_{\mathrm{i}}-3)}^2=16\times 2=80 \end{gathered}$$
$$\begin{gathered} {\mathrm{V}}_{\mathrm{B}}=\frac{\sum {(\mathrm{B}+2)}^2}{5}-{\left(\frac{\sum (\mathrm{B}+2)}{5}\right)}^2. \\ 20=\frac{\sum {(\mathrm{B}+2)}^2}{5}-100 \\ \sum {(\mathrm{B}+2)}^2=600, \\ \therefore \sum {({\mathrm{A}}_{\mathrm{i}}-3)}^2+\sum {({\mathrm{B}}_{\mathrm{i}}+2)}^2=680 \\ {\mathrm{V}}_{\mathrm{C}}=\frac{680}{10}-36=68-36=32 \\ \boxed{{\mathrm{V}}_{\mathrm{C}}=32;{\mathrm{M}}_{\mathrm{C}}=6} \end{gathered}$$