$C.V=2;\sigma =0.4\left(given\right)C.V=100\times \frac{s.d}{mean}2=\frac{100\times 0.4}{mean}Mean=\frac{40}{2}=20$

Let the mean of the last four be $A_2$. 
Then by the formula for combined mean,
$12.5=\frac{6\times 15+4\times A_2}{6+4};\text{or 125=90+4}{A}_2;$
$\therefore A_2=\frac{35}{4}$ 
Let the sixth number = x; then taking the sixth number as a collection, the combined mean of this collection and the collection of the last five is 10, by question. 
$\therefore$ By definition of combined mean
$10=\frac{1\times x+4\times \frac{35}{4}}{1+4};50=x+35;\therefore x=15$ 
$\therefore$ sixth number = 15

$\overline{x}=4,N=5\text{and}\frac{\sum (x-{\overline{x}}^2)}{N}=5.2\Rightarrow \sum (x-\overline{x}{)}^2=(5.2)5\therefore \sum (x-\overline{x}{)}^2=26\therefore (1-4{)}^2+(2-4{)}^2+(6-4{)}^2+(\alpha -4{)}^2+(\beta -4{)}^2=26\therefore (\alpha -4{)}^2+(\beta -4{)}^2=9\text{Also,}\frac{1+2+6+\alpha +\beta }{5}=4\therefore \alpha +\beta =20-9=11\text{Clearly 4, 7 only satisfy the above equation in}\alpha ,\beta .\text{Hence required numbers are 4, 7.}$

$\text{we have,}\frac{\sum x_i}{50}=38,\therefore \sum x_i=1900\text{New value of}\sum x_i=1900-55-45=1800\text{and n = 48}\therefore \text{new mean}\frac{1800}{48}=\frac{450}{12}=\frac{225}{6}=\mathrm{37.5.}$

We know that if we multiply all numbers of a set by a constant "k" then the mean of these numbers will also be a multiple of the old mean.
And the variance of these numbers will be multiple of "$k^2$".
Since standard deviation is nothing but the square root of variance, s.d will be multiple of "k" of the old standard deviation.
We are given the s.d was 4.
So after multiplying each observation by (1/4), the s.d will also be multiplied by (1/4), thus giving the value = 4.$\frac{1}{4}$ = 1 

$\text{We know that}\text{C.V =}\frac{\sigma \times 100}{\overline{x}}or\overline{x}=\frac{\sigma }{c.v}\times 100\therefore \text{Mean of first series =}\frac{21.2\times 100}{58}=\text{36.6}\text{Mean of second series =}\frac{15.6\times 100}{69}=\text{22.6}$