Let the mean and standard deviation of marks of class A of 100...

Let the mean is denoted by $\overset{¯}{{x}_{i}}$, variance by ${{\sigma}_{i}}^{2}$, so standard deviation will be ${\sigma}_{i}$, now as per given data we get,

$\begin{cases}A & & B & & A + B \\ \overset{¯}{x}1 = 40 & & \overset{¯}{x}2 = 55 & & \overset{¯}{x} = 50 \\ {\sigma}_{1} = \alpha & & {\sigma}_{2} = 30 - \alpha & & {\sigma}_{2} = 350 \\ {n}_{1} = 100 & & {n}_{2} = n & & 100 + n\end{cases}$

So, using the formula of mean in $A + B$ we get,

$\overset{¯}{x} = \dfrac{100 \times 40 + 55n}{100 + n} = 50$

$⇒ n = 200$

Now using the formula of variance we get,

${\sigma}_{1}^{2} = \dfrac{∑ {x}_{i}^{2}}{100} - {40}^{2} = {\alpha}^{2}$

And ${\sigma}_{2}^{2} = \dfrac{∑ {y}_{i}^{2}}{200} - {55}^{2} = {\left(30 - \alpha\right)}^{2}$

So, variance of $A + B$ will be,

${\sigma}^{2} = \dfrac{∑ {x}_{i}^{2} + ∑ {y}_{i}^{2}}{300} - {50}^{2} = {350}^{2}$

$⇒ {350}^{2} = \dfrac{\left(1600 + {\alpha}^{2}\right) \times 100 + \left[3025 + {\left(30 - \alpha\right)}^{2}\right] \times 200}{300} - {50}^{2}$

$⇒ {\alpha}^{2} - 40\alpha + 300 = 0$

$⇒ \alpha = 10$ or $30$

And $\alpha = 30$ is not possible

So, ${\sigma}_{1} + 2 {\sigma}_{2} = 2 {10}^{2} + {20}^{2} = 500$