1. Given data

• Class $A$:

  • Number of students $=100$
  • Mean $\mu_A=40$
  • Standard deviation $=\alpha$
  • Variance $=\alpha^2$

• Class $B$:

  • Number of students $=n$
  • Mean $\mu_B=55$
  • Standard deviation $=30-\alpha$
  • Variance $=(30-\alpha)^2$

• Combined class:

  • Number of students $=100+n$
  • Mean $\mu=50$
  • Variance $=350$

We need: $\alpha^2+(30-\alpha)^2$

2. Use combined mean formula

$\frac{100\cdot 40+n\cdot 55}{100+n}=50$

So, $4000+55n=50(100+n)=5000+50n$

$5n=1000$

$n=200$

3. Use combined variance formula

For two groups, combined variance is $\sigma^2=\frac{n_1\left(\sigma_1^2+(\mu_1-\mu)^2\right)+n_2\left(\sigma_2^2+(\mu_2-\mu)^2\right)}{n_1+n_2}$

Here,

• $n_1=100$, $\mu_1=40$, $\sigma_1^2=\alpha^2$
• $n_2=200$, $\mu_2=55$, $\sigma_2^2=(30-\alpha)^2$
• $\mu=50$, $\sigma^2=350$

Thus, $350=\frac{100\big(\alpha^2+(40-50)^2\big)+200\big((30-\alpha)^2+(55-50)^2\big)}{300}$

Now simplify: $350=\frac{100(\alpha^2+100)+200\big((30-\alpha)^2+25\big)}{300}$

Multiply by $300$: $105000=100\alpha^2+10000+200(30-\alpha)^2+5000$

$105000=100\alpha^2+200(30-\alpha)^2+15000$

$90000=100\alpha^2+200(30-\alpha)^2$

Divide by $100$: $900=\alpha^2+2(30-\alpha)^2$

Expand: $900=\alpha^2+2(900-60\alpha+\alpha^2)$

$900=\alpha^2+1800-120\alpha+2\alpha^2$

$900=3\alpha^2-120\alpha+1800$

$0=3\alpha^2-120\alpha+900$

Divide by $3$: $0=\alpha^2-40\alpha+300$

Factorize: $\alpha^2-40\alpha+300=(\alpha-10)(\alpha-30)=0$

So, $\alpha=10 \text{ or } 30$

But for class $B$, standard deviation $=30-\alpha$ and standard deviation must be non-negative. Also $\alpha>0$.

• If $\alpha=30$, then $30-\alpha=0$
• If $\alpha=10$, then $30-\alpha=20$

Both are mathematically possible, but let us compute the required sum in each case.

4. Find sum of variances

Required sum: $\alpha^2+(30-\alpha)^2$

• If $\alpha=10$: $\alpha^2+(30-\alpha)^2=10^2+20^2=100+400=500$

• If $\alpha=30$: $\alpha^2+(30-\alpha)^2=30^2+0^2=900$

Now check which one is consistent with the options and standard exam convention. Since standard deviation of class $B$ is written as $30-\alpha$ (typically taken positive), we take $30-\alpha>0 \Rightarrow \alpha<30$ so $\alpha=10$

Hence the required sum is $500$

5. Option check

• A: $450$ ❌
• B: $900$ ❌
• C: $650$ ❌
• D: $500$ ✅

Therefore, the correct answer is Option D.