JEE Mathematics 2023 Question with Solution

Given:

• Class A: mean $\overline{x_1}=40$, standard deviation $\sigma_1=\alpha$, number of students $n_1=100$
• Class B: mean $\overline{x_2}=55$, standard deviation $\sigma_2=30-\alpha$, number of students $n_2=n$
• Combined class: mean $\overline{x}=50$, variance $\sigma^2=350$

Find: The sum $\sigma_1^2+\sigma_2^2$.

Using the combined mean,

$$\overline{x}=\frac{100\times 40+55n}{100+n}$$

So,

$$50=\frac{4000+55n}{100+n}$$ $$5000+50n=4000+55n$$ $$1000=5n$$ $$n=200$$

Now use the combined variance relation shown in the solution:

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

This gives

$$8550=\alpha^2+2(30-\alpha)^2+7650$$

Hence,

$$\alpha^2+2(30-\alpha)^2=900$$

Expanding,

$$\alpha^2-40\alpha+300=0$$

So,

$$\alpha=10,30$$

From the final step in the source solution,

$$\sigma_1^2+\sigma_2^2=10^2+20^2=500$$

Therefore, the sum of variances of classes A and B is $500$. The correct option is A.