Let the Mean and Standard Deviation of Marks of Class A of 100 Students Be 40 and α, and of Class B of n Students Be 55 and 30 − α. If the Mean and Variance of the Combined Class of 100 + n Students Are 50 and 350, Then the Sum of Variances of Classes A and B Is

JEE Main 2023 — 31 January, Evening Shift. Previous Year Question.

Problem

Let the mean and standard deviation of marks of class $A$ of $100$ students be respectively $40$ and $\alpha$ ($\alpha>0$), and the mean and standard deviation of marks of class $B$ of $n$ students be respectively $55$ and $30-\alpha$. If the mean and variance of the marks of the combined class of $100+n$ students are respectively $50$ and $350$, then the sum of variances of classes $A$ and $B$ is:

Key insight. The combined variance of two groups isn’t just a weighted average of their individual variances — it also picks up an extra term for how far each group’s own mean sits from the combined mean. Writing this out gives one equation in the single unknown $\alpha$, but that equation turns out to be quadratic with two roots — only one of which describes a genuine (non-degenerate) pair of classes.

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Approach

First find $n$ using the combined mean formula. Then use the combined variance formula — accounting for both groups’ internal variance and the “distance” of each group’s mean from the combined mean — to set up one equation in $\alpha$. Solve the resulting quadratic, and reject the root that makes one class’s standard deviation degenerate (zero), before computing the sum of variances.

Solution

Step 1 — Find n using the combined mean

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

$$4000+55n = 5000+50n \implies 5n=1000 \implies n=200$$

Step 2 — Write the combined variance formula

For two groups combined, using $d_1$ and $d_2$ as each group’s mean minus the combined mean:

$$\text{Var}_{\text{combined}} = \frac{100(\alpha^2+d_1^2)+200((30-\alpha)^2+d_2^2)}{300}$$

where $d_1=40-50=-10$ and $d_2=55-50=5$.

Step 3 — Substitute the known values

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

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

Step 4 — Expand (30 − α)²

$$200(30-\alpha)^2 = 200(900-60\alpha+\alpha^2) = 180000-12000\alpha+200\alpha^2$$

Step 5 — Combine and simplify

$$105000 = 100\alpha^2+10000+180000-12000\alpha+200\alpha^2+5000$$

$$105000 = 300\alpha^2-12000\alpha+195000$$

$$300\alpha^2-12000\alpha+90000=0$$

Dividing by 300:

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

Step 6 — Solve the quadratic

$$(\alpha-10)(\alpha-30)=0 \implies \alpha=10 \text{ or } \alpha=30$$

Step 7 — Reject the degenerate root

If $\alpha=30$, then class $B$’s standard deviation is $30-\alpha=0$ — meaning every student in a 200-student class scored exactly the same mark, an unrealistic (degenerate) scenario. So $\alpha=10$ is the valid choice.

Step 8 — Compute the sum of variances

$$\text{Var}_A = \alpha^2 = 10^2=100, \qquad \text{Var}_B = (30-\alpha)^2 = 20^2=400$$

$$\text{Sum} = 100+400=500$$

Answer

Common mistakes

• Accepting both roots of the quadratic without checking for degeneracy. The algebra alone doesn’t rule out $\alpha=30$ — only the physical reasoning that a real class’s standard deviation shouldn’t collapse to zero (for a class of $200$ real students) justifies rejecting it.
• Using the simpler “weighted average of variances” formula without the correction term. Combining two datasets’ variances requires accounting for how far each sub-group’s own mean differs from the combined mean — skipping the $d_1^2$ and $d_2^2$ terms gives a significantly wrong equation.

Practise next

• Class $P$ has $50$ students with mean $30$ and SD $\beta$; class $Q$ has $m$ students with mean $45$ and SD $20-\beta$. If the combined mean is $40$ and combined variance is $200$, find the sum of variances of $P$ and $Q$, using the same approach.