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Statistics and Probability: Mathematics | JEE Main

Q: Why is rejected when the quadratic equation gives both 10 and 30?

Class B has standard deviation . In standard statistical convention, a distribution has unless all values are identical. Standard deviation is typically strictly positive in such contexts. If , . Even if were allowed, . Both 500 and 900 are options. The strict inequality (analogous to for class A) selects .

Let the mean and standard deviation of marks of class

\mathrm{A}

of 100 students be respectively 40 and

\alpha

(

> 0

), and the mean and standard deviation of marks of class

\mathrm{B}

\mathrm{n}

students be respectively 55 and 30

-\alpha

. If the mean and variance of the marks of the combined class of

100 + \mathrm{n}

studants are respectively 50 and 350 , then the sum of variances of classes

\mathrm{A}

and

\mathrm{B}

is :

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A650B450C900D500

Step-by-step solutionView

Correct answer

Using the combined mean gives $n = 200$ , and substituting into the combined variance formula yields $\alpha = 10$ , so the sum of variances is $10^2 + 20^2 = 500$ .

Option analysis

Why each option works or fails

A · 650

Calculated $\alpha$ incorrectly or misapplied deviation squared terms when setting up the combined variance formula. Ensure deviation terms $d_1^2 = (\bar{x}_1 - \bar{x})^2$ and $d_2^2 = (\bar{x}_2 - \bar{x})^2$ are correctly subtracted: $(40 - 50)^2 = 100$ and $(55 - 50)^2 = 25$ .

B · 450

Forgot to square the standard deviations, adding $\alpha^2 + (30 - \alpha)$ or committing an arithmetic error during the quadratic resolution. Variance is the square of standard deviation, so the sum of variances is $\sigma_1^2 + \sigma_2^2 = \alpha^2 + (30 - \alpha)^2$ .

C · 900

Computed $(\sigma_1 + \sigma_2)^2 = (\alpha + 30 - \alpha)^2 = 30^2 = 900$ instead of the sum of the variances $\sigma_1^2 + \sigma_2^2$ . The sum of variances is $\sigma_1^2 + \sigma_2^2$ , not the square of the sum of the standard deviations $(\sigma_1 + \sigma_2)^2$ .

D · 500

None. The value is found by finding $n = 200$ , using combined variance to find $\alpha = 10$ (since $\alpha > 0$ ), and computing $10^2 + 20^2 = 500$ . Correctly determined $n=200$ , $\alpha=10$ , and $\sigma_1^2 + \sigma_2^2 = 100 + 400 = 500$ .

Reviewed route

Solution

StepWorking

01given

For Class A: $n_1 = 100$ , mean $\bar{x}_1 = 40$ , standard deviation $\sigma_1 = \alpha > 0$ . For Class B: $n_2 = n$ , mean $\bar{x}_2 = 55$ , standard deviation $\sigma_2 = 30 - \alpha$ . Combined: $N = 100 + n$ , combined mean $\bar{x} = 50$ , combined variance $\sigma^2 = 350$ .

02goal

Find the sum of the variances of classes A and B, which is $\sigma_1^2 + \sigma_2^2 = \alpha^2 + (30 - \alpha)^2$ .

03approach

First, use the combined mean formula $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$ to solve for $n$ . Then apply the combined variance formula $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$ where $d_1 = \bar{x}_1 - \bar{x}$ and $d_2 = \bar{x}_2 - \bar{x}$ to solve for $\alpha$ , and finally compute $\sigma_1^2 + \sigma_2^2$ .

04execute

Calculate $n$ from the combined mean: $50 = \frac{100 \times 40 + n \times 55}{100 + n} \implies 5000 + 50n = 4000 + 55n \implies 5n = 1000 \implies n = 200$

05execute

Compute mean deviations: $d_1 = 40 - 50 = -10$ , $d_2 = 55 - 50 = 5$ . Substitute into combined variance formula: $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$ $350 = \frac{100(\alpha^2 + (-10)^2) + 200((30 - \alpha)^2 + 5^2)}{300}$ $350 \times 3 = (\alpha^2 + 100) + 2((30 - \alpha)^2 + 25)$ $1050 = \alpha^2 + 100 + 2(900 - 60\alpha + \alpha^2 + 25)$ $1050 = 3\alpha^2 - 120\alpha + 1950$ $3\alpha^2 - 120\alpha + 900 = 0 \implies \alpha^2 - 40\alpha + 300 = 0$ $(\alpha - 10)(\alpha - 30) = 0$ Since standard deviation of B is $30 - \alpha > 0$ , we must have $\alpha < 30$ , hence $\alpha = 10$ .

06execute

Calculate the sum of variances: $\sigma_1^2 = 10^2 = 100$ $\sigma_2^2 = (30 - 10)^2 = 20^2 = 400$ $\sigma_1^2 + \sigma_2^2 = 100 + 400 = 500$

✓verify

Check: If $\sigma_1^2 = 100, \sigma_2^2 = 400$ , combined variance is $\frac{100(100+100) + 200(400+25)}{300} = \frac{20000 + 85000}{300} = \frac{105000}{300} = 350$ . Matches exactly.

Hints that build this answer step by step

What is the number of students $n$ in class $\mathrm{B}$ , determined from the combined mean $\bar{x} = 50$ ?
$n = 200$
Using the combined variance formula $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$ , what equation in terms of $\alpha$ is obtained?
$350 = \frac{100(\alpha^2 + 100) + 200((30 - \alpha)^2 + 25)}{300}$
Simplifying the combined variance equation, what is the value of $\alpha$ given $\alpha > 0$ ?
$\alpha = 10$
What is the sum of the variances of classes $\mathrm{A}$ and $\mathrm{B}$ ?
$\sigma_1^2 + \sigma_2^2 = 10^2 + 20^2 = 500$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is

\alpha = 30

rejected when the quadratic equation gives both 10 and 30?

Class B has standard deviation $30 - \alpha$ . In standard statistical convention, a distribution has $\sigma > 0$ unless all values are identical. Standard deviation is typically strictly positive in such contexts. If $\alpha = 30$ , $\sigma_2 = 0$ . Even if $\alpha = 30$ were allowed, $\sigma_1^2 + \sigma_2^2 = 900 + 0 = 900$ . Both 500 and 900 are options. The strict inequality $30 - \alpha > 0$ (analogous to $\alpha > 0$ for class A) selects $\alpha = 10$ .

Answer

Using the combined mean gives $n = 200$ , and substituting into the combined variance formula yields $\alpha = 10$ , so the sum of variances is $10^2 + 20^2 = 500$ .

Why each option works or fails

A: 650 - Calculated $\alpha$ incorrectly or misapplied deviation squared terms when setting up the combined variance formula. Ensure deviation terms $d_1^2 = (\bar{x}_1 - \bar{x})^2$ and $d_2^2 = (\bar{x}_2 - \bar{x})^2$ are correctly subtracted: $(40 - 50)^2 = 100$ and $(55 - 50)^2 = 25$ .
B: 450 - Forgot to square the standard deviations, adding $\alpha^2 + (30 - \alpha)$ or committing an arithmetic error during the quadratic resolution. Variance is the square of standard deviation, so the sum of variances is $\sigma_1^2 + \sigma_2^2 = \alpha^2 + (30 - \alpha)^2$ .
C: 900 - Computed $(\sigma_1 + \sigma_2)^2 = (\alpha + 30 - \alpha)^2 = 30^2 = 900$ instead of the sum of the variances $\sigma_1^2 + \sigma_2^2$ . The sum of variances is $\sigma_1^2 + \sigma_2^2$ , not the square of the sum of the standard deviations $(\sigma_1 + \sigma_2)^2$ .
D · correct: 500 - None. The value is found by finding $n = 200$ , using combined variance to find $\alpha = 10$ (since $\alpha > 0$ ), and computing $10^2 + 20^2 = 500$ . Correctly determined $n=200$ , $\alpha=10$ , and $\sigma_1^2 + \sigma_2^2 = 100 + 400 = 500$ .

Step-by-step solution

given: For Class A: $n_1 = 100$ , mean $\bar{x}_1 = 40$ , standard deviation $\sigma_1 = \alpha > 0$ . For Class B: $n_2 = n$ , mean $\bar{x}_2 = 55$ , standard deviation $\sigma_2 = 30 - \alpha$ . Combined: $N = 100 + n$ , combined mean $\bar{x} = 50$ , combined variance $\sigma^2 = 350$ .
goal: Find the sum of the variances of classes A and B, which is $\sigma_1^2 + \sigma_2^2 = \alpha^2 + (30 - \alpha)^2$ .
approach: First, use the combined mean formula $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$ to solve for $n$ . Then apply the combined variance formula $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$ where $d_1 = \bar{x}_1 - \bar{x}$ and $d_2 = \bar{x}_2 - \bar{x}$ to solve for $\alpha$ , and finally compute $\sigma_1^2 + \sigma_2^2$ .
execute: Calculate $n$ from the combined mean: $50 = \frac{100 \times 40 + n \times 55}{100 + n} \implies 5000 + 50n = 4000 + 55n \implies 5n = 1000 \implies n = 200$
execute: Compute mean deviations: $d_1 = 40 - 50 = -10$ , $d_2 = 55 - 50 = 5$ . Substitute into combined variance formula: $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$ $350 = \frac{100(\alpha^2 + (-10)^2) + 200((30 - \alpha)^2 + 5^2)}{300}$ $350 \times 3 = (\alpha^2 + 100) + 2((30 - \alpha)^2 + 25)$ $1050 = \alpha^2 + 100 + 2(900 - 60\alpha + \alpha^2 + 25)$ $1050 = 3\alpha^2 - 120\alpha + 1950$ $3\alpha^2 - 120\alpha + 900 = 0 \implies \alpha^2 - 40\alpha + 300 = 0$ $(\alpha - 10)(\alpha - 30) = 0$ Since standard deviation of B is $30 - \alpha > 0$ , we must have $\alpha < 30$ , hence $\alpha = 10$ .
execute: Calculate the sum of variances: $\sigma_1^2 = 10^2 = 100$ $\sigma_2^2 = (30 - 10)^2 = 20^2 = 400$ $\sigma_1^2 + \sigma_2^2 = 100 + 400 = 500$
verify: Check: If $\sigma_1^2 = 100, \sigma_2^2 = 400$ , combined variance is $\frac{100(100+100) + 200(400+25)}{300} = \frac{20000 + 85000}{300} = \frac{105000}{300} = 350$ . Matches exactly.

Statistics and Probability: Mathematics | JEE Main

What feels right?

Using the combined mean gives $n = 200$ , and substituting into the combined variance formula yields $\alpha = 10$ , so the sum of variances is $10^2 + 20^2 = 500$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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