Maths Q9 – JEE-MAIN 2026

The mean and variance of

n

observations are

8

and

16

, respectively. If the sum of the first

(n - 1)

observations is

48

and the sum of squares of the first

(n - 1)

observations is

496

, then the value of

n

is:

Clue 1 of 3: Understanding Statistical Measures

Recall the definitions of mean and variance, and how they relate to the sum of observations and sum of squares of observations.

🧠 Full Solution Path

Step 1: Formulate equations for total sum and sum of squares✦ Active

Given the mean $\bar{x} = 8$ and variance $σ^{2} = 16$ for $n$ observations. The sum of $n$ observations is $\sum_{i = 1}^{n} x_{i} = n \bar{x} = 8 n$ . The variance formula is $σ^{2} = \frac{\sum x_{i}^{2}}{n} - (\bar{x})^{2}$ . Substituting the given values:

16 = \frac{\sum x_{i}^{2}}{n} - 8^{2} \Rightarrow 16 = \frac{\sum x_{i}^{2}}{n} - 64 \Rightarrow \frac{\sum x_{i}^{2}}{n} = 80 \Rightarrow \sum_{i = 1}^{n} x_{i}^{2} = 80 n

Step 2: Express the

n

-th observation and its square○ Expand

Let $x_{n}$ be the $n$ -th observation. We are given the sum of the first $(n - 1)$ observations as $48$ and the sum of squares of the first $(n - 1)$ observations as $496$ . We can write the total sum and sum of squares in terms of $x_{n}$ :

\sum_{i = 1}^{n} x_{i} = \sum_{i = 1}^{n - 1} x_{i} + x_{n} \Rightarrow 8 n = 48 + x_{n} \Rightarrow x_{n} = 8 n - 48

\sum_{i = 1}^{n} x_{i}^{2} = \sum_{i = 1}^{n - 1} x_{i}^{2} + x_{n}^{2} \Rightarrow 80 n = 496 + x_{n}^{2}

Step 3: Solve the quadratic equation for

n

○ Expand

Substitute the expression for $x_{n}$ from the first equation into the second equation:

80 n = 496 + (8 n - 48)^{2}

Simplify and rearrange into a quadratic equation:

80 n = 496 + 64 (n - 6)^{2}

80 n = 496 + 64 (n^{2} - 12 n + 36)

Divide by 16:

5 n = 31 + 4 (n^{2} - 12 n + 36)

5 n = 31 + 4 n^{2} - 48 n + 144

4 n^{2} - 53 n + 175 = 0

Solve the quadratic equation using the quadratic formula $n = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ :

n = \frac{53 \pm \sqrt{(- 53)^{2} - 4 (4) (175)}}{2 (4)}

n = \frac{53 \pm \sqrt{2809 - 2800}}{8}

n = \frac{53 \pm \sqrt{9}}{8}

n = \frac{53 \pm 3}{8}

This gives two possible values for $n$ : $n = \frac{53 + 3}{8} = \frac{56}{8} = 7$ or $n = \frac{53 - 3}{8} = \frac{50}{8} = \frac{25}{4}$ . Since $n$ must be an integer (number of observations), the valid value is $n = 7$ .

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Maths Question 9 – JEE-MAIN 2026

Clue 1 of 3: Understanding Statistical Measures

🧠 Full Solution Path