JEE Main 2026 — Apr 2, Shift 2 — Mathematics solution

JEE Main2026Apr 2, Shift 2Mathematics

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 :

A
$21$
B
$16$
C
$13$
D
$7$

Solution

Total sum: $\sum x_{i} = 8 n$ , so $x_{n} = 8 n - 48$ .

Variance condition: $\frac{\sum x _{i}^{2}}{n} - 64 = 16$ , giving $\sum x_{i}^{2} = 80 n$ , so $x_{n}^{2} = 80 n - 496$ .

Substituting $x_{n} = 8 n - 48$ : $(8 n - 48)^{2} = 80 n - 496$ , i.e. $64 (n - 6)^{2} = 8 (10 n - 62)$ .

Simplifying: $4 (n - 6)^{2} = 5 n - 31$ , which expands to $4 n^{2} - 53 n + 175 = 0$ .

Factoring: $(n - 7) (4 n - 25) = 0$ . Since $n$ is a positive integer, $n = 7$ .

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