JEE Main 2026 · 2 Apr · Shift 2 · Mathematics Q9

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MathematicsSingle correct+4 / -1Question 9 of 75

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 :

Answer and worked solution

Official answer: D

Totals. $\sum x=8n$ and $\sum x^2=n(16+64)=80n$ .

Last observation. $x_n=8n-48$ and $x_n^2=80n-496$ :
$(8n-48)^2=80n-496\implies4n^2-53n+175=0\implies n=7\ \left(\text{or }\tfrac{25}4\right).$

(D)

Topics:Variance and standard deviation

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JEE Main 2026 · 2 Apr · Shift 2, Mathematics question 9