JEE Main 2025 Apr 2 Shift 2, Mathematics Q11: Measure of Dispersion

JEE Main2025Apr 2, Shift 2Mathematics

If the mean and the variance of $6, 4, a, 8, b, 12, 10, 13$ are $9$ and $9.25$ respectively, then $a + b + ab$ is equal to :

Solution

From the mean condition: $6 + 4 + a + 8 + b + 12 + 10 + 13 = 8 \times 9 = 72 \Rightarrow a + b = 19$ .

For variance, use $σ^{2} = \frac{\sum x _{i}^{2}}{n} - \overset{x}{ˉ}^{2}$ :

$9.25 + 81 = \frac{\sum x _{i}^{2}}{8} \Rightarrow \sum x_{i}^{2} = 8 \times 90.25 = 722$ .

The known squares total $36 + 16 + 64 + 144 + 100 + 169 = 529$ , so $a^{2} + b^{2} = 722 - 529 = 193$ .

From $(a + b)^{2} = a^{2} + b^{2} + 2 ab$ : $361 = 193 + 2 ab \Rightarrow ab = 84$ .

$∴ a + b + ab = 19 + 84 = 103$ .

Concept behind this question

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