JEE Advanced 2026 Paper 2, Mathematics Section 3 Q3: Measure of Dispersion

JEE Advanced2026Paper 2Mathematics Section 3

Consider a data consisting of 10 observations $x_{1}, x_{2}, \dots, x_{10}$ , whose mean is 5 and variance is 7. If the mean and the variance of the first 8 observations $x_{1}, x_{2}, \dots, x_{8}$ are 4 and 3.5, respectively, and $x_{9} < x_{10}$ , then the value of $3 x_{9} + 2 x_{10}$ is ___________.

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Solution

From the means: $i = 1 \sum 10 x_{i} = 50$ and $i = 1 \sum 8 x_{i} = 32$ , so $x_{9} + x_{10} = 18$ .

From the variances: $\frac{1}{10} i = 1 \sum 10 x_{i}^{2} - 5^{2} = 7$ gives $i = 1 \sum 10 x_{i}^{2} = 320$ , and $\frac{1}{8} i = 1 \sum 8 x_{i}^{2} - 4^{2} = 3.5$ gives $i = 1 \sum 8 x_{i}^{2} = 156$ .

Therefore $x_{9}^{2} + x_{10}^{2} = 320 - 156 = 164$ .

Substituting $x_{10} = 18 - x_{9}$ : $x_{9}^{2} + (18 - x_{9})^{2} = 164$ , so $2 x_{9}^{2} - 36 x_{9} + 160 = 0$ , i.e. $x_{9}^{2} - 18 x_{9} + 80 = 0$ .

The roots are $x_{9} = 8$ or $x_{9} = 10$ . Since $x_{9} < x_{10}$ and $x_{9} + x_{10} = 18$ , we need $x_{9} < 9$ , so $x_{9} = 8$ and $x_{10} = 10$ .

$∴ 3 x_{9} + 2 x_{10} = 24 + 20 = 44$ .

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JEE Advanced 2026 Paper 2, Mathematics Section 3 Q3: Measure of Dispersion

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