JEE Main 2026 — 24 January, Morning Shift — Question 18

Let first 10 numbers are $\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots \ldots \mathrm{x}_{9}, \alpha$

$$\begin{aligned} & \Rightarrow \alpha+\sum_{\mathrm{i}=1}^{9} \mathrm{x}_{\mathrm{i}}=100 \Rightarrow \sum_{\mathrm{i}=1}^{9} \mathrm{x}_{\mathrm{i}}=100-\alpha & \end{aligned}$$

Variance $=\left(\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{\mathrm{n}}\right)-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}\right)^{2}$

$\Rightarrow \frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{\mathrm{n}}=98$

$\Rightarrow \mathrm{x}_{1}^{2}+\mathrm{x}_{2}^{2}+\ldots \ldots \mathrm{x}_{9}^{2}+\alpha^{2}=1020$

$\Rightarrow \sum \mathrm{x}_{\mathrm{i}}^{2}=1020-\alpha^{2}$

In second case, let number are $\begin{aligned} & x_{1}, x_{2}, \ldots \ldots x_{9}, \beta \end{aligned}$

$100-\alpha+\beta=101$

$\alpha-\beta+1=0$

$\frac{\sum x_{i}^{2}+\beta^{2}}{10}-(10.1)^{2}=1.99$

$\beta^{2}-\alpha^{2}=20$

$\alpha=\frac{19}{2}$

$\beta=\frac{21}{2}$

$\alpha+\beta=\frac{19+21}{2}=20$