Q26 / 75 · JEE Main 2025 (29 Jan Shift 1) · Mathematics

Let \(x_1, x_2, \ldots, x_{10}\) be ten observations such that \(\sum_{i=1}^{10}\left(x_i-2\right)=30, \sum_{i=1}^{10}\left(x_i-\beta\right)^2=98, \beta\gt2\), and their variance is \(\frac{4}{5}\). If \(\mu\) and \(\sigma^2\) are respectively the mean and the variance of \(2\left(x_1-1\right)+4 \beta\), \(2\left(x_2-1\right)+4 \beta, \ldots ., 2\left(x_{10}-1\right)+4 \beta\), then \(\frac{\beta \mu}{\sigma^2}\) is equal to :

  • Correct answer
    A
    100
  • B
    120
  • C
    110
  • D
    90

Solution

$\begin{aligned} & \sum_{l=1}^{10}\left(x_l-2\right)=30 \\ & \sum_{i=1}^{10} x_l=50 \\ & \Rightarrow \text { Mean }=5 \\ & \text { Variance }=\frac{4}{5}=\frac{\sum x_l^2}{10}-(\bar{x})^2 \\ & \frac{4}{5}=\frac{\sum x_l^2}{10}-25 \\ & \Rightarrow \sum x_l^2=258 \end{aligned}$ Now, $\sum_{l=1}^{10}\left(x_l-\beta\right)^2=98$ $\begin{aligned} & \sum_{l=1}^{10} x_l^2-2 \beta \sum_{l=1}^{10} x_l+10 \beta^2=98 \\ & \Rightarrow 258-2 \beta(50)+10 \beta^2=98 \\ & \Rightarrow 10 \beta^2-100 \beta+160=0 \\ & \Rightarrow \beta^2-10 \beta+16=0 \\ & \Rightarrow \beta=8 \text { as } \beta>2 \end{aligned}$ Now, as per the question $2\left(x_1-1\right)+4 \beta, 2\left(x_2-1\right)+4 \beta, \ldots 2\left(x_{10}-1\right)+4 \beta$ Can be simplified as $2 x_1+30,2 x_2+30, \ldots, 2 x_{10}+30$ $\begin{aligned} & \mu=2(5)+30=40 \\ & \sigma^2=2^2\left(\frac{4}{5}\right)=\frac{16}{5} \\ & \frac{\beta \mu}{\sigma^2}=\frac{8 \times 40}{\frac{16}{5}}=100\end{aligned}$

Hence, the correct answer is option: 'A'.

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