JEE Mathematics 2025 Question with Solution

Given:

• $\sum_{i=1}^{10}(x_i-2)=30$
• $\sum_{i=1}^{10}(x_i-\beta)^2=98$ with $\beta \geq 2$
• Variance of $x_1,x_2,\ldots,x_{10}$ is $\frac{4}{5}$
• Transformed observations are $y_i=2(x_i-1)+4\beta$

Find: $\frac{\beta\mu}{\sigma^2}$, where $\mu$ and $\sigma^2$ are the mean and variance of the transformed observations.

From

$$\sum_{i=1}^{10}(x_i-2)=30$$

we get

$$\sum_{i=1}^{10}x_i-20=30$$

so

$$\sum_{i=1}^{10}x_i=50$$

and hence the mean of the original observations is

$$\bar{x}=\frac{50}{10}=5$$

Now use the variance formula:

$$\frac{1}{10}\sum_{i=1}^{10}(x_i-\bar{x})^2=\frac{4}{5}$$

Since $\bar{x}=5$,

$$\frac{1}{10}\sum_{i=1}^{10}(x_i-5)^2=\frac{4}{5}$$

therefore

$$\sum_{i=1}^{10}(x_i-5)^2=8$$

Using

$$\sum_{i=1}^{10}(x_i-\beta)^2=\sum_{i=1}^{10}(x_i-5)^2+10(\beta-5)^2$$

we substitute $\sum_{i=1}^{10}(x_i-5)^2=8$ to get

$$98=8+10(\beta-5)^2$$

so

$$90=10(\beta-5)^2$$

which gives

$$(\beta-5)^2=9$$

Hence

$$\beta-5=\pm 3$$

Since $\beta \geq 2$, the admissible value taken in the solution is

$$\beta=8$$

Now the transformed observations are

$$y_i=2(x_i-1)+4\beta=2x_i-2+32=2x_i+30$$

Therefore the new mean is

$$\mu=2\bar{x}+30=2(5)+30=40$$

The variance under the linear transformation becomes

$$\sigma^2=2^2\cdot \frac{4}{5}=\frac{16}{5}$$

Finally,

$$\frac{\beta\mu}{\sigma^2}=\frac{8\times 40}{16/5}=\frac{320\times 5}{16}=100$$

Therefore, the correct option is A and the value is $100$.

Note: The solution marks option C but its own working gives $100$. Hence the worked solution has been used.