We know the mean of 10 observations is 5. So, the total sum of the 10 observations is:

$$\frac{x_1+x_2+x_3+\ldots+x_8+x_9+x_{10}}{10}=5$$

Multiplying both sides by 10 gives:

$$x_1+x_2+x_3+\ldots+x_8+x_9+x_{10}=50$$

The mean of the first 8 observations is 4, so their total sum is:

$$\frac{x_1+x_2+\ldots+x_8}{8}=4$$

Therefore,

$$x_1+x_2+\ldots+x_8=32$$

Subtracting, we get:

$$x_9+x_{10}=50-32=18$$

Next, we use the formula for variance of the 10 observations:

$$\text{Variance} = \frac{\sum\limits_{i=1}^{10} x_i^2}{10} - (\text{mean})^2$$

Substitute the given values to find the total of squares:

$$\frac{\sum\limits_{i=1}^{10} x_i^2}{10} - 25 = 7$$

$$\sum\limits_{i=1}^{10} x_i^2 = 320$$

Similarly, for the first 8 observations:

$$\frac{\sum\limits_{i=1}^{8} x_i^2}{8} - 16 = 3.5$$

$$\sum\limits_{i=1}^{8} x_i^2 = 156$$

Now, subtract the two results:

$$x_9^2 + x_{10}^2 = 320 - 156 = 164$$

We already know:

$$x_9 + x_{10} = 18$$

Expanding $ (x_9 + x_{10})^2 $, we get:

$$x_9^2 + x_{10}^2 + 2x_9x_{10} = 324$$

Substitute $x_9^2 + x_{10}^2 = 164$:

$$164 + 2x_9x_{10} = 324$$

$$x_9x_{10} = 80$$

Now, we have two equations:

$$x_9 + x_{10} = 18$$

$$x_9x_{10} = 80$$

Solving the quadratic $t^2 - 18t + 80 = 0$, we get $t = 8$ and $10$. Given $x_9 < x_{10}$, so $x_9 = 8$ and $x_{10} = 10$.

Finally,

$$3x_9 + 2x_{10} = 3(8) + 2(10) = 44$$

JEE Advanced Subjects