JEE Mathematics 2026 Question with Solution

Given: The mean of the $8$ numbers $-10,-7,-1,x,y,9,2,16$ is $2$ and the variance is $\frac{293}{4}$.

Find: The mean of the $4$ numbers $x$, $y$, $x+y+1$, $|x-y|$.

From the mean,

So,

Hence,

Using the variance formula,

This gives

So,

Therefore,

Now use

Substituting $x+y=7$ and $x^2+y^2=127$,

Thus,

and hence

To find $|x-y|$, use

So,

Thus,

The four numbers are $x$, $y$, $x+y+1=8$, and $|x-y|=\sqrt{205}$. Their mean would be

which does not simplify to any listed option. The solution states that the correct option is C and concludes the mean is $9$. Therefore, following the provided the solution, the correct option is C, though the working shown is inconsistent with that conclusion.

The key identity is

From the extracted relations,

Hence,

This confirms that the displayed intermediate working on the page does not naturally produce $9$ as the mean. Nevertheless, the source solution explicitly marks C as correct, so the answer is recorded as C.