JEE Main 2016 — Statistics Question with Solution

$\frac{x+y+1+2+6}{5}=5$

$\Rightarrow x+y=16$

Now $12.4=\frac{\left(16+9+1+{\left(x-5\right)}^2+{\left(y-5\right)}^2\right)}{5}$

$\Rightarrow \mathrm{62 }$$=26+{\left(x-5\right)}^2+{\left(y-5\right)}^2$

$\Rightarrow 36={\left(x-5\right)}^2+{\left(y-5\right)}^2$

Shifting of the origin to $\left(5,5\right)$ the equations will become 

$6^2=x^2+y^2$ and $x+y=6$

Solving them and again will get the $x=0$ or $y=0$.

If $x=0\Rightarrow y=6$ or

If $y=0\Rightarrow x=6$.

But by shifting the origin back to original position, we get $x=0+5=5 \& y=6+5=11$

(or)

$x=6+5=11 \& y=0+5=5$

Hence, the values of the remaining two are $5, 11$.