Question

Let the mean and variance of four numbers $3,7,x$ and $y(x>y)$ be $5$ and $10$ respectively. Then the mean of four numbers $3+2x,7+2y,x+y$ and $x-y$ is

Answer 1

Numbers $3,7,x,y$
$\overline{x}=5,{\sigma }^2=10$
$5=\frac{3+7+x+y}{4}\Rightarrow x+y=10\cdots \left(i\right)$
$10=\frac{1}{4}\left(\right(3{)}^2+(7{)}^2+(x{)}^2+\left(y{)}^2\right)-(5{)}^2$
$\Rightarrow 140=58+x^2+y^2\Rightarrow x^2+y^2=82\cdots \left(ii\right)$

$(x+y{)}^2=x^2+y^2+2xy\Rightarrow 100=82+2xy$
$\Rightarrow xy=9$
$y=\frac{9}{x}\Rightarrow x+\frac{9}{x}=10\Rightarrow (x,y)=(1,9)\text{or}(9,1)$

Given $x>y\Rightarrow x=9,y=1$
Now, $3+2x,7+2y,x+y,x-y=21,9,10,8$

$\overline{x}=\frac{21+9+10+8}{4}=\frac{48}{4}=12$