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 $3+2 x, 7+2 y, x+y$ and
$x-y$ is_________
Solution:
$5=\frac{3+7+x+y}{4} \Rightarrow x+y=10$
$\operatorname{Var}(x)=10=\frac{3^{2}+7^{2}+x^{2}+y^{2}}{4}-25$
$140=49+9+x^{2}+y^{2}$
$x^{2}+y^{2}=82$
$x+y=10$
$\Rightarrow(\mathrm{x}, \mathrm{y})=(9,1)$
Four numbers are $21,9,10,8$
Mean $=\frac{48}{4}=12$
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