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13.Statistics
hard
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 $3+2 \mathrm{x}, 7+2 \mathrm{y}, \mathrm{x}+\mathrm{y}$ and $x-y$ is ..... .
A
$10$
B
$11$
C
$12$
D
$48$
(JEE MAIN-2021)
Solution
$5=\frac{3+7+x+y}{4} \Rightarrow x+y=10$
$\operatorname{Var}(x)=10=\frac{3^{2}+7^{2}+x^{2}+y^{2}}{4}-25$
$140=49+9+x^{2}+y^{2}$
$x^{2}+y^{2}=82$
$x+y=10$
$\Rightarrow(x, y)=(9,1)$
Four numbers are $21,9,10,8$
$\text { Mean }=\frac{48}{4}=12$
Standard 11
Mathematics
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