Let mean and variance of four numbers 3,7, x and y(x>y) be 5 and 10 respectively. Then mean of fo

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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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