Let the mean and variance of four numbers 3, 7, x and y (xy) be 5 and 10 respectively.

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 $3+2x, 7+2y, x+y, x-y$ is

Answer: 12

Accepted Answer

Mean $=5$: $3+7+x+y=20 \implies x+y=10$

Variance $=10$: $\dfrac{9+49+x^2+y^2}{4}-25=10 \implies x^2+y^2=82$

$x^2+(10-x)^2=82 \implies 2x^2-20x+18=0 \implies x^2-10x+9=0 \implies (x-9)(x-1)=0$

Since $x>y$: $(x,y)=(9,1)$.

$$\text{Mean} = \frac{(3+18)+(7+2)+10+8}{4} = \frac{48}{4}=12$$

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