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# The mean of the numbers $a, b, 8, 5, 10$ is $6$ and the variance is $6.80$. Then which one of the following gives possible values of $a$ and $b$ ?

$\begin {array} {1 1} (A)\;a=0\: b=7 & \quad (B)\;a=5\: b=2 \\ (C)\;a=1\: b=6 & \quad (D)\;a=3\: b=4 \end {array}$

According to the given condition
$6.80 = \large\frac{(6-a)^2+(6-b)^2+(6-8)^2+(6-5)^2+(6-10)^2}{5}$
$\Rightarrow (6-a)^2+(6-b)^2+4+1+16=34$
$(6-a)^2+(6-b)^2=34-21$
$(6-a)^2+(6-b)^2=13$
$(6-a)^2+(6-b)^2=9+4$
$(6-a)^2+(6-b)^2=3^2+2^2$
$(6-a)^2 = 3^2 \: (6-b)^2 = 2^2$
$6-a = 3\: \: \:\: \: \: \: \:\; \: \: 6-b=2$
$-a=-3\: \: \: \: \: \: \: \:\: \: \: \: -b=-4$
$\: \: \: a = 3\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: b=4$
Ans : (D)