The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of six observations are 2,4,5 and 7,then the remaining two observations are:

let other two numbers be x and $(21-x)$ since, variance is 10.25
$\begin{array}{r}\therefore \left(\frac{4+16+25+49+x^2+(21-x{)}^2}{6}\right)-(6.5{)}^2=10.25\\ \Rightarrow 94+x^2+(21-x{)}^2=6(10.25)+6(6.5{)}^2\Rightarrow x^2+(21-x{)}^2=221\end{array}$

x=10 and $(21-x)=21-10=11$ Here, remaining two observations are 10, 11