Mean and variance of 10 positive observations is 7 and 8. If we add 5 to each number then sum of mean and variance of new data is

$Mean=\frac{3+12+7+a+\left(43-a\right)}{5}=13$  

Variance =  $\frac{3^2+12^2+7^2+a^2+{\left(43-a\right)}^2}{5}-{\left(13\right)}^2$  

$\Rightarrow \frac{2a^2-a+1}{5}\to Naturalnumber$  

Let 2a² – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not  $4\lambda or4\lambda +1from.$  

As each square form is  $4\lambda or4\lambda +1$  

${\sigma }^2=\frac{\sum x^2}{n}-{\left(\frac{\sum x}{n}\right)}^2=\frac{9+k^2}{10}-{\left(\frac{9+k}{10}\right)}^2<10$

$\Rightarrow k<\frac{10\sqrt[]{10}}{3}+1\Rightarrow k\le 11$        

$\therefore$ maximum value of k is 11