The sum and product of the mean and variance of a binomial...

Given, sum of mean and variance is $82 . 5$

Product of roots is $1350$

Now, Mean and variance are the roots of

${x}^{2} - \left(\text{sum of roots}\right)x + \text{product of roots} = 0$

$⇒ {x}^{2} - 82 . 5x + 1350 = 0$

$⇒ x = 60\text{or}22 . 5$

So, mean $= np = 60$ and variance $= npq = 22 . 5$

$⇒ q = \dfrac{22 . 5}{60} = \dfrac{3}{8}$

So, $p = \dfrac{5}{8}$ and $n = \dfrac{60}{\dfrac{5}{8}} = 96$