Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and $α$ (>0),and the mean and standard deviation of marks of class B of n students be respectively 55 and 30- $α$ . If the mean and variance of the marks of the combined class of 100 + n students are respectively 50 and 350, then the sum of variances of classes A and B is:

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500

450

650

900

answer is A.

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Detailed Solution

$\begin{array}{l} n_{1} = 100, \bar{x_{1}} = 40, σ_{1} = α \\ n_{2} = n, \bar{x_{2}} = 55, σ_{2} = (30 - α) \\ \bar{x_{(combined})} = 50, σ_{combined}^{2} = 350, σ_{1}^{2} + σ_{2}^{2} = ? \\ σ_{combined}^{2} = \frac{n_{1} σ_{1}^{2} + n_{2} σ_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2}}{{(n_{1} + n_{2})}^{2}} {(\bar{x_{1}} - \bar{x_{2}})}^{2} - - - - (1) \\ {\bar{x}}_{combined} = \frac{n_{1} \cdot {\bar{x}}_{1} + n_{2} \cdot {\bar{x}}_{2}}{n_{1} + n_{2}} \\ \begin{aligned} \bar{x} = \frac{100 \times 40 + 55 n}{100 + n} \\ 5000 + 50 n = 4000 + 55 n \\ 1000 = 5 n \\ n = 200 \end{aligned} \end{array}$
So substitute in (1) we get $α^{2} - 40 α + 300 = 0$
$\begin{array}{l} (α - 30) (α - 10) = 0 \\ α = 10, as (30 - α > 0) \end{array}$
So sum of squares of variances
$= 10^{2} + 20^{2} = 500$