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# The A.M of $n$ observation is $\overline x$ . If the sum of $n-5$ observations is $a$, then the mean of remaining 5 observation is

$\begin {array} {1 1} (A)\;\large\frac{n\overline x+a}{5} & \quad (B)\;\large\frac{n\overline x-a}{5} \\ (C)\;n\overline x +a & \quad (D)\;None \: of \: these \end {array}$

If $m$ is the mean of 5 observations, then
$\overline x = \large\frac{(n-5)\large\frac{a}{n-5}+5m}{n- \not{5} + \not{5}}$
$\overline x = \large\frac{(\not{n-5})\large\frac{a}{\not{n-5}}+5m}{n}$
$\Rightarrow n\overline x = a+5m$
$5m = n\overline x - a$
$m = \large\frac{n\overline x - a}{5}$
Ans : (B)