Consider a set of 3n numbers having variance 4. In this set, the mean of first 2n numbers is 6 and the mean of the remaining n numbers is 3. A new set is constructed by adding 1 into each of first 2n numbers, and subtracting 1 from each of the remaining n numbers. If the variance of the new set is k, then 9k is equal to __________.

Let first 2n observations are x₁, x₂ ...................., x_2n<br><br>and last n observations are y₁, y₂ ....................., y_n<br><br>Now, ${{\sum {{x_i}} } \over {2n}} = 6$, ${{\sum {{y_i}} } \over n} = 3$<br><br>$\Rightarrow \sum {{x_i}} = 12n,\sum {{y_i}} = 3n$ <br><br>$\therefore$ ${{\sum {{x_i}} + \sum {{y_i}} } \over {3n}} = {{15n} \over {3n}} = 5$<br><br>Now, ${{\sum {x_i^2} + \sum {y_i^2} } \over {3n}} - {5^2} = 4$<br><br>$\Rightarrow \sum {x_i^2} + \sum {y_i^2} = 29 \times 3n = 87n$<br><br>Now, mean is $${{\sum {({x_i} + 1) + \sum {({y_i} - 1)} } } \over {3n}} = {{15n + 2n - n} \over {3n}} = {{16} \over 3}$$<br><br>Now, variance is $${{{{\sum {{{({x_i} + 1)}^2} + \sum {({y_i} - 1)} } }^2}} \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$$<br><br>$$ = {{\sum {x_i^2 + \sum {y_i^2} + 2\left( {\sum {{x_i}} - \sum {{y_i}} } \right) + 3n} } \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$$<br><br>$= {{87n + 2(9n) + 3n} \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$<br><br>= $29 + 6 + 1 - {\left( {{{16} \over 3}} \right)^2}$<br><br>$= {{324 - 256} \over 9} = {{68} \over 9} = k$<br><br>$\Rightarrow$ 9k = 68<br><br>Therefore, the correct answer is 68.