Question

The mean and variance of $7$ observation are $8,16$ respectively. If $5$ of the observation are $2,4,10,12,14$, then the $LCM$ of remaining two observation is

• A. $16$
• B. $24$
• C. $20$
• D. $14$

Answer 1

The correct option is B $24$

Let other two observations are $x$and$y$.

$\therefore$our observations are $2,4,10,12,14,x,y.$ given $Mean=8$ i.e, $\frac{sumofobservation}{numberofobservation}=8$

$\therefore \frac{2+4+10+12+14+x+y}{7}=8=>x+y=56-42=>x+y=\mathrm{14...........}\left(1\right)$ Also ,given $variance=16$$\therefore \frac{1}{n}\sum {(n_i-\overline{x})}^2=16=>\frac{1}{7}\sum {(n_i-8)}^2=16\therefore \frac{1}{7}[{(2-8)}^2+{(4-8)}^2+{(10-8)}^2+{(12-8)}^2+{(14-8)}^2+{(x-8)}^2+{(y-8)}^2]=16=>\frac{1}{7}[36+16+4+16+36+x^2+8^2-2(8)x+y^2+8^2-2(8\left)y\right]=16=>236+x^2{+y}^2-224=112=>x^2{+y}^2=100........\left(2\right)\therefore x^2{+y}^2+2xy=196=>100+2xy=196=>2xy=96=>xy=48=>x=\frac{48}{y}...............\left(3\right)$puttingeq$\left(3\right)$ineq$\left(1\right)$$x+y=14=>\frac{48}{y}+y=14=>48+y^2=14y=>y^2-6y-8y+48=0=>(y-6)(y-8)=0$$y=6\&y=8$Thus remaining observations are $6$$8$$y=6\&y=8$$LCMof6\&8=2\times 3\times 4=24$