Question

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

• A. 16
• B. 24
• C. 20
• D. 14

Answer 1

Answer: (A)

16

Observation :-$(2,4,10,12,14,x,y)$

$mean=\frac{2+4+10+12+14+x+y}{7}$

$8=42+(x+y)$

$56=42+(x+y)$

$\Rightarrow x+y=14$....(1)

Variance $=\frac{\sum (mean-x;{)}^2}{N}$

$\Rightarrow 16=\frac{\left[\right(8-2{)}^2-(1-4{)}^2+(8-10{)}^2+(8-12{)}^2+(8-14{)}^2(8-x{)}^2+(8-y{)}^2]}{7}$

$\Rightarrow 112=36+16+4+16+36+(8-x{)}^2+(8-y{)}^2$

$\Rightarrow 112=108+(8-x{)}^2+(8-y{)}^2$

$(8-x{)}^2(8-y{)}^2=4$

$\therefore |8-x|=0$, $|8-y|=2$

$\therefore (x=10or6andy=8)or(x=8y=10or6)$

$x+y=14\Rightarrow (x,y)=(6,6)or(8,6)$

$LCM(x,y)=$ LCM of $6$ and $8=24$