JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Statistics

The five numbers $3,7,12,a,43-a$ always add to $65$, so their mean is $\frac{65}{5}=13$ regardless of $a$. The variance is $\frac{1}{5}\sum (x_i-13{)}^2$; the fixed entries contribute $(3-13{)}^2+(7-13{)}^2+(12-13{)}^2=100+36+1=137$, while the two variable entries contribute $(a-13{)}^2+(30-a{)}^2=2a^2-86a+1069$. Hence the variance is $\frac{2a^2-86a+1206}{5}$. For this to be a natural number the numerator must be divisible by $5$; reduced modulo $5$ it is $2a^2-a+1$, and the residues $a\equiv 0,1,2,3,4$ give $1,2,2,1,4$, none of them $0$. The numerator is therefore never a multiple of $5$, so the variance is never a natural number and no value of $a$ qualifies.