JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Statistics

Sum of the 6 observations $=6\times 55=330$, and the other four sum to $68+44+48+60=220$, so

$a+b=110$. \quad (1)

Variance condition: $\sum (x_i-55{)}^2=6\times 194=1164$.

Deviations of the known four: $13,-11,-7,5$, whose squares sum to $169+121+49+25=364$.

So $(a-55{)}^2+(b-55{)}^2=1164-364=800$.

Since $a+b=110$, write $a=55+p$, $b=55-p$ (so their sum stays 110); then

$p^2+p^2=800\Rightarrow p^2=400\Rightarrow p=\pm 20$.

Since $a>b$, take $p=20$: $a=75,b=35$.

$a+3b=75+3(35)=75+105=180$.