JEE Main 2024 — 31 January, Shift 2 — Question 12

$a, b, 68, 44, 48, 60$

Mean $=55, \quad a>b$

Variance $=194, \quad a+3 b$

$\frac{a+b+68+44+48+60}{6}=55$

$\Rightarrow 220+a+b=330$

$\therefore \mathrm{a}+\mathrm{b}=110$.

Also,$\sum \frac{\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}}{\mathrm{n}}=194$

$\Rightarrow(\mathrm{a}-55)^{2}+(\mathrm{b}-55)^{2}+(68-55)^{2}+(44-55)^{2}+(48-55)^{2}+(60-55)^{2}=194 \times 6$

$\Rightarrow(\mathrm{a}-55)^{2}+(\mathrm{b}-55)^{2}+169+121+49+25=1164$

$\Rightarrow(\mathrm{a}-55)^{2}+(\mathrm{b}-55)^{2}=1164-364=800$

$a^{2}+3025-110 a+b^{2}+3025-110 b=800$

$\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=800-6050+12100$

$\mathrm{a}^{2}+\mathrm{b}^{2}=6850$

Solve (1) & (2);

$a=75, b=35$

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