Jee Main 2024 31 01 2024 Shift2 - Question3
Question 3
Let mean and variance of 6 observations $a, b, 68$, $44,40,60$ be 55 and 194 . If $a>b$ then find $a+3 b$
(1) 211.83
(2) 201.59
(3) 189.57
(4) 198.87
Show Answer
Answer: (2)
Solution:
$\frac{a+b+68+44+40+60}{6}=55$
$212+a+b=330$
$\Rightarrow a+b=118$
$\frac{\sum x _i^{2}}{n}-(\bar{x})^{2}=194$
$\frac{a^{2}+b^{2}+(68)^{2}+(44)^{2}+(40)^{2}+(60)^{2}}{6}-(55)^{2}=194$
$=3219$
$11760+a^{2}+b^{2}=19314$
$\Rightarrow a^{2}+b^{2}=19314-11760$
$=7554$
$(a+b)^{2}-2 a b=7554$
From here $b=41.795$
$a+b=118$
$\Rightarrow a+b+2 b=118+83.59$
$=201.59$
BETA
