Q59 / 90 · JEE Main 2024 (08 Apr Shift 2) · Mathematics
Let $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{N}$ and $\mathrm{a} < \mathrm{b} < \mathrm{c}$. Let the mean, the mean deviation about the mean and the variance of the 5 observations $9,25, \mathrm{a}, \mathrm{b}, \mathrm{c}$ be 18,4 and $\frac{136}{5}$, respectively. Then $2 \mathrm{a}+\mathrm{b}-\mathrm{c}$ is equal to__________
Correct answer: 33
Solution
$a, b, c \in N \quad a < b < c$
$\overline{\mathrm{x}}=$ mean $=\frac{9+25+\mathrm{a}+\mathrm{b}+\mathrm{c}}{5}=18$
$a+b+c=56$
Mean deviation $=\frac{\Sigma\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{n}}=4$
$\begin{aligned} & =9+7+|18-\mathrm{a}|+|18-\mathrm{b}|+|18-\mathrm{c}|=20 \\ & =|18-\mathrm{a}|+|18-\mathrm{b}|+|18-\mathrm{c}|=4\end{aligned}$
Variance $=\frac{\Sigma\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|^2}{\mathrm{n}}=\frac{136}{5}$
$\begin{aligned} & =81+49+|18-a|^2+|18-b|^2+|18-c|^2=136 \\ & =(18-a)^2+(18-b)^2+(18-c)^2=6\end{aligned}$
Possible values $(18-\mathrm{a})^2=1, \quad(18-\mathrm{b})^2=1 \quad(18-\mathrm{c})^2=4$ $\mathrm{a} < \mathrm{b} < \mathrm{c}$
so $\quad \begin{array}{ccc}18-a=1 & 18-b=-1 & 18-c=-2 \\ \mathrm{a}=17 & \mathrm{~b}=19 & \mathrm{c}=20\end{array}$
$a+b+c=56$
$2 \mathrm{a}+\mathrm{b}-\mathrm{c} 34=19-20=33$
Hence, the correct answer is 33.