JEE MAIN - Mathematics (2024 - 8th April Evening Shift - No. 21)
Let $$\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbf{N}$$ and $$\mathrm{a}< \mathrm{b}< \mathrm{c}$$. Let the mean, the mean deviation about the mean and the variance of the 5 observations $$9,25, a, b, c$$ be 18, 4 and $$\frac{136}{5}$$, respectively. Then $$2 a+b-c$$ is equal to ________
Answer
33
Explanation
$$\begin{aligned} & a, b, c \in N \\ & a< b < c \\ & \text { Mean }=18 \\ & \frac{9+25+a+b+c}{5}=18 \\ & 34+a+b+c=90 \\ & a+b+c=56 \end{aligned}$$
$$\begin{aligned} & \frac{|9-18|+|25-18|+|a-18|+|b-18|+|c-18|}{5}=4 \\ & 9+7+|a-18|+|b-18|+|c-18|=20 \\ & |a-18|+|b-18|+|c-18|=4 \\ & \frac{136}{5}=\frac{706+a^2+b^2+c^2}{5}-(18)^2 \\ & \Rightarrow 136=706+a^2+b^2+c^2-1620 \\ & \Rightarrow a^2+b^2+c^2=1050 \\ & \text { Consider } a<19 < b< c \\ & \text { Solving } a=17, b=19, c=20 \\ & 2 a+b-c \\ & 34+19-20 \\ & =33 \end{aligned}$$
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