Verified Solution • 2025–26 Session moderate

QJEEM24Ma64f3: JEE Main 2024 Mathematics - Class 11 JEEM24Ma64f3 Solution: Consider 10 observations such that and , where are positive…

Consider 10 observations x1,x2,…,x10x_1, x_2, \ldots, x_{10} such that ∑i=110(xi−α)=2\sum\limits_{i=1}^{10}\left(x_i-\alpha\right)=2 and ∑i=110(xi−β)2=40\sum\limits_{i=1}^{10}\left(x_i-\beta\right)^2=40 , where α,β\alpha, \beta are positive integers. Let the mean and the variance of the observations be 65\frac{6}{5} and 8425\frac{84}{25} respectively. Then βα\frac{\beta}{\alpha} is equal to :

Multiple Choice Options

  • A 2
  • B 1
  • C 52\frac{5}{2}
  • D 32\frac{3}{2}

Step-by-Step Verified Solution

We have given xˉ(\bar{x}( mean )=65)=\frac{6}{5}

amp; Variance =8425amp;∑i=110(xi−α)=2amp;⇒x1+x2+…+x10−10α=2amp;⇒x1+x2+…+x1010−α=210amp;⇒65−α=210amp;⇒α=1\begin{aligned}& \text { Variance }=\frac{84}{25} \\\\& \sum_{i=1}^{10}\left(x_i-\alpha\right)=2 \\\\& \Rightarrow x_1+x_2+\ldots+x_{10}-10 \alpha=2 \\\\& \Rightarrow \frac{x_1+x_2+\ldots+x_{10}}{10}-\alpha=\frac{2}{10} \\\\& \Rightarrow \frac{6}{5}-\alpha=\frac{2}{10} \\\\& \Rightarrow \alpha=1\end{aligned}

amp; and ∑i=110(xi−β)2=40amp;(x1−β)2+(x2−β)2+…+(x10−β)2=40amp;x12+x22+…+x102+10β2−2β(x1+x2+…+x10)=40amp;⇒x12+x22+…+x10210+β2−2β(x1+x2+…+x10)10=4amp;⇒x12+x22+…+x10210−3625+3625+β2−2β×65=4\begin{aligned} & \text { and } \sum_{i=1}^{10}\left(x_i-\beta\right)^2=40 \\\\ & \left(x_1-\beta\right)^2+\left(x_2-\beta\right)^2+\ldots+\left(x_{10}-\beta\right)^2=40 \\\\ & x_1^2+x_2^2+\ldots+x_{10}^2+10 \beta^2-2 \beta\left(x_1+x_2+\ldots+x_{10}\right)=40 \\\\ & \Rightarrow \frac{x_1^2+x_2^2+\ldots+x_{10}^2}{10}+\beta^2-\frac{2 \beta\left(x_1+x_2+\ldots+x_{10}\right)}{10}=4 \\\\ & \Rightarrow \frac{x_1^2+x_2^2+\ldots+x_{10}^2}{10}-\frac{36}{25}+\frac{36}{25}+\beta^2-2 \beta \times \frac{6}{5}=4\end{aligned}

[\left[\right. Variance =∑i=1nxi2n−(xˉ)2]\left.=\frac{\sum_{i=1}^n x_i^2}{n}-(\bar{x})^2\right]

amp;⇒8425+3625+β2−12β5−4=0amp;⇒12025+β2−12β5−4=0amp;⇒25β2−60β+20=0amp;⇒5β2−12β+4=0amp;⇒β=2,25\begin{aligned} & \Rightarrow \frac{84}{25}+\frac{36}{25}+\beta^2-\frac{12 \beta}{5}-4=0 \\\\ & \Rightarrow \frac{120}{25}+\beta^2-\frac{12 \beta}{5}-4=0 \\\\ & \Rightarrow 25 \beta^2-60 \beta+20=0 \\\\ & \Rightarrow 5 \beta^2-12 \beta+4=0 \\\\ & \Rightarrow \beta=2, \frac{2}{5}\end{aligned}

Take β=2\beta=2

βα=21=2\frac{\beta}{\alpha}=\frac{2}{1}=2

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring