CAT 2020 Exam Shift 2 Solved Paper

Let ${\mathit{x}}_1$ be the least number
${\mathit{x}}_{10}$ be the largest number
Given $\frac{{\mathit{x}}_2+{\mathit{x}}_3+...+{\mathit{x}}_{10}}{9}=47$
${\mathit{x}}_2+{\mathit{x}}_3+\cdot \mathit{s}{\mathit{x}}_9+{\mathit{x}}_{10}=423\to \left(1\right)$
$\frac{{\mathit{x}}_1+{\mathit{x}}_2....+{\mathit{x}}_9}{9}=42$
${\mathit{x}}_1+{\mathit{x}}_2+\cdot \mathit{s}{\mathit{x}}_9=378\to \left(2\right)$
$\left(1\right)-\left(2\right)={\mathit{x}}_{10}-{\mathit{x}}_1=45$
Sum of 10 observations
${\mathit{x}}_1+{\mathit{x}}_2+{\mathit{x}}_3+\cdot \mathit{s}{\mathit{x}}_{10}=423+{\mathit{x}}_1$
Since the minimum value of ${\mathit{x}}_{10}$ is 47, the minimum value of ${\mathit{x}}_1$ is 2, minimum average
$=\frac{423+2}{10}=42.5$
The maximum value of ${\mathit{x}}_1$ is 42 ,
Maximum average $=\frac{423+42}{10}=46.5$
Required difference $=4$