CAT 2020 Quant Question [Slot 2] with Solution 26

Let \({x_1}\) be the least number

\({x_{10}}\) be the largest number

Given \(\frac{{{x_2} + {x_3} +  \ldots  + {x_{10}}}}{9} = 47\)

\({x_2} + {x_3} +  \cdots {x_9} + {x_{10}} = 423 \to (1)\)

\(\frac{{{x_1} + {x_2} \cdots  + {x_9}}}{9} = 42\)

\({x_1} + {x_2} +  \cdots {x_9} = 378\quad  \to (2)\)

\((1) - (2) = {x_{10}} - {x_1} = 45\)

Sum of 10 observations

\({x_1} + {x_2} + {x_3} +  \cdots {x_{10}} = 423 + {x_1}\)

Since the minimum value of \({x_{10}}\) is 47, the minimum value of \({x_1}\) is 2, minimum average \( = \frac{{423 + 2}}{{10}} = 42.5\)

The maximum value of \({x_1}\) is 42 ,

Maximum average \( = \frac{{423 + 42}}{{10}} = 46.5\)

Required difference =4