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CAT 2020 Quant Question [Slot 2] with Solution 26

Question

In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47. For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by

1. 4
2. 3
3. 5
4. 6
Option: 1
Solution:

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

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