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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

- $4$
- $3$
- $5$
- $6$

## 1 Answer

1 vote

Given that, in a group of $10$ students, the mean of the lowest $9$ score is $42,$ while the mean of the highest $9$ score is $47.$

Let the score of group of $10$ students be $x_{1} < x_{2} < x_{3} < x_{4} < x_{5} < x_{6} < x_{7} < x_{8} < x_{9} < x_{10}. $

Now, $ \frac{ x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9}}{9} = 42 $

$ \Rightarrow x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} = 378 \quad \longrightarrow (1) $

And, $ \frac{ x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} + x_{10}}{9} = 47 $

$ \Rightarrow x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} + x_{10} = 423 \quad \longrightarrow (2) $

Mean of $10$ students $ = \frac{ x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} + x_{10}}{10} $

Maximum mean will occur when $x_{1} = 42 :$

$\frac{x_{1} + (x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} + x_{10} )}{10} = \frac{42+423}{10} = \frac{465}{10} = 46 \cdot 5 \quad [ \because \text{From equation (2)}] $

Minimum mean will occur when $x_{10} = 47 : $

$\frac{( x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9}) + x_{10}}{10} = \frac{378+47}{10} = \frac{425}{10} = 42 \cdot 5 \quad [ \because \text{From equation (1)}] $

Difference $ = 46 \cdot 5 – 42 \cdot 5 = 4. $

$\therefore$ The maximum possible mean exceeds the minimum possible mean by $4.$

Correct Answer$: \text{A}$

Let the score of group of $10$ students be $x_{1} < x_{2} < x_{3} < x_{4} < x_{5} < x_{6} < x_{7} < x_{8} < x_{9} < x_{10}. $

Now, $ \frac{ x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9}}{9} = 42 $

$ \Rightarrow x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} = 378 \quad \longrightarrow (1) $

And, $ \frac{ x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} + x_{10}}{9} = 47 $

$ \Rightarrow x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} + x_{10} = 423 \quad \longrightarrow (2) $

Mean of $10$ students $ = \frac{ x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} + x_{10}}{10} $

Maximum mean will occur when $x_{1} = 42 :$

$\frac{x_{1} + (x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} + x_{10} )}{10} = \frac{42+423}{10} = \frac{465}{10} = 46 \cdot 5 \quad [ \because \text{From equation (2)}] $

Minimum mean will occur when $x_{10} = 47 : $

$\frac{( x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9}) + x_{10}}{10} = \frac{378+47}{10} = \frac{425}{10} = 42 \cdot 5 \quad [ \because \text{From equation (1)}] $

Difference $ = 46 \cdot 5 – 42 \cdot 5 = 4. $

$\therefore$ The maximum possible mean exceeds the minimum possible mean by $4.$

Correct Answer$: \text{A}$