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

• A. 4
• B. 3
• C. 5
• D. 6

Answer 1

The Correct Option is A

The correct answer is (A): \(4\)

Let \(x_1\)be the least number

\(x_{10}\) be the largest numbe

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

\(x_2+x_3+...x_9+x_{10} = 423 →\) (1)

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

\(x_1+x_2+....x_9 = 378 →\) (2)

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

Sum of \(10\) observations

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

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 = \(4\)

Answer 2

Let \(x(1)\) represent the smallest number and \(x(10)\) the greatest. 
Based on the condition provided in the question, we can now claim that
\(x(2)+x(3)+x(4)+........x(10)= 47\times9=423...................(1)\)
Likewise,  \(x(1)+x(2)+x(3)+x(4)................+x(9)= 42\times9=378...............(2)\)

 \(x(10)-x(1)=45\) by subtracting both equations. 
Currently, \(423+x(1)\) is the total of the 10 observations from equation (1). 
At this point, \( x(10)\) will have a minimum value of 47 and \(x(1)\) will have a minimum value of 2. 
Minimum average thus\(\frac{ 425}{10}=42.5 42\) is the maximum value of x (1). 
The highest average will therefore be \(\frac{465}{10}=46.5.\)
Therefore, the average difference will be \(46.5-42.5=4.\) which is the right response.