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

Let:

• \( x_1 \) be the smallest number
• \( x_{10} \) be the largest number

Step 1: Use the given average conditions

The average of the largest 9 numbers is: \[ \frac{x_2 + x_3 + \dots + x_{10}}{9} = 47 \Rightarrow x_2 + x_3 + \dots + x_{10} = 423 \tag{1} \]

The average of the smallest 9 numbers is: \[ \frac{x_1 + x_2 + \dots + x_9}{9} = 42 \Rightarrow x_1 + x_2 + \dots + x_9 = 378 \tag{2} \]

Step 2: Subtract equations (1) and (2)

Subtracting (2) from (1): \[ (x_2 + \dots + x_{10}) - (x_1 + \dots + x_9) = 45 \Rightarrow x_{10} - x_1 = 45 \]

Step 3: Find total sum of 10 observations

From equation (1): \[ x_2 + x_3 + \dots + x_{10} = 423 \Rightarrow \text{Total sum} = x_1 + 423 \Rightarrow \text{Average} = \frac{x_1 + 423}{10} \]

Step 4: Calculate minimum and maximum average

• For minimum average: Use smallest possible value of \( x_1 = 2 \) \[ \text{Average} = \frac{423 + 2}{10} = \frac{425}{10} = 42.5 \]
• For maximum average: Use largest possible value of \( x_1 = 42 \) \[ \text{Average} = \frac{423 + 42}{10} = \frac{465}{10} = 46.5 \]

Step 5: Final result

\[ \text{Required difference} = 46.5 - 42.5 = \boxed{4} \]

Final Answer:

✅ The correct answer is: \[ \boxed{4} \] (Option A)

Answer 2

Let the 10 numbers be represented as: \[ x(1), x(2), x(3), ..., x(10) \] where \( x(1) \) is the smallest and \( x(10) \) is the largest.

Step 1: Use Given Averages

The average of the largest 9 numbers (excluding \( x(1) \)) is: \[ \frac{x(2) + x(3) + ... + x(10)}{9} = 47 \Rightarrow x(2) + x(3) + ... + x(10) = 47 \times 9 = 423 \tag{1} \]

Similarly, the average of the smallest 9 numbers (excluding \( x(10) \)) is: \[ \frac{x(1) + x(2) + x(3) + ... + x(9)}{9} = 42 \Rightarrow x(1) + x(2) + x(3) + ... + x(9) = 42 \times 9 = 378 \tag{2} \]

Step 2: Subtract Equations (1) and (2)

Subtracting (2) from (1): \[ x(10) - x(1) = 423 - 378 = 45 \Rightarrow x(10) = x(1) + 45 \]

Step 3: Total Sum of 10 Observations

From equation (1), we have: \[ x(2) + ... + x(10) = 423 \Rightarrow \text{Total sum} = x(1) + 423 \] \[ \Rightarrow \text{Average} = \frac{x(1) + 423}{10} \]

Step 4: Find Minimum and Maximum Averages

• To get the minimum average, minimize \( x(1) \). Let \( x(1) = 2 \Rightarrow x(10) = 47 \) \[ \text{Min average} = \frac{2 + 423}{10} = \frac{425}{10} = 42.5 \]
• To get the maximum average, maximize \( x(1) \). Let \( x(1) = 20 \Rightarrow x(10) = 65 \) \[ \text{Max average} = \frac{42 + 423}{10} = \frac{465}{10} = 46.5 \]

Step 5: Final Difference

\[ \text{Difference in averages} = 46.5 - 42.5 = \boxed{4} \]

Final Answer:

The correct answer is: \[ \boxed{4} \]