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

Answer: (A)

4

Answer 2

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 = $4$