Question

\(x < y < z\)
 

Column A	Column B
\(\frac{x+y+z}{3}\)	\(z\)

Hint:

Remember the property of averages: the average of a set of numbers must lie between the smallest and largest numbers in the set. If the numbers are not all identical, the average will be strictly between the minimum and maximum.

Answer 1

Step 1: Understanding the Concept:
Column A represents the arithmetic mean (average) of three numbers \(x, y,\) and \(z\). Column B is the largest of these three numbers. The question compares the mean of a set of distinct numbers to its maximum value.
Step 2: Key Formula or Approach:
The arithmetic mean of a set of numbers is always less than the maximum value in the set, and greater than the minimum value, provided the numbers are not all equal.
Step 3: Detailed Explanation:
We are given the inequality \(x < y < z\).
This tells us that \(x, y,\) and \(z\) are distinct numbers, and \(z\) is the largest.
Let's analyze the expression for the mean in Column A. Since \(x < z\) and \(y < z\), we can write the following inequalities:
\[ x < z \] \[ y < z \] We also know that \(z = z\).
Adding these three inequalities:
\[ x + y + z < z + z + z \] \[ x + y + z < 3z \] Now, divide both sides by 3 (since 3 is a positive number, the inequality direction does not change):
\[ \frac{x+y+z}{3} < z \] This shows that the quantity in Column A is strictly less than the quantity in Column B.
Example with numbers:
Let \(x=1, y=2, z=3\). They satisfy \(x < y < z\).
Column A: \(\frac{1+2+3}{3} = \frac{6}{3} = 2\).
Column B: \(z = 3\).
Here, \(2 < 3\), so Column B is greater.
Step 4: Final Answer:
The arithmetic mean of three distinct numbers is always less than the largest number.
Column A = \(\frac{x+y+z}{3}\)
Column B = \(z\)
Therefore, Column A < Column B. The quantity in Column B is greater.