The average (arithmetic mean) of the three integers x, y, and z is 10 and x leq y leq z....

1. Translate the problem requirements: We have three integers \(\mathrm{x, y, z}\) with average 10, ordered as \(\mathrm{x ≤ y ≤ z}\), difference \(\mathrm{z - x = 3}\), and we need to find which values (9, 10, or 11) could be the median of these three numbers.
2. Establish the constraint equations: Use the given average and difference conditions to express relationships between the variables in terms of the median y.
3. Apply the ordering constraints: Use the condition \(\mathrm{x ≤ y ≤ z}\) to determine what values of y are mathematically possible.
4. Test each median candidate: For each proposed median value, verify that integer solutions exist for x and z that satisfy all constraints.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in everyday language first. We have three numbers x, y, and z that are integers (whole numbers). When we add them up and divide by 3, we get 10 - so their average is 10. The numbers are arranged in order from smallest to largest: \(\mathrm{x ≤ y ≤ z}\). The difference between the largest and smallest is 3, so \(\mathrm{z - x = 3}\).

Since the three numbers are arranged in order \(\mathrm{x ≤ y ≤ z}\), the middle number y is the median. We need to figure out which of the values 9, 10, or 11 could possibly be this median.

Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding

2. Establish the constraint equations

Now let's turn our everyday understanding into mathematical relationships. Since the average of x, y, and z is 10, we know:
\(\mathrm{(x + y + z) ÷ 3 = 10}\)

This means: \(\mathrm{x + y + z = 30}\)

We also know that \(\mathrm{z - x = 3}\), which means \(\mathrm{z = x + 3}\).

Let's substitute this into our first equation:
\(\mathrm{x + y + (x + 3) = 30}\)
\(\mathrm{2x + y + 3 = 30}\)
\(\mathrm{2x + y = 27}\)

So we can express x in terms of y: \(\mathrm{x = (27 - y) ÷ 2}\)
And z in terms of y: \(\mathrm{z = x + 3 = (27 - y) ÷ 2 + 3 = (27 - y + 6) ÷ 2 = (33 - y) ÷ 2}\)

3. Apply the ordering constraints

For our numbers to be in the correct order \(\mathrm{x ≤ y ≤ z}\), we need both conditions to be satisfied:

First condition \(\mathrm{(x ≤ y): (27 - y) ÷ 2 ≤ y}\)
\(\mathrm{27 - y ≤ 2y}\)
\(\mathrm{27 ≤ 3y}\)
\(\mathrm{y ≥ 9}\)

Second condition \(\mathrm{(y ≤ z): y ≤ (33 - y) ÷ 2}\)
\(\mathrm{2y ≤ 33 - y}\)
\(\mathrm{3y ≤ 33}\)
\(\mathrm{y ≤ 11}\)

So mathematically, y can be any value from 9 to 11 (inclusive): \(\mathrm{9 ≤ y ≤ 11}\).

But wait! We also need x and z to be integers. Since \(\mathrm{x = (27 - y) ÷ 2}\), for x to be an integer, (27 - y) must be even. Since 27 is odd, y must also be odd.

Process Skill: APPLY CONSTRAINTS - Using the ordering requirements to limit possible solutions

4. Test each median candidate

Now let's check each proposed median value:

Testing y = 9:
\(\mathrm{x = (27 - 9) ÷ 2 = 18 ÷ 2 = 9}\)
\(\mathrm{z = (33 - 9) ÷ 2 = 24 ÷ 2 = 12}\)
Check: \(\mathrm{x = 9, y = 9, z = 12}\)
Ordering: \(\mathrm{9 ≤ 9 ≤ 12}\) ✓
Average: \(\mathrm{(9 + 9 + 12) ÷ 3 = 30 ÷ 3 = 10}\) ✓
Difference: \(\mathrm{12 - 9 = 3}\) ✓

Testing y = 10:
\(\mathrm{x = (27 - 10) ÷ 2 = 17 ÷ 2 = 8.5}\)
Since x is not an integer, y = 10 is impossible.

Testing y = 11:
\(\mathrm{x = (27 - 11) ÷ 2 = 16 ÷ 2 = 8}\)
\(\mathrm{z = (33 - 11) ÷ 2 = 22 ÷ 2 = 11}\)
Check: \(\mathrm{x = 8, y = 11, z = 11}\)
Ordering: \(\mathrm{8 ≤ 11 ≤ 11}\) ✓
Average: \(\mathrm{(8 + 11 + 11) ÷ 3 = 30 ÷ 3 = 10}\) ✓
Difference: \(\mathrm{11 - 8 = 3}\) ✓

Process Skill: CONSIDER ALL CASES - Systematically checking each possibility

4. Final Answer

The possible values for the median are 9 and 11, but not 10.
This corresponds to "I and III only".

The answer is D.

Common Faltering Points

Errors while devising the approach

1. Misidentifying the median: Students often forget that the median of three numbers arranged in order \(\mathrm{x ≤ y ≤ z}\) is simply the middle value y. They might try to calculate the median using formulas or think they need to find all three values first before determining the median.

2. Missing the integer constraint: Students may overlook that x, y, and z must be integers. They focus on the average and ordering constraints but forget to check whether their solutions produce integer values for all three variables.

3. Incomplete constraint analysis: Students might set up the basic equations (average = 10, \(\mathrm{z - x = 3}\)) but fail to fully utilize the ordering constraint \(\mathrm{x ≤ y ≤ z}\) to establish the range of possible values for the median.

Errors while executing the approach

1. Algebraic manipulation errors: When substituting \(\mathrm{z = x + 3}\) into \(\mathrm{x + y + z = 30}\), students may make errors in combining like terms or solving for variables. For example, incorrectly simplifying \(\mathrm{2x + y = 27}\) or making sign errors.

2. Inequality solving mistakes: Students often struggle with converting the ordering constraints \(\mathrm{x ≤ y ≤ z}\) into proper inequalities involving y. They might flip inequality signs incorrectly or make errors when multiplying or dividing by constants.

3. Incomplete verification: When testing each median value, students may check some conditions (like the average) but forget to verify all constraints, particularly whether the resulting x and z values are integers or whether the ordering \(\mathrm{x ≤ y ≤ z}\) is satisfied.

Errors while selecting the answer

1. Misreading Roman numeral format: Students might correctly identify that medians 9 and 11 work but 10 doesn't, yet still select the wrong answer choice because they misinterpret which Roman numerals (I, II, III) correspond to which values.

2. Partial answer selection: Students may find that y = 9 works and immediately select 'A. I only' without testing the other possible median values, missing that y = 11 also works.