40 30 25 25 90 50 20 How many of the seven numbers listed are greater than the median but...

1. Translate the problem requirements: We need to find how many numbers from the given set are simultaneously greater than the median (middle value when arranged in order) and less than the arithmetic mean (average of all numbers).
2. Organize and find the median: Arrange the seven numbers in ascending order to identify the middle (4th) value.
3. Calculate the arithmetic mean: Sum all seven numbers and divide by 7 to find the average.
4. Identify the target range and count: Determine which numbers fall between the median and mean, then count how many satisfy this condition.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this question is asking us to find. We have seven numbers: 40, 30, 25, 25, 90, 50, 20.

We need to count how many of these numbers satisfy TWO conditions at the same time:

• The number must be greater than the median (the middle value)
• The number must be less than the arithmetic mean (the average)

Think of it like finding numbers that fall in a specific "sweet spot" - bigger than the middle value but smaller than the average.

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

2. Organize and find the median

To find the median, we need to arrange our seven numbers from smallest to largest:

20, 25, 25, 30, 40, 50, 90

Since we have 7 numbers (an odd number), the median is simply the middle number - the 4th number in our ordered list.

Counting from the left: \(1\mathrm{st} = 20\), \(2\mathrm{nd} = 25\), \(3\mathrm{rd} = 25\), \(4\mathrm{th} = 30\)

So our \(\mathrm{median} = 30\)

3. Calculate the arithmetic mean

The arithmetic mean is just the average of all seven numbers. We add them all up and divide by 7:

\(\mathrm{Sum} = 40 + 30 + 25 + 25 + 90 + 50 + 20 = 280\)

\(\mathrm{Arithmetic\ mean} = 280 ÷ 7 = 40\)

4. Identify the target range and count

Now we need to find numbers that are:

• Greater than the median (30), AND
• Less than the mean (40)

So we're looking for numbers in the range: \(30 < \mathrm{number} < 40\)

Let's check each of our original numbers:

• 40: Is \(40 > 30\)? Yes. Is \(40 < 40\)? No. ❌
• 30: Is \(30 > 30\)? No. ❌
• 25: Is \(25 > 30\)? No. ❌
• 25: Is \(25 > 30\)? No. ❌
• 90: Is \(90 > 30\)? Yes. Is \(90 < 40\)? No. ❌
• 50: Is \(50 > 30\)? Yes. Is \(50 < 40\)? No. ❌
• 20: Is \(20 > 30\)? No. ❌

None of our numbers fall strictly between 30 and 40.

Process Skill: APPLY CONSTRAINTS - Systematically checking each number against both conditions

Final Answer

Zero numbers from our list are greater than the median (30) but less than the mean (40).

The answer is A. None

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the compound condition

Students often miss that the question asks for numbers that satisfy BOTH conditions simultaneously (greater than median AND less than mean). They might count numbers that satisfy either condition instead of requiring both conditions to be true at the same time.

2. Confusing median calculation for even vs odd datasets

With 7 numbers, students might incorrectly think they need to average two middle values (as required for even-numbered datasets) rather than simply taking the single middle value that applies to odd-numbered datasets.

Errors while executing the approach

1. Arithmetic errors in calculating the mean

When adding the seven numbers \((40 + 30 + 25 + 25 + 90 + 50 + 20 = 280)\), students commonly make addition errors, especially with the repeated 25 or when handling the larger number 90. This leads to an incorrect mean calculation.

2. Boundary condition errors when checking inequalities

Students frequently include boundary values incorrectly. For example, they might count 30 as satisfying "greater than 30" or count 40 as satisfying "less than 40" when these strict inequalities exclude the boundary values themselves.

3. Incorrect ordering during median calculation

When arranging numbers from smallest to largest, students may incorrectly place numbers, particularly when there are repeated values like the two 25s in this dataset.

Errors while selecting the answer

No likely faltering points - the final counting step is straightforward once the range is established and each number is properly evaluated against both conditions.