Last year members of a certain professional organization for teachers consisted of teachers from 49 different school districts, with an...

1. Translate the problem requirements: We need to find the total number of teacher members. We're given: 49 districts, average 9.8 schools per district, and average 22 teacher members per school. The answer choices are orders of magnitude apart (\(10^3\) to \(10^7\)), suggesting approximation will suffice.
2. Calculate total number of schools: Multiply districts by average schools per district to find the total schools in the organization.
3. Calculate total members using approximation: Multiply total schools by average members per school, rounding appropriately since answer choices differ by powers of 10.
4. Match to closest order of magnitude: Compare the calculated result to the given answer choices to identify the closest power of 10.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're given in everyday terms. We have a professional organization for teachers, and we want to find out how many total teacher members they had last year.

Here's what we know:

• There are 49 different school districts involved
• On average, each district has 9.8 schools (nearly 10 schools per district)
• On average, each school has 22 teacher members in the organization

We need to find the total number of teacher members across all these schools and districts.

Notice that the answer choices range from \(10^3\) (1,000) to \(10^7\) (10,000,000) - they're very far apart! This tells us we can use approximation and focus on getting the right order of magnitude.

Process Skill: TRANSLATE - Converting the problem description into the mathematical components we need to work with

2. Calculate total number of schools

To find the total number of schools, we multiply the number of districts by the average number of schools per district.

Think of it this way: if you have 49 districts, and each district contributes about 9.8 schools on average, then the total number of schools is:

\(49 \text{ districts} \times 9.8 \text{ schools per district}\)

Since we're looking for an approximate answer and 9.8 is very close to 10, let's use 10 for easier calculation:
\(49 \times 10 = 490\) schools

(The exact calculation would be \(49 \times 9.8 = 480.2\) schools, so our approximation of 490 is very close)

3. Calculate total members using approximation

Now we know there are approximately 490 schools in total, and each school has an average of 22 teacher members.

To find the total number of teacher members, we multiply:
Total schools × Average members per school
\(490 \text{ schools} \times 22 \text{ members per school}\)

Let's approximate 22 as 20 to make this easier:
\(490 \times 20 = 9,800\)

If we want to be more precise:
\(490 \times 22 = 10,780\)

Both calculations give us a result around 10,000, which is \(10^4\).

4. Match to closest order of magnitude

Our calculation gives us approximately 10,000 members.

Looking at our answer choices:

1. \(10^7 = 10,000,000\) (way too big)
2. \(10^6 = 1,000,000\) (too big)
3. \(10^5 = 100,000\) (too big)
4. \(10^4 = 10,000\) (matches our calculation!)
5. \(10^3 = 1,000\) (too small)

Our result of approximately 10,000 is closest to \(10^4\).

Final Answer

The total number of members of the organization last year is closest to \(10^4\) (which equals 10,000).

The answer is D.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the hierarchy of the data structure

Students may confuse the relationship between districts, schools, and teachers. They might try to multiply districts directly by teachers (\(49 \times 22\)) instead of recognizing the three-level hierarchy: districts → schools → teachers. This leads to drastically underestimating the total.

2. Missing the significance of "average" in the problem

Students may overlook that both "9.8 schools per district" and "22 teachers per school" represent averages that need to be multiplied together. They might incorrectly think these are totals rather than per-unit averages, leading to wrong calculation setup.

Errors while executing the approach

1. Over-precision in calculations when approximation is more appropriate

Given that answer choices are orders of magnitude apart (\(10^3\) to \(10^7\)), students may waste time on exact calculations (like \(49 \times 9.8 \times 22 = 10,585.6\)) instead of using smart approximations (\(50 \times 10 \times 20 = 10,000\)), potentially making arithmetic errors in the process.

2. Arithmetic errors in multi-step multiplication

Students may correctly set up \(49 \times 9.8 \times 22\) but make computational mistakes, especially when handling decimals (9.8). Common errors include miscalculating \(49 \times 9.8\) or making errors in the final multiplication step.

1. Confusing powers of 10 with the actual numbers

Even after correctly calculating approximately 10,000, students may select \(10^3\) (thinking "it's around 10-something") instead of recognizing that \(10,000 = 10^4\). They may focus on the base number rather than counting the zeros correctly.