Four extra-large sandwiches of exactly the same size were ordered for m students, where m > 4. Three of the...

1. Translate the problem requirements: Clarify what "evenly divided" means for each scenario and identify that we need to find the total fraction Carol ate from all four sandwiches combined
2. Determine sandwich portions per student: Calculate Carol's share from the first three sandwiches (divided among m students) and from the fourth sandwich (divided among m-4 students)
3. Sum Carol's total consumption: Add the fractions from both scenarios to find her total intake as a fraction of one whole sandwich
4. Simplify the expression: Combine fractions over a common denominator and simplify to match the answer choice format

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what's happening in plain English. We have four identical extra-large sandwiches and m students (where m > 4).

For the first three sandwiches: Each sandwich is cut into m equal pieces, and each student gets one piece from each sandwich. So Carol gets 1 piece from sandwich #1, 1 piece from sandwich #2, and 1 piece from sandwich #3.

For the fourth sandwich: Here's where it gets different. Four students don't want any of this sandwich, so only (m - 4) students will share it. This sandwich is cut into (m - 4) equal pieces, and Carol gets 1 piece.

Our goal is to find what fraction of one whole extra-large sandwich Carol ate in total.

Process Skill: TRANSLATE - Converting the word problem into clear mathematical scenarios

2. Determine sandwich portions per student

Now let's figure out what fraction of a whole sandwich each piece represents.

For the first three sandwiches:

• Each sandwich is divided among m students
• So each piece = \(\frac{1}{\mathrm{m}}\) of a whole sandwich
• Carol gets one piece from each of these three sandwiches
• Carol's portion from first three sandwiches = \(\frac{1}{\mathrm{m}} + \frac{1}{\mathrm{m}} + \frac{1}{\mathrm{m}} = \frac{3}{\mathrm{m}}\)

For the fourth sandwich:

• This sandwich is divided among (m - 4) students
• So each piece = \(\frac{1}{\mathrm{m} - 4}\) of a whole sandwich
• Carol gets one piece from this sandwich
• Carol's portion from fourth sandwich = \(\frac{1}{\mathrm{m} - 4}\)

3. Sum Carol's total consumption

To find Carol's total consumption, we add up her portions from all four sandwiches:

Total = Portion from first three + Portion from fourth

Total = \(\frac{3}{\mathrm{m}} + \frac{1}{\mathrm{m} - 4}\)

To add these fractions, we need a common denominator. The common denominator is \(\mathrm{m}(\mathrm{m} - 4)\).

\(\frac{3}{\mathrm{m}} = \frac{3(\mathrm{m} - 4)}{\mathrm{m}(\mathrm{m} - 4)} = \frac{3\mathrm{m} - 12}{\mathrm{m}(\mathrm{m} - 4)}\)

\(\frac{1}{\mathrm{m} - 4} = \frac{\mathrm{m}}{\mathrm{m}(\mathrm{m} - 4)}\)

Total = \(\frac{3\mathrm{m} - 12}{\mathrm{m}(\mathrm{m} - 4)} + \frac{\mathrm{m}}{\mathrm{m}(\mathrm{m} - 4)} = \frac{3\mathrm{m} - 12 + \mathrm{m}}{\mathrm{m}(\mathrm{m} - 4)} = \frac{4\mathrm{m} - 12}{\mathrm{m}(\mathrm{m} - 4)}\)

4. Simplify the expression

Our final expression is \(\frac{4\mathrm{m} - 12}{\mathrm{m}(\mathrm{m} - 4)}\).

Let's verify this makes sense:

• The numerator \(4\mathrm{m} - 12 = 4(\mathrm{m} - 3)\), which represents Carol's total pieces
• The denominator \(\mathrm{m}(\mathrm{m} - 4)\) accounts for the different ways the sandwiches were divided

Looking at our answer choices, this matches exactly with choice E: \(\frac{4\mathrm{m} - 12}{\mathrm{m}(\mathrm{m} - 4)}\).

Final Answer

Carol ate \(\frac{4\mathrm{m} - 12}{\mathrm{m}(\mathrm{m} - 4)}\) of one whole extra-large sandwich.

This corresponds to answer choice E: \(\frac{4\mathrm{m} - 12}{\mathrm{m}(\mathrm{m} - 4)}\)

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the fourth sandwich division

Students often misread that "4 students did not want any of the fourth sandwich" and incorrectly assume the fourth sandwich is divided among 4 students instead of (m-4) students. This leads them to calculate Carol's portion from the fourth sandwich as \(\frac{1}{4}\) instead of \(\frac{1}{\mathrm{m}-4}\).

2. Confusing the constraint m > 4

Students may overlook or misunderstand why m > 4 is specified. This constraint ensures that (m-4) > 0, meaning there are actually students left to share the fourth sandwich. Without recognizing this, students might not realize the problem setup requires this condition for the scenario to make mathematical sense.

3. Misunderstanding what "one piece from each sandwich" means

Some students incorrectly interpret that Carol gets the same absolute amount from each sandwich, rather than understanding that she gets one piece from each sandwich where the piece sizes differ between the first three sandwiches and the fourth sandwich.

Errors while executing the approach

1. Common denominator errors when adding fractions

When adding \(\frac{3}{\mathrm{m}} + \frac{1}{\mathrm{m}-4}\), students frequently make mistakes in finding the common denominator \(\mathrm{m}(\mathrm{m}-4)\). Common errors include using (m-4) as the denominator or incorrectly expanding the numerators when converting to the common denominator.

2. Arithmetic mistakes in numerator calculations

Students often make calculation errors when converting \(\frac{3}{\mathrm{m}}\) to the common denominator, writing 3(m-4) as (3m-4) instead of (3m-12), or when combining the numerators (3m-12) + m, getting (4m+12) instead of (4m-12).

3. Sign errors in algebraic manipulation

When expanding 3(m-4), students sometimes get the sign wrong and write (3m+12) instead of (3m-12), which leads to a final answer of \(\frac{4\mathrm{m}+12}{\mathrm{m}(\mathrm{m}-4)}\) that doesn't match any of the given choices.

Errors while selecting the answer

1. Failing to match the exact form

Even if students arrive at the correct numerical expression \(\frac{4\mathrm{m}-12}{\mathrm{m}(\mathrm{m}-4)}\), they might not recognize that this exactly matches choice E due to differences in how the expression is written or parentheses are placed in the answer choices.

2. Selecting a similar-looking but incorrect choice

Students who make minor calculation errors might end up with expressions like \(\frac{4\mathrm{m}-4}{\mathrm{m}(\mathrm{m}-4)}\) and incorrectly select choice C, or \(\frac{4\mathrm{m}-8}{\mathrm{m}(\mathrm{m}-4)}\) and select choice D, because these look similar to their calculated result.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a strategic value for m

Since m > 4, let's choose m = 8. This value is selected because:

• It satisfies the constraint m > 4
• Both m = 8 and (m-4) = 4 are easy numbers to work with
• It will give us clean fractions that are easy to calculate

Step 2: Calculate Carol's portion from the first three sandwiches

Each of the first 3 sandwiches is divided among m = 8 students.

Carol gets \(\frac{1}{8}\) of each sandwich.

From 3 sandwiches: Carol gets \(3 \times \frac{1}{8} = \frac{3}{8}\) of a whole sandwich

Step 3: Calculate Carol's portion from the fourth sandwich

The 4th sandwich is divided among (m-4) = 8-4 = 4 students.

Carol gets \(\frac{1}{4}\) of this sandwich.

Step 4: Find Carol's total consumption

Total = \(\frac{3}{8} + \frac{1}{4}\)

Converting to common denominator: \(\frac{3}{8} + \frac{2}{8} = \frac{5}{8}\)

Step 5: Verify this matches the general formula

Using m = 8 in the correct answer choice E: \(\frac{4\mathrm{m}-12}{\mathrm{m}(\mathrm{m}-4)}\)

= \(\frac{4 \times 8-12}{8 \times (8-4)}\)

= \(\frac{32-12}{8 \times 4}\)

= \(\frac{20}{32}\)

= \(\frac{5}{8}\) ✓

Step 6: Verify with another value

Let's try m = 6 to confirm our approach:

First 3 sandwiches: Carol gets \(3 \times \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\)

Fourth sandwich: Carol gets \(\frac{1}{6-4} = \frac{1}{2}\)

Total: \(\frac{1}{2} + \frac{1}{2} = 1\)

Using formula: \(\frac{4 \times 6-12}{6 \times 2} = \frac{12}{12} = 1\) ✓

The smart numbers approach confirms that the answer is E.