Of w people in a room, x people had a sandwich for lunch and y people had soup for lunch....

1. Translate the problem requirements: We need to find what fraction of w people had neither sandwich nor soup. We're given that x people had sandwiches, y people had soup, and z people had both.
2. Apply inclusion-exclusion principle to find overlap: Determine how many people had at least one item (sandwich or soup) by avoiding double-counting those who had both.
3. Calculate the complement: Find how many people had neither item by subtracting those who had at least one item from the total.
4. Express as a fraction: Convert the count of people with neither item into a fraction of the total people in the room.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're given and what we need to find in everyday terms.

Imagine you're at a lunch gathering with w people total. Some people had sandwiches, some had soup, some had both, and some had neither. Here's what we know:

• x people had a sandwich for lunch
• y people had soup for lunch
• z people had both a sandwich AND soup

What we need to find: The fraction of people who had NEITHER a sandwich nor soup.

Think of this as: (Number of people with neither item) ÷ (Total number of people)

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

2. Apply inclusion-exclusion principle to find overlap

Now let's figure out how many people had at least one item (sandwich or soup). This is where we need to be careful about double-counting.

If we just add \(\mathrm{x + y}\), we're counting the z people who had both items twice. Here's why:

• The x people who had sandwiches includes those z people who had both
• The y people who had soup also includes those same z people who had both

So when we add \(\mathrm{x + y}\), we've counted the z people twice. To get the correct count of people who had at least one item, we need to subtract z once:

People with at least one item = \(\mathrm{x + y - z}\)

This makes intuitive sense: we add all the sandwich eaters and all the soup eaters, then subtract the overlap to avoid double-counting.

Process Skill: VISUALIZE - Using set relationships to avoid double-counting

3. Calculate the complement

Now we can find how many people had neither item. If \(\mathrm{(x + y - z)}\) people had at least one item, then the remaining people had neither.

Since there are w people total:

People with neither item = \(\mathrm{w - (x + y - z)}\)

People with neither item = \(\mathrm{w - x - y + z}\)

Let's verify this makes sense: We start with all w people, subtract those who had sandwiches (x), subtract those who had soup (y), but then we need to add back z because we subtracted those people twice (once when we subtracted x and once when we subtracted y).

4. Express as a fraction

Finally, we convert this to a fraction of the total people in the room:

Fraction with neither = (People with neither) ÷ (Total people)

Fraction with neither = \(\mathrm{(w - x - y + z) ÷ w}\)

Fraction with neither = \(\frac{\mathrm{w - x - y + z}}{\mathrm{w}}\)

Final Answer

The fraction of people who had neither a sandwich nor soup is \(\frac{\mathrm{w - x - y + z}}{\mathrm{w}}\).

Looking at our answer choices, this matches choice D: \(\frac{\mathrm{w - x - y + z}}{\mathrm{w}}\).

We can verify this makes sense: if everyone had at least one item, then \(\mathrm{w = x + y - z}\), so our numerator would be \(\mathrm{w - (x + y - z) = 0}\), giving us \(\frac{0}{\mathrm{w}} = 0\), which is correct for "no one had neither item."

Common Faltering Points

Errors while devising the approach

Students often misunderstand what "z people had both a sandwich and soup" means in the context of the overall counts. They may think z is additional to x and y, not realizing that z represents the overlap - these z people are already counted within both the x sandwich eaters and the y soup eaters. This leads to incorrect setup where they treat z as separate from x and y.

When the problem asks for people who had "neither," students need to first find those who had "at least one item." However, many students get confused and try to calculate people who had "exactly one item" instead, leading them down the wrong path entirely. The complement of "neither" is "at least one," not "exactly one."

Some students misread what fraction they need to find. Instead of finding (people with neither)/(total people), they might try to find (people with neither)/(people with at least one item) or other incorrect ratios, leading to fundamentally wrong approaches.

Errors while executing the approach

Even when students understand they need to use inclusion-exclusion, they frequently make sign errors. The most common mistake is forgetting to subtract z when calculating people with at least one item, computing \(\mathrm{x + y}\) instead of \(\mathrm{x + y - z}\). This happens because they don't carefully track that z people appear in both the x count and y count.

When calculating people with neither item as \(\mathrm{w - (people\ with\ at\ least\ one)}\), students often make algebraic mistakes. They might write \(\mathrm{w - (x + y - z)}\) as \(\mathrm{w - x - y - z}\) instead of the correct \(\mathrm{w - x - y + z}\). The double negative when distributing the subtraction is a frequent source of error.

Students sometimes correctly calculate the number of people with neither item as \(\mathrm{w - x - y + z}\), but then forget that the question asks for a fraction. They might look for this expression among the answer choices without dividing by w, missing that they need \(\frac{\mathrm{w - x - y + z}}{\mathrm{w}}\).

Errors while selecting the answer

When students see multiple fractional answer choices, they might select options like \(\frac{\mathrm{w - x - y + z}}{\mathrm{w - z}}\) from choice A, thinking that since z appears in their work, it should also appear in the denominator. They don't recognize that for a fraction of total people, the denominator must be w (total people).

Even when students arrive at the correct numerical expression \(\mathrm{w - x - y + z}\), they might second-guess themselves about the signs and select choice C: \(\frac{\mathrm{w - x - y - z}}{\mathrm{w}}\), thinking "we need to subtract everything from w." They don't trust their inclusion-exclusion logic and make a last-minute error.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose convenient smart numbers
Let's select numbers that make the relationships clear and calculations simple:
• Total people in room: \(\mathrm{w = 100}\)
• People who had sandwiches: \(\mathrm{x = 60}\)
• People who had soup: \(\mathrm{y = 40}\)
• People who had both: \(\mathrm{z = 20}\)

Step 2: Apply inclusion-exclusion principle
People who had at least one item (sandwich OR soup):
= People with sandwiches + People with soup - People with both
= \(\mathrm{60 + 40 - 20 = 80}\) people

Step 3: Find people with neither item
People with neither sandwich nor soup:
= Total people - People with at least one item
= \(\mathrm{100 - 80 = 20}\) people

Step 4: Calculate the fraction
Fraction with neither = \(\frac{20}{100} = \frac{1}{5} = 0.2\)

Step 5: Verify with answer choices using our smart numbers
Substituting \(\mathrm{w=100, x=60, y=40, z=20}\) into each choice:

1. Choice A: \(\frac{100-60-40+40}{100-20} = \frac{40}{80} = 0.5\) ❌
2. Choice B: \(\frac{60+40+20}{100-20} = \frac{120}{80} = 1.5\) ❌
3. Choice C: \(\frac{100-60-40-20}{100} = \frac{-20}{100} = -0.2\) ❌
4. Choice D: \(\frac{100-60-40+20}{100} = \frac{20}{100} = 0.2\) ✓
5. Choice E: \(\frac{100-60-40+40}{100} = \frac{40}{100} = 0.4\) ❌

Conclusion: Only Choice D gives us the correct fraction of \(\mathrm{0.2}\), confirming our answer.