A group of n students is divided into two classes in such a way that one class has one more...

1. Translate the problem requirements: We have n total students divided into exactly two classes, where one class has one more student than the other. We need to find how many students are in the larger class.
2. Set up the relationship between the two classes: If we call the smaller class size x, then the larger class has (x + 1) students, and these must sum to the total n.
3. Solve for the smaller class size: Use the equation \(\mathrm{x} + (\mathrm{x} + 1) = \mathrm{n}\) to find x in terms of n.
4. Determine the larger class size: Add 1 to the smaller class size to get the final answer and verify against the answer choices.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're being asked in everyday terms. We have a total of n students, and we need to split them into exactly two classes. The key constraint is that one class must have exactly one more student than the other class - not two more, not three more, but exactly one more.

Think of it like dividing cookies between two children where one child gets exactly one more cookie than the other. If we had 10 cookies total, one child would get 4 cookies and the other would get 6 cookies (which is 4 + 1 = 5, wait that's 4 + 5 = 9... let me reconsider: one gets 5, the other gets 6, for a total of 11... Actually, if we have 10 cookies, one child gets 5 and the other gets 5, but we need one more. So one gets 4, the other gets 6. That's 4 + 6 = 10 total, and 6 - 4 = 2, which is too much difference.)

Let me restart with a clearer example: If we have 9 students total, we could have 4 students in one class and 5 students in the other class. The larger class (5 students) has exactly one more than the smaller class (4 students).

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

2. Set up the relationship between the two classes

Now let's think about this systematically. If the smaller class has some number of students, let's call that number x. Since the larger class has exactly one more student than the smaller class, the larger class has \((\mathrm{x} + 1)\) students.

Since all n students must be divided between these two classes:
Students in smaller class + Students in larger class = Total students
\(\mathrm{x} + (\mathrm{x} + 1) = \mathrm{n}\)

This makes perfect sense - we're just saying that when we add up all the students in both classes, we get our original total of n students.

3. Solve for the smaller class size

Now we solve our equation to find x (the number of students in the smaller class):

\(\mathrm{x} + (\mathrm{x} + 1) = \mathrm{n}\)
\(\mathrm{x} + \mathrm{x} + 1 = \mathrm{n}\)
\(2\mathrm{x} + 1 = \mathrm{n}\)
\(2\mathrm{x} = \mathrm{n} - 1\)
\(\mathrm{x} = \frac{\mathrm{n} - 1}{2}\)

So the smaller class has \(\frac{\mathrm{n} - 1}{2}\) students.

Let's verify this makes sense with our earlier example: If n = 9, then \(\mathrm{x} = \frac{9 - 1}{2} = \frac{8}{2} = 4\). So the smaller class has 4 students, which matches what we found before!

4. Determine the larger class size

Since the larger class has exactly one more student than the smaller class:

Students in larger class = \(\mathrm{x} + 1\)
Students in larger class = \(\frac{\mathrm{n} - 1}{2} + 1\)

To add 1 to this fraction, we need to express 1 as a fraction with the same denominator:
Students in larger class = \(\frac{\mathrm{n} - 1}{2} + \frac{2}{2}\)
Students in larger class = \(\frac{\mathrm{n} - 1 + 2}{2}\)
Students in larger class = \(\frac{\mathrm{n} + 1}{2}\)

Let's verify with our example: If n = 9, then larger class = \(\frac{9 + 1}{2} = \frac{10}{2} = 5\) students. Perfect! This matches our earlier finding.

Looking at the answer choices, \(\frac{\mathrm{n} + 1}{2}\) corresponds exactly to choice (A).

Final Answer

The number of students in the larger class is \(\frac{\mathrm{n} + 1}{2}\).

Answer: (A) \(\frac{\mathrm{n} + 1}{2}\)

We can verify this works for both even and odd values of n:

• If n = 8: smaller class = 3.5... wait, this suggests we need to be more careful about whether n is even or odd, but the answer \(\frac{\mathrm{n} + 1}{2} = 4.5\) doesn't make sense for counting students.
• If n = 9: larger class = \(\frac{9 + 1}{2} = 5\), smaller class = 4 ✓
• If n = 10: larger class = \(\frac{10 + 1}{2} = 5.5\)... this also seems problematic.

Actually, let me reconsider. The problem states that n students CAN be divided this way, which implies that such a division is always possible. For this to work with whole numbers of students, when n is odd, we get integer values for both classes. When n is even, the problem setup wouldn't yield integer solutions, but since we're asked for the answer "in terms of n," we provide the algebraic expression \(\frac{\mathrm{n} + 1}{2}\), understanding that this applies when such a division is possible.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "one more student than the other"
Students may think this means the classes can have any difference in size, or they might set up the problem with a difference of 2 students instead of exactly 1. The phrase "one more" specifically means a difference of exactly 1 student, not "at least one more" or "a few more."

2. Incorrectly assuming which class to solve for
Students might set up their equation to solve for the larger class directly instead of working systematically. This can lead to confused variable assignments and incorrect equations. The most reliable approach is to define the smaller class as x, then express the larger class as (x + 1).

3. Not recognizing that this is a constraint problem
Some students might try to solve this by testing the answer choices with specific numbers instead of setting up algebraic relationships. While testing can be useful for verification, the systematic algebraic approach ensures we capture the general relationship for any value of n.

Errors while executing the approach

1. Arithmetic errors when combining fractions
When calculating \(\frac{\mathrm{n}-1}{2} + 1\), students often struggle with adding 1 to a fraction. They might write \(\frac{\mathrm{n}-1}{2} + 1 = \frac{\mathrm{n}-1+1}{2} = \frac{\mathrm{n}}{2}\), forgetting that 1 must be converted to \(\frac{2}{2}\) before adding. The correct calculation is \(\frac{\mathrm{n}-1}{2} + \frac{2}{2} = \frac{\mathrm{n}+1}{2}\).

2. Sign errors in algebraic manipulation
When solving \(2\mathrm{x} + 1 = \mathrm{n}\) for x, students might incorrectly write \(2\mathrm{x} = \mathrm{n} + 1\) instead of \(2\mathrm{x} = \mathrm{n} - 1\). This happens because they add 1 to both sides instead of subtracting 1 from both sides to isolate the 2x term.

3. Forgetting to find the larger class size
Students might solve correctly for \(\mathrm{x} = \frac{\mathrm{n}-1}{2}\) (the smaller class) but then forget that the question asks for the larger class. They might select answer choice (D) which represents the smaller class, instead of adding 1 to get the larger class size.

Errors while selecting the answer

1. Confusing similar-looking answer choices
Answer choices (A) \(\frac{\mathrm{n}+1}{2}\) and (B) \(\frac{\mathrm{n}}{2} + 1\) are mathematically equivalent but look different. Students might correctly derive \(\frac{\mathrm{n}+1}{2}\) but then select choice (B) thinking it looks more like their work, or vice versa. They should verify that these expressions are the same: \(\frac{\mathrm{n}}{2} + 1 = \frac{\mathrm{n}}{2} + \frac{2}{2} = \frac{\mathrm{n}+2}{2}\).

2. Selecting the wrong class size
After correctly finding that the smaller class has \(\frac{\mathrm{n}-1}{2}\) students and the larger class has \(\frac{\mathrm{n}+1}{2}\) students, students might accidentally select choice (D) which represents the smaller class instead of choice (A) which represents the larger class that the question asks for.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose convenient concrete values

Since we need to divide n students into two classes where one has exactly one more than the other, let's test with specific values of n that make the arithmetic clear.

Let's start with n = 10 students (chosen because it's even, making it easy to see the division pattern).

Step 2: Apply the constraint

With 10 total students divided into two classes where one has one more than the other:

• Smaller class: x students
• Larger class: (x + 1) students
• Total: \(\mathrm{x} + (\mathrm{x} + 1) = 10\)
• Solving: \(2\mathrm{x} + 1 = 10\), so \(2\mathrm{x} = 9\), therefore \(\mathrm{x} = 4.5\)

Wait - we can't have half students! Let's try n = 11.

Step 3: Test with n = 11

With 11 total students:

• Smaller class: x students
• Larger class: (x + 1) students
• Total: \(\mathrm{x} + (\mathrm{x} + 1) = 11\)
• Solving: \(2\mathrm{x} + 1 = 11\), so \(2\mathrm{x} = 10\), therefore \(\mathrm{x} = 5\)
• Larger class = x + 1 = 5 + 1 = 6 students

Step 4: Verify with answer choices

For n = 11, the larger class should have 6 students. Let's check each answer choice:

• (A) \(\frac{\mathrm{n}+1}{2} = \frac{11+1}{2} = \frac{12}{2} = 6\) ✓
• (B) \(\frac{\mathrm{n}}{2} + 1 = \frac{11}{2} + 1 = 5.5 + 1 = 6.5\) ✗
• (C) \(\frac{\mathrm{n}}{2} = \frac{11}{2} = 5.5\) ✗
• (D) \(\frac{\mathrm{n}-1}{2} = \frac{11-1}{2} = \frac{10}{2} = 5\) ✗
• (E) \(\frac{2\mathrm{n}-1}{2} = \frac{22-1}{2} = \frac{21}{2} = 10.5\) ✗

Step 5: Confirm with another value

Let's verify with n = 9:

• \(2\mathrm{x} + 1 = 9\), so \(\mathrm{x} = 4\)
• Larger class = 4 + 1 = 5 students
• (A) \(\frac{\mathrm{n}+1}{2} = \frac{9+1}{2} = \frac{10}{2} = 5\) ✓

The answer is (A).