An agent was billed a total of $6.00 for n reams of paper. As a result of a price increase...

1. Translate the problem requirements: We have two bills, both for $6.00. The first bill was for \(\mathrm{n}\) reams at some price per ream. The second bill was for \(\mathrm{(n-4)}\) reams at a price that's $0.25 higher per ream. We need to find how many reams were in the second bill.
2. Set up the price-quantity relationship: Use the fact that total cost equals price per ream times number of reams for both scenarios, creating two equations with the same total cost.
3. Establish the connection between scenarios: Link the two situations by expressing how the price increase and quantity decrease are related through the constant $6.00 total.
4. Solve systematically: Use substitution to find the original number of reams, then calculate how many reams were in the second bill.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what happened in plain English first. An agent bought paper twice, spending exactly $6.00 each time. Between the first and second purchase, the price per ream went up by $0.25. Because of this price increase, the agent got 4 fewer reams for the same $6.00 on the second purchase.

We need to find how many reams were in the second bill.

Let's define our variables clearly:

• \(\mathrm{n}\) = number of reams in the first bill
• \(\mathrm{(n - 4)}\) = number of reams in the second bill
• \(\mathrm{p}\) = price per ream in the first bill
• \(\mathrm{(p + 0.25)}\) = price per ream in the second bill

Process Skill: TRANSLATE - Converting the word problem into mathematical relationships

2. Set up the price-quantity relationship

Since both bills totaled $6.00, we can write two equations based on the simple relationship: Total Cost = Price per ream × Number of reams

For the first bill: \(\mathrm{p \times n = 6.00}\)
For the second bill: \(\mathrm{(p + 0.25) \times (n - 4) = 6.00}\)

From the first equation, we can see that \(\mathrm{p = \frac{6.00}{n}}\). This tells us the original price per ream in terms of how many reams were purchased.

3. Establish the connection between scenarios

Now we'll substitute the expression for \(\mathrm{p}\) from the first equation into the second equation. This connects both purchases through the price increase and quantity decrease.

Substituting \(\mathrm{p = \frac{6.00}{n}}\) into the second equation:
\(\mathrm{(\frac{6.00}{n} + 0.25) \times (n - 4) = 6.00}\)

Let's expand this step by step:
\(\mathrm{(\frac{6.00}{n} + 0.25) \times (n - 4) = 6.00}\)
\(\mathrm{\frac{6.00}{n} \times (n - 4) + 0.25 \times (n - 4) = 6.00}\)
\(\mathrm{\frac{6.00(n - 4)}{n} + 0.25(n - 4) = 6.00}\)
\(\mathrm{6.00 - \frac{24.00}{n} + 0.25n - 1.00 = 6.00}\)

Simplifying:
\(\mathrm{5.00 - \frac{24.00}{n} + 0.25n = 6.00}\)
\(\mathrm{-\frac{24.00}{n} + 0.25n = 1.00}\)

4. Solve systematically

To solve this equation, let's multiply everything by \(\mathrm{n}\) to eliminate the fraction:
\(\mathrm{-24.00 + 0.25n^2 = 1.00n}\)
\(\mathrm{0.25n^2 - 1.00n - 24.00 = 0}\)

To make this easier to work with, let's multiply by 4 to eliminate decimals:
\(\mathrm{n^2 - 4n - 96 = 0}\)

Now we can factor this quadratic equation. We need two numbers that multiply to -96 and add to -4.
Thinking about factor pairs of 96: We need factors where one is 8 more than the other.
12 × 8 = 96, and 12 - 8 = 4, so we use -12 and +8:
\(\mathrm{(n - 12)(n + 8) = 0}\)

This gives us \(\mathrm{n = 12}\) or \(\mathrm{n = -8}\). Since we can't have negative reams, \(\mathrm{n = 12}\).

Therefore:

• First bill: 12 reams
• Second bill: 12 - 4 = 8 reams

Let's verify: If the first bill was for 12 reams at $6.00, the price was \(\mathrm{\frac{\$6.00}{12} = \$0.50}\) per ream.
For the second bill: \(\mathrm{8 \times (\$0.50 + \$0.25) = 8 \times \$0.75 = \$6.00}\) ✓

5. Final Answer

The second bill was for 8 reams of paper.

The answer is A. 8

Common Faltering Points

Errors while devising the approach

1. Misidentifying what the question is asking for

Students often confuse which bill they need to find. The question asks for "how many reams of paper" in the second bill, but students might set up their variables to solve for the first bill instead. This leads to correctly finding \(\mathrm{n = 12}\) but then answering 12 instead of 8.

2. Incorrect variable setup for the price increase

Students may misinterpret the price increase relationship. Instead of understanding that the new price is (original price + $0.25), they might think the total increase is $0.25 for the entire bill, leading to incorrect equations like: \(\mathrm{p \times n = 6.00}\) and \(\mathrm{p \times (n-4) = 6.25}\).

3. Misunderstanding the "4 fewer reams" constraint

Students might incorrectly interpret "4 fewer reams" as meaning the second bill had 4 reams total, rather than understanding it as a relative decrease from the first bill. This leads them to set up the equation as \(\mathrm{(p + 0.25) \times 4 = 6.00}\) instead of \(\mathrm{(p + 0.25) \times (n - 4) = 6.00}\).

Errors while executing the approach

1. Algebraic manipulation errors when expanding

When expanding \(\mathrm{(\frac{6.00}{n} + 0.25) \times (n - 4) = 6.00}\), students frequently make distribution errors. A common mistake is incorrectly distributing to get \(\mathrm{\frac{6.00(n-4)}{n} + 0.25n - 4 = 6.00}\) instead of the correct \(\mathrm{\frac{6.00(n-4)}{n} + 0.25(n-4) = 6.00}\), missing the multiplication of 0.25 by both terms.

2. Sign errors when rearranging the equation

After simplifying to \(\mathrm{5.00 - \frac{24.00}{n} + 0.25n = 6.00}\), students often make sign errors when moving terms. A typical mistake is getting \(\mathrm{-\frac{24.00}{n} + 0.25n = -1.00}\) instead of the correct \(\mathrm{-\frac{24.00}{n} + 0.25n = 1.00}\), which leads to the wrong quadratic equation.

3. Factoring errors in the quadratic equation

When factoring \(\mathrm{n^2 - 4n - 96 = 0}\), students might incorrectly identify factor pairs. They may confuse the signs and write \(\mathrm{(n + 12)(n - 8) = 0}\) instead of \(\mathrm{(n - 12)(n + 8) = 0}\), leading them to conclude \(\mathrm{n = 8}\) instead of \(\mathrm{n = 12}\) for the first bill.

Errors while selecting the answer

1. Selecting the value for the wrong bill

Even after correctly solving to find \(\mathrm{n = 12}\), students frequently select answer choice B (12) because they forget that \(\mathrm{n}\) represents the first bill, not the second bill that the question asks for. They fail to perform the final step of calculating \(\mathrm{n - 4 = 12 - 4 = 8}\).

2. Choosing the negative solution

If students made earlier errors leading to a different quadratic equation, they might end up with solutions where both values are positive (like 8 and some other positive number). In such cases, they might incorrectly choose the larger value thinking "more reams makes more sense" rather than checking which value actually satisfies the original problem constraints.