A store currently charges the same price for each towel that it sells. If the current price of each towel...

1. Translate the problem requirements: We need to find the current price per towel. We know that if the price increases by \(\$1\), then 10 fewer towels can be purchased for \(\$120\) total.
2. Set up the quantity relationship: Express how many towels can be bought at current price versus increased price using the \(\$120\) budget.
3. Create the equation from the constraint: Use the fact that there's a difference of exactly 10 towels between the two scenarios to form an equation.
4. Solve and verify using answer choices: Solve the equation and check that our answer produces the expected 10-towel difference.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we know in plain English:

• Currently, towels cost the same price each (let's call this price \(\mathrm{P}\))
• With \(\$120\), I can buy some number of towels at the current price
• If the price goes up by \(\$1\) (so each towel costs \(\mathrm{P} + \$1\)), then with the same \(\$120\), I can only buy 10 fewer towels

What we're looking for: the current price P of each towel.

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

2. Set up the quantity relationship

Let's think about this step by step:

• At current price P: Number of towels I can buy = \(\$120 \div \mathrm{P} = \frac{120}{\mathrm{P}}\) towels
• At increased price (P + 1): Number of towels I can buy = \(\$120 \div (\mathrm{P} + 1) = \frac{120}{\mathrm{P} + 1}\) towels

Since I can buy 10 fewer towels at the higher price, this means:

\(\frac{120}{\mathrm{P}} - \frac{120}{\mathrm{P} + 1} = 10\)

3. Create the equation from the constraint

Now let's solve this equation. We have:

\(\frac{120}{\mathrm{P}} - \frac{120}{\mathrm{P} + 1} = 10\)

To make this easier to work with, let's factor out 120:

\(120 \times \left(\frac{1}{\mathrm{P}} - \frac{1}{\mathrm{P} + 1}\right) = 10\)

Dividing both sides by 10:

\(12 \times \left(\frac{1}{\mathrm{P}} - \frac{1}{\mathrm{P} + 1}\right) = 1\)

To subtract these fractions, we need a common denominator of \(\mathrm{P}(\mathrm{P} + 1)\):

\(12 \times \frac{(\mathrm{P} + 1) - \mathrm{P}}{\mathrm{P}(\mathrm{P} + 1)} = 1\)

\(12 \times \frac{1}{\mathrm{P}(\mathrm{P} + 1)} = 1\)

\(\frac{12}{\mathrm{P}(\mathrm{P} + 1)} = 1\)

Cross-multiplying:

\(12 = \mathrm{P}(\mathrm{P} + 1)\)

\(12 = \mathrm{P}^2 + \mathrm{P}\)

\(\mathrm{P}^2 + \mathrm{P} - 12 = 0\)

Process Skill: MANIPULATE - Algebraic manipulation to create a solvable equation

4. Solve and verify using answer choices

We can factor this quadratic: \(\mathrm{P}^2 + \mathrm{P} - 12 = 0\)

Looking for two numbers that multiply to -12 and add to 1: those are 4 and -3

\((\mathrm{P} + 4)(\mathrm{P} - 3) = 0\)

So \(\mathrm{P} = -4\) or \(\mathrm{P} = 3\)

Since price cannot be negative, \(\mathrm{P} = 3\).

Let's verify this makes sense:

• At current price of \(\$3\): Number of towels = \(\frac{120}{3} = 40\) towels
• At increased price of \(\$4\): Number of towels = \(\frac{120}{4} = 30\) towels
• Difference: \(40 - 30 = 10\) towels ✓

This matches our constraint perfectly!

Final Answer

The current price of each towel is \(\$3\).

Answer choice: (C) \(\$3\)

Common Faltering Points

Errors while devising the approach

• Misinterpreting the direction of the relationship: Students often confuse which scenario results in fewer towels. They might incorrectly set up the equation as \(\frac{120}{\mathrm{P}+1} - \frac{120}{\mathrm{P}} = 10\), thinking the increased price allows buying MORE towels instead of fewer.
• Confusion about what "10 fewer" refers to: Some students might think "10 fewer" means the total cost is \(\$10\) less, rather than understanding it refers to 10 fewer towels in quantity.
• Setting up the wrong variable: Students might define their variable as the increased price \((\mathrm{P}+1)\) instead of the current price \(\mathrm{P}\), leading to confusion throughout the solution and requiring adjustment at the end.

Errors while executing the approach

• Arithmetic errors when finding common denominators: When working with \(\frac{1}{\mathrm{P}} - \frac{1}{\mathrm{P}+1}\), students commonly make mistakes in the fraction subtraction, either getting the wrong numerator or denominator in the combined fraction.
• Errors in cross-multiplication: After reaching \(\frac{12}{\mathrm{P}(\mathrm{P}+1)} = 1\), students may incorrectly cross-multiply to get \(12 = \mathrm{P} + 1\) instead of the correct \(12 = \mathrm{P}(\mathrm{P}+1)\), missing the multiplication between \(\mathrm{P}\) and \((\mathrm{P}+1)\).
• Factoring mistakes in the quadratic: When solving \(\mathrm{P}^2 + \mathrm{P} - 12 = 0\), students might incorrectly factor this as \((\mathrm{P}+3)(\mathrm{P}-4) = 0\) instead of the correct \((\mathrm{P}+4)(\mathrm{P}-3) = 0\), leading to wrong values for \(\mathrm{P}\).

Errors while selecting the answer

• Choosing the negative solution: After finding \(\mathrm{P} = 3\) or \(\mathrm{P} = -4\), some students might not recognize that price cannot be negative, or they might choose -4 thinking it represents a discount or decrease.
• Forgetting to verify the solution: Students might select \(\mathrm{P} = 3\) without checking that it actually satisfies the original constraint of "10 fewer towels," potentially missing computational errors made earlier.

Alternate Solutions

Smart Numbers Approach

Instead of setting up algebraic equations, we can test the answer choices as potential current prices and see which one satisfies the given constraint.

Step 1: Understanding the constraint
We know that when the price increases by \(\$1\), exactly 10 fewer towels can be bought for \(\$120\).

Step 2: Test each answer choice systematically

Testing \(\$3\) (Choice C):
• Current price = \(\$3\) per towel
• Towels bought at current price = \(\$120 \div \$3 = 40\) towels
• New price after increase = \(\$3 + \$1 = \$4\) per towel
• Towels bought at new price = \(\$120 \div \$4 = 30\) towels
• Difference = \(40 - 30 = 10\) towels ✓

Testing other choices for verification:

Testing \(\$2\) (Choice B):
• Current: \(\$120 \div \$2 = 60\) towels
• New: \(\$120 \div \$3 = 40\) towels
• Difference = \(20\) towels ≠ 10 ✗

Testing \(\$4\) (Choice D):
• Current: \(\$120 \div \$4 = 30\) towels
• New: \(\$120 \div \$5 = 24\) towels
• Difference = \(6\) towels ≠ 10 ✗

Step 3: Conclusion
Only \(\$3\) produces exactly the 10-towel difference specified in the problem.

Answer: C