A merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators...

1. Translate the problem requirements: Clarify what 'average cost per calculator' means, how the selling price relates to this average cost, and what 'total revenue was $120 more than cost' tells us about profit
2. Set up the cost and pricing relationship: Express the average cost per calculator and determine the actual selling price for each calculator sold
3. Calculate total revenue from sales: Account for the fact that 2 calculators were kept as demonstrators, so only (x-2) calculators were actually sold
4. Apply the profit condition: Use the given information that total revenue exceeded total cost by exactly $120 to create an equation
5. Solve for the number of calculators: Solve the resulting equation and verify the answer makes practical sense

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is telling us in plain English:

• A merchant bought x calculators for a total of $300
• The "average cost per calculator" means we divide the total cost by the number of calculators: \(\$300 ÷ \mathrm{x}\)
• The merchant kept 2 calculators as demonstrators (didn't sell them)
• Each calculator that was sold had a selling price of "$5 more than the average cost"
• The total money received from sales was $120 more than what the merchant originally paid

So if the average cost per calculator is \(\$300/\mathrm{x}\), then each calculator was sold for: \((\$300/\mathrm{x}) + \$5\)

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

2. Set up the cost and pricing relationship

Now let's establish our key relationships:

• Total cost of shipment = $300
• Number of calculators in shipment = x
• Average cost per calculator = \(\$300/\mathrm{x}\)
• Selling price per calculator = \((\$300/\mathrm{x}) + \$5\)
• Number of calculators actually sold = \(\mathrm{x} - 2\) (since 2 were kept as demonstrators)

This gives us a clear picture of how the costs and prices relate to each other.

3. Calculate total revenue from sales

To find the total revenue, we multiply:

• Number of calculators sold × Selling price per calculator
• Total revenue = \((\mathrm{x} - 2) × [(\$300/\mathrm{x}) + \$5]\)

Let's expand this step by step:

Total revenue = \((\mathrm{x} - 2) × (\$300/\mathrm{x} + \$5)\)
Total revenue = \((\mathrm{x} - 2) × \$300/\mathrm{x} + (\mathrm{x} - 2) × \$5\)
Total revenue = \(\$300(\mathrm{x} - 2)/\mathrm{x} + \$5(\mathrm{x} - 2)\)
Total revenue = \(\$300 - \$600/\mathrm{x} + \$5\mathrm{x} - \$10\)
Total revenue = \(\$290 + \$5\mathrm{x} - \$600/\mathrm{x}\)

4. Apply the profit condition

We're told that total revenue was $120 more than the cost:
Total revenue = Total cost + $120
Total revenue = $300 + $120 = $420

So we can set up our equation:
\(\$290 + \$5\mathrm{x} - \$600/\mathrm{x} = \$420\)

Simplifying:
\(\$5\mathrm{x} - \$600/\mathrm{x} = \$130\)

Multiplying everything by x to eliminate the fraction:
\(\$5\mathrm{x}^2 - \$600 = \$130\mathrm{x}\)
\(\$5\mathrm{x}^2 - \$130\mathrm{x} - \$600 = 0\)

Dividing by 5:
\(\mathrm{x}^2 - 26\mathrm{x} - 120 = 0\)

Process Skill: MANIPULATE - Algebraic manipulation to solve for the unknown

5. Solve for the number of calculators

We can factor this quadratic equation:
\(\mathrm{x}^2 - 26\mathrm{x} - 120 = 0\)

Looking for two numbers that multiply to -120 and add to -26:
We need factors of 120: 1×120, 2×60, 3×40, 4×30, 5×24, 6×20, 8×15, 10×12

Trying 30 and -4: 30 × (-4) = -120 and 30 + (-4) = 26 ✓

So: \((\mathrm{x} - 30)(\mathrm{x} + 4) = 0\)

This gives us x = 30 or x = -4

Since we can't have a negative number of calculators, x = 30.

4. Final Answer

Let's verify our answer makes sense:

• 30 calculators cost $300 total, so $10 per calculator on average
• Selling price = $10 + $5 = $15 per calculator
• Number sold = 30 - 2 = 28 calculators
• Total revenue = \(28 × \$15 = \$420\)
• Profit = $420 - $300 = $120 ✓

The answer is E. 30

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what "average cost" refers to

Students may confuse the average cost per calculator (which is \(\$300/\mathrm{x}\)) with some other cost metric. They might think the selling price is $5 more than the cost of the entire shipment ($300) rather than $5 more than the cost per individual calculator.

2. Incorrectly setting up the revenue calculation

Students often forget that only (x-2) calculators were sold, not all x calculators. They may set up their revenue equation as \(\mathrm{x} × [(\$300/\mathrm{x}) + \$5]\) instead of the correct \((\mathrm{x}-2) × [(\$300/\mathrm{x}) + \$5]\), missing the fact that 2 calculators were kept as demonstrators.

3. Misinterpreting the profit condition

Students may misunderstand "$120 more than the cost" and set up the equation as Revenue - Cost = $120 but then use the wrong values. Some might think the $120 refers to profit per calculator rather than total additional revenue above the original cost.

Errors while executing the approach

1. Algebraic manipulation errors when expanding

When expanding \((\mathrm{x}-2) × [(\$300/\mathrm{x}) + \$5]\), students commonly make distribution errors. They might incorrectly expand to get \(\$300(\mathrm{x}-2) + \$5\mathrm{x}\) instead of the correct \(\$300(\mathrm{x}-2)/\mathrm{x} + \$5(\mathrm{x}-2)\), forgetting that the \(\$300/\mathrm{x}\) term needs to be multiplied by (x-2) and then divided by x.

2. Errors when clearing fractions

When multiplying the equation \(\$5\mathrm{x} - \$600/\mathrm{x} = \$130\) by x to eliminate fractions, students may forget to multiply every term by x, leading to incorrect equations like \(\$5\mathrm{x}^2 - \$600 = \$130\) instead of the correct \(\$5\mathrm{x}^2 - \$600 = \$130\mathrm{x}\).

3. Factoring the quadratic equation incorrectly

For the equation \(\mathrm{x}^2 - 26\mathrm{x} - 120 = 0\), students may struggle to find the correct factor pairs. They might try factors that multiply to 120 but forget about the negative sign, or find factors that multiply correctly but don't add to -26.

Errors while selecting the answer

1. Choosing the wrong root of the quadratic

When solving \((\mathrm{x} - 30)(\mathrm{x} + 4) = 0\), some students might not immediately recognize that x = -4 is impossible in this context (can't have negative calculators) and may second-guess their answer or make computational errors trying to "fix" the negative solution.

2. Failing to verify the answer

Students may arrive at x = 30 but not check their work by substituting back into the original conditions. Without verification, they might not catch earlier errors and could select an incorrect answer choice that seemed reasonable during their calculations.