At the opening of business yesterday, there were 10 cereal boxes weighing 20 ounces each and 20 cereal boxes weighing...

1. Translate the problem requirements: We start with 10 boxes of 20 oz each and 20 boxes of 15 oz each. After selling 18 boxes total, the remaining boxes weigh 220 oz. We need to find how many of the sold boxes were 15 oz boxes.
2. Set up the constraint equations: Express the total boxes sold and the weight of unsold boxes using variables for the unknown quantities.
3. Solve using substitution: Use the weight constraint to find the number of each type of unsold box, then determine how many of each type were sold.
4. Verify against answer choices: Check that our solution makes logical sense and matches one of the given options.

Execution of Strategic Approach

1. Translate the problem requirements

Let's first understand what we're working with in plain English:

• We started with 10 boxes weighing 20 ounces each
• We also started with 20 boxes weighing 15 ounces each
• So we had 30 total boxes at the beginning
• 18 boxes were sold yesterday
• This means \(30 - 18 = 12\) boxes remain unsold
• These 12 unsold boxes weigh exactly 220 ounces total
• We need to find how many 15-ounce boxes were among the 18 sold boxes

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

2. Set up the constraint equations

Let's think about this step by step using variables for what we don't know:

Let's say:

• x = number of 20-ounce boxes that remain unsold
• y = number of 15-ounce boxes that remain unsold

Now we can write down what we know in simple equations:

• Total unsold boxes: \(\mathrm{x} + \mathrm{y} = 12\)
• Total weight of unsold boxes: \(20\mathrm{x} + 15\mathrm{y} = 220\)

These two equations capture all the constraints from the problem.

3. Solve using substitution

From the first equation, we know that \(\mathrm{y} = 12 - \mathrm{x}\)

Let's substitute this into the weight equation:

\(20\mathrm{x} + 15(12 - \mathrm{x}) = 220\)

Expanding: \(20\mathrm{x} + 180 - 15\mathrm{x} = 220\)

Simplifying: \(5\mathrm{x} + 180 = 220\)

Solving: \(5\mathrm{x} = 40\), so \(\mathrm{x} = 8\)

This means \(\mathrm{y} = 12 - 8 = 4\)

So we have:

• 8 twenty-ounce boxes remain unsold
• 4 fifteen-ounce boxes remain unsold

Now we can figure out how many were sold:

• Twenty-ounce boxes sold: \(10 - 8 = 2\)
• Fifteen-ounce boxes sold: \(20 - 4 = 16\)

Let's verify: \(2 + 16 = 18\) total boxes sold ✓

Process Skill: APPLY CONSTRAINTS - Using both the count and weight constraints simultaneously

4. Verify against answer choices

We found that 16 fifteen-ounce boxes were sold.

Let's double-check our work:

• Unsold boxes: 8 boxes of 20 oz + 4 boxes of 15 oz = \(8(20) + 4(15) = 160 + 60 = 220\) oz ✓
• Total boxes sold: \(2 + 16 = 18\) ✓
• All constraints satisfied

Final Answer: E. 16

The number of sold boxes weighing 15 ounces was 16.

Common Faltering Points

Errors while devising the approach

• Misinterpreting what the question is asking for: Students may confuse "number of sold boxes weighing 15 ounces" with "number of unsold boxes weighing 15 ounces." This leads them to set up the problem correctly but then provide the wrong final answer (4 instead of 16).
• Setting up variables for sold boxes instead of unsold boxes: Some students may define variables as "number of 20-ounce boxes sold" and "number of 15-ounce boxes sold," making the constraint equations more complex and error-prone, rather than using the simpler approach of defining variables for unsold boxes.
• Missing the connection between sold and unsold quantities: Students may fail to recognize that if they find how many boxes of each type remain unsold, they can easily determine how many were sold by subtracting from the original quantities (10 and 20 respectively).

Errors while executing the approach

• Arithmetic errors during substitution: When substituting \(\mathrm{y} = 12 - \mathrm{x}\) into the weight equation, students commonly make errors like: expanding \(15(12 - \mathrm{x})\) as \(180 + 15\mathrm{x}\) instead of \(180 - 15\mathrm{x}\), or incorrectly combining like terms (\(20\mathrm{x} - 15\mathrm{x} = 5\mathrm{x}\)).
• Calculation errors when finding sold quantities: After correctly finding that 8 twenty-ounce boxes and 4 fifteen-ounce boxes remain unsold, students may make subtraction errors: calculating \(10 - 8 = 3\) or \(20 - 4 = 15\) instead of the correct values 2 and 16.

Errors while selecting the answer

• Selecting the number of unsold boxes instead of sold boxes: The most common error is choosing 4 (the number of 15-ounce boxes that remained unsold) instead of 16 (the number of 15-ounce boxes that were sold), because students lose track of what the question is actually asking for.