Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and average (arithmetic mean)...

1. Translate the problem requirements: We need to understand what "average weight" means in terms of total weight, and clarify that we're removing boxes to change this average from 18 pounds to 14 pounds.
2. Determine the initial composition: Find how many 10-pound and 20-pound boxes we start with using the given average of 18 pounds.
3. Set up the target scenario: Calculate what the total weight and composition should be after removing some 20-pound boxes to achieve the new 14-pound average.
4. Find the difference: Compare the initial and final scenarios to determine exactly how many 20-pound boxes must be removed.

Execution of Strategic Approach

1. Translate the problem requirements

Let's think about what we're dealing with here. We have 30 boxes total, and they come in only two weights: 10 pounds and 20 pounds. The current average weight is 18 pounds per box.

Now, what does "average weight" actually mean? If we multiply the average weight by the number of boxes, we get the total weight of everything. So right now:

\(\mathrm{Total\,weight} = 30\,\mathrm{boxes} \times 18\,\mathrm{pounds\,per\,box} = 540\,\mathrm{pounds}\)

Our goal is to remove some 20-pound boxes so that the remaining boxes have an average weight of 14 pounds. This means we need to figure out how many boxes we'll have left and what their total weight will be.

Process Skill: TRANSLATE - Converting the average weight concept into concrete total weights

2. Determine the initial composition

Let's figure out how many 10-pound boxes and 20-pound boxes we start with. Let's say we have:

• x boxes that weigh 10 pounds each
• y boxes that weigh 20 pounds each

We know two things:

The total number of boxes: \(x + y = 30\)

The total weight: \(10x + 20y = 540\,\mathrm{pounds}\)

From the first equation: \(x = 30 - y\)

Substituting into the second equation:

\(10(30 - y) + 20y = 540\)

\(300 - 10y + 20y = 540\)

\(300 + 10y = 540\)

\(10y = 240\)

\(y = 24\)

So \(x = 30 - 24 = 6\)

This means we start with 6 boxes of 10 pounds and 24 boxes of 20 pounds.

3. Set up the target scenario

Now we need to figure out what happens when we remove some 20-pound boxes. The key insight is that we're only removing 20-pound boxes, so all the 10-pound boxes stay.

Let's say we remove n boxes that weigh 20 pounds each. After removal:

• We still have 6 boxes of 10 pounds
• We have (24 - n) boxes of 20 pounds
• Total boxes remaining: \(6 + (24 - n) = 30 - n\)
• Total weight remaining: \(6 \times 10 + (24 - n) \times 20 = 60 + 480 - 20n = 540 - 20n\)

We want the new average to be 14 pounds, so:

Average = Total weight ÷ Number of boxes

\(14 = \frac{540 - 20n}{30 - n}\)

Process Skill: INFER - Recognizing that only 20-pound boxes are removed, so 10-pound boxes remain constant

4. Find the difference

Now let's solve for n:

\(14 = \frac{540 - 20n}{30 - n}\)

\(14(30 - n) = 540 - 20n\)

\(420 - 14n = 540 - 20n\)

\(420 - 540 = 14n - 20n\)

\(-120 = -6n\)

\(n = 20\)

Let's verify this makes sense:

• Boxes remaining: \(30 - 20 = 10\,\mathrm{boxes}\)
• Weight remaining: \(540 - (20 \times 20) = 540 - 400 = 140\,\mathrm{pounds}\)
• New average: \(140 \div 10 = 14\,\mathrm{pounds}\) ✓

What boxes remain? We have 6 boxes of 10 pounds and 4 boxes of 20 pounds (since we removed 20 out of the original 24). Total: \(6 \times 10 + 4 \times 20 = 60 + 80 = 140\,\mathrm{pounds}\), with an average of \(140 \div 10 = 14\,\mathrm{pounds}\).

5. Final Answer

We must remove 20 of the 20-pound boxes.

The answer is D. 20

Common Faltering Points

Errors while devising the approach

• Misunderstanding what "average weight reduced to 14 pounds" means: Students might think they need to reduce each individual box's weight rather than understanding that removing heavier boxes will lower the overall average. They may try to somehow change the weights of existing boxes instead of removing boxes entirely.
• Overlooking the constraint that only 20-pound boxes can be removed: The problem specifically states that 20-pound boxes will be removed to reduce the average, but students might set up equations assuming both types of boxes are removed or that 10-pound boxes are removed instead.
• Confusing total weight with average weight relationships: Students may struggle to connect that average weight × number of boxes = total weight, and instead try to work directly with averages throughout the problem, making the solution much more complex than necessary.

Errors while executing the approach

• Arithmetic errors in the system of equations: When solving \(10x + 20y = 540\) and \(x + y = 30\), students commonly make mistakes like incorrectly distributing \(10(30 - y)\) or making sign errors when combining like terms, leading to wrong initial compositions.
• Setting up the final average equation incorrectly: Students may incorrectly write the equation as \(14 = \frac{540 - 20n}{30}\) instead of \(14 = \frac{540 - 20n}{30 - n}\), forgetting that the total number of boxes also decreases when boxes are removed.
• Algebraic manipulation errors: When solving \(14(30 - n) = 540 - 20n\), students frequently make mistakes in expanding, combining terms, or isolating n, such as getting the signs wrong when moving terms across the equals sign.

Errors while selecting the answer

• Selecting the number of boxes remaining instead of boxes removed: After calculating that n = 20, students might see that 10 boxes remain and mistakenly choose answer choice C (10) instead of the correct answer D (20), confusing what the question is actually asking for.