The moisture content of wet wood is the ratio of the weight of the water in the wood to the...

1. Translate the problem requirements: Understand that moisture content is defined as \(\frac{\mathrm{weight\,of\,water}}{\mathrm{weight\,of\,dry\,wood}} \times 100\%\). We need to find the weight of water in 12,000 pounds of wet wood with 20% moisture content.
2. Set up the relationship equation: Express the moisture content formula using the given information to create an equation with one unknown variable.
3. Solve for the weight of water: Use algebraic manipulation to isolate and calculate the weight of water from the moisture content equation.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what moisture content really means in everyday terms. Think of wet wood as having two parts: the actual wood material (dry wood) and the water absorbed into it.

The problem tells us that moisture content is calculated as: the weight of water divided by the weight of dry wood, then expressed as a percentage.

Let's verify this with the given example: 60 pounds of wet wood contains 30 pounds of water. This means the dry wood weighs \(60 - 30 = 30\) pounds. So moisture content = \(\frac{30\,\mathrm{pounds\,of\,water}}{30\,\mathrm{pounds\,of\,dry\,wood}} \times 100\% = 100\%\).

Now for our problem: We have 12,000 pounds of wet wood with 20% moisture content. We need to find how much of that 12,000 pounds is water.

Process Skill: TRANSLATE - Converting the problem's definition into a workable understanding

2. Set up the relationship equation

Let's think through this step by step using plain language:

If we call the weight of water "W" pounds, then:

• Weight of wet wood = 12,000 pounds
• Weight of water = W pounds
• Weight of dry wood = 12,000 - W pounds

Since moisture content is 20%, we know that:

The weight of water divided by the weight of dry wood equals 0.20

In mathematical terms: \(\mathrm{W} \div (12,000 - \mathrm{W}) = 0.20\)

3. Solve for the weight of water

Now let's solve this equation step by step:

Starting with: \(\mathrm{W} \div (12,000 - \mathrm{W}) = 0.20\)

Multiply both sides by (12,000 - W):

\(\mathrm{W} = 0.20 \times (12,000 - \mathrm{W})\)

Distribute the 0.20:

\(\mathrm{W} = 0.20 \times 12,000 - 0.20 \times \mathrm{W}\)

\(\mathrm{W} = 2,400 - 0.20\mathrm{W}\)

Add 0.20W to both sides:

\(\mathrm{W} + 0.20\mathrm{W} = 2,400\)

\(1.20\mathrm{W} = 2,400\)

Divide both sides by 1.20:

\(\mathrm{W} = 2,400 \div 1.20 = 2,000\)

Let's verify: If there are 2,000 pounds of water, then dry wood weighs \(12,000 - 2,000 = 10,000\) pounds. Moisture content = \(2,000 \div 10,000 = 0.20 = 20\%\) ✓

4. Final Answer

The weight of water in the wet wood is 2,000 pounds.

Looking at our answer choices, this corresponds to choice B: 2,000.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the moisture content definition: Students often confuse the definition and think moisture content = (weight of water) ÷ (total weight of wet wood) instead of the correct definition: moisture content = (weight of water) ÷ (weight of dry wood). This leads them to set up the equation as \(\mathrm{W} \div 12,000 = 0.20\), giving an incorrect answer of 2,400.

2. Incorrectly identifying what the 12,000 pounds represents: Some students may think the 12,000 pounds refers to the dry wood weight rather than the total wet wood weight, leading to a completely wrong setup where they treat 12,000 as the denominator in the moisture content ratio.

Errors while executing the approach

1. Algebraic manipulation errors: When solving \(\mathrm{W} \div (12,000 - \mathrm{W}) = 0.20\), students commonly make mistakes when cross-multiplying or distributing. For example, they might incorrectly write \(\mathrm{W} = 0.20 + (12,000 - \mathrm{W})\) instead of \(\mathrm{W} = 0.20 \times (12,000 - \mathrm{W})\).

2. Arithmetic errors in the final calculation: Students may correctly set up \(1.20\mathrm{W} = 2,400\) but then make calculation errors when dividing \(2,400 \div 1.20\), potentially getting 2,000 wrong or not recognizing that \(1.20 = \frac{6}{5}\), leading to incorrect final answers.

1. Selecting the wrong computed value: If students made the initial setup error (using total weight instead of dry weight), they would get 2,400 and select choice C instead of the correct choice B (2,000). They might feel confident because 2,400 appears as an answer choice, not realizing their conceptual error from the beginning.