A 10text{-kilogram} soil mixture containing 30% sand and 70% humus, by weight, is thoroughly mixed together. In order to create...

1. Translate the problem requirements: We start with 10 kg of soil (\(30\%\) sand, \(70\%\) humus) and need to end with 10 kg of soil (\(50\%\) sand, \(50\%\) humus). We remove some amount of the original mixture and replace it with pure sand.
2. Calculate current and target component amounts: Determine how much sand and humus we currently have versus what we need in the final mixture.
3. Set up the replacement scenario: When we remove x kg of mixture and add x kg of pure sand, track how the sand and humus amounts change.
4. Create equation based on target composition: Use the fact that we need exactly 5 kg of sand in the final mixture to solve for the amount to replace.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we have and what we want in simple terms.

What we start with: We have a 10-kilogram bag of soil mixture. Think of it like a bag of trail mix, but instead of nuts and raisins, we have sand and humus (decomposed organic matter). Right now, this bag contains:

• \(30\%\) sand by weight
• \(70\%\) humus by weight

What we want to end up with: We want the same 10-kilogram bag, but with a different composition:

• \(50\%\) sand by weight
• \(70\%\) humus by weight

How we're going to do it: We'll scoop out some amount of the current mixture and replace that exact same weight with pure sand. Think of it like removing some trail mix and adding pure nuts to change the overall ratio.

Process Skill: TRANSLATE - Converting the word problem into a clear understanding of what needs to happen

2. Calculate current and target component amounts

Now let's figure out the actual amounts in kilograms, since that's easier to work with than percentages.

Current mixture (10 kg total):

• Sand: \(30\% \text{ of } 10 \text{ kg} = 3 \text{ kg of sand}\)
• Humus: \(70\% \text{ of } 10 \text{ kg} = 7 \text{ kg of humus}\)

Target mixture (10 kg total):

• Sand: \(50\% \text{ of } 10 \text{ kg} = 5 \text{ kg of sand}\)
• Humus: \(50\% \text{ of } 10 \text{ kg} = 5 \text{ kg of humus}\)

So we need to go from having 3 kg of sand to having 5 kg of sand. That means we need 2 more kilograms of sand in our final mixture.

3. Set up the replacement scenario

Let's say we remove x kilograms of the original mixture and replace it with x kilograms of pure sand.

When we remove x kg of original mixture, what are we taking out?

• Sand removed: \(30\% \text{ of } \mathrm{x} \text{ kg} = 0.3\mathrm{x} \text{ kg}\)
• Humus removed: \(70\% \text{ of } \mathrm{x} \text{ kg} = 0.7\mathrm{x} \text{ kg}\)

When we add x kg of pure sand, what are we adding?

• Sand added: \(\mathrm{x} \text{ kg}\) (since it's pure sand)
• Humus added: \(0 \text{ kg}\) (since it's pure sand)

So the net change is:

• Net sand change: \(+\mathrm{x} \text{ kg (added)} - 0.3\mathrm{x} \text{ kg (removed)} = +0.7\mathrm{x} \text{ kg}\)
• Net humus change: \(0 \text{ kg (added)} - 0.7\mathrm{x} \text{ kg (removed)} = -0.7\mathrm{x} \text{ kg}\)

4. Create equation based on target composition

We know we currently have 3 kg of sand, and we need to end up with 5 kg of sand.

Starting sand + Net sand change = Target sand
\(3 + 0.7\mathrm{x} = 5\)

Solving for x:
\(0.7\mathrm{x} = 5 - 3\)
\(0.7\mathrm{x} = 2\)
\(\mathrm{x} = 2 \div 0.7\)
\(\mathrm{x} = 2 \div \left(\frac{7}{10}\right)\)
\(\mathrm{x} = 2 \times \left(\frac{10}{7}\right)\)
\(\mathrm{x} = \frac{20}{7}\)
\(\mathrm{x} \approx 2.86\)

Let's verify this makes sense for humus too:
Starting humus + Net humus change = Target humus
\(7 + (-0.7\mathrm{x}) = 7 - 0.7(2.86) = 7 - 2 = 5 \text{ kg}\) ✓

5. Final Answer

We need to remove and replace approximately 2.86 kg of the soil mixture with pure sand.

Looking at our answer choices: ["2.0","2.9","3.4","4.0","5.0"]

2.86 is closest to 2.9, so the answer is B.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what "replace" means in the context

Students often think they need to add pure sand to the existing mixture rather than removing some mixture first and then adding sand. This leads to setting up equations where they simply add sand to make an 11kg or 12kg mixture instead of maintaining the 10kg total weight constraint.

2. Confusing the composition of what gets removed

When removing x kg of the original mixture, students might incorrectly assume they're removing pure sand or pure humus, rather than removing x kg of the \(30\%\)-\(70\%\) mixture. This leads to incorrect calculations of how much sand and humus are actually being taken out.

3. Setting up the wrong target equation

Students may focus on the humus changes instead of sand changes, or try to set up separate equations for both sand and humus without realizing that solving for one automatically gives the answer (since the total weight is constrained to 10kg).

Errors while executing the approach

1. Arithmetic errors with percentages and decimals

Common mistakes include calculating \(30\%\) of x as \(0.03\mathrm{x}\) instead of \(0.3\mathrm{x}\), or making errors when converting between fractions and decimals (like \(2 \div 0.7 = 2 \div \frac{7}{10} = 2 \times \frac{10}{7}\)).

2. Sign errors in net change calculations

Students often get confused about whether changes should be positive or negative, especially when calculating "Net sand change = +x - 0.3x = +0.7x". They might incorrectly write this as -0.7x or make other sign errors.

3. Incorrect fraction-to-decimal conversion

When solving \(\mathrm{x} = \frac{20}{7}\), students might make calculation errors, getting values like 2.6 or 3.2 instead of the correct 2.86, leading them to select the wrong answer choice.

Errors while selecting the answer

1. Rounding to the wrong answer choice

With \(\mathrm{x} \approx 2.86\), students might round to 3.4 (choice C) instead of 2.9 (choice B), especially if they're not careful about which value is actually closest to their calculated result.

2. Forgetting the "approximately" instruction

Students might try to get an exact answer and become frustrated when their calculation doesn't match any answer choice perfectly, not realizing that 2.86 is meant to be approximated to the nearest given option.