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A \(10\text{-kilogram}\) soil mixture containing \(30\%\) sand and \(70\%\) humus, by weight, is thoroughly mixed together. In order to create a \(10\text{-kilogram}\) mixture that contains \(50\%\) of each component, approximately how many kilograms of the soil mixture should be removed and replaced by pure sand?
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:
What we want to end up with: We want the same 10-kilogram bag, but with a different composition:
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
Now let's figure out the actual amounts in kilograms, since that's easier to work with than percentages.
Current mixture (10 kg total):
Target mixture (10 kg total):
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.
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?
When we add x kg of pure sand, what are we adding?
So the net change is:
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
\(7 + (-0.7\mathrm{x}) = 7 - 0.7(2.86) = 7 - 2 = 5 \text{ kg}\) ✓
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.
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.
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.
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).
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 \(2 \div 0.7 = 2 \div \frac{7}{10} = 2 \times \frac{10}{7}\)).
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.
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.
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.
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.