David used part of $100,000 to purchase a house. Of the remaining portion, he invested 1/3 of it at 4%...

1. Translate the problem requirements: David starts with \(\$100,000\), buys a house for an unknown amount, then invests the remaining money in two parts - \(\frac{1}{3}\) at \(4\%\) and \(\frac{2}{3}\) at \(6\%\) simple interest. After one year, total interest earned is \(\$320\). We need to find the house purchase price.
2. Set up the money flow: Define the remaining amount after house purchase and express the two investment portions in terms of this amount.
3. Calculate total interest earned: Use simple interest formula to find the total interest from both investments and set it equal to \(\$320\).
4. Solve for house purchase price: Work backwards from the remaining amount to find how much was spent on the house.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what's happening with David's money step by step:

David starts with \(\$100,000\). He uses some unknown amount to buy a house. Let's call the amount he spent on the house '\(\mathrm{H}\)' dollars.

This means the remaining money he has for investing is: \(\$100,000 - \mathrm{H}\)

Now, with this remaining money, he splits it into two investment portions:

• \(\frac{1}{3}\) of the remaining money goes into a \(4\%\) simple interest investment
• \(\frac{2}{3}\) of the remaining money goes into a \(6\%\) simple interest investment

After one year, the total interest he earned from both investments combined is \(\$320\).

Our goal is to find \(\mathrm{H}\), the purchase price of the house.

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

2. Set up the money flow

Let's define our remaining amount clearly. If \(\mathrm{R}\) represents the money remaining after buying the house, then:

\(\mathrm{R} = \$100,000 - \mathrm{H}\)

Now let's express each investment portion in simple terms:

• First investment amount = \(\frac{1}{3} \times \mathrm{R}\)
• Second investment amount = \(\frac{2}{3} \times \mathrm{R}\)

We can verify this makes sense: \(\frac{1}{3}\mathrm{R} + \frac{2}{3}\mathrm{R} = \mathrm{R}\) ✓

So the two investments are:

• \(\frac{1}{3} \times \mathrm{R}\) dollars invested at \(4\%\) simple interest
• \(\frac{2}{3} \times \mathrm{R}\) dollars invested at \(6\%\) simple interest

3. Calculate total interest earned

Now let's figure out how much interest each investment earned after one year.

For simple interest, the interest earned = Principal × Rate × Time

Since we're looking at one year, Time = 1, so:
Interest = Principal × Rate

Interest from first investment = \(\frac{1}{3} \times \mathrm{R} \times 0.04\)
Interest from second investment = \(\frac{2}{3} \times \mathrm{R} \times 0.06\)

The total interest earned is the sum of both:
Total interest = \(\frac{1}{3} \times \mathrm{R} \times 0.04 + \frac{2}{3} \times \mathrm{R} \times 0.06\)

We know this total equals \(\$320\):
\(\frac{1}{3} \times \mathrm{R} \times 0.04 + \frac{2}{3} \times \mathrm{R} \times 0.06 = 320\)

Let's factor out \(\mathrm{R}\) to simplify:
\(\mathrm{R} \times [\frac{1}{3} \times 0.04 + \frac{2}{3} \times 0.06] = 320\)
\(\mathrm{R} \times [\frac{0.04}{3} + \frac{0.12}{3}] = 320\)
\(\mathrm{R} \times [\frac{0.16}{3}] = 320\)
\(\mathrm{R} \times \frac{0.16}{3} = 320\)

Therefore: \(\mathrm{R} = 320 \div \frac{0.16}{3} = 320 \times \frac{3}{0.16} = 320 \times 18.75 = 6,000\)

So the remaining amount after buying the house was \(\$6,000\).

4. Solve for house purchase price

Now we can find the house purchase price using our relationship:

\(\mathrm{R} = \$100,000 - \mathrm{H}\)
\(\$6,000 = \$100,000 - \mathrm{H}\)
\(\mathrm{H} = \$100,000 - \$6,000\)
\(\mathrm{H} = \$94,000\)

Let's verify this makes sense by checking our interest calculation:

• Remaining money: \(\$6,000\)
• First investment: \(\frac{1}{3} \times \$6,000 = \$2,000\) at \(4\%\)
• Second investment: \(\frac{2}{3} \times \$6,000 = \$4,000\) at \(6\%\)
• Interest from first: \(\$2,000 \times 0.04 = \$80\)
• Interest from second: \(\$4,000 \times 0.06 = \$240\)
• Total interest: \(\$80 + \$240 = \$320\) ✓

Final Answer

The purchase price of the house was \(\$94,000\).

Looking at our answer choices, this corresponds to choice (B) \(\$94,000\), which matches the given correct answer.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting what the fractions apply to: Students often think that \(\frac{1}{3}\) and \(\frac{2}{3}\) refer to fractions of the original \(\$100,000\) rather than fractions of the remaining money after the house purchase. This leads to setting up equations with \(\$100,000 \times \frac{1}{3}\) and \(\$100,000 \times \frac{2}{3}\) instead of \(\mathrm{R} \times \frac{1}{3}\) and \(\mathrm{R} \times \frac{2}{3}\).

2. Confusing simple interest formula: Students may forget that simple interest for one year simplifies to just Principal × Rate, and instead try to use more complex formulas or include unnecessary time calculations, making the setup more complicated than needed.

3. Setting up the wrong variable: Some students define their variable as the remaining amount \(\mathrm{R}\) instead of the house price \(\mathrm{H}\), or vice versa, leading to confusion about what they're actually solving for and requiring unnecessary extra steps.

Errors while executing the approach

1. Arithmetic errors in fraction calculations: When calculating \(\mathrm{R} \times [\frac{1}{3} \times 0.04 + \frac{2}{3} \times 0.06]\), students often make mistakes with decimal-fraction operations, especially when computing \(\frac{0.04}{3} + \frac{0.12}{3} = \frac{0.16}{3}\), or when dividing \(320\) by \(\frac{0.16}{3}\).

2. Incorrect percentage conversions: Students may forget to convert percentages to decimals, using \(4\) and \(6\) instead of \(0.04\) and \(0.06\) in their calculations, leading to dramatically wrong intermediate results.

3. Sign errors in the final step: When solving \(\mathrm{R} = \$100,000 - \mathrm{H}\) for \(\mathrm{H}\), students sometimes write \(\mathrm{H} = \$6,000 - \$100,000\) instead of \(\mathrm{H} = \$100,000 - \$6,000\), resulting in a negative house price.

Errors while selecting the answer

1. Reporting the wrong quantity: Students may correctly calculate that \(\mathrm{R} = \$6,000\) (the remaining investment amount) but then select this value or a related calculation as their final answer instead of the house purchase price of \(\$94,000\).

2. Failing to verify the answer: Students might arrive at one of the wrong answer choices through calculation errors but not check their work by substituting back into the original interest equation, missing the opportunity to catch their mistake.