An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period. In terms...

1. Translate the problem requirements: We need to clarify what "simple annual interest" means (interest calculated only on principal), identify the given scenario (d dollars for 2 years yields \(\$600\)), and understand what we're solving for (investment amount for different time and yield targets).
2. Establish the relationship between investment parameters: Use the first scenario to determine how the interest rate, principal, and time relate to the total interest earned.
3. Apply proportional reasoning to the new scenario: Compare the two investment scenarios by analyzing how changes in time period and desired yield affect the required principal amount.
4. Verify through direct comparison: Check that our answer makes intuitive sense by comparing the relative changes in all variables between the two scenarios.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're dealing with in everyday terms. When we say "simple annual interest," we mean that the interest is calculated only on the original amount you invested (the principal) - not on any interest that accumulates over time.

Here's what we know from the first scenario:

• Someone invested d dollars
• At k percent simple annual interest
• For 2 years
• This earned exactly \(\$600\) in interest

And here's what we need to find:

• How much money (in terms of d) should be invested
• At the same k percent rate
• For 3 years
• To earn \(\$2,400\) in interest

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

2. Establish the relationship between investment parameters

Let's think about how simple interest works in plain English first. The total interest you earn depends on three things:

• How much you invest (the principal)
• What interest rate you get
• How long you leave the money invested

Specifically, if you double the amount you invest, you double the interest. If you double the time, you double the interest. If you double the rate, you double the interest.

From our first scenario, we can say:
"d dollars, invested at rate k, for 2 years, produces \(\$600\) interest"

This gives us: \(\mathrm{Interest} = \mathrm{Principal} \times \mathrm{Rate} \times \mathrm{Time}\)
So: \(\$600 = \mathrm{d} \times \mathrm{k} \times 2\)

This means: \(\mathrm{d} \times \mathrm{k} = \$600 \div 2 = \$300\)

So we now know that \(\mathrm{d} \times \mathrm{k} = \$300\). This is our key relationship!

3. Apply proportional reasoning to the new scenario

Now let's figure out the second scenario using the same logic. We want to know:
"How much money (let's call it P), invested at rate k, for 3 years, produces \(\$2,400\) interest?"

Using the same relationship:
\(\$2,400 = \mathrm{P} \times \mathrm{k} \times 3\)

We can solve for P:
\(\mathrm{P} \times \mathrm{k} = \$2,400 \div 3 = \$800\)

So we need \(\mathrm{P} \times \mathrm{k} = \$800\)

But remember, we already found that \(\mathrm{d} \times \mathrm{k} = \$300\)

Since both expressions involve "\(\times \mathrm{k}\)", we can compare them directly:

• We need \(\mathrm{P} \times \mathrm{k} = \$800\)
• We know \(\mathrm{d} \times \mathrm{k} = \$300\)

Therefore: \(\mathrm{P} = \$800 \div \mathrm{k}\) and \(\mathrm{d} = \$300 \div \mathrm{k}\)

This means: \(\frac{\mathrm{P}}{\mathrm{d}} = \frac{\$800}{\$300} = \frac{8}{3}\)

So: \(\mathrm{P} = \frac{8}{3} \times \mathrm{d} = \frac{8\mathrm{d}}{3}\)

Process Skill: INFER - Drawing the non-obvious conclusion that we can compare the two scenarios through their d×k relationships

4. Verify through direct comparison

Let's check that our answer makes sense by comparing what changed between the two scenarios:

First scenario: d dollars → 2 years → \(\$600\) interest
Second scenario: \(\frac{8\mathrm{d}}{3}\) dollars → 3 years → \(\$2,400\) interest

Let's see if the proportions work out:

• Time increased from 2 years to 3 years (factor of \(\frac{3}{2} = 1.5\))
• Desired interest increased from \(\$600\) to \(\$2,400\) (factor of \(\frac{2,400}{600} = 4\))
• Investment amount increased from d to \(\frac{8\mathrm{d}}{3}\) (factor of \(\frac{8}{3} \approx 2.67\))

Checking: \(1.5 \times 2.67 = 4.00\) ✓

This confirms our answer is correct!

5. Final Answer

The dollar amount that needs to be invested is \(\frac{8\mathrm{d}}{3}\).

Looking at our answer choices, this corresponds to Answer Choice E: \(\frac{8\mathrm{d}}{3}\).

Common Faltering Points

Errors while devising the approach

1. Misunderstanding Simple vs. Compound Interest

Students often confuse simple interest with compound interest. In this problem, they might try to apply compound interest formulas where interest earns interest over time, rather than recognizing that simple interest is calculated only on the principal amount. This leads to unnecessarily complicated approaches involving exponential calculations.

2. Thinking They Need to Find the Actual Interest Rate

Many students get stuck trying to solve for the exact value of k (the interest rate percentage) before proceeding. They don't realize that this problem can be solved using proportional relationships without ever determining the specific interest rate value.

3. Missing the Proportional Relationship Between Scenarios

Students may attempt to solve each scenario completely separately instead of recognizing that both scenarios share the same interest rate k, which creates a powerful proportional relationship that can be exploited to find the answer efficiently.

Errors while executing the approach

1. Incorrectly Setting Up the Interest Formula

Even when students know to use \(\mathrm{I} = \mathrm{P} \times \mathrm{r} \times \mathrm{t}\), they often make errors in identifying which values correspond to which variables. For example, they might confuse the time period (2 years vs 3 years) or mix up the principal amounts between the two scenarios.

2. Arithmetic Errors in Proportional Calculations

When calculating the ratio \(\$2,400 \div 3 = \$800\) and \(\$600 \div 2 = \$300\), students frequently make division errors. Additionally, when finding the final ratio \(\frac{800}{300} = \frac{8}{3}\), they often simplify incorrectly or get confused with fraction operations.

3. Mixing Up the Direction of the Proportion

Students sometimes set up the proportion backwards, writing \(\frac{\mathrm{d}}{\mathrm{P}} = \frac{8}{3}\) instead of \(\frac{\mathrm{P}}{\mathrm{d}} = \frac{8}{3}\), which leads them to calculate \(\mathrm{P} = \frac{3\mathrm{d}}{8}\) instead of the correct \(\frac{8\mathrm{d}}{3}\).

Errors while selecting the answer

1. Choosing the Reciprocal Answer

Due to proportion setup errors during execution, students often arrive at \(\frac{3\mathrm{d}}{8}\), which isn't among the choices. They then might incorrectly select choice A: \(\frac{2\mathrm{d}}{3}\) or choice B: \(\frac{3\mathrm{d}}{4}\), thinking these are close to their incorrect calculation.

2. Confusing Time Period Ratios with the Final Answer

Since the time periods change from 2 to 3 years (ratio \(\frac{3}{2}\)), some students mistakenly think this ratio should directly appear in their answer and incorrectly select choice D: \(\frac{3\mathrm{d}}{2}\), not realizing they also need to account for the change in desired interest amount.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose convenient values that satisfy the constraint

From the given information: d dollars at k percent simple interest yields \(\$600\) over 2 years

Using simple interest: \(\mathrm{Interest} = \mathrm{Principal} \times \mathrm{Rate} \times \mathrm{Time}\)

So: \(\mathrm{d} \times \frac{\mathrm{k}}{100} \times 2 = 600\)

This gives us: \(\mathrm{d} \times \mathrm{k} = 30,000\)

Let's choose smart numbers: \(\mathrm{d} = 1,000\) and \(\mathrm{k} = 30\)

Verification: \(1,000 \times \frac{30}{100} \times 2 = 1,000 \times 0.30 \times 2 = 600\) ✓

Step 2: Apply the same interest rate to the new scenario

We need to find the investment amount that yields \(\$2,400\) over 3 years at the same 30% annual rate

Let \(\mathrm{x}\) = the unknown investment amount

Using simple interest: \(\mathrm{x} \times \frac{30}{100} \times 3 = 2,400\)

\(\mathrm{x} \times 0.30 \times 3 = 2,400\)

\(\mathrm{x} \times 0.90 = 2,400\)

\(\mathrm{x} = 2,400 \div 0.90 = \frac{8,000}{3} \approx 2,666.67\)

Step 3: Express in terms of d

We found \(\mathrm{x} = \frac{8,000}{3}\) and we chose \(\mathrm{d} = 1,000\)

So \(\mathrm{x} = \frac{8,000}{3} = \frac{8 \times 1,000}{3} = \frac{8\mathrm{d}}{3}\)

Step 4: Verify with answer choices

Our result \(\frac{8\mathrm{d}}{3}\) matches answer choice E.

Step 5: Double-check our work

Original scenario: \(1,000 \times 30\% \times 2\text{ years} = 600\) ✓

New scenario: \(\frac{8,000}{3} \times 30\% \times 3\text{ years} = \frac{8,000}{3} \times 0.90 = 2,400\) ✓