Math Problem Statement

Melissa deposits $10,000 into an account that pays 3% interest per year, compounded annually. Greg deposits $10,000 into an account that also pays 3% per year. But it is simple interest. Find the interest Melissa and Greg earn during each of the first three years. Then decide who earns more interest for each year. Assume there are no withdrawals and no additional deposits.

Solution

This question involves comparing the interest earned by Melissa and Greg over three years, with the interest formulas differing for each. Melissa's account earns compound interest, while Greg’s earns simple interest. Let's work through the problem:

Given:

  • Melissa deposits $10,000 in an account with 3% interest compounded annually.
  • Greg deposits $10,000 in an account with 3% simple interest.
  • We need to find how much interest they each earn in the first three years and then compare who earns more interest.

For Melissa (Compound Interest Formula):

The formula for compound interest is: A=P(1+r)tA = P(1 + r)^t Where:

  • AA is the amount of money accumulated after interest.
  • PP is the principal (initial deposit), $10,000.
  • rr is the interest rate (0.03 for 3%).
  • tt is the time in years.

The interest earned is APA - P.

For Greg (Simple Interest Formula):

The formula for simple interest is: I=P×r×tI = P \times r \times t Where:

  • II is the interest earned.
  • PP is the principal, $10,000.
  • rr is the interest rate, 0.03.
  • tt is the time in years.

Now, let’s calculate the interest for both over the three years.

First Year:

For Melissa (Compound Interest): A1=10000(1+0.03)1=10000×1.03=10300A_1 = 10000(1 + 0.03)^1 = 10000 \times 1.03 = 10300 Interest earned = $10300 - $10000 = $300

For Greg (Simple Interest): I1=10000×0.03×1=300I_1 = 10000 \times 0.03 \times 1 = 300 Interest earned = $300

Second Year:

For Melissa (Compound Interest): A2=10000(1+0.03)2=10000×1.0609=10609A_2 = 10000(1 + 0.03)^2 = 10000 \times 1.0609 = 10609 Interest earned = $10609 - $10000 = $609

For Greg (Simple Interest): I2=10000×0.03×2=600I_2 = 10000 \times 0.03 \times 2 = 600 Interest earned = $600

Third Year:

For Melissa (Compound Interest): A3=10000(1+0.03)3=10000×1.092727=10927.27A_3 = 10000(1 + 0.03)^3 = 10000 \times 1.092727 = 10927.27 Interest earned = $10927.27 - $10000 = $927.27

For Greg (Simple Interest): I3=10000×0.03×3=900I_3 = 10000 \times 0.03 \times 3 = 900 Interest earned = $900

Summary of Interest Earned:

YearInterest Melissa Earns (Compound)Interest Greg Earns (Simple)Who Earns More?
First$300$300Same
Second$609$600Melissa
Third$927.27$900Melissa

Conclusion:

  • In the first year, both earn the same amount.
  • In the second year, Melissa earns slightly more.
  • In the third year, Melissa again earns slightly more.

Would you like further explanations or have any other questions about the problem?

Here are a few related questions for further exploration:

  1. How would the result change if the interest rate was higher, say 5%?
  2. How does compound interest grow more significantly over time compared to simple interest?
  3. How does the time (t) in the formula affect the final interest earned in both cases?
  4. What would happen if the principal for Greg was doubled?
  5. Can you explain the difference between compound interest and simple interest in a real-world scenario?

Tip: Compound interest works by earning interest on both the original deposit and any interest already earned, making it more beneficial as time goes on compared to simple interest, which only calculates interest on the initial deposit.

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Math Problem Analysis

Mathematical Concepts

Compound Interest
Simple Interest
Interest Accumulation

Formulas

A = P(1 + r)^t
I = P × r × t

Theorems

Interest Calculation

Suitable Grade Level

Grades 9-12