Math Problem Statement

Question content area top Part 1 Person A deposits ​$2600 in an account that pays 7​% interest compounded once a year. Person B deposits ​$2450 in an account that pays 8​% interest compounded monthly. Complete parts​ (a) through​ (c) below.

Solution

To solve this problem, we will use the compound interest formula:

A=P(1+rn)ntA = P \left( 1 + \frac{r}{n} \right)^{nt}

Where:

  • AA is the amount of money accumulated after interest.
  • PP is the principal amount (the initial deposit).
  • rr is the annual interest rate (decimal form).
  • nn is the number of times interest is compounded per year.
  • tt is the number of years the money is invested or borrowed for.

Given Data:

  • Person A:

    • Principal PA=2600P_A = 2600
    • Interest rate rA=7%=0.07r_A = 7\% = 0.07
    • Compounded annually nA=1n_A = 1 (since it's compounded once a year).
  • Person B:

    • Principal PB=2450P_B = 2450
    • Interest rate rB=8%=0.08r_B = 8\% = 0.08
    • Compounded monthly nB=12n_B = 12 (since it's compounded monthly).

Now, let's proceed with solving the parts of the question.

(a) Find the amount of money for Person A after 1 year.

For Person A, compounded once a year:

AA=PA(1+rAnA)nAtA_A = P_A \left( 1 + \frac{r_A}{n_A} \right)^{n_A \cdot t}

Substituting the values for t=1t = 1 year:

AA=2600(1+0.071)11A_A = 2600 \left( 1 + \frac{0.07}{1} \right)^{1 \cdot 1} AA=2600×(1.07)1=2600×1.07=2782A_A = 2600 \times (1.07)^1 = 2600 \times 1.07 = 2782

So, Person A's amount after 1 year is $2782.

(b) Find the amount of money for Person B after 1 year.

For Person B, compounded monthly:

AB=PB(1+rBnB)nBtA_B = P_B \left( 1 + \frac{r_B}{n_B} \right)^{n_B \cdot t}

Substituting the values for t=1t = 1 year:

AB=2450(1+0.0812)121A_B = 2450 \left( 1 + \frac{0.08}{12} \right)^{12 \cdot 1} AB=2450(1+0.006667)12A_B = 2450 \left( 1 + 0.006667 \right)^{12} AB=2450×(1.006667)122450×1.083282=2654.04A_B = 2450 \times (1.006667)^{12} \approx 2450 \times 1.083282 = 2654.04

So, Person B's amount after 1 year is $2654.04.

(c) Compare the results for Person A and Person B after 1 year.

After 1 year:

  • Person A has $2782.
  • Person B has $2654.04.

Conclusion: Person A has earned more money than Person B after 1 year, even though Person B's interest rate is higher. This is because Person A's interest is compounded annually, while Person B’s interest is compounded monthly. The frequency of compounding plays a significant role in the final amount.


Would you like further explanations or details on any part? Feel free to ask!

Here are some related questions to deepen your understanding:

  1. How does the frequency of compounding affect the final amount in an interest-bearing account?
  2. What would the amounts for both individuals be if the interest was compounded quarterly instead of annually or monthly?
  3. How would the amounts change if the interest rates were reversed (7% for Person B and 8% for Person A)?
  4. How does the initial principal amount affect the compound interest formula over time?
  5. What would happen if the money was left for more than one year (e.g., 5 years)?

Tip: When comparing different interest rates and compounding frequencies, always calculate the effective annual rate (EAR) to see the actual impact of compounding.

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Math Problem Analysis

Mathematical Concepts

Compound Interest
Exponential Growth
Interest Rate Calculations

Formulas

A = P(1 + r/n)^(nt)

Theorems

Compound Interest Theorem

Suitable Grade Level

Grades 9-12