Math Problem Statement

Jenny deposits $60,000 into an account that pays 4% interest per year, compounded annually.

Frank deposits $60,000 into an account that also pays 4% per year. But it is simple interest.

Find the interest Jenny and Frank 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

Let's break down the problem step by step to find the interest Jenny and Frank earn during each of the first three years and compare their earnings.

Jenny's Account: Compound Interest

Jenny's interest is compounded annually at a rate of 4%. The formula for compound interest is:

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

Where:

  • AA is the amount of money accumulated after nn years, including interest.
  • PP is the principal amount ($60,000).
  • rr is the annual interest rate (4% or 0.04).
  • nn is the number of times interest is compounded per year (1 in this case).
  • tt is the number of years.

We need to find the interest earned at the end of each year. The interest earned each year is the difference between the accumulated amount at the end of the year and the initial amount at the start of the year.

Year 1:

A1=60,000×(1+0.04)1=60,000×1.04=62,400A_1 = 60{,}000 \times \left(1 + 0.04\right)^1 = 60{,}000 \times 1.04 = 62{,}400 Interest earned in Year 1: Interest1=A1P=62,40060,000=2,400\text{Interest}_1 = A_1 - P = 62{,}400 - 60{,}000 = 2{,}400

Year 2:

A2=60,000×(1+0.04)2=60,000×1.0816=64,896A_2 = 60{,}000 \times \left(1 + 0.04\right)^2 = 60{,}000 \times 1.0816 = 64{,}896 Interest earned in Year 2: Interest2=A2A1=64,89662,400=2,496\text{Interest}_2 = A_2 - A_1 = 64{,}896 - 62{,}400 = 2{,}496

Year 3:

A3=60,000×(1+0.04)3=60,000×1.124864=67,491.84A_3 = 60{,}000 \times \left(1 + 0.04\right)^3 = 60{,}000 \times 1.124864 = 67{,}491.84 Interest earned in Year 3: Interest3=A3A2=67,491.8464,896=2,595.84\text{Interest}_3 = A_3 - A_2 = 67{,}491.84 - 64{,}896 = 2{,}595.84

Frank's Account: Simple Interest

Frank's interest is simple interest, so the formula is:

I=P×r×tI = P \times r \times t

Where:

  • II is the interest.
  • PP is the principal amount ($60,000).
  • rr is the annual interest rate (4% or 0.04).
  • tt is the number of years.

The interest Frank earns each year remains constant.

Year 1:

I1=60,000×0.04×1=2,400I_1 = 60{,}000 \times 0.04 \times 1 = 2{,}400

Year 2:

I2=60,000×0.04×1=2,400I_2 = 60{,}000 \times 0.04 \times 1 = 2{,}400

Year 3:

I3=60,000×0.04×1=2,400I_3 = 60{,}000 \times 0.04 \times 1 = 2{,}400

Comparing Interest Earned:

  • Year 1: Both Jenny and Frank earn $2,400.
  • Year 2: Jenny earns $2,496, while Frank earns $2,400.
  • Year 3: Jenny earns $2,595.84, while Frank earns $2,400.

Conclusion:

  • Year 1: Jenny and Frank earn the same interest.
  • Year 2: Jenny earns more interest ($2,496 vs. $2,400).
  • Year 3: Jenny earns significantly more interest ($2,595.84 vs. $2,400).

Jenny earns more interest each year starting from Year 2 due to the compounding effect.


Would you like more details, or do you have any questions? Here are some related questions:

  1. What would happen if Frank had a higher interest rate but still used simple interest?
  2. How would the results change if the interest was compounded semi-annually?
  3. Can we calculate the total amount each person has at the end of 5 years?
  4. How does compounding frequency affect the total interest earned?
  5. What if Jenny deposited an additional amount each year?

Tip: Compound interest tends to outperform simple interest over time because it earns "interest on interest," leading to faster growth.

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Math Problem Analysis

Mathematical Concepts

Compound Interest
Simple Interest

Formulas

Compound interest formula: A = P(1 + r/n)^(nt)
Simple interest formula: I = P * r * t

Theorems

Compound interest grows faster due to interest on interest over time.

Suitable Grade Level

Grades 9-12