QuestionCalculate and compare interest for Laura's compound and Eric's simple interest on \$70,000 at 3\% for 3 years.

Studdy Solution

STEP 1

Assumptions1. Laura's account compounds interest annually. . Eric's account earns simple interest.
3. Both Laura and Eric deposit $70,000.
4. The annual interest rate for both accounts is3%.
5. There are no withdrawals and no additional deposits.
6. We need to calculate the interest earned for each of the first three years.

STEP 2

First, we need to understand the formulas for calculating simple and compound interest.
For simple interest $Interest = Principal \times Rate \times Time$ For compound interest $Amount = Principal \times (1 + Rate)^{Time}$ Then, to find the interest, subtract the principal from the amount.

STEP 3

Let's calculate the simple interest for Eric for each of the first three years.
For year1 $Interest = \$70,000 \times3\% \times1$ For year2 $Interest = \$70,000 \times3\% \times1$ For year3 $Interest = \$70,000 \times3\% \times1$

STEP 4

Convert the percentage to a decimal value.
$3\% =0.03$ Then, calculate the interest for each year.

STEP 5

Calculate the simple interest for Eric for each of the first three years.
For year1 $Interest = \$70,000 \times0.03 \times1 = \$2,100$ For year2 $Interest = \$70,000 \times0.03 \times1 = \$2,100$ For year3 $Interest = \$70,000 \times0.03 \times1 = \$2,100$

STEP 6

Now, let's calculate the compound interest for Laura for each of the first three years.
For year1 $Amount = \$70,000 \times (1 +0.03)^1$ $Interest = Amount - Principal$ For year2 $Amount = \$70,000 \times (1 +0.03)^2$ $Interest = Amount - Principal$ For year3 $Amount = \$70,000 \times (1 +0.03)^3$ $Interest = Amount - Principal$

STEP 7

Calculate the compound interest for Laura for each of the first three years.
For year1 $Amount = \$70,000 \times (1 +0.03)^1 = \$72,100$ $Interest = \$72,100 - \$70,000 = \$2,100$ For year2 $Amount = \$70,000 \times (1 +0.03)^2 = \$74,263$ $Interest = \$74,263 - \$70,000 = \$4,263$ For year3 $Amount = \$70,000 \times (1 +0.03)^3 = \$76,490.89$ $Interest = \$76,490.89 - \$70,000 = \$6,490.89$

STEP 8

Now, let's compare the interest earned by Laura and Eric for each of the first three years.
For year1, both Laura and Eric earn \$2,100.
For year2, Laura earns \$4,263 and Eric earns \$2,100. Laura earns more.
For year3, Laura earns \$6,490.89 and Eric earns \$2,100. Laura earns more.
So, Laura earns more interest than Eric for each year after the first year.

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