Question

read each of the problems below, determine which formula should be used, and then find the solution.
problem

1. mrs. baxter deposits $2,000 in an account that earns 5% simple interest. how much is mrs. baxter’s investment worth after 8 years?
2. joey made a deposit into an account that earns 6% simple interest. after 3 years, joey had earned $400. how much was joey’s initial deposit?
3. coach cliffman made a deposit of $1,800 into an account that earns 2% annual simple interest. find the amount of interest that coach cliffman earned after 3 years.

use your understanding of simple interest to answer the questions below.

1. peter is calculating the interest earned on a deposit of $275 in an account that earns 8% simple interest after 12 years.

i = prt
i = 275(0.8)(12)
i = 2,640
a. what did peter do incorrectly?
b. what is the correct amount of interest?

1. isabella is calculating the interest earned on a deposit of $3,000 in an account that earns 4% simple interest after 6 years.

i = prt
i = 3,000(0.04)(6)
i = 3,720
a. isabella determines that her deposit will then be worth $6,720.00. explain what isabella did incorrectly.
b. what is the correct amount of interest?

Explanation:

Problem 1

Step1: Recall simple interest total formula

The total value \( A \) of an investment with simple interest is \( A = P(1 + rt) \), where \( P=\$2000 \), \( r=0.05 \), \( t=8 \).

Step2: Substitute values into formula

\( A = 2000(1 + 0.05 \times 8) \)

Step3: Calculate inside parentheses first

\( 0.05 \times 8 = 0.4 \), so \( 1 + 0.4 = 1.4 \)

Step4: Compute final total value

\( A = 2000 \times 1.4 = 2800 \)

Problem 2

Step1: Recall simple interest formula

\( I = Prt \), solve for \( P \): \( P = \frac{I}{rt} \), where \( I=\$400 \), \( r=0.06 \), \( t=3 \).

Step2: Substitute values into formula

\( P = \frac{400}{0.06 \times 3} \)

Step3: Calculate denominator

\( 0.06 \times 3 = 0.18 \)

Step4: Compute initial deposit

\( P = \frac{400}{0.18} \approx 2222.22 \)

Problem 3

Step1: Recall simple interest formula

\( I = Prt \), where \( P=\$1800 \), \( r=0.02 \), \( t=3 \).

Step2: Substitute values into formula

\( I = 1800 \times 0.02 \times 3 \)

Step3: Calculate step-by-step

\( 1800 \times 0.02 = 36 \), then \( 36 \times 3 = 108 \)

Problem 4

Part a

Step1: Identify Peter's error

Peter used \( 0.8 \) instead of \( 0.08 \) for 8% interest.

Part b

Step1: Use correct simple interest formula

\( I = Prt \), where \( P=\$275 \), \( r=0.08 \), \( t=12 \).

Step2: Substitute values into formula

\( I = 275 \times 0.08 \times 12 \)

Step3: Calculate step-by-step

\( 275 \times 0.08 = 22 \), then \( 22 \times 12 = 264 \)

Problem 5

Part a

Step1: Identify Isabella's error

Isabella added the calculated interest to the principal incorrectly: she used \( 3000 + 3720 \) instead of the correct interest value, and miscalculated \( 3000 \times 0.04 \times 6 \).

Part b

Step1: Use correct simple interest formula

\( I = Prt \), where \( P=\$3000 \), \( r=0.04 \), \( t=6 \).

Step2: Substitute values into formula

\( I = 3000 \times 0.04 \times 6 \)

Step3: Calculate step-by-step

\( 3000 \times 0.04 = 120 \), then \( 120 \times 6 = 720 \)

Answer:

1. Total value after 8 years: $\$2800$
2. Joey's initial deposit: $\$2222.22$ (rounded to nearest cent)
3. Interest earned: $\$108$
4. a. Peter converted 8% to 0.8 instead of 0.08.

b. Correct interest: $\$264$

1. a. Isabella miscalculated the interest value ($3000×0.04×6=720$, not 3720) and incorrectly added this wrong value to the principal.

b. Correct interest: $\$720$