I'll help you solve these problems using the GRESA method (Given, Required, Equation, Solution, Answer) and the simple and compound interest formulas.

Let's analyze each question in the image:

Problem 1

Given:

• Principal ($P$): $2,000
• Rate ($r$): 12% per annum (simple interest)
• Time ($t$): 3 years

Required:

• The amount of interest earned after 3 years

Equation: The formula for simple interest is: $I = P \times r \times t$

Solution: $I = 2000 \times 0.12 \times 3 = 720$

Answer: The interest earned after 3 years is $720.

Problem 2

Given:

• Principal ($P$): $10,000
• Rate ($r$): 7% per annum
• Time ($t$): 2 years
• Compounding Frequency: Quarterly (4 times a year)

Required:

• The amount in the account after 2 years with compound interest

Equation: The formula for compound interest is: $A = P \left(1 + \frac{r}{n}\right)^{nt}$ where $n$ is the number of times interest is compounded per year.

Solution: $A = 10000 \left(1 + \frac{0.07}{4}\right)^{4 \times 2} = 10000 \left(1 + 0.0175\right)^8$ $A = 10000 \left(1.0175\right)^8 \approx 10000 \times 1.1489 = 11489$

Answer: The amount in the account after 2 years is approximately $11,489.

Problem 3

Given:

• Initial amount ($P$): 15% of $20,000 = $3,000
• Annual rate of decrease ($r$): 3%
• Time ($t$): 8 years

Required:

• The value after 8 years with a 3% annual decrease

Equation: The formula for depreciation with annual decrease is similar to compound interest: $A = P \left(1 - r\right)^t$

**Solution