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Solve using GRESA and simple,compound interest formula 

Let's go through the problems in the image using the **GRESA** method (Given, Required, Equation, Solution, Answer) and the appropriate formulas for simple and compound interest.

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### Problem 1

**G**:  
- Principal (P) = $5,750
- Rate (R) = 7%
- Time (T) = 16 months (convert to years: $ \frac{16}{12} $ years)

**R**:  
Find the Simple Interest (I).

**E**:  
The formula for Simple Interest is:  
\[ I = P \times R \times T \]

**S**:  
\[ I = 5750 \times \frac{7}{100} \times \frac{16}{12} \]
\[ I = 5750 \times 0.07 \times \frac{4}{3} \]
\[ I = 5750 \times 0.07 \times 1.3333 \]
\[ I \approx 5750 \times 0.093333 \]
\[ I \approx 537.67 \]

**A**:  
The interest due is **$537.67**.

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### Problem 2

**G**:  
- Principal (P) = $50,000
- Rate (R) = 8%
- Time (T) = 37 days (convert to years: $ \frac{37}{365} $ years)

**R**:  
Find the Simple Interest (I).

**E**:  
\[ I = P \times R \times T \]

**S**:  
\[ I = 50000 \times \frac{8}{100} \times \frac{37}{365} \]
\[ I = 50000 \times 0.08 \times \frac{37}{365} \]
\[ I \approx 50000 \times 0.08 \times 0.10137 \]
\[ I \approx 50000 \times 0.0081096 \]
\[ I \approx 405.48 \]

**A**:  
The interest is **$405.48**.

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### Problem 3

**G**:  
- Simple Interest (I) = $1,250
- Rate (R) = 5%
- Time (T) = 6 months (convert to years: $ \frac{6}{12} $ years)

**R**:  
Find the Principal (P).

**E**:  
Rearrange the simple interest formula to solve for P:  
\[ P = \frac{I}{R \times T} \]

**S**:  
\[ P = \frac{1250}{0.05 \times \frac{6}{12}} \]
\[ P = \frac{1250}{0.05 \times 0.5} \]
\[ P = \frac{1250}{0.025} \]
\[ P = 50000 \]

**A**:  
The principal is **$50,000**.

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### Problem 4

**G**:  
- Principal (P) = $6,000
- Rate (R) = 8% compounded quarterly
- Time (T) = 1 year

**R**:  
Find the Compound Interest (CI).

**E**:  
The formula for Compound Interest is:  
\[ A = P \left(1 + \frac{R}{n}\right)^{nT} \]  
Where:
- A is the amount after interest
- n is the number of compounding periods per year

Then:  
\[ CI = A - P \]

**S**:  
Given that R = 8% or 0.08, n = 4 (quarterly compounding):  
\[ A = 6000 \left(1 + \frac{0.08}{4}\right)^{4 \times 1} \]
\[ A = 6000 \left(1 + 0.02\right)^{4} \]
\[ A = 6000 \times 1.08243216 \]
\[ A \approx 6494.59 \]

\[ CI = 6494.59 - 6000 \]
\[ CI \approx 494.59 \]

**A**:  
The compound interest is **$494.59**.

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### Problem 5

**G**:  
- Principal (P) = $12,500
- Rate (R) = 7.5% per annum
- Time (T) = 6 months (convert to years: $ \frac{6}{12} $ years)

**R**:  
Find the Simple Interest (I) and the Amount (A).

**E**:  
\[ I = P \times R \times T \]  
\[ A = P + I \]

**S**:  
\[ I = 12500 \times \frac{7.5}{100} \times \frac{6}{12} \]
\[ I = 12500 \times 0.075 \times 0.5 \]
\[ I = 12500 \times 0.0375 \]
\[ I = 468.75 \]

\[ A = 12500 + 468.75 \]
\[ A = 12968.75 \]

**A**:  
The interest is **$468.75**, and the total amount due is **$12,968.75**.

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Do you want more details or have any questions?

Here are five related questions for further practice:

1. What would be the interest for Problem 1 if the rate was 5% instead of 7%?
2. How does the compound interest change in Problem 4 if the compounding is done monthly?
3. What is the time required for the principal in Problem 3 to double at the same interest rate?
4. What would be the amount after 2 years in Problem 5, if the rate remains 7.5% per annum?
5. What would be the difference in interest between Problem 2 and Problem 5 if both had the same time duration?

**Tip:** For problems involving interest, always convert time to years if the rate is given per annum, and be mindful of compounding periods for compound interest calculations.

Solve using GRESA and simple, compound interest formula.

Learn how to solve problems involving simple and compound interest using the GRESA method. This page covers key mathematical concepts such as simple interest, compound interest, and percentage calculations. Suitable for students in Grades 7-10.

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