To solve this, we use the formula for Simple Interest (SI):

$A = P + SI$

where:

• $A$ is the amount,
• $P$ is the principal,
• $SI$ is the simple interest,
• $SI = \frac{P \cdot r \cdot t}{100}$,
• $r$ is the rate of interest, and
• $t$ is the time.

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Step 1: Setup equations from the given data

• For 2 years, the amount is $5600$: $5600 = P + \frac{P \cdot r \cdot 2}{100}$ $5600 = P (1 + \frac{2r}{100})$

• For 5 years, the amount is $6500$: $6500 = P + \frac{P \cdot r \cdot 5}{100}$ $6500 = P (1 + \frac{5r}{100})$

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Step 2: Eliminate $P$ to find $r$

Divide the second equation by the first equation: $\frac{6500}{5600} = \frac{1 + \frac{5r}{100}}{1 + \frac{2r}{100}}$ Simplify: $\frac{6500}{5600} = \frac{100 + 5r}{100 + 2r}$ $\frac{13}{11} = \frac{100 + 5r}{100 + 2r}$

Cross-multiply: $13(100 + 2r) = 11(100 + 5r)$ $1300 + 26r = 1100 + 55r$ $200 = 29r$ $r = \frac{200}{29} \approx 6.9\% \, \text{(Rate of interest)}.$

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Step 3: Find $P$

Substitute $r = \frac{200}{29}$ back into one of the original equations. Using the first equation: $5600 = P (1 + \frac{2r}{100})$ $5600 = P (1 + \frac{2 \cdot \frac{200}{29}}{100})$ $5600 = P (1 + \frac{400}{2900})$ $5600 = P (1 + \frac{4}{29})$ $5600 = P \cdot \frac{33}{29}$ $P = \frac{5600 \cdot 29}{33} = 4927.27 \, \text{(Principal)}.$

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Final Answers:

• Principal: $P = 4927.27$
• Rate of Interest: $r = 6.9\%$ (approximately).

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Would you like detailed explanations for any part of this?

Here are 5 related questions for further exploration:

1. What happens to the rate if the time periods change significantly?
2. How do you solve for $P$ and $r$ using compound interest instead of simple interest?
3. What is the total interest earned in the 5-year period?
4. How can this calculation be visualized graphically for better understanding?
5. Can the formula be adapted for varying rates of interest over different periods?

Tip: Always verify your final values by plugging them back into the original equations to ensure