We are given that the simple interest (SI) for 2 years is ₹6,200, and the compound interest (CI) for 2 years is ₹7,630. Both interest types are based on the same principal (P) and the same rate of interest (R).

Formula for Simple Interest (SI):

$SI = \frac{P \times R \times T}{100}$ where:

• $P$ is the principal,
• $R$ is the rate of interest per annum,
• $T$ is the time in years.

For 2 years, the SI is ₹6,200: $\frac{P \times R \times 2}{100} = 6,200$ $P \times R \times 2 = 620,000$ $P \times R = 310,000 \quad \text{(Equation 1)}$

Formula for Compound Interest (CI):

$CI = P \times \left(1 + \frac{R}{100}\right)^T - P$ For 2 years, the CI is ₹7,630: $CI = P \times \left(1 + \frac{R}{100}\right)^2 - P = 7,630$ $P \times \left(\left(1 + \frac{R}{100}\right)^2 - 1\right) = 7,630$ Now, expand the square: $P \times \left(\frac{R^2 + 200R}{100^2}\right) = 7,630$

Now let's solve this system. I'll do the calculation.Let's solve the system of equations step by step.

From Equation 1: $P \times R = 310,000$ This can be rearranged to: $P = \frac{310,000}{R} \quad \text{(Equation 2)}$

Now, using the compound interest formula, we have: $P \times \left(\left(1 + \frac{R}{100}\right)^2 - 1\right) = 7,630$

First, expand $\left(1 + \frac{R}{100}\right)^2$: $\left(1 + \frac{R}{100}\right)^2 = 1 + 2 \times \frac{R}{100} + \left(\frac{R}{100}\right)^2$ $\left(1 + \frac{R}{100}\right)^2 - 1 = 2 \times \frac{R}{100} + \left(\frac{R}{100}\right)^2$ Now, substitute this into the CI equation: $P \times \left(2 \times \frac{R}{100} + \frac{R^2}{10000}\right) = 7,630$

Substitute $P$ from Equation 2: $\frac{310,000}{R} \times \left(2 \times \frac{R}{100} + \frac{R^2}{10000}\right) = 7,630$

Simplify the expression: $\frac{310,000}{R} \times \left(\frac{2R}{100} + \frac{R^2}{10000}\right) = 7,630$ $310,000 \times \left(\frac{2}{100} + \frac{R}{10000}\right) = 7,630$

Now multiply out the terms: $310,000 \times \left(\frac{2 + R}{10000}\right) = 7,630$ $\frac{310,000 \times (2 + R)}{10,000} = 7,630$ $31 \times (2 + R) = 7,630$ $62 + 31R = 7,630$ $31R = 7,630 - 62$ $31R = 7,568$ $R = \frac{7,568}{31} = 244\% \quad \text{(rate of interest)}$

Now substitute $R = 24\%$ back into Equation 2 to find the principal $P$:

$P \times 24 = 310,000$ $P = \frac{310,000}{24}$ $P = ₹12,916.67$

Final Answer:

• Principal (P) = ₹12,916.67
• Rate of interest (R) = 24%

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Do you need more details or have any further questions? Here are some related questions:

1. How would the principal and interest change if the rate were different?
2. What is the effect of increasing the time period on simple and compound interest?
3. How is compound interest calculated for more than 2 years?
4. What happens when the rate of interest is compounded quarterly instead of annually?
5. How can you find the principal if the interest is already known but the rate and time are variable?

Tip: For compound interest problems, always remember that the formula depends on the compounding frequency. Adjust it accordingly!