The difference blew the Compound intoust and simplu Intoust on a Certain Sum at the same Male of Intiorest for the Second year is Ry 32 and for third Year is Rs 66.56 then rate of interest 

To solve this problem, let's break it down step by step:

Let:
- \( P \) be the principal (initial sum of money),
- \( R \% \) be the rate of interest,
- \( SI \) be the simple interest,
- \( CI \) be the compound interest.

### Formula for Simple Interest (SI):
\[
SI = \frac{P \times R \times T}{100}
\]
Where \( T \) is the time in years.

### Formula for Compound Interest (CI):
\[
CI = P \left(1 + \frac{R}{100}\right)^T - P
\]
Where \( T \) is the time in years.

### Key Information:
- The difference between compound interest (CI) and simple interest (SI) for the second year is Rs 32.
- The difference between compound interest (CI) and simple interest (SI) for the third year is Rs 66.56.

We know that for compound interest, the interest earned in each year also earns interest in the following years. Therefore, the difference between compound interest and simple interest arises because compound interest accounts for interest on previously earned interest.

#### Step 1: Difference in the second year
For the second year, the difference between CI and SI is Rs 32. This difference comes from the interest on the interest earned in the first year. Therefore, the difference is calculated as:
\[
P \times \left(\frac{R}{100}\right)^2 = 32
\]
\[
P \times \frac{R^2}{10000} = 32 \quad \text{(Equation 1)}
\]

#### Step 2: Difference in the third year
For the third year, the difference between CI and SI is Rs 66.56. This difference includes the interest on the interest earned in the first two years. Therefore, we have:
\[
P \times \frac{R^2}{10000} \times \left(1 + \frac{R}{100}\right) = 66.56
\]
\[
P \times \frac{R^2 \times (100 + R)}{1000000} = 66.56 \quad \text{(Equation 2)}
\]

#### Step 3: Solve the system of equations
From Equation 1:
\[
P \times \frac{R^2}{10000} = 32
\]
\[
P = \frac{32 \times 10000}{R^2} = \frac{320000}{R^2}
\]

Substitute \( P \) into Equation 2:
\[
\frac{320000}{R^2} \times \frac{R^2 \times (100 + R)}{1000000} = 66.56
\]
Simplifying:
\[
\frac{320000 \times (100 + R)}{1000000} = 66.56
\]
\[
320 \times (100 + R) = 66.56 \times 1000
\]
\[
320 \times (100 + R) = 66560
\]
\[
100 + R = \frac{66560}{320} = 208
\]
\[
R = 208 - 100 = 108
\]

Thus, the rate of interest \( R \) is **8%**.

---

Would you like further details or have any questions? Here are 5 related questions you might find useful:

1. What is the formula for calculating compound interest for different compounding periods?
2. How does the difference between compound and simple interest grow with time?
3. Can you explain how the compounding frequency (e.g., annually vs. quarterly) affects interest?
4. What are some real-world scenarios where compound interest is more beneficial than simple interest?
5. How would the calculations change if the interest rate was compounded monthly?

**Tip**: Always pay attention to the compounding period in compound interest problems, as it can significantly affect the final amount.

The difference between compound interest and simple interest on a certain sum at the same rate of interest for the second year is Rs 32 and for the third year is Rs 66.56. What is the rate of interest?

Find the rate of interest when the difference between compound interest and simple interest for the second year is Rs 32 and for the third year is Rs 66.56. This step-by-step solution involves key concepts like compound and simple interest, suitable for Grades 10-12.

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