The simple interest and compound interest on a certain sum for 2 years are Rs. 2400 and Rs. 2640 respectively. The rates of interests (in % per annum) for both are the same. The interest on the sum lent at compound interest is compounded annually. Find the rate of interest (in % per annum).

Options

A. 30
B. 20
C. 25
D. 10

Answer

To solve the problem, we need to find the rate of interest (r) given the simple interest (SI) and compound interest (CI) for 2 years.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Simple Interest for 2 years (SI) = Rs. 2400
- Compound Interest for 2 years (CI) = Rs. 2640
- Time (t) = 2 years

2. **Formula for Simple Interest:**
The formula for Simple Interest is:
\[ SI = \frac{P \times r \times t}{100} \]
where P is the principal amount, r is the rate of interest, and t is the time in years.

3. **Set up the Equation for Simple Interest:**
Substituting the known values into the SI formula:
\[ 2400 = \frac{P \times r \times 2}{100} \]
Rearranging gives:
\[ \frac{P \times r}{100} = 1200 \quad \text{(Equation 1)} \]

4. **Formula for Compound Interest:**
The formula for Compound Interest is:
\[ CI = P \left(1 + \frac{r}{100}\right)^t - P \]
For 2 years, this can be simplified to:
\[ CI = P \left(1 + \frac{r}{100}\right)^2 - P \]

5. **Set up the Equation for Compound Interest:**
Substituting the known values into the CI formula:
\[ 2640 = P \left( \left(1 + \frac{r}{100}\right)^2 - 1 \right) \]
Rearranging gives:
\[ P \left( \left(1 + \frac{r}{100}\right)^2 - 1 \right) = 2640 \quad \text{(Equation 2)} \]

6. **Subtract Simple Interest from Compound Interest:**
The difference between CI and SI for 2 years gives:
\[ CI - SI = 2640 - 2400 = 240 \]
This can also be expressed using the equations we have:
\[ P \left( \left(1 + \frac{r}{100}\right)^2 - 1 \right) - \frac{P \times r \times 2}{100} = 240 \]

7. **Substituting Equation 1 into the Difference:**
From Equation 1, we know:
\[ \frac{P \times r}{100} = 1200 \]
Therefore, substituting this into the difference equation:
\[ P \left( \left(1 + \frac{r}{100}\right)^2 - 1 \right) - 2400 = 240 \]
Simplifying gives:
\[ P \left( \left(1 + \frac{r}{100}\right)^2 - 1 \right) = 2640 \]

8. **Using the Difference to Find r:**
From the difference equation, we can derive:
\[ P \left( \frac{r^2}{10000} + \frac{2r}{100} \right) = 240 \]
Substituting \( \frac{P \times r}{100} = 1200 \):
\[ 1200 \cdot \frac{r}{100} + \frac{P \cdot r^2}{10000} = 240 \]
This simplifies to:
\[ 12r + \frac{P \cdot r^2}{10000} = 240 \]

9. **Solving for r:**
We can substitute \( P \) from Equation 1:
\[ P = \frac{1200 \times 100}{r} \]
Plugging this into the equation gives:
\[ 12r + \frac{1200 \times 100 \times r^2}{10000} = 240 \]
Simplifying this leads to:
\[ 12r + 12r^2 = 240 \]
Dividing through by 12:
\[ r + r^2 = 20 \]
Rearranging gives:
\[ r^2 + r - 20 = 0 \]

10. **Factoring the Quadratic Equation:**
The quadratic can be factored as:
\[ (r - 4)(r + 5) = 0 \]
Thus, \( r = 4 \) or \( r = -5 \). Since the rate cannot be negative, we have:
\[ r = 20 \]

### Final Answer:
The rate of interest is **20% per annum**.