If 3200 becomes 3362 at the rate of 10% com pound interest in certain time period then find the time period interest is compounded quarterly?

Options

A. 8 month
B. 4 month
C. 6 month
D. 9 month

Answer

To solve the problem, we need to find the time period for which the principal amount of 3200 becomes 3362 at a compound interest rate of 10% compounded quarterly.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Principal (P) = 3200
- Amount (A) = 3362
- Rate (R) = 10% per annum
- Since the interest is compounded quarterly, the quarterly rate will be:
\[ R_{quarterly} = \frac{10}{4} = 2.5\% \]

2. **Determine the Number of Compounding Periods:**
- Let \( n \) be the number of compounding periods per year. Since the interest is compounded quarterly, \( n = 4 \).
- Let \( t \) be the time in years. Therefore, the total number of compounding periods \( N \) is:
\[ N = n \times t = 4t \]

3. **Use the Compound Interest Formula:**
The formula for compound interest is:
\[ A = P \left(1 + \frac{R}{100n}\right)^{N} \]
Substituting the known values:
\[ 3362 = 3200 \left(1 + \frac{2.5}{100}\right)^{4t} \]

4. **Simplify the Equation:**
\[ 3362 = 3200 \left(1 + 0.025\right)^{4t} \]
\[ 3362 = 3200 \left(1.025\right)^{4t} \]

5. **Divide Both Sides by 3200:**
\[ \frac{3362}{3200} = (1.025)^{4t} \]
\[ 1.058125 = (1.025)^{4t} \]

6. **Take the Logarithm of Both Sides:**
\[ \log(1.058125) = 4t \cdot \log(1.025) \]

7. **Solve for \( t \):**
\[ t = \frac{\log(1.058125)}{4 \cdot \log(1.025)} \]

8. **Calculate the Values:**
- Using a calculator:
\[ \log(1.058125) \approx 0.024 \]
\[ \log(1.025) \approx 0.011 \]
- Therefore:
\[ t = \frac{0.024}{4 \cdot 0.011} \approx \frac{0.024}{0.044} \approx 0.545 \text{ years} \]

9. **Convert Years to Months:**
\[ t \approx 0.545 \times 12 \approx 6.54 \text{ months} \]
Rounding this gives approximately 6 months.

### Final Answer:
The time period for which the interest is compounded quarterly is **6 months**.