Neeraj lent Rs 65536 for 2 years at `12(1)/(2)%` per annum, compounded annually. How much more could he earn if the interest were compounded half-yearly?

Answer

To solve the problem step by step, we need to calculate the compound interest for both annual and half-yearly compounding, and then find the difference between the two amounts.

### Step 1: Understand the Given Information
- Principal (P) = Rs 65,536
- Rate of Interest (R) = 12.5% per annum (which is \(12\frac{1}{2}\%\))
- Time (T) = 2 years

### Step 2: Calculate the Amount with Annual Compounding
The formula for the amount (A) when interest is compounded annually is:
\[ A = P \left(1 + \frac{R}{100}\right)^T \]

Substituting the values:
\[ A = 65536 \left(1 + \frac{12.5}{100}\right)^2 \]
\[ A = 65536 \left(1 + 0.125\right)^2 \]
\[ A = 65536 \left(1.125\right)^2 \]
\[ A = 65536 \times 1.265625 \]
Calculating this gives:
\[ A \approx 83148.4375 \]

### Step 3: Calculate the Amount with Half-Yearly Compounding
When interest is compounded half-yearly, the rate and time need to be adjusted:
- Half-yearly rate = \( \frac{12.5}{2} = 6.25\% \)
- Number of half-year periods in 2 years = \( 2 \times 2 = 4 \)

Using the formula for the amount:
\[ A = P \left(1 + \frac{R}{100}\right)^{n} \]
where \( n \) is the number of compounding periods.

Substituting the values:
\[ A = 65536 \left(1 + \frac{6.25}{100}\right)^4 \]
\[ A = 65536 \left(1 + 0.0625\right)^4 \]
\[ A = 65536 \left(1.0625\right)^4 \]
Calculating \( (1.0625)^4 \):
\[ (1.0625)^4 \approx 1.28368 \]
Now substituting back:
\[ A \approx 65536 \times 1.28368 \]
Calculating this gives:
\[ A \approx 84000.00 \]

### Step 4: Calculate the Difference
Now, we find the difference between the amounts earned through annual and half-yearly compounding:
\[ \text{Difference} = A_{\text{half-yearly}} - A_{\text{annually}} \]
\[ \text{Difference} = 84000 - 83148.4375 \]
\[ \text{Difference} \approx 851.5625 \]

### Final Answer
Neeraj could earn approximately Rs 851.56 more if the interest were compounded half-yearly instead of annually.

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