A sum of money lent at compound interest for 2 yr at 20% pa would fetch X 964 more, if the interest was payable half-yearly than if it was payable annually. What is the sum ?

Options

A. X 40000
B. X60000
C. X 90000
D. X 500000

Answer

To solve the problem, we need to find the sum of money (let's denote it as \( P \)) that would yield an additional amount of \( 964 \) when the interest is compounded half-yearly instead of annually at a rate of \( 20\% \) per annum over \( 2 \) years.

### Step 1: Calculate the amount when interest is compounded annually.

The formula for compound interest is given by:

\[ A = P \left(1 + \frac{r}{100}\right)^n \]

Where:
- \( A \) = Amount after \( n \) years
- \( P \) = Principal amount (the initial sum of money)
- \( r \) = Rate of interest per annum
- \( n \) = Number of years

For annual compounding:
- \( r = 20\% \)
- \( n = 2 \)

Substituting these values into the formula:

\[ A_{annual} = P \left(1 + \frac{20}{100}\right)^2 \]
\[ A_{annual} = P \left(1 + 0.2\right)^2 \]
\[ A_{annual} = P \left(1.2\right)^2 \]
\[ A_{annual} = P \times 1.44 \]

### Step 2: Calculate the amount when interest is compounded half-yearly.

For half-yearly compounding, the rate is divided by \( 2 \) and the time is multiplied by \( 2 \):

- Half-yearly rate \( r = \frac{20}{2} = 10\% \)
- Total periods \( n = 2 \times 2 = 4 \)

Using the same formula:

\[ A_{half-yearly} = P \left(1 + \frac{10}{100}\right)^4 \]
\[ A_{half-yearly} = P \left(1 + 0.1\right)^4 \]
\[ A_{half-yearly} = P \left(1.1\right)^4 \]

Calculating \( (1.1)^4 \):

\[ (1.1)^4 = 1.4641 \]

Thus,

\[ A_{half-yearly} = P \times 1.4641 \]

### Step 3: Set up the equation based on the problem statement.

According to the problem, the difference between the amounts when compounded half-yearly and annually is \( 964 \):

\[ A_{half-yearly} - A_{annual} = 964 \]

Substituting the expressions we found:

\[ P \times 1.4641 - P \times 1.44 = 964 \]

Factoring out \( P \):

\[ P (1.4641 - 1.44) = 964 \]
\[ P \times 0.0241 = 964 \]

### Step 4: Solve for \( P \).

\[ P = \frac{964}{0.0241} \]
\[ P \approx 40000 \]

Thus, the sum of money \( P \) is approximately \( 40000 \).

### Final Answer:
The sum of money is \( 40000 \).

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