Understanding the Compound Interest Problem

This problem involves comparing compound interest calculated under two different frequencies for the same principal sum, rate, and time period. We are given the difference in the interest earned when compounding is done annually versus half-yearly and need to find the initial sum lent, which is the principal.

Key Concepts: Compound Interest

Compound interest is the interest calculated on the principal amount and the accumulated interest from previous periods. The formula for compound interest is:

\( A = P \left(1 + \frac{R}{100}\right)^T \)

Where:

• \( A \) is the amount after T years
• \( P \) is the principal amount
• \( R \) is the annual interest rate in percent
• \( T \) is the time period in years

The Compound Interest (CI) is \( CI = A - P \).

Calculating Compound Interest for Different Compounding Frequencies

When compounding is done more than once a year, the formula is adjusted:

\( A = P \left(1 + \frac{R/(n)}{100}\right)^{T \times n} \)

Where:

• \( n \) is the number of times interest is compounded per year (e.g., \( n=1 \) for annually, \( n=2 \) for half-yearly, \( n=4 \) for quarterly).

Scenario 1: Annual Compounding

Given:

• Rate (R) = 16% per annum
• Time (T) = 1 year
• Compounding frequency = Annually (n=1)

The amount after 1 year is:

\( A_{annual} = P \left(1 + \frac{16}{100}\right)^1 = P (1 + 0.16)^1 = P \times 1.16 \)

The compound interest for annual compounding is:

\( CI_{annual} = A_{annual} - P = 1.16P - P = 0.16P \)

Scenario 2: Half-Yearly Compounding

Given:

• Rate (R) = 16% per annum
• Time (T) = 1 year
• Compounding frequency = Half-yearly (n=2)

The rate per period is \( \frac{16\%}{2} = 8\% \). The number of periods in 1 year is \( 1 \times 2 = 2 \).

The amount after 1 year with half-yearly compounding is:

\( A_{half-yearly} = P \left(1 + \frac{8}{100}\right)^2 = P (1 + 0.08)^2 = P (1.08)^2 \)

Calculating \((1.08)^2\):

\( 1.08 \times 1.08 = 1.1664 \)

So,

\( A_{half-yearly} = P \times 1.1664 \)

The compound interest for half-yearly compounding is:

\( CI_{half-yearly} = A_{half-yearly} - P = 1.1664P - P = 0.1664P \)

Finding the Principal Sum

We are given that the interest would have increased by Rs. 64 if compounding had been done half-yearly instead of annually. This means the difference between the half-yearly compound interest and the annual compound interest is Rs. 64.

\( CI_{half-yearly} - CI_{annual} = 64 \)

Substitute the expressions for \( CI_{half-yearly} \) and \( CI_{annual} \):

\( 0.1664P - 0.16P = 64 \)

Simplify the equation:

\( (0.1664 - 0.16)P = 64 \)

\( 0.0064P = 64 \)

Now, solve for P:

\( P = \frac{64}{0.0064} \)

To divide by 0.0064, we can multiply both the numerator and denominator by 10000 to remove the decimal:

\( P = \frac{64 \times 10000}{0.0064 \times 10000} = \frac{640000}{64} \)

\( P = 10000 \)

Thus, the sum lent was Rs. 10000.

Verification

• CI (Annual) = 16% of 10000 = \( 0.16 \times 10000 = 1600 \)
• CI (Half-yearly) = \( 10000 \times (1.08)^2 - 10000 = 10000 \times 1.1664 - 10000 = 11664 - 10000 = 1664 \)
• Difference = \( 1664 - 1600 = 64 \)

The calculated difference matches the given difference (Rs. 64), confirming the principal is Rs. 10000.

The difference in CI is \( 1664 - 1600 = 64 \), which matches the problem statement.

Revision Table: Compound Interest Basics

Additional Information: Effect of Compounding Frequency

Increasing the frequency of compounding (from annually to half-yearly, quarterly, etc.) while keeping the nominal annual rate the same results in a higher effective annual rate and thus higher compound interest over the same time period. This is because interest is calculated and added to the principal more often, allowing subsequent interest calculations to be based on a larger sum.

• For a given nominal annual rate, the effective annual rate increases as the compounding frequency increases.
• Annual compounding: \( n=1 \)
• Half-yearly compounding: \( n=2 \)
• Quarterly compounding: \( n=4 \)
• Monthly compounding: \( n=12 \)
• Daily compounding: \( n=365 \)

The effective annual rate (EAR) for a nominal rate R compounded n times per year is given by:

\( EAR = \left(1 + \frac{R/n}{100}\right)^n - 1 \)

In our problem, the nominal annual rate is 16%:

• EAR (Annual) = \( \left(1 + \frac{16/1}{100}\right)^1 - 1 = (1.16)^1 - 1 = 1.16 - 1 = 0.16 \), or 16%
• EAR (Half-yearly) = \( \left(1 + \frac{16/2}{100}\right)^2 - 1 = (1.08)^2 - 1 = 1.1664 - 1 = 0.1664 \), or 16.64%

The difference in interest arises from the difference in these effective rates applied to the principal over the year.