A bank offers 10% per annum compound interest calculated on half-yearly basis. A customer deposits ₹3,60,000 each on 1st January and 1st July of a year. At the end of the year, find the amount he would have gained by way of interest.

Correct option is C

Given:

Deposit made on 1st January = ₹3,60,000

Deposit made on 1st July = ₹3,60,000

Rate of interest = 10% per annum, compounded half-yearly

Formula Used:
The compound interest formula is given by:

A = $$P \left(1 + \frac{r}{n}\right)^{nt}$$​

Where:

A is the amount (principal + interest)

P is the principal

r is the annual interest rate (in decimal form)

n is the number of times the interest is compounded per year

t is the time the money is invested for (in years)

Solution:

Since the interest is compounded half-yearly, n = 2

For the deposit made on 1st January:

Principal, P = 3,60,000

Rate, r = 10% or 0.10

Compounding frequency, n = 2 (half-yearly)

Time, t = 1 year

Using the compound interest formula:

​$$A_1 = 3,60,000 \left(1 + \frac{0.10}{2}\right)^{2 \times 1} = 3,60,000 \left(1 + 0.05\right)^2 = 3,60,000 \times 1.1025 = 3,96,900$$​

Interest earned from the 1st January deposit:

​$$\text{Interest}_1 = A_1 - P = 3,96,900 - 3,60,000 = 36,900$$​

For the deposit made on 1st July:

Principal, P = 3,60,000

Rate, r = 10% or 0.10

Compounding frequency, n = 2 (half-yearly)

Time, t =$$\frac{1}{2}$$​ year (6 months)

Using the compound interest formula:

​$$A_2 = 3,60,000 \left(1 + \frac{0.10}{2}\right)^{2 \times \frac{1}{2}} = 3,60,000 \left(1 + 0.05\right)^1 = 3,60,000 \times 1.05 = 3,78,000$$​

Interest earned from the 1st July deposit:

​$$\text{Interest}_2 = A_2 - P = 3,78,000 - 3,60,000 = 18,000$$​

Total interest earned:

​$$\text{Total Interest} = \text{Interest}_1 + \text{Interest}_2 = 36,900 + 18,000 = 54,900$$​

Thus, the total interest gained at the end of the year is ₹54,900