6. Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after \(1\frac { 1 }{ 2 }\) years if the interest is
(i) compounded annually.

(ii) compounded half yearly.

(i) We are given: P = ₹ 80,000 , R = 10% p.a. and n = \(1\frac { 1 }{ 2 }\) years Since the interest is compounded annually so we have

Simple Interest=\(\frac{P \times R \times n}{100}\)

=\(\frac{80,000\times 10\times 1}{100}\)

=\(₹ \;8,000 \)

Principal for the second year=P+SI

=₹ 80,000 + ₹ 8,000 = ₹ 88,000

Now for \(\frac{1}2\) year we have interest==\(\frac{88,000\times 10\times 1}{100\times 2}=₹ \;4,400\)

Thus, amount=₹ 88,000 + ₹ 4,400=₹ 92,400

(ii)So for compounded interest half yearly we have:

R=\(\frac{10}2=5\)% half yearly.

n=\(1\frac{1}2 \)years=\(\frac{3}2\times 2=3\) half years

∴ Amount=\(P(1+\frac{R}{100})^n\)

=\(80,000(1+\frac{5}{100})^3\)

=\(80,000(\frac{21}{20})^3\)

=\(80,000\times(\frac{21}{20}\times \frac{21}{20}\times \frac{21}{20})\)

=\(10\times 9261\)

=\(₹ \; 92610\)

Difference between the amounts = ₹ 92,610 – ₹ 92,400 = ₹ 210