8. Find the amount and the compound interest on ₹ 10,000 for \(1\frac { 1 }{ 2 }\) years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

We are given: P = ₹ 10,000, n = \(1\frac { 1 }{ 2 }\)years and R = 10% per annum
Since the interest is compounded half yearly:

∴ n=\(1\frac{1}2 \) years

=\(\frac{3}2\times 2 = 3\) half yearly

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

Amount=\(P(1+\frac{R}{100})^n\)

=\(10,000(1+\frac{5}{100})^3\)

=\(10,000(\frac{21}{20})^3\)

=\(10,000\times(\frac{21}{20}\times \frac{21}{20}\times \frac{21}{20})\)

=\(\frac{5}4\times 9261 = \frac{46305}4\)

=\(₹ \; 11576.25\)

Thus, we have: Compound interest=A - P

=₹ 11576.25 - ₹ 10,000 = ₹ 1576.25

If the interest is compounded annually, then we have:

n=\(1\frac{1}2\) year and R=10 %

Therefore, we have SI= \(\frac{P \times R \times n}{100}\)

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

Principal for the second year:

=₹ 10,000 + ₹ 1,000

=₹ 11,000

∴ Interest for \(\frac{1}2\) year=

=\(\frac{11,000\times 10 \times 1}{100\times 2}=₹ 550\)

Total interest = ₹ 1,000 + ₹ 550 = ₹ 1,550 Difference between the two interests = ₹ 1,576.25 – ₹ 1,550 = ₹ 26.25

Hence, the interest will be ₹ 26.25 more when compounded half yearly than the interest when compounded annually.