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Answer :
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.