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Answer :
(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