1. Calculate the amount and compound interest on
(a)₹ 10,800 for 3 years at \(12\frac { 1 }{ 2 }\) % per annum compounded annually.

(b)₹ 18,000 for \(2\frac { 1 }{ 2 }\) years at 10% per annum compounded annually.

(c)₹ 62,500 for \(1\frac { 1 }{ 2 }\) years at 8% per annum compounded half yearly.

(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify).

(e)₹ 10,000 for 1 year at 8% per annum compounded half yearly.

(a) Given, P=₹ 10,800,n=3 years.
We have, R=\(12\frac{1}2\)% = \(\frac{25}2\)%p.a.

So, we know that \(Amount=P(1+\frac{R}{100})^n\)

\(=10,800(1+\frac{25}{100})^3\)

\(=10,800(\frac{9}{8})^3\)

\(=10,800\times(\frac{9}{8})\times(\frac{9}{8})\times(\frac{9}{8})\)

\(=\frac{4,92,075}{32}=₹ \; 15,377.34\)
Thus, we have Compound interest=A - P

=₹ 15,377.34-₹ 10,800 = ₹ 4,577.35

Hence we have, amount = ₹ 15,377.34 and CI = ₹ 4,577.34

(b)We have: P = ₹ 18,000, n = \(2\frac { 1 }{ 2 }\) years and R = 10% p.a.
The amount for \(2\frac { 1 }{ 2 }\) years, can be calculated by first calculating the amount to 2 years using CI formula and then calculating the simple interest by using SI formula.

The amount for 2 years:

Amount = \(18000(1+\frac{10}{100})^2\)

= \(18000\times(\frac{110}{100})^2\)

= \(18000\times(\frac{11}{10})\times(\frac{11}{10})=₹ \; 21,780\)

∴ Interest after 2 years = Amount - P

=21,780-18,000=₹ 3,780

So, now taking principal amount as ₹ 21,780, the SI for the next \(\frac{1}2\)years will be as:

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

=\(\frac{21,780\times10\times1}{100\times2}\)

=\(₹\; 1,089\)

Therefore,Total CI = ₹ 3780 +₹ 1,089= ₹ 4,869

Amount = P + I = ₹ 21,780 + ₹ 1,089 =₹ 22,869

Hence, the amount = ₹ 22,869 and CI = ₹ 4,869
(c)Given, P=₹ 62,500,n=\(1\frac{1}2=\frac{3}2\) years per annum compounded half yearly i.e., \(\frac{3}2\times 2\) years=3 half years.
We have, R=\(8\)%=\(\frac{8}2\)% = \(4\)%half yearly

So, we know that \(Amount=P(1+\frac{R}{100})^n\)

\(=62,500(1+\frac{4}{100})^3\)

\(=62,500(\frac{26}{25})^3\)

\(=62,500\times(\frac{26}{25})\times(\frac{26}{25})\times(\frac{26}{25})\)

\(=4 \times 26\times 26\times 26=₹ \; 70,304\)
Thus, we have Compound interest=A - P

=₹ 70,304-₹ 62,500 = ₹ 7,804

Hence we have, amount = ₹ 70,304 and CI = ₹ 7804
(d)Given, P=₹ 8,000,n=1 years and R = 9% per annum compounded half yearly
Since, the interest is compounded half yearly n =\(1 \times 2 = 2\) half years.So, R=\(\frac{9}2\)% per half years.

So, we know that \(Amount=P(1+\frac{R}{100})^n\)

\(=8,000(1+\frac{9}{2\times100})^2\)

\(=8,000(\frac{209}{200})^2\)

\(=8,000\times(\frac{209}{200})\times(\frac{209}{200})\)

\(=₹ \;(\frac{8,7362}{10})\)

\(=₹ \; 8,736.20\)

Thus, we have Compound interest=A - P

=₹ 8,736.20-₹ 8,000 = ₹ 736.20

Hence we have, amount = ₹ 8736.20 and CI = ₹ 736.20
(e)Given, P=₹ 10,000,n=1 years and R = 8% per annum compounded half yearly
Since, the interest is compounded half yearly n =\(1 \times 2 = 2\) half years.So, R=\(\frac{8}2=4\)% per half years.

So, we know that \(Amount=P(1+\frac{R}{100})^n\)

\(=10,000(1+\frac{4}{100})^2\)

\(=10,000(\frac{26}{25})^2\)

\(=10,000\times(\frac{26}{25})\times(\frac{26}{25})\)

\(=16\times 26\times 26\)

\(=₹ \; 10,816\)

Thus, we have Compound interest=A - P

=₹ 10,816 -₹ 10,000 = ₹ 816

Hence we have, amount = ₹ 10,816 and CI = ₹ 816