Amount for compound interest:

If P amount of money is invested for t years at r% rate of interest which compounded n times annually then total amount of money to be received is calculated as per the formula given below:

\(\displaystyle{A}={P}{\left({1}+\frac{r}{{n}}\right)}^{r}{t}\)

(a)

It is given that the principal amount of money is $10,000, rate of interest is \(\displaystyle{2.4}\%\), time is 10 years and the sum is compounded annually. This implies that the value of P is 10000, value of r is 0.024, value of t is 10 and value of n is 1.

\(\displaystyle{A}={10000}{\left({1}+\frac{0.024}{{1}}\right)}^{{{1}{\left({10}\right)}}}\)

\(\displaystyle={10000}{\left({1}+{0.024}\right)}^{10}\)

\(\displaystyle\approx{12676.51}\)

Thus, is the future value will be $12,676.51 in the account after 10 years

(b)

It is given that the principal amount of money is $10,000, rate of interest is \(\displaystyle{2.4}\%\), time is 10 years and the sum is compounded annually. This implies that the value of P is 10000, value of r is 0.024, value of t is 10 and value of n is 4.

\(\displaystyle{A}={10000}{\left({1}+\frac{0.024}{{4}}\right)}^{{{4}{\left({10}\right)}}}\)

\(\displaystyle={10000}{\left({1}+\frac{0.024}{{4}}\right)}{40}\)

\(\displaystyle\approx{12703.38}\)

Thus, is the future value will be $12703.37 in the account after 10 years

(c)

It is given that the principal amount of money is $10,000, rate of interest is \(\displaystyle{2.4}\%\), time is 10 years and the sum is compounded annually. This implies that the value of P is 10000, value of r is 0.024, value of t is 10 and value of n is 12.

\(\displaystyle{A}={10000}{\left({1}+\frac{{{0.024}}}{{12}}\right)}^{{{12}{\left({10}\right)}}}\)

\(\displaystyle={10000}{\left({1}+\frac{{{0.024}}}{{12}}\right)}{120}\)

\(\displaystyle\approx{12709.44}\)

Thus is the future value will be $12,709.44 in the account after 10 years

(d) It is given that the principal amount of money is $10,000, rate of interest is \(\displaystyle{2.4}\%\), time is 10 years and the sum is compounded annually. This implies that the value of P is 10000, value of r is 0.024, value of t is 10 and value of n is 365.

\(\displaystyle{A}={10000}{\left({1}+\frac{{{0.024}}}{{365}}\right)}^{{{365}{\left({10}\right)}}}\)

\(\displaystyle={10000}{\left({1}+\frac{{{0.024}}}{{365}}\right)}{3650}\)

\(\displaystyle\approx{12712.39}\)

Thus is the future value will be $12712.39 in the account after 10 years

If P amount of money is invested for t years at r% rate of interest which compounded n times annually then total amount of money to be received is calculated as per the formula given below:

\(\displaystyle{A}={P}{\left({1}+\frac{r}{{n}}\right)}^{r}{t}\)

(a)

It is given that the principal amount of money is $10,000, rate of interest is \(\displaystyle{2.4}\%\), time is 10 years and the sum is compounded annually. This implies that the value of P is 10000, value of r is 0.024, value of t is 10 and value of n is 1.

\(\displaystyle{A}={10000}{\left({1}+\frac{0.024}{{1}}\right)}^{{{1}{\left({10}\right)}}}\)

\(\displaystyle={10000}{\left({1}+{0.024}\right)}^{10}\)

\(\displaystyle\approx{12676.51}\)

Thus, is the future value will be $12,676.51 in the account after 10 years

(b)

It is given that the principal amount of money is $10,000, rate of interest is \(\displaystyle{2.4}\%\), time is 10 years and the sum is compounded annually. This implies that the value of P is 10000, value of r is 0.024, value of t is 10 and value of n is 4.

\(\displaystyle{A}={10000}{\left({1}+\frac{0.024}{{4}}\right)}^{{{4}{\left({10}\right)}}}\)

\(\displaystyle={10000}{\left({1}+\frac{0.024}{{4}}\right)}{40}\)

\(\displaystyle\approx{12703.38}\)

Thus, is the future value will be $12703.37 in the account after 10 years

(c)

It is given that the principal amount of money is $10,000, rate of interest is \(\displaystyle{2.4}\%\), time is 10 years and the sum is compounded annually. This implies that the value of P is 10000, value of r is 0.024, value of t is 10 and value of n is 12.

\(\displaystyle{A}={10000}{\left({1}+\frac{{{0.024}}}{{12}}\right)}^{{{12}{\left({10}\right)}}}\)

\(\displaystyle={10000}{\left({1}+\frac{{{0.024}}}{{12}}\right)}{120}\)

\(\displaystyle\approx{12709.44}\)

Thus is the future value will be $12,709.44 in the account after 10 years

(d) It is given that the principal amount of money is $10,000, rate of interest is \(\displaystyle{2.4}\%\), time is 10 years and the sum is compounded annually. This implies that the value of P is 10000, value of r is 0.024, value of t is 10 and value of n is 365.

\(\displaystyle{A}={10000}{\left({1}+\frac{{{0.024}}}{{365}}\right)}^{{{365}{\left({10}\right)}}}\)

\(\displaystyle={10000}{\left({1}+\frac{{{0.024}}}{{365}}\right)}{3650}\)

\(\displaystyle\approx{12712.39}\)

Thus is the future value will be $12712.39 in the account after 10 years