question_answer

Calculate the amount and compound interest on.

(a) Rs. 10,800 for 3 years at \[12\frac{1}{2}%\] per annum compounded annually.

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

Answer:

Here, Principal (P) = Rs. 10,800

Rate (R) = \[12\frac{1}{2}%=\frac{25}{2}%\]

Number of years (n) = 3

We have, A = P\[{{\left( 1\times \frac{R}{100} \right)}^{n}}\]

= 10,800 \[{{\left( 1+\frac{25}{2\times 100} \right)}^{3}}\]

= 10,800 \[{{\left( 1+\frac{1}{2\times 4} \right)}^{3}}\]         = 10,800 \[{{\left( 1+\frac{1}{8} \right)}^{3}}\]

= 10,800 \[{{\left( \frac{9}{8} \right)}^{3}}\]

\[=10,800\times \frac{9}{8}\times \frac{9}{8}\times \frac{9}{8}\]

= Rs. 15377.34

C. I. = A - P = Rs. 15377.36 \[-\] Rs. 10800

= Rs. 4577.34                                                                                           (b) Here, Principal (P) = Rs.  18000

Rate (R) = 10%, Time (n) =\[\frac{1}{2}\] years

A = P \[{{\left( 1+\frac{R}{100} \right)}^{n}}\]

\[=18000{{\left( 1+\frac{10}{100} \right)}^{2}}\]

\[=18000{{\left( 1+\frac{1}{10} \right)}^{2}}\]

\[=18000{{\left( \frac{11}{10} \right)}^{2}}\]

\[=18000\times \frac{11}{10}\times \frac{11}{10}\]

= Rs. 21780

Interest for \[\frac{1}{2}\] on Rs. 21,780 rate of 10%

\[=\frac{1}{2}\times \frac{21780\times 10\times 1}{100}=\]Rs. 1089

Total amount for \[2\frac{1}{2}\]years = Rs. 21780 + Rs. 1089

= Rs. 22869

\[C.\text{ }I.\text{ }=\text{ }A\text{ }\text{ }P\text{ }=\]Rs. 22869 - Rs 18000 = Rs. 4869