question_answer 24)

                Find the amount and the compound interest on Rs 10,000 for \[1\frac{1}{2}\] years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Answer:

                When compounded half yearly P = Rs 10,000 R = 10% per annum \[=\frac{10}{2}%\]per half year = 5% per half year \[n=1\frac{1}{2}\] years \[=\frac{3}{2}\times 2\] half years = 3 half years \[\therefore \]  \[A=P{{\left( 1+\frac{R}{100} \right)}^{n}}\] \[=10,000{{\left( 1+\frac{5}{1000} \right)}^{3}}\] \[=10,000\,{{\left( 1+\frac{1}{20} \right)}^{3}}\] \[=10,000\,{{\left( \frac{21}{20} \right)}^{3}}\] \[=10,000\times \frac{21}{20}\times \frac{21}{20}\times \frac{21}{20}\] = Rs 11,576.25 This is the required amount. Now, CI = A ? P = Rs 11,576.25 ? Rs 10,000 = Rs 1,576.25 This is the required CI. When compounded annually P = Rs 10,000 R = 10% per annum \[n=1\] year \[\therefore \]  \[A=P{{\left( 1+\frac{R}{100} \right)}^{n}}\] \[=10,000{{\left( 1+\frac{10}{100} \right)}^{1}}\] \[=10,000\left( 1+\frac{1}{10} \right)\] \[=10,000\times \frac{11}{10}\,=\text{Rs}\,11,000\] = Principal for next \[\frac{1}{2}\] year \[\therefore \] Interest for first year = A ? p = Rs 11, 000 ? Rs 10, 000 = Rs 1000 SI on Rs 11,000 at 10% per annum for \[\frac{1}{2}\] year \[=\frac{11,000\times 10\times 1}{2\times 100}=\text{Rs}550\] \[\therefore \] Total compound interest = Rs 1,000 + Rs 550 = Rs 1,550 Hence, the interest when compounded half yearly would be more than the interest when compounded annually.