question_answer 22)

                Arif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after \[1\frac{1}{2}\] years if the interest is (i) compounded annually (ii) compounded half yearly.

Answer:

                (i) Compounded annually P = Rs 80,000 R = 10% per annum \[n=1\]year \[\therefore \]  \[A=P{{\left( 1+\frac{R}{100} \right)}^{n}}\] \[=80,000{{\left( 1+\frac{10}{100} \right)}^{1}}\] \[=80,000\left( 1+\frac{1}{10} \right)\] \[=80,000\times \frac{11}{10}\] = Rs 88,000 SI on Rs 88,000 at 10% per annum for \[\frac{1}{2}\] year \[=\frac{88,000\times 10\times 1}{2\times 100}=\text{Rs}4,400\] \[\therefore \] Required amount = Rs 88,000 + Rs 4,400 = Rs 92,400 (ii) Compounded half yearly P = Rs 80,000 R = 10% per annum \[=\frac{10}{2}%\] per half year 5% per half year \[n=1\frac{1}{2}\] years \[=1\frac{1}{2}\times 2\] half years = 3 half years \[\therefore \]  \[A=P\,{{\left( 1+\frac{R}{100} \right)}^{n}}\] \[=80,000{{\left( 1+\frac{5}{100} \right)}^{3}}\]  \[=80,000{{\left( 1+\frac{1}{20} \right)}^{3}}\] \[=80,000{{\left( \frac{21}{20} \right)}^{3}}\] \[=80,000\times \,\frac{21}{20}\times \frac{21}{20}\times \frac{21}{20}\] =  Rs 92,610 This is the required amount. Difference in amounts = Rs 92,610 ? Rs 92,400 = Rs 210. Hence, the difference in amounts is Rs 210.