DAV Class 8 Maths Chapter 6 Worksheet 3

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DAV Class 8 Maths Chapter 6 Worksheet 3

Compound Interest Worksheet 3


1. Find the amount for ₹ 15,000 at 8% per annum compounded annually for two years.

Solution

\[ \begin{aligned} P &= \text{₹ }15000 \\ R &= 8\% \\ n &= 2\ \text{years} \end{aligned} \]\[ \begin{aligned} A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 15000\left(1 + \frac{8}{100}\right)^2 \\[6pt] &= 15000\left(\frac{108}{100}\right)^2 \\[6pt] &= 15000\left(\frac{27}{25}\right)^2 \\[6pt] &= \cancel{15000}^{\cancel{600}^{ \ 24}} \times \frac{27}{\cancel{25}_1} \times \frac{27}{\cancel{25}_1} \\[6pt] &= 24 \times 27 \times 27 \\[6pt] &= 24 \times 729 \\[6pt] A &= \text{₹ }17496 \end{aligned} \]

Answer Amount after 2 years \(=\; \color{red}{\text{₹ }17{,}496}\)

2. Find the compound interest on ₹ 11,200 at \(17\dfrac{1}{2}\%\) per annum for two years.

Solution

\[ \begin{aligned} P &= \text{₹ }11200 \\[6pt] R &= 17\dfrac{1}{2}\% \implies \frac{35}{2}\% \\[6pt] n &= 2\ \text{years} \end{aligned} \]\[ \begin{aligned} A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 11200\left(1 + \frac{35}{200}\right)^2 \\[6pt] &= 11200\left(1 + \frac{7}{40}\right)^2 \\[6pt] &= 11200\left(\frac{47}{40}\right)^2 \\[6pt] &= 11200 \times \frac{47}{40} \times \frac{47}{40} \\[6pt] &= \cancel{112}^{\cancel{28}^7} \times \frac{47}{\cancel4_1} \times \frac{47}{\cancel4_1} \\[6pt] &= 7 \times 47 \times 47 \\[6pt] &= 7 \times 2209 \\[6pt] A &= \text{₹ }15463 \\[6pt] \text{CI} &= A - P \\[4pt] &= 15463 - 11200 \\[4pt] &= 4263 \end{aligned} \]

Answer Compound Interest \(=\; \color{red}{\text{₹ }4{,}263}\)

3. Ram borrowed a sum of ₹ 30,000 from Shyam for three years. If the rate of interest is 6% per annum compounded annually, find the interest paid by Ram to Shyam after three years.

Solution

\[ \begin{aligned} P &= \text{₹ }30000 \\[6pt] R &= 6\% \\[6pt] n &= 3\ \text{years} \end{aligned} \]\[ \begin{aligned} A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 30000\left(1 + \frac{6}{100}\right)^3 \\[6pt] &= 30000\left(1 + \frac{3}{50}\right)^3 \\[6pt] &= 30000\left(\frac{53}{50}\right)^3 \\[6pt] &= 30000 \times \frac{53}{50} \times \frac{53}{50} \times \frac{53}{50} \\[6pt] &= \cancel{30}^{\ 6} \times \frac{53}{\cancel5_1} \times \frac{53}{5} \times \frac{53}{5} \\[6pt] &= \frac{6 \times 148877}{25} \\[6pt] &= \frac{893262}{25} \\[6pt] A &= \text{₹ }35730.48\\\\ C.I. &= A - P \\[4pt] &= 35730.48 - 30000 \\[4pt] &= 5730.48 \end{aligned} \]

Answer Interest paid by Ram after 3 years \(=\; \color{red}{\text{₹ }5{,}730.48}\)

4. Nidhi deposited ₹ 7,500 in a bank which pays her 4% interest per annum compounded annually. Find the amount and the interest received by her after three years.

Solution

\[ \begin{aligned} P &= \text{₹ }7500 \\ R &= 4\% \\ n &= 3 \text{ years} \\[8pt] A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 7500\left(1 + \frac{4}{100}\right)^3 \\[6pt] &= 7500\left(1 + \frac{1}{25}\right)^3 \\[6pt] &= 7500\left(\frac{26}{25}\right)^3 \\[6pt] &= \cancel{7500}^{\cancel{300}^{12}} \times \frac{26}{\cancel{25}_1} \times \frac{26}{\cancel{25}_1} \times \frac{26}{25} \\[6pt] &= \frac{12 \times 17576}{25} \\[6pt] &= \frac{210912}{25} \\[6pt] A &= \text{₹ }8436.48 \end{aligned} \]\[ \begin{aligned} \text{C.I. for 3 years} &= A - P \\[4pt] &= 8436.48 - 7500 \\[4pt] &= \text{₹ }936.48 \end{aligned} \]

Answer
Amount after 3 years \(= \color{red}{\text{₹ }8{,}436.48 } \) , Interest received \(=\; \color{red}{\text{₹ }936.48}\)

5. Find the difference between the compound interest and the simple interest on ₹ 30,000 at 7% per annum for three years.

Solution

\[ \begin{aligned} P &= \text{₹ }30000 \\ R &= 7\% \\ n &= 3 \text{ years} \end{aligned} \] \[ \begin{aligned} &\color{magenta}\textbf{Compound Interest } \\[6pt] A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 30000\left(1 + \frac{7}{100}\right)^3 \\[6pt] &= 30000\left(\frac{107}{100}\right)^3 \\[6pt] &= 30000 \times \frac{107 \times 107 \times 107}{100 \times 100 \times 100} \\[6pt] &= \frac{3 \times 1225043}{100} \\[6pt] &= \frac{3675129}{100} \\[6pt] A&= \text{₹ }36751.29 \\[6pt] \text{C.I.} &= A - P \\[4pt] &= 36751.29 - 30000 \\[4pt] \text{C.I.} &= \text{₹ }6751.29 \\ \end{aligned} \] \[ \begin{aligned} &\color{magenta}\textbf{Simple Interest } \\[6pt] \text{S.I.} &= \frac{P \times R \times T}{100} \\[6pt] &= \frac{30000 \times 7 \times 3}{100} \\[6pt] &= 300 \times 21 \\[6pt] \text{S.I.} &= \text{₹ }6300 \end{aligned} \]\[ \begin{aligned} \text{Difference} &= \text{C.I.} - \text{S.I.} \\[4pt] &= 6751.29 - 6300 \\[4pt] &= \text{₹ }451.29 \end{aligned} \]

Answer Difference between C.I. and S.I. \(= \color{red}{\text{₹ }451.29}\)

6. Aman borrows ₹ 14,500 at 11% per annum for three years at simple interest and Tarun borrows the same amount at 10% per annum for the same time compounded annually. Who pays more interest and by how much?

Solution

\[ \begin{aligned} &\color{magenta}\textbf{Aman (Simple Interest)} \\[6pt] P &= \text{₹ }14500 \\ R &= 11\% \\ T &= 3\ \text{years} \\[6pt] \text{S.I.} &= \frac{P \times R \times T}{100} \\[6pt] &= \frac{14500 \times 11 \times 3}{100} \\[6pt] &= 145 \times 11 \times 3 \\[6pt] &= 1595 \times 3 \\[6pt] \text{S.I.} &= \text{₹ }4785 \end{aligned} \]\[ \begin{aligned} &\color{magenta}\textbf{Tarun (Compound Interest)} \\[6pt] P &= \text{₹ }14500 \\ R &= 10\% \\ n &= 3\ \text{years} \\[6pt] A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 14500\left(1 + \frac{10}{100}\right)^3 \\[6pt] &= 14500\left(1 + \frac{1}{10}\right)^3 \\[6pt] &= 14500\left(\frac{11}{10}\right)^3 \\[6pt] &= 14500 \times \frac{1331}{1000} \\[6pt] &= 14.5 \times 1331 \\[6pt] A &= \text{₹ }19299.50 \\[6pt] \text{C.I.} &= A - P \\[4pt] &= 19299.50 - 14500 \\[4pt] \text{C.I.} &= \text{₹ }4799.50 \\ \end{aligned} \] \[ \begin{aligned} \text{Extra interest paid by Tarun} &= 4799.50 - 4785 \\[4pt] &= \text{₹ }14.50 \end{aligned} \]

AnswerTarun pays more interest by \( \color{red}{\text{₹ }14.50} \)

7. The simple interest on a certain sum of money for two years at \(5\dfrac{1}{2}\%\) is ₹ 6,600. What will be the compound interest on that sum at the same rate for the same time period?

Solution

\[ \begin{aligned} P & = P \\[4pt] R &= 5\dfrac{1}{2}\% \implies \frac{11}{2}\% \\[6pt] T &= 2\ \text{years} \\[6pt] \text{S.I.} &= \text{₹ }6600 \\[6pt] \frac{P \times R \times T}{100} &= \text{₹ }6600 \\[8pt] \frac{P \times \frac{11}{2} \times 2}{100} &= 6600 \\[8pt] \frac{P \times 11}{100} &= 6600 \\[8pt] P &= \frac{\cancel{6600}^{ \ 600} \times 100}{\cancel{11}} \\[6pt] P &= \text{₹ }60000 \end{aligned} \]\[ \begin{aligned} &\color{magenta}\textbf{Compound Interest} \\[6pt] P &= \text{₹ }60000 \\[6pt] R &= 5\dfrac{1}{2}\% \implies \frac{11}{2}\% \\[6pt] n &= 2 \\[6pt] A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 60000\left(1 + \frac{\frac{11}{2}}{100}\right)^2 \\[6pt] &= 60000\left(1 + \frac{11}{200}\right)^2 \\[6pt] &= 60000\left(\frac{200 + 11}{200}\right)^2 \\[6pt] &= 60000\left(\frac{211}{200}\right)^2 \\[6pt] &= 60000 \times \frac{211 \times 211}{200 \times 200} \\[6pt] &= \cancel6^3 \times \frac{211 \times 211}{\cancel2 \times 2} \\[6pt] &= \frac{3 \times 44521}{2} \\[6pt] &= \frac{133563}{2} \\[6pt] &= \text{₹ }66781.50 \\[6pt] \text{C.I.} &= A - P \\[4pt] &= 66781.50 - 60000 \\[4pt] &= \text{₹ }6781.50 \end{aligned} \]

Answer Compound Interest for 2 years \(=\; \color{red}{\text{₹ }6{,}781.50}\)

8. A certain sum amounts to ₹ 2,970.25 in two years at 9% per annum compounded annually. Find the sum.

Solution

\[ \begin{aligned} P & = P \\[6pt] R &= 9\% \\[6pt] n &= 2\ \text{years} \\[6pt] Amount &= \text{₹ }2970.25 \\[8pt] P\left(1 + \frac{R}{100}\right)^n &= \text{₹ }2970.25 \\[6pt] P\left(1 + \frac{9}{100}\right)^2 &= \text{₹ }2970.25 \\[6pt] P\left(\frac{109}{100}\right)^2 &= \text{₹ }2970.25 \\[6pt] P \times \frac{109 \times 109}{100 \times 100} &= \text{₹ }2970.25 \\[8pt] P &= \frac{2970.25 \times 100 \times 100}{109 \times 109} \\ \\ P &= \frac{297025 \times 100}{109 \times 109} \\ \\ P &= \frac{{\cancel{297025}^{\cancel{2725}}}^{25} \times 100}{\cancel{109}_1 \times \cancel{109}_1} \\[8pt] P &= \text{₹ }2500 \end{aligned} \]

AnswerThe required sum \(= \color{red}{\text{₹ }2{,}500}\)

9. On what sum will the compound interest at \(7\dfrac{1}{2}\%\) per annum for three years, compounded annually, be ₹ 3,101.40?

Solution

\[ \begin{aligned} P &= P \\[4pt] R &= 7\dfrac{1}{2}\% \implies \frac{15}{2}\% \\[4pt] n &= 3 \text{ years} \\[8pt] \text{C.I.} &= \text{₹ }3101.40 \\[8pt] P\left[\left(1 + \frac{R}{100}\right)^n - 1\right] &= \text{₹ }3101.40 \\[8pt] P\left[\left(1 + \frac{15}{200}\right)^3 - 1\right] &= \text{₹ }3101.40 \\[8pt] P\left[\left(1 + \frac{3}{40}\right)^3 - 1\right] &= \text{₹ }3101.40 \\[8pt] P\left[\left(\frac{40 + 3}{40}\right)^3 - 1\right] &= \text{₹ }3101.40 \\[8pt] P\left[\left(\frac{43}{40}\right)^3 - 1\right] &= \text{₹ }3101.40 \\[8pt] P\left[\frac{79507}{64000} - 1\right] &= \text{₹ }3101.40 \\[8pt] P\left[\frac{79507 - 64000}{64000} \right] &= \text{₹ }3101.40 \\[8pt] P\left[\frac{15507}{64000} \right] &= \text{₹ }3101.40 \\[8pt] P &= \frac{3101.40 \times 64000}{15507}\\[8pt] P &= \frac{31014 \times 6400}{15507}\\[8pt] P &= \frac{\cancel{31014}^{ \ 2} \times 6400}{\cancel{15507}_1}\\[8pt] P &= 2 \times 6400 \\[8pt] P &= \text{₹ }12800 \end{aligned} \]

Answer Required sum \(= \color{red}{\text{₹ }12{,}800}\)

10. At what rate per cent will a sum of ₹ 640 be compounded to ₹ 774.40 in two years?

Solution

\[ \begin{aligned} P &= \text{₹ }640 \\[6pt] n &= 2 \text{ years} \\[6pt] R &= R \\[6pt] A &= \text{₹ }774.40 \\[8pt] P\left(1 + \frac{R}{100}\right)^n &= 774.40 \\[8pt] 640 \left(1 + \frac{R}{100}\right)^2 &= 774.40 \\[8pt] \left(\frac{100 + R}{100}\right)^2 &= \frac{774.40 \times 10}{640 \times 10} \\[8pt] \left(\frac{100 + R}{100}\right)^2 &= \frac{7744}{6400} \\[8pt] \left(\frac{100 + R}{100}\right)^2 &= \frac{\cancel{7744}^{121}}{\cancel{6400}_{100}} \\[8pt] \left(\frac{100 + R}{100}\right)^2 &= \frac{121}{100} \\[8pt] \frac{100 + R}{100} &= \sqrt{\frac{121}{100}} \\[8pt] \frac{100 + R}{100} &= \frac{11}{10} \\[8pt] 100 + R &= \frac{11}{10} \times 100 \\[8pt] 100 + R &= 110 \\[8pt] R &= 110 - 100 \\[8pt] R &= 10\% \end{aligned} \]

Answer Rate of interest \(=\; \color{red}{10\% \text{ per annum}}\)

11. At what rate per cent will a sum of ₹ 64,000 be compounded to ₹ 68,921 in three years?

Solution

\[ \begin{aligned} P &= \text{₹ }64000 \\[6pt] n &= 3 \text{ years} \\[6pt] R &= R \\[8pt] A &= \text{₹ }68921 \\[8pt] P\left(1 + \frac{R}{100}\right)^n &= 68921 \\[8pt] 64000\left(1 + \frac{R}{100}\right)^3 &= 68921 \\[8pt] \left(1 + \frac{R}{100}\right)^3 &= \frac{68921}{64000} \\[8pt]1 + \frac{R}{100} &= \sqrt[3]{\frac{68921}{64000}} \\[8pt] 1 + \frac{R}{100} &= \frac{41}{40} \\[8pt] \frac{R}{100} &= \frac{41}{40} - 1 \\[8pt] \frac{R}{100} &= \frac{41 - 40}{40} \\[8pt] \frac{R}{100} &= \frac{1}{40} \\[8pt] R &= \frac{100}{40} \\[8pt] R &= \frac{10}{4} \\[8pt] R &= 2\dfrac{1}{2}\% \end{aligned} \]

Answer Rate of interest \(=\; \color{red}{2\dfrac{1}{2}\% \text{ per annum}}\)

12. In how many years will ₹ 8,000 amount to ₹ 9,261 at 5% per annum compounded annually?

Solution

\[ \begin{aligned} P &= \text{₹ }8000 \\[6pt] R &= 5\% \\[6pt] n &= n \\[6pt] A &= \text{₹ }9261 \\[8pt] P\left(1 + \frac{R}{100}\right)^n &= \text{₹ }9261 \\[8pt] 8000\left(1 + \frac{5}{100}\right)^n &= 9261 \\[8pt] \left(1 + \frac{1}{20}\right)^n &= \frac{9261}{8000} \\[8pt] \left(\frac{20+1}{20}\right)^n &= \frac{9261}{8000} \\[8pt] \left(\frac{21}{20}\right)^n &= \frac{9261}{8000} \\[8pt] \left(\frac{21}{20}\right)^n &= \left(\frac{21}{20}\right)^3 \\[8pt] \Rightarrow n &= 3 \end{aligned} \]

Answer Required time \(=\; \color{red}{3 \text{ years}}\)

13. In what time will a sum of ₹ 3,750 at 20% per annum compounded annually amount to ₹ 6,480?

Solution

\[ \begin{aligned} P &= \text{₹ }3750 \\ R &= 20\% \\ n &= n \\ A &= \text{₹ }6480 \\[8pt] P\left(1 + \frac{R}{100}\right)^n &= \text{₹ }6480 \\[8pt] 3750\left(1 + \frac{20}{100}\right)^n &= 6480 \\[8pt] \left(1 + \frac{20}{100}\right)^n &= \frac{6480}{3750} \\[8pt] \left(1 + \frac{1}{5}\right)^n &= \frac{648}{375} \\[8pt] \left(\frac{5 + 1}{5}\right)^n &= \frac{216}{125} \\[8pt] \left(\frac{6}{5}\right)^n &= \frac{216}{125} \\[8pt] \left(\frac{6}{5}\right)^n &= \left(\frac{6}{5}\right)^3 \\[8pt] \Rightarrow n &= 3 \end{aligned} \]

Answer Required time \(=\; \color{red}{3 \text{ years}}\)

14. The difference between the compound interest and the simple interest on a certain sum of money at \(6\dfrac{2}{3}\%\) per annum for three years is ₹ 46. Find the sum.

Solution

\[ \begin{aligned} P &= P \\[4pt] R &= 6\dfrac{2}{3}\% \implies \frac{20}{3}\% \\[4pt] n &= 3 \text{ years} \\[4pt] CI - SI & = \text{₹ }46 \\[8pt] \end{aligned} \] \[ \begin{aligned} P\left[\left(1 + \frac{R}{100}\right)^n - 1\right] - \frac{P \times R \times T}{100} & = 46 \\[8pt] P\left[\left(1 + \frac{20}{300}\right)^3 - 1\right] - \frac{P \times 20 \times 3}{300} & = 46 \\[8pt] P\left[\left(1 + \frac{\cancel{20}^1}{\cancel{300}_{15}}\right)^3 - 1\right] - \frac{P \times \cancel{60}^1}{\cancel{300}_5} & = 46 \\[8pt] P\left[\left(1 + \frac{1}{15}\right)^3 - 1\right] - \frac{1}{5}P & = 46 \\[8pt] P\left[\left(\frac{15+1}{15}\right)^3 - 1\right] - \frac{1}{5}P & = 46 \\[8pt] P\left[\left(\frac{16}{15}\right)^3 - 1\right] - \frac{1}{5}P & = 46 \\[8pt] P\left[\frac{4096}{3375} - 1\right] - \frac{1}{5}P & = 46 \\[8pt] P\left[\frac{4096-3375}{3375} \right] - \frac{1}{5}P & = 46 \\[8pt] \frac{721}{3375}P - \frac{1}{5}P & = 46 \\[8pt] P\left(\frac{721}{3375} - \frac{1}{5}\right) & = 46 \\[8pt] P\left(\frac{721 - 675}{3375} \right) & = 46 \\[8pt] P\left(\frac{46}{3375} \right) & = 46 \\[8pt] P &= \frac{46 \times 3375}{46} \\[8pt] P &= \text{₹ } 3375 \end{aligned} \]

Answer Sum of money \(=\; \color{red}{\text{₹ }3{,}375}\)

15. The difference between the compound interest and the simple interest on a certain sum of money at 15% per annum for three years is ₹ 283.50. Find the sum.

Solution

\[ \begin{aligned} P &= P \\[4pt] R &= 15\% \\[4pt] n &= 3 \text{ years} \\[4pt] CI - SI & = \text{₹ }283.50 \end{aligned} \]\[ \begin{aligned} P\left[\left(1 + \frac{R}{100}\right)^n - 1\right] - \frac{P \times R \times T}{100} &= 283.50 \\[8pt] P\left[\left(1 + \frac{\cancel{15}^3}{\cancel{100}_{20}}\right)^3 - 1\right] - \frac{P \times \cancel{15}^3 \times 3}{\cancel{100}_{20}} &= 283.50 \\[8pt] P\left[\left(1 + \frac{3}{20}\right)^3 - 1\right] - \frac{9}{20}P &= 283.50 \\[8pt] P\left[\left(\frac{20+3}{20}\right)^3 - 1\right] - \frac{9}{20}P &= 283.50 \\[8pt] P\left[\left(\frac{23}{20}\right)^3 - 1\right] - \frac{9}{20}P &= 283.50 \\[8pt] P\left[\frac{12167}{8000} - 1\right] - \frac{9}{20}P &= 283.50 \\[8pt] P\left[\frac{12167 - 8000}{8000}\right] - \frac{9}{20}P &= 283.50 \\[8pt] P\left(\frac{4167}{8000}\right) - \frac{9}{20}P &= 283.50 \\[8pt] \frac{4167}{8000}P - \frac{9}{20}P &= 283.50 \\[8pt] \left(\frac{4167}{8000} - \frac{9}{20}\right) P &= 283.50 \\[8pt] \left(\frac{4167 - 3600}{8000} \right) P &= 283.50 \\[8pt] \left(\frac{567}{8000} \right) P &= 283.50 \\[8pt] P &= \frac{283.50 \times 8000}{567} \\[8pt] P &= \frac{\cancel{2835}^5 \times 800}{\cancel{567}_1} \\[8pt] P &= \text{₹ }4000 \end{aligned} \]

Answer Sum of money \(=\; \color{red}{\text{₹ }4{,}000}\)

16. Find the amount and the compound interest on ₹ 12,800 for one year at \(7\dfrac{1}{2}\%\) per annum, compounded semi-annually.

Solution

\[ \begin{aligned} P &= \text{₹ }12800 \\[4pt] R &= 7\dfrac{1}{2}\% \implies \frac{15}{2}\% \text{ per annum} \\[4pt] n &= 1 \\[4pt] &\color{magenta}\textbf{CI - half yearly} \\[6pt] A &= P\left(1 + \frac{R}{200}\right)^{2n} \\[8pt] &= 12800 \left(1 + \frac{\frac{15}{2}}{200}\right)^{2 \times 1} \\[8pt] &= 12800\left(1 + \frac{15}{400}\right)^2 \\[8pt] &= 12800\left(\frac{400 + 15}{400}\right)^2 \\[8pt] &= 12800\left(\frac{415}{400}\right)^2 \\[8pt] &= 12800\left(\frac{83}{80}\right)^2 \\[8pt] &= 12800 \times \frac{83 \times 83}{80 \times 80} \\[8pt] &= \frac{{\cancel{128}^{\cancel{16}}}^{\ 2} \times 6889}{\cancel8_1 \times \cancel8_1} \\[8pt] &= 2 \times 6889 \\[8pt] A &= \text{₹ }13778 \\[8pt] \text{C.I.} &= A - P \\[4pt] &= 13778 - 12800 \\[4pt] &= \text{₹ }978 \end{aligned} \]

Answer Amount after 1 year \(= \color{red}{\text{₹ }13{,}778}\) , Compound Interest \(= \color{red}{\text{₹ }978}\)

17. Mr Arora borrowed ₹ 40,960 from a bank to start a play school. If the bank charges \(12\dfrac{1}{2}\%\) per annum compounded half-yearly, what amount will he have to pay after \(1\dfrac{1}{2}\) years?

Solution

\[ \begin{aligned} P &= \text{₹ }40960 \\[4pt] R &= 12\dfrac{1}{2}\% \implies \frac{25}{2}\% \\[4pt] n &= 1\dfrac{1}{2}\ \text{years} \end{aligned} \]\[ \begin{aligned} &\color{magenta}\textbf{CI - half yearly} \\[6pt] A &= P\left(1 + \frac{R}{200}\right)^{2n} \\[6pt] &= 40960\left(1 + \frac{\frac{25}{2}}{200}\right)^{2 \times \frac{3}{2}} \\[6pt] &= 40960\left(1 + \frac{25}{400}\right)^3 \\[6pt] &= 40960\left(1 + \frac{1}{16}\right)^3 \\[6pt] &= 40960\left(\frac{16+1}{16}\right)^3 \\[6pt] &= 40960\left(\frac{17}{16}\right)^3 \\[6pt] &= 40960 \times \frac{4913}{4096} \\[6pt] &= \frac{\cancel{40960}^{\ 10} \times 4913}{\cancel{4096}_1} \\[6pt] A &= \text{₹ }49130 \\[6pt] \text{C.I.} &= A - P \\[4pt] &= 49130 - 40960 \\[4pt] &= \text{₹ }8170 \end{aligned} \]

Answer Amount to be paid after \(1\dfrac{1}{2}\) years \(= \color{red}{\text{₹ }49{,}130}\) , Compound Interest \(= \color{red}{\text{₹ }8{,}170}\)

18. Meera lent out ₹ 20,000 for nine months at 20% per annum compounded quarterly to Mrs Sharma. What amount will she get after the expiry of the period?

Solution

\[ \begin{aligned} P &= \text{₹ }20000 \\[4pt] R &= 20\% \\[4pt] n &= 9 \text{ months } = \frac{9}{12} \implies \frac{3}{4}\ \text{years} \end{aligned} \]\[ \begin{aligned} &\color{magenta}\textbf{CI - quarterly} \\[6pt] A &= P\left(1 + \frac{R}{400}\right)^{4n} \\[6pt] &= 20000\left(1 + \frac{20}{400}\right)^{4 \times \frac{3}{4}} \\[6pt] &= 20000\left(1 + \frac{1}{20}\right)^3 \\[6pt] &= 20000\left(\frac{20+1}{20}\right)^3 \\[6pt] &= 20000\left(\frac{21}{20}\right)^3 \\[6pt] &= \frac{20000 \times 21 \times 21 \times 21}{20 \times 20 \times 20} \\[6pt] &= \frac{20 \times 9261}{2 \times 2 \times 2} \\[6pt] &= \frac{5 \times 9261}{2} \\[6pt] &= \frac{46305}{2} \\[6pt] A &= \text{₹ }23152.50 \end{aligned} \]

Answer Amount after 9 months \(= \color{red}{\text{₹ }23{,}152.50}\)

19. Find the amount and the compound interest on ₹ 24,000 for six months if the interest is payable quarterly at the rate of 20 paise a rupee per annum.

Solution

\[ \begin{aligned} 1 \text{ rupee} &= 100 \text{ paise} \\[8pt] 20 \text{ paise per rupee} & = \dfrac{20}{100} \\[8pt] & = \dfrac{20}{100} \times 100 \%\\[8pt] R &= 20\% \\[4pt] P &= \text{₹ }24000 \\[4pt] n &= 6 \text{ months } = \frac{6}{12} \implies \frac{1}{2}\ \text{year} \end{aligned} \] \[ \begin{aligned} &\color{magenta}\textbf{CI - quarterly} \\[8pt] A &= P\left(1 + \frac{R}{400}\right)^{4n} \\[8pt] &= 24000\left(1 + \frac{20}{400}\right)^{4 \times \frac{1}{2}} \\[8pt] &= 24000\left(1 + \frac{1}{20}\right)^2 \\[8pt] &= 24000\left(\frac{20+1}{20}\right)^2 \\[8pt] &= 24000\left(\frac{21}{20}\right)^2 \\[8pt] &= \frac{24000 \times 21 \times 21}{20 \times 20} \\[8pt] &= \frac{\cancel{240}^{60} \times 441}{\cancel{4}_1} \\[8pt] &= 60 \times 441 \\[8pt] A &= \text{₹ }26460 \\[8pt] \text{C.I.} &= A - P \\[4pt] &= 26460 - 24000 \\[4pt] &= \text{₹ }2460 \end{aligned} \]

Answer Amount after 6 months \(=\; \color{red}{\text{₹ }26{,}460}\) , Compound Interest \(=\; \color{red}{\text{₹ }2{,}460}\)

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