DAV Class 8 Maths Chapter 6 Practice Worksheet

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DAV Class 8 Maths Chapter 6 Practice Worksheet

Compound Interest Practice Worksheet


1. Reeta received a sum of ₹ 40,000 as a loan from a finance company at the rate of 7% per annum. Find the compound interest paid by Reeta after 2 years.

Solution

\[ \begin{aligned} P &= \text{₹ }40000 \\ R &= 7\% \\ n &= 2\ \text{years} \end{aligned} \] \[ \begin{aligned} A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 40000\left(1 + \frac{7}{100}\right)^2 \\[6pt] &= 40000\left(\frac{107}{100}\right)^2 \\[6pt] &= 40000 \times \frac{107}{100} \times \frac{107}{100} \\[6pt] &= 4 \times 107 \times 107 \\[6pt] &= 4 \times 11449 \\[6pt] A &= \text{₹ }45796 \\[6pt] \text{C.I.} &= A - P \\[4pt] &= 45796 - 40000 \\[4pt] &= 5796 \end{aligned} \]

Answer Compound Interest \(=\; \color{red}{\text{₹ }5{,}796}\)

2. The simple interest on a certain sum of money for 3 years at 5% per annum is ₹ 540. What will be compound interest on that sum at the same rate for the same period?

Solution

\[ \begin{aligned} \text{S.I.} &= \text{₹ }540 \\ R &= 5\% \\ T &= 3\ \text{years} \\[6pt] \text{S.I.} &= \frac{P \times R \times T}{100} \\[6pt] 540 &= \frac{P \times \cancel5^1 \times 3}{\cancel{100}_{20}} \\[6pt] 540 &= \frac{3P}{20} \\[6pt] P &= \frac{\cancel{540}^{180} \times 20}{\cancel3_1} \\[6pt] P &= 180 \times 20 \\[6pt] P &= \text{₹ }3600 \end{aligned} \] \[ \begin{aligned} &\color{magenta}\textbf{Compound Interest} \\[6pt] P &= \text{₹ }3600 \\ R &= 5\% \\ n &= 3\ \text{years} \\[6pt] A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 3600\left(1 + \frac{5}{100}\right)^3 \\[6pt] &= 3600\left(\frac{21}{20}\right)^3 \\[6pt] &= 3600 \times \frac{9261}{8000} \\[6pt] &= \frac{36 \times 9261}{80} \\[6pt] &= \frac{9 \times 9261}{20} \\[6pt] &= \frac{83349}{20} \\[6pt] A &= \text{₹ }4167.45 \\[6pt] \text{C.I.} &= A - P \\[4pt] &= 4167.45 - 3600 \\[4pt] &= 567.45 \end{aligned} \]

Answer Compound Interest \(=\; \color{red}{\text{₹ }567.45}\)

3. The annual rate of growth in population of a certain city is 8%. If its present population is 196830, then what was the population 3 years ago?

Solution

\[ \begin{aligned} \text{Present Population } (A) &= 196830 \\ R &= 8\% \\ n &= 3\ \text{years} \\ \end{aligned} \] \[ \begin{aligned} &\text{Let P be the population 3 years ago} \\\\ A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] 196830 &= P\left(1 + \frac{8}{100}\right)^3 \\[6pt] 196830 &= P\left(\frac{27}{25}\right)^3 \\[6pt] P &= 196830 \times \left(\frac{25}{27}\right)^3 \\[6pt] P &= 196830 \times \frac{15625}{19683} \\[6pt] P &= 10 \times 15625 \\[6pt] P &= 156250 \end{aligned} \]

Answer Population 3 years ago \(=\; \color{red}{1{,}56{,}250}\)

4. At what rate percent will a sum of ₹ 3750 amount to ₹ 4374 in 2 years, if the interest is compounded annually?

Solution

\[ \begin{aligned} P &= \text{₹ }3750 \\ A &= \text{₹ }4374 \\ n &= 2\ \text{years} \end{aligned} \] \[ \begin{aligned} A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] 4374 &= 3750\left(1 + \frac{R}{100}\right)^2 \\[6pt] \frac{4374}{3750} &= \left(1 + \frac{R}{100}\right)^2 \\[6pt] \frac{729}{625} &= \left(1 + \frac{R}{100}\right)^2 \\[6pt] \left(\frac{27}{25}\right)^2 &= \left(1 + \frac{R}{100}\right)^2 \\[6pt] \frac{27}{25} &= 1 + \frac{R}{100} \\[6pt] \frac{R}{100} &= \frac{27}{25} - 1 \\[6pt] \frac{R}{100} &= \frac{2}{25} \\[6pt] R &= \frac{200}{25} \\[6pt] R &= 8\% \end{aligned} \]

Answer Rate of interest \(=\; \color{red}{8\%}\)

5. The difference between the compound interest and the simple interest on a certain sum of money at 15% per annum for 3 years is ₹ 283.50. Find the sum.

Solution

\[ \begin{aligned} R &= 15\% \\ n &= 3\ \text{years} \\ \text{CI - SI} &= \text{₹ }283.50 \end{aligned} \] \[ \begin{aligned} P\left[\left(1 + \frac{15}{100}\right)^3 - 1\right] - \frac{P \times 15 \times 3}{100} & = 283.50\\[6pt] P\left[\left(\frac{23}{20}\right)^3 - 1\right]- \frac{45P}{100}& = 283.50\\[6pt] P\left[\frac{12167}{8000} - 1\right] -\frac{9P}{20} & = 283.50\\[6pt] P\left[\frac{12167 - 8000}{8000}\right] - \frac{9P}{20}& = 283.50\\[6pt] P\left[\frac{4167}{8000}\right] - \frac{9P}{20}& = 283.50\\[6pt] \frac{4167P}{8000} - \frac{9P}{20} & = 283.50 \\[6pt] P\left(\frac{4167}{8000} - \frac{9}{20} \right) & = 283.50 \\[6pt] P\left(\frac{4167 - 3600}{8000}\right) & = 283.50 \\[6pt] P\left(\frac{567}{8000}\right) & = 283.50 \\[6pt] P &= \frac{283.50 \times 8000}{567} \\[6pt] P &= \frac{2835 \times 800}{567} \\[6pt] P &= 5 \times 800 \\[6pt] P &= 4000 \end{aligned} \]

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

6. Compute the amount for ₹ 65,536 invested for \(1\dfrac{1}{2}\) years at \(12\dfrac{1}{2}\%\) per annum, the interest being compounded semi-annually.

Solution

\[ \begin{aligned} P &= \text{₹ }65536 \\[6pt] n &= 1\frac{1}{2}\ \text{years} \implies \frac{3}{2}\ \text{years} \\[6pt] R &= 12\dfrac{1}{2}\% \implies \frac{25}{2}\% \text{ per annum} \\[6pt] \end{aligned} \] \[ \begin{aligned} A &= P\left(1 + \frac{R}{200}\right)^{2n} \\[6pt] &= 65536\left(1 + \frac{25}{400}\right)^3 \\[6pt] &= 65536\left(1 + \frac{1}{16}\right)^3 \\[6pt] &= 65536\left(\frac{17}{16}\right)^3 \\[6pt] &= 65536 \times \frac{4913}{4096} \\[6pt] &= 16 \times 4913 \\[6pt] A &= \text{₹ }78608 \end{aligned} \]

Answer Amount \(=\; \color{red}{\text{₹ }78{,}608}\)

7. In a factory, the production of scooters has risen to 48400 from 40000 in 2 years. Find the rate of growth per annum.

Solution

\[ \begin{aligned} P &= 40000 \\ A &= 48400 \\ n &= 2\ \text{years} \end{aligned} \] \[ \begin{aligned} A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] 48400 &= 40000\left(1 + \frac{R}{100}\right)^2 \\[6pt] \frac{48400}{40000} &= \left(1 + \frac{R}{100}\right)^2 \\[6pt] \frac{121}{100} &= \left(1 + \frac{R}{100}\right)^2 \\[6pt] \left(\frac{11}{10}\right)^2 &= \left(1 + \frac{R}{100}\right)^2 \\[6pt] \frac{11}{10} &= 1 + \frac{R}{100} \\[6pt] \frac{R}{100} &= \frac{11}{10} - 1 \\[6pt] \frac{R}{100} &= \frac{1}{10} \\[6pt] R &= 10\% \end{aligned} \]

Answer Rate of growth \(=\; \color{red}{10\%}\)

8. In what time will a sum of ₹ 3750 amount to ₹ 6480 at 20% per annum compound interest, if the interest is compounded annually.

Solution

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

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

9. The value of a machine depreciates at the rate of 10% per annum It was purchased 3 years ago. If its present value is ₹ 43740, then at what price was it purchased?

Solution

\[ \begin{aligned} \text{Present Value } (A) &= \text{₹ }43740 \\ R &= 10\% \text{ (depreciation)} \\ n &= 3\ \text{years} \\ \end{aligned} \] \[ \begin{aligned} &\text{Let P be the price 3 years ago} \\ \\ A &= P\left(1 - \frac{R}{100}\right)^n \\[6pt] 43740 &= P\left(1 - \frac{10}{100}\right)^3 \\[6pt] 43740 &= P\left(\frac{9}{10}\right)^3 \\[6pt] 43740 &= P \times \frac{729}{1000} \\[6pt] P &= 43740 \times \frac{1000}{729} \\[6pt] P &= 60 \times 1000 \\[6pt] P &= 60000 \end{aligned} \]

Answer Purchase price \(=\; \color{red}{\text{₹ }60{,}000}\)

10. Ramesh borrowed from Suresh certain sum for 2 years at simple interest. He lent this sum to Dinesh at the same rate for 2 years compound interest. At the end of 2 years, Ramesh received ₹ 110 as compound interest from Dinesh but paid ₹ 100 as simple interest to Suresh. Find the sum and the rate of interest.

Solution

\[ \begin{aligned} \text{Let the principal }(P) &= P \\[4pt] \text{Rate of interest }(R) &= R\% \\[4pt] n &= 2\ \text{years} \\[6pt] \text{S.I. paid by Ramesh} &= \text{₹ }100 \\[4pt] \text{C.I. received by Ramesh} &= \text{₹ }110 \end{aligned} \]\[ \begin{aligned} &\color{magenta}\textbf{Simple Interest (paid by Ramesh)} \\[6pt] \end{aligned} \]\[ \begin{aligned} \text{S.I.} &= \text{₹ }100 \\[6pt] \frac{P \times R \times T}{100} &= 100 \\[6pt] \frac{P \times R \times 2}{100} &= 100 \\[6pt] \frac{PR}{50} &= 100 \\[6pt] PR &= 100 \times 50 \\[6pt] PR &= 5000 \qquad \implies \text{(i)} \end{aligned} \]\[ \begin{aligned} &\color{magenta}\textbf{Compound Interest (received by Ramesh)} \\[6pt] \end{aligned} \]\[ \begin{aligned} \text{C.I.} &= \text{₹ }110 \\[6pt] P\left[\left(1 + \frac{R}{100}\right)^2 - 1\right] &= 110 \\[6pt] P\left[\left(\frac{100 + R}{100}\right)^2 - 1\right] &= 110 \\[6pt] P\left[\frac{(100 + R)^2}{10000} - 1\right] &= 110 \\[6pt] P\left[\frac{(100 + R)^2 - 10000}{10000}\right] &= 110 \\[6pt] P\left[\frac{10000 + 200R + R^2 - 10000}{10000}\right] &= 110 \\[6pt] P\left[\frac{200R + R^2}{10000}\right] &= 110 \\[6pt] P\left[\frac{R(200 + R)}{10000}\right] &= 110 \\[6pt] \frac{PR(200 + R)}{10000} &= 110 \\[6pt] PR(200 + R) &= 110 \times 10000 \\[6pt] PR(200 + R) &= 1100000 \qquad \implies \text{(ii)} \end{aligned} \]

Substituting \(PR = 5000\) from (i) in (ii):

\[ \begin{aligned} PR(200 + R) &= 1100000 \\[6pt] 5000(200 + R) &= 1100000 \\[6pt] 200 + R &= \frac{1100000}{5000} \\[6pt] 200 + R &= 220 \\[6pt] R &= 220 - 200 \\[4pt] \color{green}R &= \color{green}20\% \end{aligned} \]\[ \begin{aligned} \therefore\ PR &= 5000 \\[6pt] P &= \frac{5000}{R} \\[6pt] P &= \frac{5000}{20} \\[6pt] P &= \text{₹ }250 \end{aligned} \]

Answer Sum \(=\; \color{red}{\text{₹ }250}\), Rate \(=\; \color{red}{20\% \text{ per annum}}\)

11. The difference between the compound interest and simple interest on a certain sum of money at 10% per annum for 2 years is ₹ 500. Find the sum.

Solution

\[ \begin{aligned} R &= 10\% \\ n &= 2\ \text{years} \\ \text{CI - SI} &= \text{₹ }500 \end{aligned} \] \[ \begin{aligned} P\left[\left(1 + \frac{10}{100}\right)^2 - 1\right] - \frac{P \times 10 \times 2}{100} & = 500 \\[6pt] P\left[\left(\frac{11}{10}\right)^2 - 1\right] - \frac{20P}{100} & = 500 \\[6pt] P\left[\frac{121}{100} - 1\right] - \frac{P}{5} & = 500 \\[6pt] P\left[\frac{121 - 100}{100}\right] - \frac{P}{5} & = 500 \\[6pt] P\left(\frac{21}{100}\right) - \frac{P}{5} & = 500 \\[6pt] P\left(\frac{21}{100} - \frac{1}{5}\right) & = 500 \\[6pt] P\left(\frac{21}{100} - \frac{20}{100}\right) & = 500 \\[6pt] P\left(\frac{1}{100}\right) & = 500 \\[6pt] P &= 500 \times 100 \\[6pt] P &= 50{,}000 \end{aligned} \]

Answer Sum \(=\; \color{red}{\text{₹ }50{,}000}\)

12. Find the compound interest on ₹ 18000 for \(1\dfrac{1}{2}\) years at 10% per annum, if interest being compounded half yearly.

Solution

\[ \begin{aligned} P &= \text{₹ }18000 \\[6pt] n &= 1\dfrac{1}{2}\ \text{years} \implies \dfrac{3}{2}\ \text{years} \\[6pt] R &= 10\% \text{ per annum} \end{aligned} \] \[ \begin{aligned} A &= P\left(1 + \frac{R}{200}\right)^{2n} \\[6pt] &= 18000\left(1 + \frac{10}{200}\right)^{2 \times \frac{3}{2}} \\[6pt] &= 18000\left(1 + \frac{1}{20}\right)^3 \\[6pt] &= 18000\left(\frac{21}{20}\right)^3 \\[6pt] &= 18000 \times \frac{9261}{8000} \\[6pt] &= \frac{18 \times 9261}{8} \\[6pt] &= \frac{9 \times 9261}{4} \\[6pt] &= \frac{83349}{4} \\[6pt] A &= 20837.25 \\[6pt] \text{C.I.} &= A - P \\[4pt] &= 20837.25 - 18000 \\[4pt] &= \text{₹ }2837.25 \end{aligned} \]

Answer Compound Interest \(=\; \color{red}{\text{₹ }2{,}837.25}\)

13. Megha and her friends started a firm to manufacture natural fertilizer to promote organic farming. The profits of the firm were ₹ 1,84,000 in the year 2014. During the next year, it increased by 5% and it decreased by 2% in the following year. What is the profit of the firm after 2 years?

Solution

\[ \begin{aligned} \text{Initial Profit } (P) &= \text{₹ }184000 \\ a &= 5\% \\ b &= 2\% \end{aligned} \] \[ \begin{aligned} \text{Profit after 2 years} &= P\left(1 + \frac{a}{100}\right)\left(1 - \frac{b}{100}\right) \\[6pt] &= 184000\left(1 + \frac{5}{100}\right)\left(1 - \frac{2}{100}\right) \\[6pt] &= 184000\left(1 + \frac{1}{20}\right)\left(1 - \frac{1}{50}\right) \\[6pt] &= 184000\left(\frac{21}{20}\right)\left(\frac{49}{50}\right) \\[6pt] &= \frac{184000 \times 21 \times 49}{1000} \\[6pt] &= 184 \times 1029 \\[6pt] &= 189336 \end{aligned} \]

Answer Profit after 2 years \(=\; \color{red}{\text{₹ }1{,}89{,}336}\)

14. At what rate percent will a sum of ₹ 640 be compounded to ₹ 774.40 in 2 years, when the interest is compounded annually?

Solution

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

Answer Rate percent \(=\; \color{red}{10\%}\)

15. A sum gives simple interest of ₹ 800 for 2 years at the rate of 4% per annum. Find the compound interest on the same sum at the same rate of interest and same time period when interest is compounded annually.

Solution

\[ \begin{aligned} \text{S.I.} &= \text{₹ }800 \\ T &= 2\ \text{years} \\ R &= 4\% \\[6pt] \text{S.I.} &= \frac{P \times R \times T}{100} \\[6pt] 800 &= \frac{P \times 4 \times 2}{100} \\[6pt] 800 &= \frac{8P}{100} \\[6pt] P &= \frac{800 \times 100}{8} \\[6pt] P &= \text{₹ }10000 \end{aligned} \] \[ \begin{aligned} &\color{magenta}\textbf{Compound Interest} \\[6pt] A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 10000\left(1 + \frac{4}{100}\right)^2 \\[6pt] &= 10000\left(\frac{26}{25}\right)^2 \\[6pt] &= 10000 \times \frac{676}{625} \\[6pt] &= 16 \times 676 \\[6pt] A &= 10816 \\[6pt] \text{C.I.} &= A - P \\[4pt] &= 10816 - 10000 \\[4pt] &= 816 \end{aligned} \]

Answer Compound Interest \(=\; \color{red}{\text{₹ }816}\)

16. The value of a machine depreciates at 12.5% per annum which was purchased 3 years ago. If its present value is ₹ 13720, then find the original value of the machine 3 years ago.

Solution

\[ \begin{aligned} \text{Present Value } (A) &= \text{₹ }13720 \\ n &= 3\ \text{years} \\ R &= 12.5\% \end{aligned} \] \[ \begin{aligned} A &= P\left(1 - \frac{R}{100}\right)^n \\[6pt] 13720 &= P\left(1 - \frac{12.5}{100}\right)^3 \\[6pt] 13720 &= P\left(1 - \frac{125}{1000}\right)^3 \\[6pt] 13720 &= P\left(1 - \frac{1}{8}\right)^3 \\[6pt] 13720 &= P\left(\frac{7}{8}\right)^3 \\[6pt] 13720 &= P \times \frac{343}{512} \\[6pt] P &= \frac{13720 \times 512}{343} \\[6pt] P &= 40 \times 512 \\[6pt] P &= 20480 \end{aligned} \]

Answer Original value \(=\; \color{red}{\text{₹ }20{,}480}\)

17. In the year 2001, the number of malaria patients admitted in the hospitals of the state was 4375. Every year this number decreases by 8%. Find the number of patients in 2003.

Solution

\[ \begin{aligned} P &= 4375 \\ R &= 8\% \text{ (decrease)} \\ n &= 2\ \text{years (2001 to 2003)} \end{aligned} \] \[ \begin{aligned} A &= P\left(1 - \frac{R}{100}\right)^n \\[6pt] &= 4375\left(1 - \frac{8}{100}\right)^2 \\[6pt] &= 4375\left(\frac{92}{100}\right)^2 \\[6pt] &= 4375\left(\frac{23}{25}\right)^2 \\[6pt] &= 4375 \times \frac{529}{625} \\[6pt] &= 7 \times 529 \\[6pt] A &= 3703 \end{aligned} \]

Answer Number of patients \(=\; \color{red}{3703}\)

18. At what rate percent will a sum of ₹ 8000 amounts to ₹ 15625 in 3 years, if the interest is compounded annually?

Solution

\[ \begin{aligned} P &= \text{₹ }8000 \\ A &= \text{₹ }15625 \\ n &= 3\ \text{years} \end{aligned} \] \[ \begin{aligned} A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] 15625 &= 8000\left(1 + \frac{R}{100}\right)^3 \\[6pt] \frac{15625}{8000} &= \left(1 + \frac{R}{100}\right)^3 \\[6pt] \frac{125}{64} &= \left(1 + \frac{R}{100}\right)^3 \\[6pt] \left(\frac{5}{4}\right)^3 &= \left(1 + \frac{R}{100}\right)^3 \\[6pt] \frac{5}{4} &= 1 + \frac{R}{100} \\[6pt] \frac{5}{4} &= \frac{100 + R}{100} \\[6pt] 100 + R &= \frac{5}{4} \times 100 \\[6pt] 100 + R &= 5 \times 25 \\[6pt] 100 + R &= 125 \\[6pt] R &= 125 - 100 \\[6pt] R &= 25\% \end{aligned} \]

Answer Rate of interest \(=\; \color{red}{25\%}\)

19. The simple interest on a certain sum of money for 2 years at 8% per annum is ₹ 2400. What will be the compound interest on that sum at the same rate for the same time, when the interest is compounded annually?

Solution

\[ \begin{aligned} \text{S.I.} &= \text{₹ }2400 \\ n &= 2\ \text{years} \\ R &= 8\% \\[6pt] \text{S.I.} &= \frac{P \times R \times T}{100} \\[6pt] 2400 &= \frac{P \times 8 \times 2}{100} \\[6pt] 2400 &= \frac{16P}{100} \\[6pt] P &= \frac{2400 \times 100}{16} \\[6pt] P &= 150 \times 100 \\[6pt] P &= \text{₹ }15000 \end{aligned} \] \[ \begin{aligned} &\color{magenta}\textbf{Compound Interest} \\[6pt] A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 15000\left(1 + \frac{8}{100}\right)^2 \\[6pt] &= 15000\left(\frac{27}{25}\right)^2 \\[6pt] &= 15000 \times \frac{729}{625} \\[6pt] &= 24 \times 729 \\[6pt] A &= 17496 \\[6pt] \text{C.I.} &= A - P \\[4pt] &= 17496 - 15000 \\[4pt] &= \text{₹ }2496 \end{aligned} \]

Answer Compound Interest \(=\; \color{red}{\text{₹ }2{,}496}\)

20. The difference between the compound interest and simple interest on a certain sum of money at 5% per annum for 3 years is ₹ 27.45. Find the sum.

Solution

\[ \begin{aligned} R &= 5\% \\ n &= 3\ \text{years} \\ \text{CI - SI} &= \text{₹ }27.45 \end{aligned} \] \[ \begin{aligned} P\left[\left(1 + \frac{5}{100}\right)^3 - 1\right] - \frac{P \times 5 \times 3}{100} & = 27.45\\[6pt] P\left[\left(\frac{21}{20}\right)^3 - 1\right]- \frac{15P}{100}& = 27.45\\[6pt] P\left[\frac{9261}{8000} - 1\right] -\frac{3P}{20} & = 27.45\\[6pt] P\left[\frac{9261 - 8000}{8000}\right] - \frac{3P}{20}& = 27.45\\[6pt] P\left[\frac{1261}{8000}\right] - \frac{3P}{20}& = 27.45\\[6pt] \frac{1261P}{8000} - \frac{3P}{20} & = 27.45 \\[6pt] P\left(\frac{1261}{8000} - \frac{3}{20} \right) & = 27.45 \\[6pt] P\left(\frac{1261 - 1200}{8000}\right) & = 27.45 \\[6pt] P\left(\frac{61}{8000}\right) & = 27.45 \\[6pt] P &= \frac{27.45 \times 8000}{61} \\[6pt] P &= \frac{2745 \times 80}{61} \\[6pt] P &= 45 \times 80 \\[6pt] P &= 3600 \end{aligned} \]

Answer Sum \(=\; \color{red}{\text{₹ }3{,}600}\)