\displaystyle \textbf{Question 1. }\text{A person invests Rs }10000\text{ for three years at a certain}
\displaystyle \text{rate of interest compounded annually. At the end of one year this sum}
\displaystyle \text{amounts to Rs }11200.\text{ Calculate : [ICSE 2006]}
\displaystyle \text{(i) the rate of interest per annum.}
\displaystyle \text{(ii) the amount at the end of the second year.}
\displaystyle \text{(iii) the amount at the end of the third year.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) For the first year :}
\displaystyle \text{Principal}=\text{Rs }10000\text{ and amount}=\text{Rs }11200
\displaystyle \therefore I=A-P=\text{Rs }11200-\text{Rs }10000=\text{Rs }1200
\displaystyle \text{Rate}=\frac{I\times100}{P\times T}\%
\displaystyle \therefore \text{Rate of interest p.a.}=\frac{1200\times100}{10000\times1}\%=12\%
\displaystyle \text{(ii) For the second year :}
\displaystyle P=\text{Rs }11200,\ R=12\%\text{ and }T=1\text{ year}
\displaystyle I=\text{Rs }\frac{P\times R\times T}{100}=\text{Rs }\frac{11200\times12\times1}{100}
\displaystyle =\text{Rs }1344
\displaystyle \therefore \text{Amount}=P+I=\text{Rs }11200+\text{Rs }1344=\text{Rs }12544
\displaystyle \text{(iii) For the third year :}
\displaystyle P=\text{Rs }12544,\ R=12\%\text{ and }T=1\text{ year}
\displaystyle I=\text{Rs }\frac{12544\times12\times1}{100}=\text{Rs }1505.28
\displaystyle \therefore \text{Amount}=P+I=\text{Rs }12544+\text{Rs }1505.28=\text{Rs }14049.28
\displaystyle \\

\displaystyle \textbf{Question 2. }\text{A sum of Rs }9600\text{ is invested for }3\text{ years at }10\%
\displaystyle \text{per annum at compound interest. [ICSE 1996]}
\displaystyle \text{(i) What is the sum due at the end of the first year?}
\displaystyle \text{(ii) What is the sum due at the end of the second year?}
\displaystyle \text{(iii) Find the difference between the answers in (ii) and (i) and find}
\displaystyle \text{the interest on this sum (difference) for one year.}
\displaystyle \text{(iv) Hence, write down the compound interest for the third year.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Interest for the 1st year}=\text{Rs }\frac{9600\times10\times1}{100}=\text{Rs }960
\displaystyle \therefore \text{The sum due at the end of the 1st year}
\displaystyle =\text{Rs }9600+\text{Rs }960=\text{Rs }10560
\displaystyle \text{(ii) Interest for the 2nd year}=\text{Rs }\frac{10560\times10\times1}{100}=\text{Rs }1056
\displaystyle \therefore \text{The sum due at the end of the 2nd year}
\displaystyle =\text{Rs }10560+\text{Rs }1056=\text{Rs }11616
\displaystyle \text{(iii) Required difference}=\text{Rs }11616-\text{Rs }10560=\text{Rs }1056
\displaystyle \text{Rs }1056\text{ is C.I. for the 2nd year}
\displaystyle \therefore \text{Interest for one year on this difference}
\displaystyle =\text{Rs }\frac{1056\times10\times1}{100}=\text{Rs }105.60
\displaystyle \text{(iv) C.I. for the 3rd year}=\text{C.I. of the 2nd year}+\text{Interest on it for 1 year}
\displaystyle =\text{Rs }1056+\text{Rs }105.60
\displaystyle =\text{Rs }1161.60
\displaystyle \\

\displaystyle \textbf{Question 3. }\text{What sum of money will amount to Rs }9261\text{ in}
\displaystyle \text{3 years at }5\%\text{ per annum compound interest? [ICSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Given : }A=\text{Rs }9261,\ n=3\text{ years and }r=5\%
\displaystyle \therefore 9261=P\left(1+\frac{5}{100}\right)^3
\displaystyle \therefore P=9261\times\left(\frac{20}{21}\right)^3
\displaystyle =\text{Rs }8000
\displaystyle \therefore \text{Required sum}=\text{Rs }8000
\displaystyle \\

\displaystyle \textbf{Question 4. }\text{In what period of time will Rs }12000\text{ yield Rs }3972
\displaystyle \text{as compound interest at }10\%\text{ per cent, if compounded on a yearly}
\displaystyle \text{basis. [ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given : }P=\text{Rs }12000,\ \text{C.I.}=\text{Rs }3972\text{ and }r=10\%
\displaystyle \text{Amount }A=P+I
\displaystyle =\text{Rs }12000+\text{Rs }3972=\text{Rs }15972
\displaystyle \therefore 15972=12000\left(1+\frac{10}{100}\right)^n
\displaystyle \therefore \frac{15972}{12000}=\left(\frac{11}{10}\right)^n
\displaystyle \therefore \left(\frac{11}{10}\right)^3=\left(\frac{11}{10}\right)^n
\displaystyle \therefore n=3
\displaystyle \therefore \text{Required time}=3\text{ years}
\displaystyle \\

\displaystyle \textbf{Question 5. }\text{On what sum of money will the difference between}
\displaystyle \text{the compound interest and the simple interest for }2\text{ years be equal}
\displaystyle \text{to Rs }25,\text{ if the rate of interest charged for both is }5\%\text{ p.a.?}
\displaystyle \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{C.I.}=P\left[\left(1+\frac{r}{100}\right)^n-1\right]
\displaystyle =P\left[\left(1+\frac{5}{100}\right)^2-1\right]=\frac{41P}{400}
\displaystyle \text{S.I.}=\frac{P\times5\times2}{100}=\frac{P}{10}
\displaystyle \text{Given, C.I.}-\text{S.I.}=\text{Rs }25
\displaystyle \therefore \frac{41P}{400}-\frac{P}{10}=25
\displaystyle \therefore P=\text{Rs }10000
\displaystyle \therefore \text{Required sum}=\text{Rs }10000
\displaystyle \text{Alternative method :}
\displaystyle \text{Let sum (principal)}=\text{Rs }100
\displaystyle \therefore \text{C.I.}=\text{Rs }100\left(1+\frac{5}{100}\right)^2-\text{Rs }100
\displaystyle =\text{Rs }10.25
\displaystyle \text{S.I.}=\text{Rs }\frac{100\times5\times2}{100}=\text{Rs }10
\displaystyle \text{Difference between C.I. and S.I.}=\text{Rs }10.25-\text{Rs }10
\displaystyle =\text{Rs }0.25
\displaystyle \text{When difference between C.I. and S.I.}=\text{Rs }0.25,\ \text{sum}=\text{Rs }100
\displaystyle \therefore \text{When difference between C.I. and S.I.}=\text{Rs }25,\ \text{sum}
\displaystyle =\text{Rs }\frac{100}{0.25}\times25
\displaystyle =\text{Rs }10000
\displaystyle \therefore \text{Required sum}=\text{Rs }10000
\displaystyle \\

\displaystyle \textbf{Question 6. }\text{The simple interest on a sum of money for }2\text{ years}
\displaystyle \text{at }4\%\text{ per annum is Rs }340.\text{ Find : [ICSE 2008]}
\displaystyle \text{(i) the sum of money and}
\displaystyle \text{(ii) the compound interest on this sum for one year payable}
\displaystyle \text{half-yearly at the same rate.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given : }I=\text{Rs }340,\ T=2\text{ years and }R=4\%
\displaystyle \therefore P=\frac{I\times100}{R\times T}=\frac{340\times100}{4\times2}
\displaystyle =\text{Rs }4250
\displaystyle \text{(ii) C.I.}=P\left(1+\frac{r}{2\times100}\right)^{n\times2}-P
\displaystyle =\text{Rs }4250\left(1+\frac{4}{2\times100}\right)^{1\times2}-\text{Rs }4250
\displaystyle =\text{Rs }4421.70-\text{Rs }4250=\text{Rs }171.70
\displaystyle \\

\displaystyle \textbf{Question 7. }\text{Simple interest for 2 years at }4\%\text{ p.a. is Rs. }450.\text{ Find the} \\ \text{compound interest for 1 year, compounded half-yearly. [ICSE Board 1997]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the principal be Rs. }P.
\displaystyle \frac{P\times4\times2}{100}=450
\displaystyle \frac{2P}{25}=450
\displaystyle P=\frac{450\times25}{2}=5625
\displaystyle \text{Rate per half-year}=\frac{4}{2}=2\%
\displaystyle \text{Amount after 1 year}=5625\left(1+\frac{2}{100}\right)^2
\displaystyle =5625\left(\frac{102}{100}\right)^2=5852.25
\displaystyle \therefore \text{Compound Interest}=5852.25-5625
\displaystyle =\text{Rs. }227.25
\displaystyle \text{Hence, the required compound interest is Rs. }227.25.
\\

\displaystyle \textbf{Question 8. }\text{Rohit borrows Rs. }86000\text{ at simple interest and lends it at compound} \\ \text{interest at }5\%\text{ p.a. for }2\text{ years. Find his profit. [ICSE Board 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Simple Interest}=86000\times\frac{5}{100}\times2=\text{Rs. }8600
\displaystyle \text{Amount at compound interest}=86000\left(1+\frac{5}{100}\right)^2=\text{Rs. }94815
\displaystyle \text{Compound Interest}=94815-86000=\text{Rs. }8815
\displaystyle \text{Profit}=8815-8600=\text{Rs. }215
\displaystyle \text{Hence, Rohit's profit is Rs. }215.
\\

\displaystyle \textbf{Question 9. }\text{Nikita invests Rs. }6000.\text{ After }1\text{ year it becomes Rs. }6720. \\ \text{ Find the rate of interest and the amount after }2\text{ years. [ICSE Board 2010]}
\displaystyle \text{Answer:}
\displaystyle 6000\left(1+\frac{r}{100}\right)=6720
\displaystyle 1+\frac{r}{100}=\frac{6720}{6000}=1.12
\displaystyle \therefore r=12\%
\displaystyle \text{Amount after }2\text{ years}=6000\left(1+\frac{12}{100}\right)^2
\displaystyle =6000(1.12)^2=\text{Rs. }7526.40
\displaystyle \text{Hence, the rate of interest is }12\%\text{ p.a. and the amount after }2\text{ years is Rs. }7526.40.
\\

\displaystyle \textbf{Question 10. }\text{Find the amount on Rs. }7400\text{ at }5\%\text{ p.a., compounded} \\ \text{half-yearly for }1\text{ year. [ICSE Board 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Rate per half-year}=\frac{5}{2}\%=2.5\%
\displaystyle \text{Amount}=7400\left(1+\frac{5}{200}\right)^2
\displaystyle =7400(1.025)^2
\displaystyle =\text{Rs. }7774.63
\displaystyle \text{Hence, the required amount is Rs. }7774.63.
\\

\displaystyle \textbf{Question 11. }\text{At what rate will Rs. }4000\text{ yield a compound interest of Rs. }1324 \\ \text{ in }3\text{ years? [ICSE Board 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Amount}=4000+1324=5324
\displaystyle 4000\left(1+\frac{r}{100}\right)^3=5324
\displaystyle \left(1+\frac{r}{100}\right)^3=\frac{5324}{4000}=1.331
\displaystyle 1+\frac{r}{100}=1.1
\displaystyle \therefore r=10\%
\displaystyle \text{Hence, the required rate of interest is }10\%\text{ p.a.}
\\

\displaystyle \textbf{Question 12. }\text{The difference between compound interest and simple interest for } \\ 1\text{ year at }10\%\text{ p.a., compounded half-yearly, is Rs. }15.\text{ Find the sum. [ICSE Board 1998]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the principal be Rs. }P.
\displaystyle \text{Simple Interest}=P\times\frac{10}{100}=0.1P
\displaystyle \text{Compound Interest}=P\left(1+\frac{10}{200}\right)^2-P
\displaystyle =P\left(\frac{21}{20}\right)^2-P
\displaystyle \left[P\left(\frac{21}{20}\right)^2-P\right]-0.1P=15
\displaystyle P\left(\frac{441}{400}-1-\frac{1}{10}\right)=15
\displaystyle P\left(\frac{441-400-40}{400}\right)=15
\displaystyle \frac{P}{400}=15
\displaystyle P=\text{Rs. }6000
\displaystyle \text{Hence, the required sum is Rs. }6000.
\\
\displaystyle \textbf{Question 13. }\text{The compound interest for the second year is Rs. }1320\text{ and for the} \\ \text{third year is Rs. }1452.\text{ Find the rate and the sum. [ICSE Board 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Increase in interest}=1452-1320=\text{Rs. }132
\displaystyle \therefore \text{Rate}=\frac{132}{1320}\times100=10\%
\displaystyle \text{Interest for first year}=\frac{1320}{1.10}=1200
\displaystyle \text{Principal}=\frac{1200\times100}{10}=\text{Rs. }12000
\displaystyle \text{Hence, the rate is }10\%\text{ and the sum is Rs. }12000.
\\

\displaystyle \textbf{Question 14. }\text{Ramesh invests Rs. }12800\text{ at }10\%\text{ p.a. compound interest.} \\ \text{Find the amounts. [ICSE Board 2007]}
\displaystyle \text{Answer:}
\displaystyle A_1=12800\left(1+\frac{10}{100}\right)=\text{Rs. }14080
\displaystyle A_2=14080\left(1+\frac{10}{100}\right)=\text{Rs. }15488
\displaystyle A_3=15488\left(1+\frac{10}{100}\right)=\text{Rs. }17036.80
\displaystyle \text{Hence, the amounts after }1,\ 2\text{ and }3\text{ years are Rs. }14080,\text{ Rs. }15488 \\ \text{ and Rs. }17036.80\text{ respectively.}
\\

\displaystyle \textbf{Question 15. }\text{The compound interest for the second year is Rs. }880\text{ and for the} \\ \text{third year is Rs. }968.\text{ Find the rate and the sum. [ICSE Board 1995]}
\displaystyle \text{Answer:}
\displaystyle \text{Increase in interest}=968-880=\text{Rs. }88
\displaystyle \therefore \text{Rate}=\frac{88}{880}\times100=10\%
\displaystyle \text{Interest for first year}=\frac{880}{1.10}=800
\displaystyle \text{Principal}=\frac{800\times100}{10}=\text{Rs. }8000
\displaystyle \text{Hence, the rate is }10\%\text{ and the sum is Rs. }8000.
\\

\displaystyle \textbf{Question 16. }\text{Mr. Kumar borrowed Rs. }15000.\text{ The rates of compound interest for} \\ \text{the first and second years are }8\%\text{ and }10\%\text{ respectively. He paid Rs. }6200 \\ \text{ at the end of the first year. Find the outstanding amount. [ICSE Board 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Amount after first year}=15000\left(1+\frac{8}{100}\right)=\text{Rs. }16200
\displaystyle \text{Amount paid}=\text{Rs. }6200
\displaystyle \text{New principal}=16200-6200=\text{Rs. }10000
\displaystyle \text{Amount after second year}=10000\left(1+\frac{10}{100}\right)=\text{Rs. }11000
\displaystyle \text{Hence, the outstanding amount is Rs. }11000.
\\

\displaystyle \textbf{Question 17. }\text{Ranbir borrows Rs }20000\text{ at }12\%\text{ per cent C.I.}
\displaystyle \text{If he repays Rs }8400\text{ at the end of first year and Rs }9680\text{ at}
\displaystyle \text{the end of second year, find the amount of loan outstanding at the}
\displaystyle \text{beginning of the third year. [ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{For 1st year : }P=\text{Rs }20000,\ R=12\%\text{ and }T=1\text{ year.}
\displaystyle \therefore I=\text{Rs }\frac{20000\times12\times1}{100}=\text{Rs }2400
\displaystyle \text{And, amount }(A)=P+I
\displaystyle =\text{Rs }20000+\text{Rs }2400=\text{Rs }22400
\displaystyle \text{Since, the man pays Rs }8400\text{ at the end of 1st year}
\displaystyle \therefore \text{Principal for 2nd year}=\text{Rs }22400-\text{Rs }8400=\text{Rs }14000
\displaystyle \text{For 2nd year : }P=\text{Rs }14000,\ R=12\%\text{ and }T=1\text{ year.}
\displaystyle \therefore I=\text{Rs }\frac{14000\times12\times1}{100}=\text{Rs }1680
\displaystyle \text{And, amount }(A)=P+I
\displaystyle =\text{Rs }14000+\text{Rs }1680=\text{Rs }15680
\displaystyle \text{Since, the man pays Rs }9680\text{ at the end of 2nd year}
\displaystyle \therefore \text{Principal for 3rd year}=\text{Rs }15680-\text{Rs }9680=\text{Rs }6000
\displaystyle \therefore \text{The amount of loan outstanding at the beginning of the}
\displaystyle \text{3rd year}=\text{Rs }6000
\displaystyle \\

\displaystyle \textbf{Question 18. }\text{A sum of Rs. }9600\text{ is invested for }3\text{ years at }10\%\text{ per} \\ \text{annum at compound interest.}
\displaystyle \text{(i) What is the sum due at the end of the first year?}
\displaystyle \text{(ii) What is the sum due at the end of the second year?}
\displaystyle \text{(iii) Find the difference between the answers in (ii) and (i) and find the interest on this sum} \\ \text{(difference) for one year.}
\displaystyle \text{(iv) Hence, write down the compound interest for the third year. [ICSE Board 1996]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Interest for the first year}=\frac{9600\times10\times1}{100}=\text{Rs. }960
\displaystyle \text{Amount at the end of the first year}=9600+960=\text{Rs. }10560
\displaystyle \text{(ii) Interest for the second year}=\frac{10560\times10\times1}{100}=\text{Rs. }1056
\displaystyle \text{Amount at the end of the second year}=10560+1056=\text{Rs. }11616
\displaystyle \text{(iii) Difference}=11616-10560=\text{Rs. }1056
\displaystyle \text{Interest on this difference for one year}=\frac{1056\times10\times1}{100}=\text{Rs. }105.60
\displaystyle \text{(iv) Compound interest for the third year}=1056+105.60=\text{Rs. }1161.60
\displaystyle \text{Hence, the compound interest for the third year is Rs. }1161.60.
\\

\displaystyle \textbf{Question 19. }\text{What sum of money will amount to Rs. }9261\text{ in }3\text{ years} \\ \text{at }5\%\text{ per annum compound interest? [ICSE Board 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Amount }(A)=\text{Rs. }9261,\qquad r=5\%,\qquad n=3\text{ years}
\displaystyle A=P\left(1+\frac{5}{100}\right)^3
\displaystyle 9261=P\left(\frac{21}{20}\right)^3
\displaystyle P=9261\left(\frac{20}{21}\right)^3
\displaystyle P=\text{Rs. }8000
\displaystyle \text{Hence, the required sum is Rs. }8000.
\\

\displaystyle \textbf{Question 20. }\text{On a certain sum, the compound interest in }3\text{ years at }10\%\text{ per} \\ \text{annum amounts to Rs. }2317.\text{ Find the sum. [ICSE Board 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{Compound Interest}=\text{Rs. }2317,\qquad r=10\%,\qquad n=3\text{ years}
\displaystyle \text{Compound Interest}=P\left[\left(1+\frac{10}{100}\right)^3-1\right]
\displaystyle 2317=P\left[\left(\frac{11}{10}\right)^3-1\right]
\displaystyle 2317=P\left(\frac{1331-1000}{1000}\right)
\displaystyle 2317=P\left(\frac{331}{1000}\right)
\displaystyle P=\frac{2317\times1000}{331}=\text{Rs. }7000
\displaystyle \text{Hence, the required sum is Rs. }7000.
\\

\displaystyle \textbf{Question 21. }\text{On a certain sum, the compound interest in two years amounts to Rs. }2256. \\ \text{ If the rates of interest for successive years are }8\%\text{ and }10\%\text{ respectively, find} \\ \text{the sum. [ICSE Board 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the principal be Rs. }P.
\displaystyle \text{Amount after two years}=P\left(1+\frac{8}{100}\right)\left(1+\frac{10}{100}\right)
\displaystyle =P\times\frac{108}{100}\times\frac{110}{100}=P\times\frac{297}{250}
\displaystyle \text{Given Compound Interest}=\text{Rs. }2256
\displaystyle \frac{297}{250}P-P=2256
\displaystyle \frac{297P-250P}{250}=2256
\displaystyle \frac{47P}{250}=2256
\displaystyle P=\frac{2256\times250}{47}=\text{Rs. }12000
\displaystyle \text{Hence, the required sum is Rs. }12000.
\\

\displaystyle \textbf{Question 22. }\text{A person lends Rs. }15000\text{ for }2\text{ years at a certain rate of} \\ \text{compound interest. If after }2\text{ years it amounts to Rs. }16224,\text{ find the rate of} \\ \text{interest. [ICSE Board 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{Principal }(P)=\text{Rs. }15000,\qquad \text{Amount }(A)=\text{Rs. }16224,\qquad n=2\text{ years}
\displaystyle A=P\left(1+\frac{r}{100}\right)^2
\displaystyle 16224=15000\left(1+\frac{r}{100}\right)^2
\displaystyle \left(1+\frac{r}{100}\right)^2=\frac{16224}{15000}=\frac{676}{625}=\left(\frac{26}{25}\right)^2
\displaystyle 1+\frac{r}{100}=\frac{26}{25}
\displaystyle \frac{r}{100}=\frac{1}{25}
\displaystyle r=4\%
\displaystyle \text{Hence, the required rate of interest is }4\%\text{ per annum.}
\\

\displaystyle \textbf{Question 23. }\text{In what period of time will Rs. }12000\text{ yield Rs. }3972\text{ as} \\ \text{compound interest at }10\%\text{ per annum, compounded yearly? [ICSE Board 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Principal }(P)=\text{Rs. }12000,\qquad \text{Compound Interest}=\text{Rs. }3972,\qquad r=10\%
\displaystyle \text{Amount}=P+\text{Compound Interest}=12000+3972=\text{Rs. }15972
\displaystyle A=P\left(1+\frac{10}{100}\right)^n
\displaystyle 15972=12000\left(\frac{11}{10}\right)^n
\displaystyle \frac{15972}{12000}=\frac{1331}{1000}=\left(\frac{11}{10}\right)^3
\displaystyle \therefore \left(\frac{11}{10}\right)^n=\left(\frac{11}{10}\right)^3
\displaystyle \therefore n=3
\displaystyle \text{Hence, the required time is }3\text{ years.}
\\

\displaystyle \textbf{Question 24. }\text{On what sum of money will the difference between the compound interest} \\ \text{and the simple interest for }2\text{ years be equal to Rs. }25,\text{ if the rate of interest charged} \\ \text{for both is }5\%\text{ p.a.? [ICSE Board 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the principal be Rs. }P.
\displaystyle \text{Compound Interest}=P\left[\left(1+\frac{5}{100}\right)^2-1\right]
\displaystyle =P\left[\left(\frac{21}{20}\right)^2-1\right]
\displaystyle =P\left(\frac{441-400}{400}\right)=\frac{41P}{400}
\displaystyle \text{Simple Interest}=\frac{P\times5\times2}{100}=\frac{P}{10}
\displaystyle \text{Given, Compound Interest}-\text{Simple Interest}=25
\displaystyle \frac{41P}{400}-\frac{P}{10}=25
\displaystyle \frac{41P-40P}{400}=25
\displaystyle \frac{P}{400}=25
\displaystyle P=\text{Rs. }10000
\displaystyle \text{Hence, the required sum is Rs. }10000.
\\

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