Class 10: Compound Interest – ICSE Board Problems

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

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

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

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

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

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

Question 7: S.I. for 2 years at 4\% is Rs. 450. Find C.I. for 1 year (half-yearly) [ICSE Board 1997]
$\displaystyle \text{Answer:}$
$\displaystyle 450=\frac{2}{25}x \Rightarrow x=5625$
$\displaystyle A=5625 \Big(1+ \frac{4}{200} \Big)^2=5852.25$
$\displaystyle \text{C.I.}=227.25$
$\\$
Question 8: Rohit borrows Rs. 86000 at S.I. and lends at C.I. Find profit [ICSE Board 2010]
$\displaystyle \text{Answer:}$
$\displaystyle \text{S.I.}=86000 \times \frac{5}{100} \times 2=8600$
$\displaystyle A=86000 \Big(1+ \frac{5}{100} \Big)^2=94815$
$\displaystyle \text{C.I.}=94815-86000=4815$
$\displaystyle \text{Profit}=4815-8600=215$
$\\$
Question 9: Nikita invests Rs.6000. After 1 year it becomes Rs.6720. Find rate and amount after 2 years [ICSE Board 2010]
$\displaystyle \text{Answer:}$
$\displaystyle 6000 \Big(1+ \frac{x}{100} \Big)=6720 \Rightarrow x=12\%$
$\displaystyle A=6000 \Big(1+ \frac{12}{100} \Big)^2=7526.40$
$\\$
Question 10: Find amount for Rs.7400 at 5\% half-yearly for 1 year [ICSE Board 2005]
$\displaystyle \text{Answer:}$
$\displaystyle A=7400 \Big(1+ \frac{5}{200} \Big)^2=7774.63$
$\\$
Question 11: At what rate will Rs.4000 yield Rs.1324 in 3 years? [ICSE Board 2013]
$\displaystyle \text{Answer:}$
$\displaystyle 4000 \Big(1+ \frac{x}{100} \Big)^3-4000=1324 \Rightarrow x=10\%$
$\\$
Question 12: Difference between C.I. and S.I. (half-yearly) is Rs.15 at 10\%. Find sum [ICSE Board 1998]
$\displaystyle \text{Answer:}$
$\displaystyle \text{S.I.}=0.1x,\; \text{C.I.}=x \Big(\frac{21}{20} \Big)^2-x$
$\displaystyle x \Big(\frac{21}{20} \Big)^2-x-0.1x=15 \Rightarrow x=6000$
$\\$
Question 13: C.I. for 2nd year Rs.1320 and 3rd year Rs.1452. Find rate and sum [ICSE Board 2014]
$\displaystyle \text{Answer:}$
$\displaystyle \text{Rate}=10\%$
$\displaystyle \text{Sum}=12000$
$\\$
Question 14: Ramesh invests Rs.12800 at 10\%. Find amounts [ICSE Board 2007]
$\displaystyle \text{Answer:}$
$\displaystyle A_1=14080,\; I_2=1408,\; A_3=17036.80$
$\\$
Question 15: C.I. for 2nd year Rs.880 and 3rd year Rs.968. Find rate and sum [ICSE Board 1995]
$\displaystyle \text{Answer:}$
$\displaystyle \text{Rate}=10\%,\; \text{Sum}=8000$
$\\$
Question 16: Mr. Kumar borrowed Rs.15000. Rates 8\% and 10\%. Paid Rs.6200. Find outstanding [ICSE Board 2011]
$\displaystyle \text{Answer:}$
$\displaystyle A_1=16200$
$\displaystyle \text{New principal}=16200-6200=10000$
$\displaystyle A_2=10000 \Big(1+ \frac{10}{100} \Big)=11000$
$\\$

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

Question 18: A sum of Rs. 9600 is invested for 3 years at 10% per annum at compound interest.
(i) What is the sum due at the end of the first year?
(ii) What is the sum due at the end of the second year.
(iii) Find the difference between the answers in (ii) and (i) and find the interest on this sum (difference) for one year.
(iv) Hence, write down the compound interest for the third year [ICSE Board 1996]
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Interest for the 1st year } = \frac{9600 \times 10 \times 1}{100} = \text{Rs. } 960$
$\displaystyle \text{The sum due at the end of the 1st year } = 9600 + 960 = \text{Rs. } 10560$
$\displaystyle \text{(ii) Interest for the 2nd year } = \frac{10560 \times 10 \times 1}{100} = \text{Rs. } 1056$
$\displaystyle \text{The sum due at the end of the 2nd year } = 10560 + 1056 = \text{Rs. } 11616$
$\displaystyle \text{(iii) Required difference } = 11616 - 10560 = \text{Rs. } 1056$
$\displaystyle \text{Therefore interest for 1 year on this difference } = \frac{1056 \times 10 \times 1}{100} = \text{Rs. } 105.60$
$\displaystyle \text{(iv) C.I. for the 3rd year } = 1056 + 105.60 = \text{Rs. } 1161.60$
$\\$
Question 19: What sum of money will amount to Rs. 9261 in 3 years at 5% per annum compound interest? [ICSE Board 2009]
$\displaystyle \text{Answer:}$
$\displaystyle A=9261 \text{ Rs.; } \hspace{1.0cm} r=5\%; \hspace{1.0cm} n=3 \text{ years }$
$\displaystyle \text{ Amount: } 9261=P \Big(1+ \frac{5}{100} \Big)^3$
$\displaystyle P = 9261 \times \Big( \frac{20}{21} \Big)^3 = \text{Rs. } 8000$
$\displaystyle \text{ Required sum = Rs. } 8000$
$\\$
Question 20: On a certain sum, the compound interest in 3 years and at 10 per cent per annum amounts to Rs. 2317. Find the sum. [ICSE Board 2006]
$\displaystyle \text{Answer:}$
$\displaystyle C.I.=2317 \text{ Rs.; } \hspace{1.0cm} r=10\%; \hspace{1.0cm} n=3 \text{ years }$
$\displaystyle \text{ Amount: } 2317=P \Bigg[ \Big(1+ \frac{10}{100} \Big)^3 - 1 \Bigg]$
$\displaystyle 2317 = P(1.331 - 1)$
$\displaystyle P = \frac{2317}{0.331} = \text{Rs. } 7000$
$\displaystyle \text{Required sum = Rs. } 7000$
$\\$
Question 21: On a certain sum, the compound interest in two years amounts to Rs. 2256. If the rates of interest for successive years are 8% and 10% respectively, find the sum. [ICSE Board 2008]
$\displaystyle \text{Answer:}$
$\displaystyle \text{ Amount: } A=P \Big(1+ \frac{8}{100} \Big)\Big(1+ \frac{10}{100} \Big) = \frac{297}{250} P$
$\displaystyle \text{ Given C.I. = Rs. } 2256$
$\displaystyle \Rightarrow \frac{297}{250} P - P = 2256$
$\displaystyle \Rightarrow \frac{297P-250P}{250} = 2256$
$\displaystyle \Rightarrow P = \frac{2256 \times 250}{47} = \text{Rs. } 12000$
$\displaystyle \text{ Required sum = Rs. } 12000$
$\\$
Question 22: A person lends Rs. 15000 for 2 years at a certain rate of compound interest. If after 2 years, it amounts to Rs.16224; find the rate of interest. [ICSE Board 2007]
$\displaystyle \text{Answer:}$
$\displaystyle P=15000 \text{ Rs.; } \hspace{1.0cm} A=16224 \text{ Rs. }; \hspace{1.0cm} n=2 \text{ years }$
$\displaystyle \Rightarrow 16224=15000 \Big(1+ \frac{r}{100} \Big)^2$
$\displaystyle \Big(1+ \frac{r}{100} \Big)^2 = \Big( \frac{26}{25} \Big)^2$
$\displaystyle 1 + \frac{r}{100} = \frac{26}{25}$
$\displaystyle r = 4\%$
$\\$
Question 23: In what period of time will Rs. 12,000 yield Rs. 3972 as compound interest at 10 per cent, if compounded on an yearly basis. [ICSE Board 2011]
$\displaystyle \text{Answer:}$
$\displaystyle P=12000 \text{ Rs.; } \hspace{1.0cm} C.I.=3972 \text{ Rs. }; \hspace{1.0cm} r=10\% \text{ years }$
$\displaystyle \text{Amount: } A = P + I = 12000 + 3972 = 15972$
$\displaystyle \therefore 15972 = 12000 \Big( 1 + \frac{10}{100} \Big)^n$
$\displaystyle \Rightarrow \frac{15972}{12000} = \Big( \frac{11}{10} \Big)^3$
$\displaystyle \Rightarrow n = 3$
$\displaystyle \text{Therefore Required Time = 3 years }$
$\\$
Question 24: On what sum of money will the difference between the compound interest and the simple interest for 2 years be equal to Rs. 25, if the rate of interest charged for both is 5% p.a.? [ICSE Board 2012]
$\displaystyle \text{Answer:}$
$\displaystyle C.I. = P \Bigg[ \Big( 1 + \frac{r}{100} \Big)^n - 1 \Bigg] = P \Bigg[ \Big( 1 + \frac{5}{100} \Big)^2 - 1 \Bigg] = \frac{41P}{400}$
$\displaystyle S.I. = \frac{P \times 5 \times 2}{100} = \frac{P}{10}$
$\displaystyle \text{Given }C.I. - S.I. = 25$
$\displaystyle \Rightarrow \frac{41P}{400} - \frac{P}{10} = 25$
$\displaystyle \Rightarrow P = 10000$
$\displaystyle \text{Therefore Required sum = Rs. 10000 }$
$\\$