Class 9: Compound Interest

$\displaystyle \textbf{Exercise 3(A)}$

$\displaystyle \textbf{Question 1: }\text{Find the amount and the compound interest}$
$\displaystyle \text{on Rs. }12,000\text{ in 3 years at }5\%\text{; interest being compounded annually.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 12,000,\ \text{Rate }(r)=5\%\ \text{p.a., Time }(n)=3\ \text{years}$
$\displaystyle \text{Amount}=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle =12,000\left(1+\frac{5}{100}\right)^3$
$\displaystyle =12,000\left(\frac{21}{20}\right)^3$
$\displaystyle =12,000\times\frac{9261}{8000}$
$\displaystyle =Rs.\ 13,891.50$
$\displaystyle \therefore \text{Amount}=Rs.\ 13,891.50$
$\displaystyle \text{Compound Interest}=13,891.50-12,000$
$\displaystyle =Rs.\ 1,891.50$
$\displaystyle \therefore \text{Compound Interest}=Rs.\ 1,891.50$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Calculate the amount, if Rs. }15,000\text{ is lent at}$
$\displaystyle \text{compound interest for 2 years and the rates for the successive years are}$
$\displaystyle 8\%\text{ p.a. and }10\%\text{ p.a. respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 15,000$
$\displaystyle \text{Rate for first year}=8\%,\ \text{Rate for second year}=10\%$
$\displaystyle \text{Amount}=P\left(1+\frac{8}{100}\right)\left(1+\frac{10}{100}\right)$
$\displaystyle =15,000\times\frac{108}{100}\times\frac{110}{100}$
$\displaystyle =15,000\times1.188$
$\displaystyle =Rs.\ 17,820$
$\displaystyle \therefore \text{Required amount}=Rs.\ 17,820$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Calculate the compound interest accrued on}$
$\displaystyle \text{Rs. }6,000\text{ in 3 years, compounded yearly, if the rates for the}$
$\displaystyle \text{successive years are }5\%,\ 8\%\text{ and }10\%\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 6,000$
$\displaystyle \text{Rates of interest for successive years are }5\%,\ 8\%\text{ and }10\%.$
$\displaystyle \text{Amount}=P\left(1+\frac{5}{100}\right)\left(1+\frac{8}{100}\right)\left(1+\frac{10}{100}\right)$
$\displaystyle =6,000\times\frac{105}{100}\times\frac{108}{100}\times\frac{110}{100}$
$\displaystyle =Rs.\ 7,484.40$
$\displaystyle \text{Compound Interest}=7,484.40-6,000$
$\displaystyle =Rs.\ 1,484.40$
$\displaystyle \therefore \text{Required compound interest}=Rs.\ 1,484.40$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{What sum of money will amount to Rs. }5,445$
$\displaystyle \text{in 2 years at }10\%\text{ per annum compound interest?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Amount }(A)=Rs.\ 5,445,\ \text{Rate }(r)=10\%,\ \text{Time }(n)=2\ \text{years}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle 5,445=P\left(\frac{11}{10}\right)^2$
$\displaystyle P=5,445\times\frac{10}{11}\times\frac{10}{11}$
$\displaystyle =Rs.\ 4,500$
$\displaystyle \therefore \text{Required sum}=Rs.\ 4,500$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{On what sum of money will the compound}$
$\displaystyle \text{interest for 2 years at }5\%\text{ per annum amount to Rs. }768.75\text{?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Compound Interest}=Rs.\ 768.75,\ \text{Rate }(r)=5\%,\ \text{Time }(n)=2\ \text{years}$
$\displaystyle \text{C.I.}=P\left[\left(1+\frac{r}{100}\right)^n-1\right]$
$\displaystyle 768.75=P\left[\left(\frac{21}{20}\right)^2-1\right]$
$\displaystyle 768.75=P\left(\frac{441-400}{400}\right)$
$\displaystyle 768.75=P\times\frac{41}{400}$
$\displaystyle P=768.75\times\frac{400}{41}$
$\displaystyle =Rs.\ 7,500$
$\displaystyle \therefore \text{Required sum}=Rs.\ 7,500$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Find the sum on which the compound interest}$
$\displaystyle \text{for 3 years at }10\%\text{ per annum amounts to Rs. }1,655.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Compound Interest}=Rs.\ 1,655,\ \text{Rate }(r)=10\%,\ \text{Time }(n)=3\ \text{years}$
$\displaystyle \text{C.I.}=P\left[\left(1+\frac{r}{100}\right)^n-1\right]$
$\displaystyle 1,655=P\left[\left(1+\frac{10}{100}\right)^3-1\right]$
$\displaystyle 1,655=P\left[\left(\frac{11}{10}\right)^3-1\right]$
$\displaystyle 1,655=P\left(\frac{1331-1000}{1000}\right)$
$\displaystyle 1,655=P\times\frac{331}{1000}$
$\displaystyle P=1,655\times\frac{1000}{331}$
$\displaystyle =Rs.\ 5,000$
$\displaystyle \therefore \text{Required sum}=Rs.\ 5,000$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{What principal will amount to Rs. }9,856\text{ in two}$
$\displaystyle \text{years, if the rates of interest for successive years are }10\%\text{ and }12\%$
$\displaystyle \text{respectively?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Amount }(A)=Rs.\ 9,856$
$\displaystyle \text{Rate for the first year}=10\%,\ \text{Rate for the second year}=12\%$
$\displaystyle A=P\left(1+\frac{10}{100}\right)\left(1+\frac{12}{100}\right)$
$\displaystyle 9,856=P\times\frac{110}{100}\times\frac{112}{100}$
$\displaystyle 9,856=P\times\frac{11}{10}\times\frac{28}{25}$
$\displaystyle P=9,856\times\frac{10}{11}\times\frac{25}{28}$
$\displaystyle =Rs.\ 8,000$
$\displaystyle \therefore \text{Required principal}=Rs.\ 8,000$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{On a certain sum, the compound interest in}$
$\displaystyle 2\text{ years amounts to Rs. }4,240.\text{ If the rates of interest for successive}$
$\displaystyle \text{years are }10\%\text{ and }15\%\text{ respectively, find the sum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Compound Interest}=Rs.\ 4,240$
$\displaystyle \text{Rate for the first year}=10\%,\ \text{Rate for the second year}=15\%$
$\displaystyle \text{Amount}=P\left(1+\frac{10}{100}\right)\left(1+\frac{15}{100}\right)$
$\displaystyle =P\times\frac{110}{100}\times\frac{115}{100}$
$\displaystyle =P\times\frac{253}{200}$
$\displaystyle \text{C.I.}=\text{Amount}-\text{Principal}$
$\displaystyle 4,240=P\left(\frac{253}{200}-1\right)$
$\displaystyle 4,240=P\left(\frac{53}{200}\right)$
$\displaystyle P=4,240\times\frac{200}{53}$
$\displaystyle =Rs.\ 16,000$
$\displaystyle \therefore \text{Required sum}=Rs.\ 16,000$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{At what rate per cent per annum will Rs. }6,000$
$\displaystyle \text{amount to Rs. }6,615\text{ in 2 years when interest is compounded annually?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 6,000,\ \text{Amount }(A)=Rs.\ 6,615,\ \text{Time }(n)=2\text{ years}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle 6,615=6,000\left(1+\frac{r}{100}\right)^2$
$\displaystyle \left(1+\frac{r}{100}\right)^2=\frac{6,615}{6,000}$
$\displaystyle =\frac{441}{400}$
$\displaystyle =\left(\frac{21}{20}\right)^2$
$\displaystyle \therefore 1+\frac{r}{100}=\frac{21}{20}$
$\displaystyle \frac{r}{100}=\frac{21}{20}-1$
$\displaystyle =\frac{1}{20}$
$\displaystyle r=5$
$\displaystyle \therefore \text{The required rate of interest is }5\%\text{ per annum.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{At what rate per cent compound interest does}$
$\displaystyle \text{a sum of money become }1.44\text{ times itself in 2 years?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the principal be }Rs.\ P.$
$\displaystyle \therefore \text{Amount}=Rs.\ 1.44P$
$\displaystyle \text{Time }(n)=2\text{ years}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle 1.44P=P\left(1+\frac{r}{100}\right)^2$
$\displaystyle \left(1+\frac{r}{100}\right)^2=1.44$
$\displaystyle =\left(1.2\right)^2$
$\displaystyle \therefore 1+\frac{r}{100}=1.2$
$\displaystyle \frac{r}{100}=0.2$
$\displaystyle r=20$
$\displaystyle \therefore \text{The required rate of interest is }20\%\text{ per annum.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{At what rate per cent will a sum of Rs. }4,000$
$\displaystyle \text{yield Rs. }1,324\text{ as compound interest in 3 years?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 4,000,\ \text{Compound Interest}=Rs.\ 1,324$
$\displaystyle \text{Time }(n)=3\text{ years}$
$\displaystyle \text{Amount}=\text{Principal}+\text{Compound Interest}$
$\displaystyle =4,000+1,324$
$\displaystyle =Rs.\ 5,324$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle 5,324=4,000\left(1+\frac{r}{100}\right)^3$
$\displaystyle \left(1+\frac{r}{100}\right)^3=\frac{5,324}{4,000}$
$\displaystyle =\frac{1331}{1000}$
$\displaystyle =\left(\frac{11}{10}\right)^3$
$\displaystyle \therefore 1+\frac{r}{100}=\frac{11}{10}$
$\displaystyle \frac{r}{100}=\frac{11}{10}-1$
$\displaystyle =\frac{1}{10}$
$\displaystyle r=10$
$\displaystyle \therefore \text{The required rate of interest is }10\%\text{ per annum.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{A person invests Rs. }5,000\text{ for three years at a}$
$\displaystyle \text{certain rate of interest compounded annually. At the end of two years}$
$\displaystyle \text{this sum amounts to Rs. }6,272.\text{ Calculate:}$
$\displaystyle \text{(i) the rate of interest per annum.}$
$\displaystyle \text{(ii) the amount at the end of the third year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 5,000,\ \text{Amount after 2 years}=Rs.\ 6,272$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^2$
$\displaystyle 6,272=5,000\left(1+\frac{r}{100}\right)^2$
$\displaystyle \left(1+\frac{r}{100}\right)^2=\frac{6,272}{5,000}$
$\displaystyle =\frac{784}{625}=\left(\frac{28}{25}\right)^2$
$\displaystyle \therefore 1+\frac{r}{100}=\frac{28}{25}$
$\displaystyle \frac{r}{100}=\frac{28}{25}-1=\frac{3}{25}$
$\displaystyle r=12\%$
$\displaystyle \therefore \text{(i) Rate of interest per annum}=12\%$
$\displaystyle \\$

$\displaystyle \text{(ii) Amount at the end of the third year}$
$\displaystyle =6,272\left(1+\frac{12}{100}\right)$
$\displaystyle =6,272\times\frac{112}{100}$
$\displaystyle =Rs.\ 7,024.64$
$\displaystyle \therefore \text{Amount at the end of the third year}=Rs.\ 7,024.64$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{In how many years will Rs. }7,000\text{ amount to}$
$\displaystyle \text{Rs. }9,317\text{ at }10\%\text{ per cent per annum compound interest?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 7,000,\ \text{Amount }(A)=Rs.\ 9,317,\ \text{Rate }(r)=10\%$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle 9,317=7,000\left(\frac{11}{10}\right)^n$
$\displaystyle \frac{9,317}{7,000}=\left(\frac{11}{10}\right)^n$
$\displaystyle \frac{1331}{1000}=\left(\frac{11}{10}\right)^n$
$\displaystyle \left(\frac{11}{10}\right)^3=\left(\frac{11}{10}\right)^n$
$\displaystyle \therefore n=3\text{ years}$
$\displaystyle \therefore \text{Required time}=3\text{ years}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Find the time, in years, in which Rs. }4,000\text{ will}$
$\displaystyle \text{produce Rs. }630.50\text{ as compound interest at }5\%\text{ per cent p.a.}$
$\displaystyle \text{interest being compounded annually.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 4,000,\ \text{Compound Interest}=Rs.\ 630.50$
$\displaystyle \text{Amount}=4,000+630.50=Rs.\ 4,630.50$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle 4,630.50=4,000\left(\frac{21}{20}\right)^n$
$\displaystyle \frac{4,630.50}{4,000}=\left(\frac{21}{20}\right)^n$
$\displaystyle \frac{9261}{8000}=\left(\frac{21}{20}\right)^n$
$\displaystyle \left(\frac{21}{20}\right)^3=\left(\frac{21}{20}\right)^n$
$\displaystyle \therefore n=3\text{ years}$
$\displaystyle \therefore \text{Required time}=3\text{ years}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Divide Rs. }28,730\text{ between A and B so that when their}$
$\displaystyle \text{shares are lent out at }10\%\text{ compound interest compounded per year,}$
$\displaystyle \text{the amount that A receives in 3 years is the same as that B receives in 5 years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the share of A}=Rs.\ x$
$\displaystyle \therefore \text{The share of B}=Rs.\ (28,730-x)$
$\displaystyle \text{Amount received by A after 3 years}$
$\displaystyle =x\left(1+\frac{10}{100}\right)^3$
$\displaystyle =x\left(\frac{11}{10}\right)^3$
$\displaystyle \text{Amount received by B after 5 years}$
$\displaystyle =(28,730-x)\left(1+\frac{10}{100}\right)^5$
$\displaystyle =(28,730-x)\left(\frac{11}{10}\right)^5$
$\displaystyle \text{Since both the amounts are equal,}$
$\displaystyle x\left(\frac{11}{10}\right)^3=(28,730-x)\left(\frac{11}{10}\right)^5$
$\displaystyle x=(28,730-x)\left(\frac{11}{10}\right)^2$
$\displaystyle x=(28,730-x)\times\frac{121}{100}$
$\displaystyle 100x=121(28,730-x)$
$\displaystyle 100x=34,76,330-121x$
$\displaystyle 221x=34,76,330$
$\displaystyle x=Rs.\ 15,730$
$\displaystyle \therefore \text{The share of A}=Rs.\ 15,730$
$\displaystyle \text{The share of B}=28,730-15,730$
$\displaystyle =Rs.\ 13,000$
$\displaystyle \therefore \text{The shares of A and B are Rs. }15,730\text{ and Rs. }13,000\text{ respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{A sum of Rs. }44,200\text{ is divided between John and Smith,}$
$\displaystyle 12\text{ years and }14\text{ years old respectively, in such a way that if their}$
$\displaystyle \text{portions are invested at }10\%\text{ per annum compound interest, they will}$
$\displaystyle \text{receive equal amounts on reaching }16\text{ years of age.}$
$\displaystyle \text{(i) What is the share of each out of Rs. }44,200\text{?}$
$\displaystyle \text{(ii) What will each receive, when }16\text{ years old?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{John's age}=12\text{ years and Smith's age}=14\text{ years}$
$\displaystyle \therefore \text{John's money will be invested for }16-12=4\text{ years}$
$\displaystyle \text{Smith's money will be invested for }16-14=2\text{ years}$
$\displaystyle \text{Let John's share}=Rs.\ x$
$\displaystyle \therefore \text{Smith's share}=Rs.\ (44,200-x)$
$\displaystyle \text{Amount received by John at the age of }16$
$\displaystyle =x\left(1+\frac{10}{100}\right)^4$
$\displaystyle =x\left(\frac{11}{10}\right)^4$
$\displaystyle \text{Amount received by Smith at the age of }16$
$\displaystyle =(44,200-x)\left(1+\frac{10}{100}\right)^2$
$\displaystyle =(44,200-x)\left(\frac{11}{10}\right)^2$
$\displaystyle \text{Since both the amounts are equal,}$
$\displaystyle x\left(\frac{11}{10}\right)^4=(44,200-x)\left(\frac{11}{10}\right)^2$
$\displaystyle x\left(\frac{11}{10}\right)^2=44,200-x$
$\displaystyle \frac{121x}{100}=44,200-x$
$\displaystyle 121x=44,20,000-100x$
$\displaystyle 221x=44,20,000$
$\displaystyle x=Rs.\ 20,000$
$\displaystyle \therefore \text{John's share}=Rs.\ 20,000$
$\displaystyle \text{Smith's share}=44,200-20,000$
$\displaystyle =Rs.\ 24,200$
$\displaystyle \therefore \text{(i) John's share is Rs. }20,000\text{ and Smith's share is Rs. }24,200.$
$\displaystyle \\$

$\displaystyle \text{(ii) Amount received by John at the age of }16$
$\displaystyle =20,000\left(\frac{11}{10}\right)^4$
$\displaystyle =20,000\times\frac{14641}{10000}$
$\displaystyle =Rs.\ 29,282$
$\displaystyle \text{Amount received by Smith at the age of }16$
$\displaystyle =24,200\left(\frac{11}{10}\right)^2$
$\displaystyle =24,200\times\frac{121}{100}$
$\displaystyle =Rs.\ 29,282$
$\displaystyle \therefore \text{Each will receive Rs. }29,282\text{ on reaching the age of }16\text{ years.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{The simple interest on a certain sum of money}$
$\displaystyle \text{at }10\%\text{ per annum is Rs. }6,000\text{ in 2 years. Find:}$
$\displaystyle \text{(i) the sum.}$
$\displaystyle \text{(ii) the amount due at the end of 3 years at the same rate of interest}$
$\displaystyle \text{compounded annually.}$
$\displaystyle \text{(iii) the compound interest earned in 3 years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Simple Interest}=Rs.\ 6,000,\ \text{Rate }(r)=10\%\text{ p.a., Time }(t)=2\text{ years}$
$\displaystyle \text{S.I.}=\frac{P\times r\times t}{100}$
$\displaystyle 6,000=\frac{P\times10\times2}{100}$
$\displaystyle P=\frac{6,000\times100}{10\times2}$
$\displaystyle =Rs.\ 30,000$
$\displaystyle \therefore \text{(i) The required sum}=Rs.\ 30,000$
$\displaystyle \\$

$\displaystyle \text{(ii) Amount due at the end of 3 years}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle =30,000\left(1+\frac{10}{100}\right)^3$
$\displaystyle =30,000\left(\frac{11}{10}\right)^3$
$\displaystyle =30,000\times\frac{1331}{1000}$
$\displaystyle =Rs.\ 39,930$
$\displaystyle \therefore \text{The amount due at the end of 3 years}=Rs.\ 39,930$
$\displaystyle \\$

$\displaystyle \text{(iii) Compound Interest}=\text{Amount}-\text{Principal}$
$\displaystyle =39,930-30,000$
$\displaystyle =Rs.\ 9,930$
$\displaystyle \therefore \text{The compound interest earned in 3 years}=Rs.\ 9,930$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{Find the difference between compound interest}$
$\displaystyle \text{and simple interest on Rs. }8,000\text{ in 2 years at }5\%\text{ per annum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 8,000,\ \text{Rate }(r)=5\%\text{ p.a., Time }(t)=2\text{ years}$
$\displaystyle \text{Simple Interest}=\frac{P\times r\times t}{100}$
$\displaystyle =\frac{8,000\times5\times2}{100}$
$\displaystyle =Rs.\ 800$
$\displaystyle \text{Amount at compound interest}=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle =8,000\left(1+\frac{5}{100}\right)^2$
$\displaystyle =8,000\left(\frac{21}{20}\right)^2$
$\displaystyle =8,000\times\frac{441}{400}$
$\displaystyle =Rs.\ 8,820$
$\displaystyle \text{Compound Interest}=8,820-8,000$
$\displaystyle =Rs.\ 820$
$\displaystyle \text{Difference between C.I. and S.I.}=820-800$
$\displaystyle =Rs.\ 20$
$\displaystyle \therefore \text{The required difference}=Rs.\ 20$
$\displaystyle \\$

$\displaystyle \textbf{Exercise 3(B)}$

$\displaystyle \textbf{Question 1: }\text{The difference between simple interest and}$
$\displaystyle \text{compound interest on a certain sum is Rs. }54.40\text{ for 2 years}$
$\displaystyle \text{at }8\%\text{ per annum. Find the sum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required sum be Rs. }P.$
$\displaystyle \text{For 2 years, the difference between C.I. and S.I.}$
$\displaystyle =P\left(\frac{r}{100}\right)^2$
$\displaystyle 54.40=P\left(\frac{8}{100}\right)^2$
$\displaystyle 54.40=P\times\frac{64}{10,000}$
$\displaystyle P=\frac{54.40\times10,000}{64}$
$\displaystyle =Rs.\ 8,500$
$\displaystyle \therefore \text{The required sum}=Rs.\ 8,500$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{A sum of money, invested at compound interest,}$
$\displaystyle \text{amounts to Rs. }19,360\text{ in 2 years and to Rs. }23,425.60$
$\displaystyle \text{in 4 years. Find the rate per cent and the original sum of money.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original sum be Rs. }P\text{ and the rate of interest be }r\%\text{ p.a.}$
$\displaystyle \text{Amount after 2 years}=P\left(1+\frac{r}{100}\right)^2=19,360\qquad\ldots\text{(i)}$
$\displaystyle \text{Amount after 4 years}=P\left(1+\frac{r}{100}\right)^4=23,425.60\qquad\ldots\text{(ii)}$
$\displaystyle \text{Dividing (ii) by (i), we get}$
$\displaystyle \left(1+\frac{r}{100}\right)^2=\frac{23,425.60}{19,360}$
$\displaystyle =\frac{121}{100}$
$\displaystyle =\left(\frac{11}{10}\right)^2$
$\displaystyle \therefore 1+\frac{r}{100}=\frac{11}{10}$
$\displaystyle \frac{r}{100}=\frac{1}{10}$
$\displaystyle r=10\%$
$\displaystyle \therefore \text{The rate of interest is }10\%\text{ per annum.}$
$\displaystyle \\$

$\displaystyle \text{Substituting }r=10\%\text{ in (i), we get}$
$\displaystyle 19,360=P\left(1+\frac{10}{100}\right)^2$
$\displaystyle 19,360=P\left(\frac{11}{10}\right)^2$
$\displaystyle 19,360=P\times\frac{121}{100}$
$\displaystyle P=19,360\times\frac{100}{121}$
$\displaystyle =Rs.\ 16,000$
$\displaystyle \therefore \text{The original sum of money}=Rs.\ 16,000$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{A sum of money lent out at C.I. at a certain}$
$\displaystyle \text{rate per annum becomes three times itself in 8 years. Find in how many}$
$\displaystyle \text{years the money will become twenty-seven times itself at the same rate}$
$\displaystyle \text{of interest per annum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the principal be Rs. }P\text{ and the rate of interest be }r\%\text{ p.a.}$
$\displaystyle \text{Since the sum becomes three times itself in 8 years,}$
$\displaystyle 3P=P\left(1+\frac{r}{100}\right)^8$
$\displaystyle \therefore \left(1+\frac{r}{100}\right)^8=3\qquad\ldots\text{(i)}$
$\displaystyle \text{Let the sum become twenty-seven times itself in }n\text{ years.}$
$\displaystyle 27P=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle \therefore \left(1+\frac{r}{100}\right)^n=27$
$\displaystyle =3^3$
$\displaystyle =\left[\left(1+\frac{r}{100}\right)^8\right]^3$
$\displaystyle =\left(1+\frac{r}{100}\right)^{24}$
$\displaystyle \therefore n=24$
$\displaystyle \therefore \text{The money will become twenty-seven times itself in }24\text{ years.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{On what sum of money will compound interest}$
$\displaystyle \text{(payable annually) for 2 years be the same as simple interest on}$
$\displaystyle \text{Rs. }9,430\text{ for 10 years, both at the rate of }5\%\text{ per annum?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Simple interest on Rs. }9,430\text{ for 10 years at }5\%\text{ per annum}$
$\displaystyle =\frac{P\times r\times t}{100}$
$\displaystyle =\frac{9,430\times5\times10}{100}$
$\displaystyle =Rs.\ 4,715$
$\displaystyle \text{Let the required sum be Rs. }P.$
$\displaystyle \text{Compound interest on Rs. }P\text{ for 2 years at }5\%\text{ per annum}$
$\displaystyle =P\left[\left(1+\frac{5}{100}\right)^2-1\right]$
$\displaystyle =P\left[\left(\frac{21}{20}\right)^2-1\right]$
$\displaystyle =P\left(\frac{441-400}{400}\right)$
$\displaystyle =\frac{41P}{400}$
$\displaystyle \text{Since the compound interest and simple interest are equal,}$
$\displaystyle \frac{41P}{400}=4,715$
$\displaystyle P=\frac{4,715\times400}{41}$
$\displaystyle =Rs.\ 46,000$
$\displaystyle \therefore \text{The required sum}=Rs.\ 46,000$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Kamal and Anand each lent the same sum of money}$
$\displaystyle \text{for 2 years at }5\%\text{ at simple interest and compound interest}$
$\displaystyle \text{respectively. Anand received Rs. }15\text{ more than Kamal. Find the}$
$\displaystyle \text{amount of money lent by each and the interest received.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the amount of money lent by each be Rs. }P.$
$\displaystyle \text{For 2 years, the difference between C.I. and S.I.}$
$\displaystyle =P\left(\frac{r}{100}\right)^2$
$\displaystyle 15=P\left(\frac{5}{100}\right)^2$
$\displaystyle 15=P\times\frac{25}{10,000}$
$\displaystyle P=\frac{15\times10,000}{25}$
$\displaystyle =Rs.\ 6,000$
$\displaystyle \therefore \text{The amount of money lent by each}=Rs.\ 6,000$
$\displaystyle \\$

$\displaystyle \text{Simple interest received by Kamal}$
$\displaystyle =\frac{P\times r\times t}{100}$
$\displaystyle =\frac{6,000\times5\times2}{100}$
$\displaystyle =Rs.\ 600$
$\displaystyle \text{Compound interest received by Anand}$
$\displaystyle =6,000\left[\left(1+\frac{5}{100}\right)^2-1\right]$
$\displaystyle =6,000\left[\left(\frac{21}{20}\right)^2-1\right]$
$\displaystyle =6,000\times\frac{41}{400}$
$\displaystyle =Rs.\ 615$
$\displaystyle \therefore \text{Kamal received Rs. }600\text{ and Anand received Rs. }615\text{ as interest.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Simple interest on a sum of money for 2 years}$
$\displaystyle \text{at }4\%\text{ is Rs. }450.\text{ Find compound interest on the same sum}$
$\displaystyle \text{and at the same rate for 2 years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Simple Interest}=Rs.\ 450,\ \text{Rate }(r)=4\%,\ \text{Time }(t)=2\text{ years}$
$\displaystyle \text{S.I.}=\frac{P\times r\times t}{100}$
$\displaystyle 450=\frac{P\times4\times2}{100}$
$\displaystyle P=\frac{450\times100}{4\times2}$
$\displaystyle =Rs.\ 5,625$
$\displaystyle \text{Compound Interest}=P\left[\left(1+\frac{r}{100}\right)^2-1\right]$
$\displaystyle =5,625\left[\left(1+\frac{4}{100}\right)^2-1\right]$
$\displaystyle =5,625\left[\left(\frac{26}{25}\right)^2-1\right]$
$\displaystyle =5,625\left(\frac{676-625}{625}\right)$
$\displaystyle =5,625\times\frac{51}{625}$
$\displaystyle =Rs.\ 459$
$\displaystyle \therefore \text{The required compound interest}=Rs.\ 459$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Simple interest on a certain sum of money for}$
$\displaystyle 4\text{ years at }4\%\text{ per annum exceeds the compound interest on the}$
$\displaystyle \text{same sum for 3 years at }5\%\text{ per annum by Rs. }228.\text{ Find the sum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required sum be Rs. }P.$
$\displaystyle \text{Simple interest for 4 years at }4\%\text{ per annum}$
$\displaystyle =\frac{P\times4\times4}{100}$
$\displaystyle =\frac{4P}{25}$
$\displaystyle \text{Compound interest for 3 years at }5\%\text{ per annum}$
$\displaystyle =P\left[\left(1+\frac{5}{100}\right)^3-1\right]$
$\displaystyle =P\left[\left(\frac{21}{20}\right)^3-1\right]$
$\displaystyle =P\left(\frac{9261-8000}{8000}\right)$
$\displaystyle =\frac{1261P}{8000}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{4P}{25}-\frac{1261P}{8000}=228$
$\displaystyle \frac{1280P-1261P}{8000}=228$
$\displaystyle \frac{19P}{8000}=228$
$\displaystyle P=\frac{228\times8000}{19}$
$\displaystyle =Rs.\ 96,000$
$\displaystyle \therefore \text{The required sum}=Rs.\ 96,000$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Compound interest on a certain sum of money}$
$\displaystyle \text{at }5\%\text{ per annum for two years is Rs. }246.\text{ Calculate simple}$
$\displaystyle \text{interest on the same sum for 3 years at }6\%\text{ per annum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required sum be Rs. }P.$
$\displaystyle \text{Compound interest for 2 years at }5\%\text{ per annum}$
$\displaystyle =P\left[\left(1+\frac{5}{100}\right)^2-1\right]$
$\displaystyle =P\left[\left(\frac{21}{20}\right)^2-1\right]$
$\displaystyle =P\left(\frac{441-400}{400}\right)$
$\displaystyle =\frac{41P}{400}$
$\displaystyle \therefore \frac{41P}{400}=246$
$\displaystyle P=\frac{246\times400}{41}$
$\displaystyle =Rs.\ 2,400$
$\displaystyle \text{Simple interest on Rs. }2,400\text{ for 3 years at }6\%\text{ per annum}$
$\displaystyle =\frac{P\times r\times t}{100}$
$\displaystyle =\frac{2,400\times6\times3}{100}$
$\displaystyle =Rs.\ 432$
$\displaystyle \therefore \text{The required simple interest}=Rs.\ 432$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{A certain sum of money amounts to Rs. }23,400$
$\displaystyle \text{in 3 years at }10\%\text{ per annum simple interest. Find the amount}$
$\displaystyle \text{of the same sum in 2 years and at }10\%\text{ p.a. compound interest.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required sum be Rs. }P.$
$\displaystyle \text{Amount at simple interest}=P+\frac{P\times r\times t}{100}$
$\displaystyle 23,400=P+\frac{P\times10\times3}{100}$
$\displaystyle 23,400=P+\frac{3P}{10}$
$\displaystyle 23,400=\frac{13P}{10}$
$\displaystyle P=\frac{23,400\times10}{13}$
$\displaystyle =Rs.\ 18,000$
$\displaystyle \text{Amount at compound interest for 2 years at }10\%\text{ per annum}$
$\displaystyle =P\left(1+\frac{r}{100}\right)^n$
$\displaystyle =18,000\left(1+\frac{10}{100}\right)^2$
$\displaystyle =18,000\left(\frac{11}{10}\right)^2$
$\displaystyle =18,000\times\frac{121}{100}$
$\displaystyle =Rs.\ 21,780$
$\displaystyle \therefore \text{The required amount}=Rs.\ 21,780$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Mohit borrowed a certain sum at }5\%\text{ per}$
$\displaystyle \text{annum compound interest and cleared this loan by paying Rs. }12,600$
$\displaystyle \text{at the end of the first year and Rs. }17,640\text{ at the end of the second}$
$\displaystyle \text{year. Find the sum borrowed.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum borrowed be Rs. }P.$
$\displaystyle \text{Amount due at the end of the first year}=P\left(1+\frac{5}{100}\right)$
$\displaystyle =P\times\frac{21}{20}$
$\displaystyle \text{After paying Rs. }12,600,\text{ the outstanding balance}$
$\displaystyle =\frac{21P}{20}-12,600$
$\displaystyle \text{This balance amounts to Rs. }17,640\text{ at the end of the second year.}$
$\displaystyle \left(\frac{21P}{20}-12,600\right)\left(\frac{21}{20}\right)=17,640$
$\displaystyle \frac{21P}{20}-12,600=17,640\times\frac{20}{21}$
$\displaystyle =Rs.\ 16,800$
$\displaystyle \frac{21P}{20}=16,800+12,600$
$\displaystyle =29,400$
$\displaystyle P=29,400\times\frac{20}{21}$
$\displaystyle =Rs.\ 28,000$
$\displaystyle \therefore \text{The sum borrowed}=Rs.\ 28,000$
$\displaystyle \\$

$\displaystyle \textbf{Exercise 3(C)}$

$\displaystyle \textbf{Question 1: }\text{If the interest is compounded half-yearly, calculate the amount}$
$\displaystyle \text{when the principal is Rs. }7,400\text{; the rate of interest is }5\%\text{ per annum}$
$\displaystyle \text{and the duration is one year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 7,400,\ \text{Rate }(r)=5\%\text{ p.a., Time }(n)=1\text{ year}$
$\displaystyle \text{Since the interest is compounded half-yearly,}$
$\displaystyle \text{Rate per half-year}=\frac{5}{2}\%=2.5\%$
$\displaystyle \text{Number of half-years}=1\times2=2$
$\displaystyle A=P\left(1+\frac{r}{2\times100}\right)^{2n}$
$\displaystyle =7,400\left(1+\frac{5}{2\times100}\right)^2$
$\displaystyle =7,400\left(\frac{41}{40}\right)^2$
$\displaystyle =7,400\times\frac{1681}{1600}$
$\displaystyle =Rs.\ 7,774.625$
$\displaystyle \therefore \text{The required amount}=Rs.\ 7,774.63\text{ (approximately).}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the difference between the compound interest compounded}$
$\displaystyle \text{yearly and half-yearly on Rs. }10,000\text{ for 18 months at }10\%$
$\displaystyle \text{per annum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 10,000,\ \text{Rate }(r)=10\%\text{ p.a.}$
$\displaystyle \text{Time}=18\text{ months}=1\frac{1}{2}\text{ years}$
$\displaystyle \text{When interest is compounded yearly,}$
$\displaystyle A_1=10,000\left(1+\frac{10}{100}\right)\left(1+\frac{10}{2\times100}\right)$
$\displaystyle =10,000\times\frac{11}{10}\times\frac{21}{20}$
$\displaystyle =Rs.\ 11,550$
$\displaystyle \text{C.I. when compounded yearly}=11,550-10,000$
$\displaystyle =Rs.\ 1,550$

$\displaystyle \text{When interest is compounded half-yearly,}$
$\displaystyle \text{Rate per half-year}=\frac{10}{2}\%=5\%$
$\displaystyle \text{Number of half-years}=18\div6=3$
$\displaystyle A_2=10,000\left(1+\frac{5}{100}\right)^3$
$\displaystyle =10,000\left(\frac{21}{20}\right)^3$
$\displaystyle =10,000\times\frac{9261}{8000}$
$\displaystyle =Rs.\ 11,576.25$
$\displaystyle \text{C.I. when compounded half-yearly}=11,576.25-10,000$
$\displaystyle =Rs.\ 1,576.25$
$\displaystyle \text{Required difference}=1,576.25-1,550$
$\displaystyle =Rs.\ 26.25$
$\displaystyle \therefore \text{The required difference}=Rs.\ 26.25$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{A man borrowed Rs. }16,000\text{ for 3 years under the}$
$\displaystyle \text{following terms: }20\%\text{ simple interest for the first 2 years and}$
$\displaystyle 20\%\text{ C.I. for the remaining one year on the amount due after 2 years,}$
$\displaystyle \text{the interest being compounded half-yearly. Find the total amount to be}$
$\displaystyle \text{paid at the end of three years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 16,000$
$\displaystyle \text{Simple interest for the first 2 years}$
$\displaystyle =\frac{P\times r\times t}{100}$
$\displaystyle =\frac{16,000\times20\times2}{100}$
$\displaystyle =Rs.\ 6,400$
$\displaystyle \text{Amount due after 2 years}=16,000+6,400$
$\displaystyle =Rs.\ 22,400$

$\displaystyle \text{For the remaining one year, }P=Rs.\ 22,400,\ r=20\%\text{ p.a.}$
$\displaystyle \text{Since the interest is compounded half-yearly,}$
$\displaystyle \text{Rate per half-year}=\frac{20}{2}\%=10\%$
$\displaystyle \text{Number of half-years}=1\times2=2$
$\displaystyle A=22,400\left(1+\frac{20}{2\times100}\right)^2$
$\displaystyle =22,400\left(1+\frac{10}{100}\right)^2$
$\displaystyle =22,400\left(\frac{11}{10}\right)^2$
$\displaystyle =22,400\times\frac{121}{100}$
$\displaystyle =Rs.\ 27,104$
$\displaystyle \therefore \text{The total amount to be paid at the end of three years}=Rs.\ 27,104.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{What sum of money will amount to Rs. }27,783\text{ in}$
$\displaystyle \text{one and a half years at }10\%\text{ per annum compounded half-yearly?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Amount }(A)=Rs.\ 27,783,\ \text{Rate }(r)=10\%\text{ p.a.}$
$\displaystyle \text{Time}=1\frac{1}{2}\text{ years}$
$\displaystyle \text{Since the interest is compounded half-yearly,}$
$\displaystyle \text{Rate per half-year}=\frac{10}{2}\%=5\%$
$\displaystyle \text{Number of half-years}=1\frac{1}{2}\times2=3$
$\displaystyle A=P\left(1+\frac{r}{2\times100}\right)^{2n}$
$\displaystyle 27,783=P\left(1+\frac{10}{2\times100}\right)^3$
$\displaystyle 27,783=P\left(\frac{21}{20}\right)^3$
$\displaystyle 27,783=P\times\frac{9261}{8000}$
$\displaystyle P=27,783\times\frac{8000}{9261}$
$\displaystyle =Rs.\ 24,000$
$\displaystyle \therefore \text{The required sum}=Rs.\ 24,000$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Ashok invests a certain sum of money at }20\%$
$\displaystyle \text{per annum, compounded yearly. Geeta invests an equal amount of money}$
$\displaystyle \text{at the same rate of interest per annum compounded half-yearly. If Geeta}$
$\displaystyle \text{gets Rs. }33\text{ more than Ashok in 18 months, calculate the money invested.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the money invested by each be Rs. }P.$
$\displaystyle \text{Time}=18\text{ months}=1\frac{1}{2}\text{ years}$
$\displaystyle \text{For Ashok, the interest is compounded yearly.}$
$\displaystyle \text{Amount after the first year}=P\left(1+\frac{20}{100}\right)$
$\displaystyle =P\times\frac{6}{5}$
$\displaystyle \text{For the remaining half-year, the interest is calculated at }10\%.$
$\displaystyle \therefore \text{Ashok's amount}=P\times\frac{6}{5}\left(1+\frac{10}{100}\right)$
$\displaystyle =P\times\frac{6}{5}\times\frac{11}{10}$
$\displaystyle =\frac{33P}{25}$
$\displaystyle \\$

$\displaystyle \text{For Geeta, the interest is compounded half-yearly.}$
$\displaystyle \text{Rate per half-year}=\frac{20}{2}\%=10\%$
$\displaystyle \text{Number of half-years}=1\frac{1}{2}\times2=3$
$\displaystyle \therefore \text{Geeta's amount}=P\left(1+\frac{10}{100}\right)^3$
$\displaystyle =P\left(\frac{11}{10}\right)^3$
$\displaystyle =\frac{1331P}{1000}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{1331P}{1000}-\frac{33P}{25}=33$
$\displaystyle \frac{1331P-1320P}{1000}=33$
$\displaystyle \frac{11P}{1000}=33$
$\displaystyle P=\frac{33\times1000}{11}$
$\displaystyle =Rs.\ 3,000$
$\displaystyle \therefore \text{The money invested by each}=Rs.\ 3,000$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{At what rate of interest per annum will a sum}$
$\displaystyle \text{of Rs. }62,500\text{ earn a compound interest of Rs. }5,100\text{ in one year?}$
$\displaystyle \text{The interest is to be compounded half-yearly.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 62,500,\ \text{Compound Interest}=Rs.\ 5,100$
$\displaystyle \text{Amount}=62,500+5,100$
$\displaystyle =Rs.\ 67,600$
$\displaystyle \text{Let the rate of interest be }r\%\text{ per annum.}$
$\displaystyle \text{Since the interest is compounded half-yearly,}$
$\displaystyle \text{Rate per half-year}=\frac{r}{2}\%$
$\displaystyle \text{Number of half-years}=1\times2=2$
$\displaystyle A=P\left(1+\frac{r}{2\times100}\right)^2$
$\displaystyle 67,600=62,500\left(1+\frac{r}{200}\right)^2$
$\displaystyle \left(1+\frac{r}{200}\right)^2=\frac{67,600}{62,500}$
$\displaystyle =\frac{676}{625}$
$\displaystyle =\left(\frac{26}{25}\right)^2$
$\displaystyle \therefore 1+\frac{r}{200}=\frac{26}{25}$
$\displaystyle \frac{r}{200}=\frac{26}{25}-1$
$\displaystyle =\frac{1}{25}$
$\displaystyle r=8$
$\displaystyle \therefore \text{The required rate of interest is }8\%\text{ per annum.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{In what time will Rs. }1,500\text{ yield Rs. }496.50$
$\displaystyle \text{as compound interest at }20\%\text{ per year compounded half-yearly?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 1,500,\ \text{Compound Interest}=Rs.\ 496.50$
$\displaystyle \text{Amount}=1,500+496.50$
$\displaystyle =Rs.\ 1,996.50$
$\displaystyle \text{Rate per half-year}=\frac{20}{2}\%=10\%$
$\displaystyle \text{Let the number of half-years be }n.$
$\displaystyle A=P\left(1+\frac{10}{100}\right)^n$
$\displaystyle 1,996.50=1,500\left(\frac{11}{10}\right)^n$
$\displaystyle \left(\frac{11}{10}\right)^n=\frac{1,996.50}{1,500}$
$\displaystyle =1.331$
$\displaystyle =\left(\frac{11}{10}\right)^3$
$\displaystyle \therefore n=3\text{ half-years}$
$\displaystyle \text{Required time}=\frac{3}{2}\text{ years}$
$\displaystyle =1\frac{1}{2}\text{ years}$
$\displaystyle \therefore \text{The required time is }1\frac{1}{2}\text{ years.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Calculate the C.I. on Rs. }3,500\text{ at }6\%\text{ per}$
$\displaystyle \text{annum for 3 years, the interest being compounded half-yearly.}$
$\displaystyle \text{Do not use mathematical tables. Use the necessary information:}$
$\displaystyle (1.06)^3=1.191016,\quad(1.03)^3=1.092727$
$\displaystyle (1.06)^6=1.418519,\quad(1.03)^6=1.194052$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 3,500,\ \text{Rate }(r)=6\%\text{ p.a., Time}=3\text{ years}$
$\displaystyle \text{Since the interest is compounded half-yearly,}$
$\displaystyle \text{Rate per half-year}=\frac{6}{2}\%=3\%$
$\displaystyle \text{Number of half-years}=3\times2=6$
$\displaystyle A=P\left(1+\frac{r}{2\times100}\right)^{2n}$
$\displaystyle =3,500\left(1+\frac{6}{2\times100}\right)^6$
$\displaystyle =3,500(1.03)^6$
$\displaystyle =3,500\times1.194052$
$\displaystyle =Rs.\ 4,179.182$
$\displaystyle \text{Compound Interest}=4,179.182-3,500$
$\displaystyle =Rs.\ 679.182$
$\displaystyle \therefore \text{The required compound interest}=Rs.\ 679.18\text{ (approximately).}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Find the difference between compound interest}$
$\displaystyle \text{and simple interest on Rs. }12,000\text{ in }1\frac{1}{2}\text{ years at }10\%$
$\displaystyle \text{per annum, compounded yearly.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 12,000,\ \text{Rate }(r)=10\%\text{ p.a.},\ \text{Time}=1\frac{1}{2}\text{ years}$
$\displaystyle \text{Simple Interest}=\frac{12,000\times10\times\frac{3}{2}}{100}$
$\displaystyle =Rs.\ 1,800$
$\displaystyle \text{Amount at compound interest}$
$\displaystyle =12,000\left(1+\frac{10}{100}\right)\left(1+\frac{10}{2\times100}\right)$
$\displaystyle =12,000\times\frac{11}{10}\times\frac{21}{20}$
$\displaystyle =Rs.\ 13,860$
$\displaystyle \text{Compound Interest}=13,860-12,000$
$\displaystyle =Rs.\ 1,860$
$\displaystyle \text{Difference}=1,860-1,800$
$\displaystyle =Rs.\ 60$
$\displaystyle \therefore \text{The required difference}=Rs.\ 60$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Find the difference between compound interest}$
$\displaystyle \text{and simple interest on Rs. }12,000\text{ in }1\frac{1}{2}\text{ years at }10\%$
$\displaystyle \text{per annum, compounded half-yearly.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 12,000,\ \text{Rate }(r)=10\%\text{ p.a.},\ \text{Time}=1\frac{1}{2}\text{ years}$
$\displaystyle \text{Simple Interest}=\frac{12,000\times10\times\frac{3}{2}}{100}$
$\displaystyle =Rs.\ 1,800$
$\displaystyle \text{Since the interest is compounded half-yearly,}$
$\displaystyle \text{Rate per half-year}=\frac{10}{2}\%=5\%$
$\displaystyle \text{Number of half-years}=1\frac{1}{2}\times2=3$
$\displaystyle \text{Amount at compound interest}$
$\displaystyle =12,000\left(1+\frac{5}{100}\right)^3$
$\displaystyle =12,000\left(\frac{21}{20}\right)^3$
$\displaystyle =12,000\times\frac{9261}{8000}$
$\displaystyle =Rs.\ 13,891.50$
$\displaystyle \text{Compound Interest}=13,891.50-12,000$
$\displaystyle =Rs.\ 1,891.50$
$\displaystyle \text{Difference}=1,891.50-1,800$
$\displaystyle =Rs.\ 91.50$
$\displaystyle \therefore \text{The required difference}=Rs.\ 91.50$
$\displaystyle \\$

$\displaystyle \textbf{Exercise 3(D)}$

$\displaystyle \textbf{Question 1: }\text{The cost of a machine is supposed to depreciate each year}$
$\displaystyle \text{by }12\%\text{ of its value at the beginning of the year. If the machine is}$
$\displaystyle \text{valued at Rs. }44,000\text{ at the beginning of 2008, find its value:}$
$\displaystyle \text{(i) at the end of 2009.}$
$\displaystyle \text{(ii) at the beginning of 2007.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Value of the machine at the beginning of 2008}=Rs.\ 44,000$
$\displaystyle \text{Rate of depreciation}=12\%\text{ per annum}$
$\displaystyle \text{(i) From the beginning of 2008 to the end of 2009, the time is }2\text{ years.}$
$\displaystyle \text{Value after }n\text{ years}=P\left(1-\frac{r}{100}\right)^n$
$\displaystyle \text{Value at the end of 2009}=44,000\left(1-\frac{12}{100}\right)^2$
$\displaystyle =44,000\left(\frac{88}{100}\right)^2$
$\displaystyle =44,000\left(\frac{22}{25}\right)^2$
$\displaystyle =44,000\times\frac{484}{625}$
$\displaystyle =Rs.\ 34,073.60$
$\displaystyle \therefore \text{The value of the machine at the end of 2009 is Rs. }34,073.60.$
$\displaystyle \\$

$\displaystyle \text{(ii) Let the value of the machine at the beginning of 2007 be Rs. }P.$
$\displaystyle \text{After one year's depreciation, its value at the beginning of 2008 is Rs. }44,000.$
$\displaystyle 44,000=P\left(1-\frac{12}{100}\right)$
$\displaystyle 44,000=P\times\frac{88}{100}$
$\displaystyle P=44,000\times\frac{100}{88}$
$\displaystyle =Rs.\ 50,000$
$\displaystyle \therefore \text{The value of the machine at the beginning of 2007 was Rs. }50,000.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{The value of an article decreased for two years at the rate}$
$\displaystyle \text{of }10\%\text{ per year and then in the third year it increased by }10\%.\text{ Find}$
$\displaystyle \text{the original value of the article, if its value at the end of 3 years is Rs. }40,095.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original value of the article be Rs. }P.$
$\displaystyle \text{Value after a decrease of }10\%\text{ in the first year}$
$\displaystyle =P\left(1-\frac{10}{100}\right)$
$\displaystyle \text{Value after a decrease of }10\%\text{ in the second year}$
$\displaystyle =P\left(1-\frac{10}{100}\right)^2$
$\displaystyle \text{In the third year, the value increases by }10\%.$
$\displaystyle \therefore 40,095=P\left(1-\frac{10}{100}\right)^2\left(1+\frac{10}{100}\right)$
$\displaystyle 40,095=P\left(\frac{9}{10}\right)^2\left(\frac{11}{10}\right)$
$\displaystyle 40,095=P\times\frac{81}{100}\times\frac{11}{10}$
$\displaystyle 40,095=P\times\frac{891}{1000}$
$\displaystyle P=40,095\times\frac{1000}{891}$
$\displaystyle =Rs.\ 45,000$
$\displaystyle \therefore \text{The original value of the article was Rs. }45,000.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{According to a census taken towards the end of the year}$
$\displaystyle 2009,\text{ the population of a rural town was found to be }64,000.\text{ The census}$
$\displaystyle \text{authority also found that the population of this particular town had a}$
$\displaystyle \text{growth of }5\%\text{ per annum. In how many years after 2009 did the population}$
$\displaystyle \text{of this town reach }74,088\text{?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Present population }(P)=64,000$
$\displaystyle \text{Population after }n\text{ years }(A)=74,088$
$\displaystyle \text{Rate of growth }(r)=5\%\text{ per annum}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle 74,088=64,000\left(1+\frac{5}{100}\right)^n$
$\displaystyle \left(\frac{21}{20}\right)^n=\frac{74,088}{64,000}$
$\displaystyle =\frac{9261}{8000}$
$\displaystyle =\left(\frac{21}{20}\right)^3$
$\displaystyle \therefore n=3$
$\displaystyle \therefore \text{The population reached }74,088\text{ after }3\text{ years, i.e., at the end of }2012.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{The population of a town decreased by }12\%\text{ during}$
$\displaystyle 1998\text{ and then increased by }8\%\text{ during }1999.\text{ Find the population}$
$\displaystyle \text{of the town at the beginning of }1998,\text{ if at the end of }1999\text{ its}$
$\displaystyle \text{population was }2,85,120.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the population at the beginning of }1998\text{ be }P.$
$\displaystyle \text{Population after a decrease of }12\%\text{ in }1998=P\left(1-\frac{12}{100}\right)$
$\displaystyle =P\times\frac{88}{100}$
$\displaystyle \text{Population after an increase of }8\%\text{ in }1999=P\times\frac{88}{100}\times\frac{108}{100}$
$\displaystyle =P\times\frac{2376}{2500}$
$\displaystyle \therefore 2,85,120=P\times\frac{2376}{2500}$
$\displaystyle P=2,85,120\times\frac{2500}{2376}$
$\displaystyle =Rs.\ 3,00,000$
$\displaystyle \therefore \text{The population at the beginning of }1998\text{ was }3,00,000.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{The difference between C.I. and S.I. on Rs. }7,500$
$\displaystyle \text{for two years is Rs. }12\text{ at the same rate of interest per annum.}$
$\displaystyle \text{Find the rate of interest.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 7,500$
$\displaystyle \text{Difference between C.I. and S.I.}=Rs.\ 12$
$\displaystyle \text{For 2 years, }\text{C.I.}-\text{S.I.}=P\left(\frac{r}{100}\right)^2$
$\displaystyle 12=7,500\left(\frac{r}{100}\right)^2$
$\displaystyle \left(\frac{r}{100}\right)^2=\frac{12}{7,500}$
$\displaystyle =\frac{1}{625}$
$\displaystyle \frac{r}{100}=\frac{1}{25}$
$\displaystyle r=\frac{100}{25}=4$
$\displaystyle \therefore \text{The required rate of interest is }4\%\text{ per annum.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Mr. Sharma borrowed a certain sum of money}$
$\displaystyle \text{at }10\%\text{ per annum compounded annually. If by paying Rs. }19,360$
$\displaystyle \text{at the end of the second year and Rs. }31,944\text{ at the end of the third}$
$\displaystyle \text{year, he clears the debt; find the sum borrowed by him.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum borrowed by Mr. Sharma be Rs. }P.$
$\displaystyle \text{The present value of the payment of Rs. }19,360\text{ due at the end of 2 years}$
$\displaystyle =\frac{19,360}{\left(1+\frac{10}{100}\right)^2}$
$\displaystyle =19,360\left(\frac{10}{11}\right)^2$
$\displaystyle =19,360\times\frac{100}{121}$
$\displaystyle =Rs.\ 16,000$
$\displaystyle \text{The present value of the payment of Rs. }31,944\text{ due at the end of 3 years}$
$\displaystyle =\frac{31,944}{\left(1+\frac{10}{100}\right)^3}$
$\displaystyle =31,944\left(\frac{10}{11}\right)^3$
$\displaystyle =31,944\times\frac{1000}{1331}$
$\displaystyle =Rs.\ 24,000$
$\displaystyle \therefore P=16,000+24,000$
$\displaystyle =Rs.\ 40,000$
$\displaystyle \therefore \text{The sum borrowed by Mr. Sharma was Rs. }40,000.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{The ages of Pramod and Rohit are }16\text{ years}$
$\displaystyle \text{and }18\text{ years respectively. In what ratio must they invest money}$
$\displaystyle \text{at }5\%\text{ p.a. compounded yearly so that both get the same sum on}$
$\displaystyle \text{attaining the age of }25\text{ years?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Pramod's money will remain invested for }25-16=9\text{ years.}$
$\displaystyle \text{Rohit's money will remain invested for }25-18=7\text{ years.}$
$\displaystyle \text{Let Pramod invest Rs. }x\text{ and Rohit invest Rs. }y.$
$\displaystyle \text{Since both receive the same amount at the age of }25,$
$\displaystyle x\left(1+\frac{5}{100}\right)^9=y\left(1+\frac{5}{100}\right)^7$
$\displaystyle x\left(\frac{21}{20}\right)^2=y$
$\displaystyle \frac{x}{y}=\left(\frac{20}{21}\right)^2$
$\displaystyle =\frac{400}{441}$
$\displaystyle \therefore \text{The required ratio of the investments of Pramod and Rohit is }400:441.$
$\displaystyle \\$

$\displaystyle \textbf{Exercise 3(E)}$

$\displaystyle \textbf{Question 1: }\text{Simple interest on a sum of money for 2 years}$
$\displaystyle \text{at }4\%\text{ is Rs. }450.\text{ Find compound interest on the same sum}$
$\displaystyle \text{and at the same rate for 1 year, if the interest is reckoned half-yearly.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Simple Interest}=Rs.\ 450,\ \text{Rate }(r)=4\%\text{ p.a., Time }(t)=2\text{ years}$
$\displaystyle \text{S.I.}=\frac{P\times r\times t}{100}$
$\displaystyle 450=\frac{P\times4\times2}{100}$
$\displaystyle P=\frac{450\times100}{4\times2}$
$\displaystyle =Rs.\ 5,625$
$\displaystyle \text{For compound interest, the interest is reckoned half-yearly.}$
$\displaystyle \text{Rate per half-year}=\frac{4}{2}\%=2\%$
$\displaystyle \text{Number of half-years}=1\times2=2$
$\displaystyle A=P\left(1+\frac{r}{2\times100}\right)^2$
$\displaystyle =5,625\left(1+\frac{4}{2\times100}\right)^2$
$\displaystyle =5,625\left(\frac{51}{50}\right)^2$
$\displaystyle =5,625\times\frac{2601}{2500}$
$\displaystyle =Rs.\ 5,852.25$
$\displaystyle \text{Compound Interest}=5,852.25-5,625$
$\displaystyle =Rs.\ 227.25$
$\displaystyle \therefore \text{The required compound interest}=Rs.\ 227.25$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the compound interest to the nearest rupee}$
$\displaystyle \text{on Rs. }10,800\text{ for }2\frac{1}{2}\text{ years at }10\%\text{ per annum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 10,800,\ \text{Rate }(r)=10\%\text{ p.a.}$
$\displaystyle \text{Time}=2\frac{1}{2}\text{ years}$
$\displaystyle \text{Amount after 2 complete years}$
$\displaystyle =P\left(1+\frac{r}{100}\right)^2$
$\displaystyle =10,800\left(1+\frac{10}{100}\right)^2$
$\displaystyle =10,800\left(\frac{11}{10}\right)^2$
$\displaystyle =Rs.\ 13,068$
$\displaystyle \text{For the remaining half-year, interest is calculated at }5\%\text{ on Rs. }13,068.$
$\displaystyle \text{Amount after }2\frac{1}{2}\text{ years}=13,068\left(1+\frac{5}{100}\right)$
$\displaystyle =13,068\times\frac{21}{20}$
$\displaystyle =Rs.\ 13,721.40$
$\displaystyle \text{Compound Interest}=13,721.40-10,800$
$\displaystyle =Rs.\ 2,921.40$
$\displaystyle \therefore \text{The compound interest to the nearest rupee}=Rs.\ 2,921$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{The value of a machine, purchased two years ago,}$
$\displaystyle \text{depreciates at the annual rate of }10\%.\text{ If its present value is}$
$\displaystyle \text{Rs. }97,200,\text{ find:}$
$\displaystyle \text{(i) its value after 2 years.}$
$\displaystyle \text{(ii) its value when it was purchased.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Present value of the machine}=Rs.\ 97,200$
$\displaystyle \text{Rate of depreciation}=10\%\text{ per annum}$

$\displaystyle \text{(i) Value after 2 years}=P\left(1-\frac{r}{100}\right)^2$
$\displaystyle =97,200\left(1-\frac{10}{100}\right)^2$
$\displaystyle =97,200\left(\frac{9}{10}\right)^2$
$\displaystyle =97,200\times\frac{81}{100}$
$\displaystyle =Rs.\ 78,732$
$\displaystyle \therefore \text{Its value after 2 years will be Rs. }78,732.$

$\displaystyle \text{(ii) Let the value of the machine when purchased be Rs. }P.$
$\displaystyle \text{After 2 years of depreciation, its present value is Rs. }97,200.$
$\displaystyle 97,200=P\left(1-\frac{10}{100}\right)^2$
$\displaystyle 97,200=P\left(\frac{9}{10}\right)^2$
$\displaystyle 97,200=P\times\frac{81}{100}$
$\displaystyle P=97,200\times\frac{100}{81}$
$\displaystyle =Rs.\ 1,20,000$
$\displaystyle \therefore \text{The value of the machine when purchased was Rs. }1,20,000.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Anuj and Rajesh each lent the same sum of money}$
$\displaystyle \text{for 2 years at }8\%\text{ simple interest and compound interest respectively.}$
$\displaystyle \text{Rajesh received Rs. }64\text{ more than Anuj. Find the money lent by each}$
$\displaystyle \text{and the interest received.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum of money lent by each be Rs. }P.$
$\displaystyle \text{For 2 years, the difference between C.I. and S.I.}$
$\displaystyle =P\left(\frac{r}{100}\right)^2$
$\displaystyle 64=P\left(\frac{8}{100}\right)^2$
$\displaystyle 64=P\times\frac{64}{10,000}$
$\displaystyle P=\frac{64\times10,000}{64}$
$\displaystyle =Rs.\ 10,000$
$\displaystyle \therefore \text{The money lent by each}=Rs.\ 10,000$

$\displaystyle \text{Simple interest received by Anuj}$
$\displaystyle =\frac{P\times r\times t}{100}$
$\displaystyle =\frac{10,000\times8\times2}{100}$
$\displaystyle =Rs.\ 1,600$
$\displaystyle \text{Compound interest received by Rajesh}$
$\displaystyle =10,000\left[\left(1+\frac{8}{100}\right)^2-1\right]$
$\displaystyle =10,000\left[\left(\frac{27}{25}\right)^2-1\right]$
$\displaystyle =10,000\left(\frac{729-625}{625}\right)$
$\displaystyle =10,000\times\frac{104}{625}$
$\displaystyle =Rs.\ 1,664$
$\displaystyle \therefore \text{Anuj received Rs. }1,600\text{ and Rajesh received Rs. }1,664\text{ as interest.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Calculate the sum of money on which the compound}$
$\displaystyle \text{interest (payable annually) for 2 years is four times the simple interest}$
$\displaystyle \text{on Rs. }4,715\text{ for 5 years, both at the rate of }5\%\text{ per annum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Simple interest on Rs. }4,715\text{ for 5 years at }5\%\text{ per annum}$
$\displaystyle =\frac{P\times r\times t}{100}$
$\displaystyle =\frac{4,715\times5\times5}{100}$
$\displaystyle =Rs.\ 1,178.75$
$\displaystyle \therefore \text{Four times the simple interest}=4\times1,178.75$
$\displaystyle =Rs.\ 4,715$
$\displaystyle \text{Let the required sum be Rs. }P.$
$\displaystyle \text{Compound interest for 2 years at }5\%\text{ per annum}$
$\displaystyle =P\left[\left(1+\frac{5}{100}\right)^2-1\right]$
$\displaystyle =P\left[\left(\frac{21}{20}\right)^2-1\right]$
$\displaystyle =P\left(\frac{441-400}{400}\right)$
$\displaystyle =\frac{41P}{400}$
$\displaystyle \therefore \frac{41P}{400}=4,715$
$\displaystyle P=\frac{4,715\times400}{41}$
$\displaystyle =Rs.\ 46,000$
$\displaystyle \therefore \text{The required sum}=Rs.\ 46,000$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{A sum of money was invested for 3 years, interest}$
$\displaystyle \text{being compounded annually. The rates for successive years were }10\%,$
$\displaystyle 15\%\text{ and }18\%\text{ respectively. If the compound interest for the second}$
$\displaystyle \text{year amounted to Rs. }4,950,\text{ find the sum invested.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum invested be Rs. }P.$
$\displaystyle \text{Amount at the end of the first year}$
$\displaystyle =P\left(1+\frac{10}{100}\right)$
$\displaystyle =\frac{11P}{10}$
$\displaystyle \text{The interest for the second year is }15\%\text{ of the amount at the end of the first year.}$
$\displaystyle \therefore \frac{15}{100}\times\frac{11P}{10}=4,950$
$\displaystyle \frac{33P}{200}=4,950$
$\displaystyle P=4,950\times\frac{200}{33}$
$\displaystyle =Rs.\ 30,000$
$\displaystyle \therefore \text{The sum invested}=Rs.\ 30,000$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{A sum of money is invested at }10\%\text{ per annum}$
$\displaystyle \text{compounded half-yearly. If the difference of amounts at the end of}$
$\displaystyle 6\text{ months and }12\text{ months is Rs. }189,\text{ find the sum of money invested.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum of money invested be Rs. }P.$
$\displaystyle \text{Rate per half-year}=\frac{10}{2}\%=5\%$
$\displaystyle \text{Amount at the end of }6\text{ months}$
$\displaystyle =P\left(1+\frac{5}{100}\right)$
$\displaystyle =P\times\frac{21}{20}$
$\displaystyle \text{Amount at the end of }12\text{ months}$
$\displaystyle =P\left(1+\frac{5}{100}\right)^2$
$\displaystyle =P\left(\frac{21}{20}\right)^2$
$\displaystyle \text{According to the question,}$
$\displaystyle P\left(\frac{21}{20}\right)^2-P\left(\frac{21}{20}\right)=189$
$\displaystyle P\left(\frac{441}{400}-\frac{420}{400}\right)=189$
$\displaystyle \frac{21P}{400}=189$
$\displaystyle P=189\times\frac{400}{21}$
$\displaystyle =Rs.\ 3,600$
$\displaystyle \therefore \text{The sum of money invested}=Rs.\ 3,600$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Rohit borrows Rs. }86,000\text{ from Arun for two}$
$\displaystyle \text{years at }5\%\text{ per annum simple interest. He immediately lends out}$
$\displaystyle \text{this money to Akshay at }5\%\text{ compound interest compounded annually}$
$\displaystyle \text{for the same period. Calculate Rohit's profit in the transaction at the end}$
$\displaystyle \text{of two years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 86,000,\ \text{Rate }(r)=5\%\text{ p.a., Time }(t)=2\text{ years}$
$\displaystyle \text{Simple interest payable by Rohit to Arun}$
$\displaystyle =\frac{P\times r\times t}{100}$
$\displaystyle =\frac{86,000\times5\times2}{100}$
$\displaystyle =Rs.\ 8,600$
$\displaystyle \text{Amount payable by Rohit to Arun}=86,000+8,600$
$\displaystyle =Rs.\ 94,600$

$\displaystyle \text{Amount received by Rohit from Akshay}$
$\displaystyle =P\left(1+\frac{r}{100}\right)^2$
$\displaystyle =86,000\left(1+\frac{5}{100}\right)^2$
$\displaystyle =86,000\left(\frac{21}{20}\right)^2$
$\displaystyle =86,000\times\frac{441}{400}$
$\displaystyle =Rs.\ 94,815$
$\displaystyle \text{Rohit's profit}=94,815-94,600$
$\displaystyle =Rs.\ 215$
$\displaystyle \therefore \text{Rohit's profit at the end of two years}=Rs.\ 215$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{The simple interest on a certain sum of money}$
$\displaystyle \text{for 3 years at }5\%\text{ per annum is Rs. }1,200.\text{ Find the amount due}$
$\displaystyle \text{and the compound interest on this sum of money at the same rate and after}$
$\displaystyle 2\text{ years, interest is reckoned annually.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Simple Interest}=Rs.\ 1,200,\ \text{Rate }(r)=5\%\text{ p.a., Time }(t)=3\text{ years}$
$\displaystyle \text{S.I.}=\frac{P\times r\times t}{100}$
$\displaystyle 1,200=\frac{P\times5\times3}{100}$
$\displaystyle P=\frac{1,200\times100}{5\times3}$
$\displaystyle =Rs.\ 8,000$
$\displaystyle \text{Amount due after 2 years at compound interest}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^2$
$\displaystyle =8,000\left(1+\frac{5}{100}\right)^2$
$\displaystyle =8,000\left(\frac{21}{20}\right)^2$
$\displaystyle =8,000\times\frac{441}{400}$
$\displaystyle =Rs.\ 8,820$
$\displaystyle \therefore \text{The amount due after 2 years}=Rs.\ 8,820$
$\displaystyle \text{Compound Interest}=8,820-8,000$
$\displaystyle =Rs.\ 820$
$\displaystyle \therefore \text{The compound interest}=Rs.\ 820$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Nikita invests Rs. }6,000\text{ for two years at a certain}$
$\displaystyle \text{rate of interest compounded annually. At the end of first year it amounts}$
$\displaystyle \text{to Rs. }6,720.\text{ Calculate:}$
$\displaystyle \text{(a) the rate of interest.}$
$\displaystyle \text{(b) the amount at the end of the second year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Principal }(P)=Rs.\ 6,000,\ \text{Amount after 1 year}=Rs.\ 6,720$
$\displaystyle \text{Interest for the first year}=6,720-6,000$
$\displaystyle =Rs.\ 720$
$\displaystyle \text{(a) Rate of interest}=\frac{720}{6,000}\times100\%$
$\displaystyle =12\%$
$\displaystyle \therefore \text{The rate of interest is }12\%\text{ per annum.}$

$\displaystyle \text{(b) Amount at the end of the second year}$
$\displaystyle =6,720\left(1+\frac{12}{100}\right)$
$\displaystyle =6,720\times\frac{112}{100}$
$\displaystyle =Rs.\ 7,526.40$
$\displaystyle \therefore \text{The amount at the end of the second year}=Rs.\ 7,526.40$
$\displaystyle \\$