Class 10: Compound Interest

$\displaystyle \textbf{Question 1: }\text{The height of a plant is }80\text{ cm and it is expected to grow}$
$\displaystyle \text{at the rate of }20\%\text{ every month. What will be its height at the end of }3\text{ months?}$
$\displaystyle \text{Answer:}$
$\displaystyle P=80\text{ cm},\ r=20\%,\ n=3\text{ months}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle =80\left(1+\frac{20}{100}\right)^3$
$\displaystyle =80(1.2)^3$
$\displaystyle =80\times1.728=138.24\text{ cm}$
$\displaystyle \therefore\ \text{Height after }3\text{ months}=138.24\text{ cm}$
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$\displaystyle \textbf{Question 2: }\text{The cost of a machine depreciates each year by }12\%\text{ of its value.}$
$\displaystyle \text{If it is valued at Rs. }44000\text{ at the beginning of }2008,\text{ find its value:}$
$\displaystyle \text{(i) at the end of }2009\qquad\text{(ii) at the beginning of }2007$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }44000,\ r=12\%$
$\displaystyle \textbf{(i) Value at the end of }2009$
$\displaystyle A=P\left(1-\frac{r}{100}\right)^2$
$\displaystyle =44000(0.88)^2$
$\displaystyle =\text{Rs. }34073.60$
$\displaystyle \textbf{(ii) Value at the beginning of }2007$
$\displaystyle 44000=P(0.88)$
$\displaystyle P=\frac{44000}{0.88}=\text{Rs. }50000$
$\displaystyle \therefore\ \text{Required values are Rs. }34073.60\text{ and Rs. }50000$
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$\displaystyle \textbf{Question 3: }\text{The value of a machine is Rs. }27000\text{ at the end of }2004$
$\displaystyle \text{and Rs. }21870\text{ at the beginning of }2007.\text{ Calculate the rate of depreciation}$
$\displaystyle \text{and its value at the beginning of }2004.$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }27000,\ A=\text{Rs. }21870,\ n=2\text{ years}$
$\displaystyle \text{Let the rate of depreciation be }r\%\text{ per annum.}$
$\displaystyle 21870=27000\left(1-\frac{r}{100}\right)^2$
$\displaystyle \left(1-\frac{r}{100}\right)^2=\frac{21870}{27000}=0.81=(0.9)^2$
$\displaystyle 1-\frac{r}{100}=0.9$
$\displaystyle \frac{r}{100}=0.1$
$\displaystyle r=10\%$
$\displaystyle \therefore\ \text{Rate of depreciation}=10\%\text{ p.a.}$
$\displaystyle \text{Let the value at the beginning of }2004\text{ be Rs. }x$
$\displaystyle 27000=x\left(1-\frac{10}{100}\right)$
$\displaystyle 27000=0.9x$
$\displaystyle x=\frac{27000}{0.9}=\text{Rs. }30000$
$\displaystyle \therefore\ \text{Value at the beginning of }2004=\text{Rs. }30000$
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$\displaystyle \textbf{Question 4: }\text{The value of an article decreased for }2\text{ years at }10\%\text{ per year}$
$\displaystyle \text{and then increased by }10\%\text{ in the third year. Find its original value,}$
$\displaystyle \text{if its value at the end of }3\text{ years is Rs. }40095.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original value of the article be Rs. }x$
$\displaystyle \text{After }2\text{ years of depreciation at }10\%\text{ per year,}$
$\displaystyle \text{value}=x\left(1-\frac{10}{100}\right)^2=x(0.9)^2=0.81x$
$\displaystyle \text{In the third year, the value increased by }10\%$
$\displaystyle \therefore\ \text{value after }3\text{ years}=0.81x\left(1+\frac{10}{100}\right)$
$\displaystyle 40095=0.81x(1.1)$
$\displaystyle 40095=0.891x$
$\displaystyle x=\frac{40095}{0.891}=\text{Rs. }45000$
$\displaystyle \therefore\ \text{Original value of the article}=\text{Rs. }45000$
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$\displaystyle \textbf{Question 5: }\text{According to a census taken towards the end of }2009,$
$\displaystyle \text{the population of a rural town was }64000.\text{ If it grew at }5\%\text{ p.a.,}$
$\displaystyle \text{in how many years after }2009\text{ did the population reach }74088?$
$\displaystyle \text{Answer:}$
$\displaystyle P=64000,\ A=74088,\ r=5\%,\ n=n\text{ years}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle 74088=64000\left(1+\frac{5}{100}\right)^n$
$\displaystyle \left(1.05\right)^n=\frac{74088}{64000}=1.157625=(1.05)^3$
$\displaystyle \therefore\ n=3\text{ years}$
$\displaystyle \therefore\ \text{The population reached }74088\text{ after }3\text{ years after }2009.$
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$\displaystyle \textbf{Question 6: }\text{The population of a town decreased by }12\%\text{ during }1998$
$\displaystyle \text{and then increased by }8\%\text{ during }1999.\text{ Find the population at the beginning}$
$\displaystyle \text{of }1998,\text{ if at the end of }1999\text{ it was }285120.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the population at the beginning of }1998\text{ be }x$
$\displaystyle \text{After }12\%\text{ decrease during }1998,\text{ population}=x\left(1-\frac{12}{100}\right)$
$\displaystyle =0.88x$
$\displaystyle \text{After }8\%\text{ increase during }1999,\text{ population}=0.88x\left(1+\frac{8}{100}\right)$
$\displaystyle =0.88x\times1.08=0.9504x$
$\displaystyle 0.9504x=285120$
$\displaystyle x=\frac{285120}{0.9504}=300000$
$\displaystyle \therefore\ \text{Population at the beginning of }1998=300000$
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$\displaystyle \textbf{Question 7: }\text{A sum of money invested at compound interest amounts to}$
$\displaystyle \text{Rs. }16500\text{ in }1\text{ year and Rs. }19965\text{ in }3\text{ years. Find the rate}$
$\displaystyle \text{per cent and the original sum of money.}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }16500,\ A=\text{Rs. }19965,\ n=2\text{ years}$
$\displaystyle 19965=16500\left(1+\frac{r}{100}\right)^2$
$\displaystyle \left(1+\frac{r}{100}\right)^2=\frac{19965}{16500}=1.21=(1.1)^2$
$\displaystyle 1+\frac{r}{100}=1.1$
$\displaystyle r=10\%$
$\displaystyle \text{Let the original sum be Rs. }x$
$\displaystyle 16500=x\left(1+\frac{10}{100}\right)$
$\displaystyle x=\frac{16500}{1.1}=\text{Rs. }15000$
$\displaystyle \therefore\ \text{Rate}=10\%\text{ p.a. and Original sum}=\text{Rs. }15000$
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$\displaystyle \textbf{Question 8: }\text{The difference between compound interest and simple interest}$
$\displaystyle \text{on Rs. }7500\text{ for }2\text{ years is Rs. }12.\text{ Find the rate of interest.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the rate of interest be }x\%\text{ per annum.}$
$\displaystyle \text{Simple Interest}=7500\times\frac{x}{100}\times2=150x$
$\displaystyle \text{Compound Interest}=7500\left(1+\frac{x}{100}\right)^2-7500$
$\displaystyle \text{Given, Compound Interest}-\text{Simple Interest}=12$
$\displaystyle 7500\left(1+\frac{x}{100}\right)^2-7500-150x=12$
$\displaystyle 7500\left(1+\frac{2x}{100}+\frac{x^2}{10000}\right)-7500-150x=12$
$\displaystyle 150x+0.75x^2-150x=12$
$\displaystyle 0.75x^2=12$
$\displaystyle x^2=16$
$\displaystyle x=4\%$
$\displaystyle \therefore\ \text{Required rate}=4\%\text{ p.a.}$
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$\displaystyle \textbf{Question 9: }\text{A sum of money lent out at C.I. becomes three times itself}$
$\displaystyle \text{in }10\text{ years. In how many years will it become twenty-seven times itself}$
$\displaystyle \text{at the same rate of interest?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the principal be Rs. }P\text{ and the rate be }r\%\text{ p.a.}$
$\displaystyle \text{After }10\text{ years, the amount becomes }3P$
$\displaystyle 3P=P\left(1+\frac{r}{100}\right)^{10}$
$\displaystyle \left(1+\frac{r}{100}\right)^{10}=3$
$\displaystyle \therefore\ 1+\frac{r}{100}=3^{\frac{1}{10}}\qquad\ldots(i)$
$\displaystyle \text{Let the money become }27\text{ times in }n\text{ years.}$
$\displaystyle 27P=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle 27=\left(1+\frac{r}{100}\right)^n$
$\displaystyle \text{Using }(i),$
$\displaystyle 27=\left(3^{\frac{1}{10}}\right)^n$
$\displaystyle 3^3=3^{\frac{n}{10}}$
$\displaystyle \frac{n}{10}=3$
$\displaystyle n=30\text{ years}$
$\displaystyle \therefore\ \text{The money will become }27\text{ times itself in }30\text{ years.}$
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$\displaystyle \textbf{Question 10: }\text{Sharma borrowed a certain sum at }10\%\text{ p.a., compounded annually.}$
$\displaystyle \text{If he pays Rs. }19360\text{ at the end of the second year and Rs. }31944$
$\displaystyle \text{at the end of the third year to clear the debt, find the sum borrowed.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum borrowed be Rs. }x$
$\displaystyle \text{Amount due at the end of }2\text{ years}=x\left(1+\frac{10}{100}\right)^2$
$\displaystyle =1.21x$
$\displaystyle \text{After paying Rs. }19360,\text{ balance}=1.21x-19360$
$\displaystyle \text{This balance amounts to Rs. }31944\text{ at the end of the third year.}$
$\displaystyle 31944=(1.21x-19360)\left(1+\frac{10}{100}\right)$
$\displaystyle 31944=(1.21x-19360)(1.1)$
$\displaystyle 29040=1.21x-19360$
$\displaystyle 1.21x=48400$
$\displaystyle x=40000$
$\displaystyle \therefore\ \text{Sum borrowed by Sharma}=\text{Rs. }40000$
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$\displaystyle \textbf{Question 11: }\text{The difference between compound interest for a year payable}$
$\displaystyle \text{half-yearly and simple interest on a certain sum at }10\%\text{ for a year is Rs. }15.$
$\displaystyle \text{Find the sum of money lent out.}\hfill\text{[ICSE 1998]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum of money lent out be Rs. }x$
$\displaystyle \textbf{Simple Interest}$
$\displaystyle S.I.=x\times\frac{10}{100}\times1=0.1x$
$\displaystyle \textbf{Compound Interest}$
$\displaystyle \text{Rate per half-year}=5\%,\ \text{Number of half-years}=2$
$\displaystyle A=x\left(1+\frac{10}{200}\right)^2=x\left(\frac{21}{20}\right)^2$
$\displaystyle C.I.=x\left(\frac{21}{20}\right)^2-x$
$\displaystyle \text{Given, }C.I.-S.I.=15$
$\displaystyle x\left(\frac{21}{20}\right)^2-x-0.1x=15$
$\displaystyle x\left(\frac{441}{400}-1-\frac{1}{10}\right)=15$
$\displaystyle x\left(\frac{441-400-40}{400}\right)=15$
$\displaystyle \frac{x}{400}=15$
$\displaystyle x=\text{Rs. }6000$
$\displaystyle \therefore\ \text{Sum of money lent out}=\text{Rs. }6000$
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$\displaystyle \textbf{Question 12: }\text{The ages of Person 1 and Person 2 are }16\text{ years and }18\text{ years}$
$\displaystyle \text{respectively. In what ratio must they invest money at }5\%\text{ p.a., compounded yearly,}$
$\displaystyle \text{so that both get the same sum on attaining the age of }25\text{ years?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Person 1 invests for }25-16=9\text{ years}$
$\displaystyle \text{Person 2 invests for }25-18=7\text{ years}$
$\displaystyle \text{Let the investments of Person 1 and Person 2 be Rs. }x\text{ and Rs. }y$
$\displaystyle \text{Since both receive the same amount,}$
$\displaystyle x\left(1+\frac{5}{100}\right)^9=y\left(1+\frac{5}{100}\right)^7$
$\displaystyle x(1.05)^2=y$
$\displaystyle \frac{x}{y}=\frac{1}{(1.05)^2}=\frac{1}{1.1025}=\frac{400}{441}$
$\displaystyle \therefore\ x:y=400:441$
$\displaystyle \therefore\ \text{Required ratio of investment}=400:441$
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