Class 10: Compound Interest

$\displaystyle \textbf{Question 1: } \text{Find the amount and the compound interest on Rs. }12000$
$\displaystyle \text{in }3\text{ years at }5\%\text{ per annum, compounded annually.}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }12000,\ r=5\%,\ n=3\text{ years}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle =12000\left(1+\frac{5}{100}\right)^3$
$\displaystyle =12000\left(\frac{21}{20}\right)^3$
$\displaystyle =\text{Rs. }13891.50$
$\displaystyle \therefore\ \text{Amount}=\text{Rs. }13891.50$
$\displaystyle \text{Compound Interest}=13891.50-12000=\text{Rs. }1891.50$
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$\displaystyle \textbf{Question 2: } \text{Calculate the amount, if Rs. }15000\text{ is lent at compound}$
$\displaystyle \text{interest for }2\text{ years at }8\%\text{ p.a. and }10\%\text{ p.a. respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{First year: }P=\text{Rs. }15000,\ r=8\%,\ n=1\text{ year}$
$\displaystyle A=15000\left(1+\frac{8}{100}\right)=\text{Rs. }16200$
$\displaystyle \text{Second year: }P=\text{Rs. }16200,\ r=10\%,\ n=1\text{ year}$
$\displaystyle A=16200\left(1+\frac{10}{100}\right)=\text{Rs. }17820$
$\displaystyle \therefore\ \text{Amount}=\text{Rs. }17820$
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$\displaystyle \textbf{Question 3: } \text{Calculate the compound interest accrued on Rs. }9000$
$\displaystyle \text{in }3\text{ years at }5\%,\ 8\%\text{ and }10\%\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{First year: }P=\text{Rs. }9000,\ r=5\%,\ n=1\text{ year}$
$\displaystyle A=9000\left(1+\frac{5}{100}\right)=\text{Rs. }9450$
$\displaystyle \text{Second year: }P=\text{Rs. }9450,\ r=8\%,\ n=1\text{ year}$
$\displaystyle A=9450\left(1+\frac{8}{100}\right)=\text{Rs. }10206$
$\displaystyle \text{Third year: }P=\text{Rs. }10206,\ r=10\%,\ n=1\text{ year}$
$\displaystyle A=10206\left(1+\frac{10}{100}\right)=\text{Rs. }11226.60$
$\displaystyle \text{Compound Interest}=11226.60-9000=\text{Rs. }2226.60$
$\\$

$\displaystyle \textbf{Question 4: } \text{What sum of money will amount to Rs. }5445\text{ in }2\text{ years}$
$\displaystyle \text{at }10\%\text{ per annum compound interest?}$
$\displaystyle \text{Answer:}$
$\displaystyle A=\text{Rs. }5445,\ P=\text{Rs. }x,\ r=10\%,\ n=2\text{ years}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle 5445=x\left(1+\frac{10}{100}\right)^2$
$\displaystyle 5445=x\left(\frac{11}{10}\right)^2$
$\displaystyle x=\frac{5445\times100}{121}=\text{Rs. }4500$
$\displaystyle \therefore\ \text{Required sum}=\text{Rs. }4500$
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$\displaystyle \textbf{Question 5: } \text{On what sum of money will the compound interest for }2\text{ years}$
$\displaystyle \text{at }5\%\text{ per annum amount to Rs. }768.75?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the principal be }P=\text{Rs. }x,\ r=5\%,\ n=2\text{ years}$
$\displaystyle A=x\left(1+\frac{5}{100}\right)^2=x\left(\frac{21}{20}\right)^2$
$\displaystyle \text{Compound Interest}=A-P=768.75$
$\displaystyle x\left(\frac{21}{20}\right)^2-x=768.75$
$\displaystyle x\left(\frac{441}{400}-1\right)=768.75$
$\displaystyle x\left(\frac{41}{400}\right)=768.75$
$\displaystyle x=\frac{768.75\times400}{41}=\text{Rs. }7500$
$\displaystyle \therefore\ \text{Required sum}=\text{Rs. }7500$
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$\displaystyle \textbf{Question 6: } \text{Find the sum on which the compound interest for }3\text{ years}$
$\displaystyle \text{at }10\%\text{ per annum amounts to Rs. }1655.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the principal be }P=\text{Rs. }x,\ r=10\%,\ n=3\text{ years}$
$\displaystyle A=x\left(1+\frac{10}{100}\right)^3=x\left(\frac{11}{10}\right)^3$
$\displaystyle \text{Compound Interest}=A-P=1655$
$\displaystyle x\left(\frac{11}{10}\right)^3-x=1655$
$\displaystyle x\left(\frac{1331}{1000}-1\right)=1655$
$\displaystyle x\left(\frac{331}{1000}\right)=1655$
$\displaystyle x=\frac{1655\times1000}{331}=\text{Rs. }5000$
$\displaystyle \therefore\ \text{Required sum}=\text{Rs. }5000$
$\\$

$\displaystyle \textbf{Question 7: } \text{What principal will amount to Rs. }9856\text{ in }2\text{ years, if the}$
$\displaystyle \text{rates of interest for successive years are }10\%\text{ and }12\%\text{ respectively?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the principal be }P=\text{Rs. }x$
$\displaystyle \text{After the first year at }10\%,\ A=1.1x$
$\displaystyle \text{After the second year at }12\%,\ A=1.1x\left(1+\frac{12}{100}\right)$
$\displaystyle =1.1x\times1.12=1.232x$
$\displaystyle 1.232x=9856$
$\displaystyle x=\frac{9856}{1.232}=\text{Rs. }8000$
$\displaystyle \therefore\ \text{Required principal}=\text{Rs. }8000$
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$\displaystyle \textbf{Question 8: } \text{On a certain sum, the compound interest in }2\text{ years amounts}$
$\displaystyle \text{to Rs. }4240.\text{ If the rates for successive years are }10\%\text{ and }15\%,$
$\displaystyle \text{find the sum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the principal be }P=\text{Rs. }x$
$\displaystyle \text{After the first year at }10\%,\ A=1.1x$
$\displaystyle \text{After the second year at }15\%,\ A=1.1x\left(1+\frac{15}{100}\right)$
$\displaystyle =1.1x\times1.15=1.265x$
$\displaystyle \text{Compound Interest}=1.265x-x=4240$
$\displaystyle 0.265x=4240$
$\displaystyle x=\frac{4240}{0.265}=\text{Rs. }16000$
$\displaystyle \therefore\ \text{Required sum}=\text{Rs. }16000$
$\\$

$\displaystyle \textbf{Question 9: } \text{At what rate per cent per annum will Rs. }6000\text{ amount to}$
$\displaystyle \text{Rs. }6615\text{ in }2\text{ years, compounded annually?}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }6000,\ A=\text{Rs. }6615,\ r=x\%,\ n=2\text{ years}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle 6615=6000\left(1+\frac{x}{100}\right)^2$
$\displaystyle \left(1+\frac{x}{100}\right)^2=\frac{6615}{6000}=\frac{441}{400}$
$\displaystyle 1+\frac{x}{100}=\frac{21}{20}$
$\displaystyle \frac{x}{100}=\frac{1}{20}$
$\displaystyle x=5\%$
$\displaystyle \therefore\ \text{Required rate}=5\%\text{ p.a.}$
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$\displaystyle \textbf{Question 10: } \text{At what rate per cent compound interest does a sum of money}$
$\displaystyle \text{become }1.44\text{ times of itself in }2\text{ years?}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }x,\ A=1.44x,\ r=r\%,\ n=2\text{ years}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^2$
$\displaystyle 1.44x=x\left(1+\frac{r}{100}\right)^2$
$\displaystyle \left(1+\frac{r}{100}\right)^2=1.44=\left(1.2\right)^2$
$\displaystyle 1+\frac{r}{100}=1.2$
$\displaystyle \frac{r}{100}=0.2$
$\displaystyle r=20\%$
$\displaystyle \therefore\ \text{Required rate}=20\%\text{ p.a.}$
$\\$

$\displaystyle \textbf{Question 11: } \text{At what rate per cent will a sum of Rs. }4000\text{ yield}$
$\displaystyle \text{Rs. }1324\text{ as compound interest in }3\text{ years?}\hfill\text{[ICSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }4000,\ r=x\%,\ n=3\text{ years}$
$\displaystyle \text{Compound Interest}=\text{Rs. }1324$
$\displaystyle \therefore\ A=4000+1324=\text{Rs. }5324$
$\displaystyle A=P\left(1+\frac{x}{100}\right)^3$
$\displaystyle 5324=4000\left(1+\frac{x}{100}\right)^3$
$\displaystyle \left(1+\frac{x}{100}\right)^3=\frac{5324}{4000}=1.331=(1.1)^3$
$\displaystyle 1+\frac{x}{100}=1.1$
$\displaystyle x=10\%$
$\displaystyle \therefore\ \text{Required rate}=10\%\text{ p.a.}$
$\\$

$\displaystyle \textbf{Question 12: } \text{A person invests Rs. }5000\text{ for }3\text{ years at a certain}$
$\displaystyle \text{rate of compound interest. At the end of }2\text{ years the amount is}$
$\displaystyle \text{Rs. }6272.\text{ Calculate: (i) the rate of interest (ii) the amount after }3\text{ years.}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }5000,\ A=\text{Rs. }6272,\ r=x\%,\ n=2\text{ years}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle 6272=5000\left(1+\frac{x}{100}\right)^2$
$\displaystyle \left(1+\frac{x}{100}\right)^2=\frac{6272}{5000}=1.2544=(1.12)^2$
$\displaystyle 1+\frac{x}{100}=1.12$
$\displaystyle x=12\%$
$\displaystyle \text{Amount at the end of the third year}$
$\displaystyle A=5000\left(1+\frac{12}{100}\right)^3$
$\displaystyle =5000(1.12)^3=\text{Rs. }7024.64$
$\displaystyle \therefore\ \text{Rate}=12\%\text{ p.a. and Amount after }3\text{ years}=\text{Rs. }7024.64$
$\\$

$\displaystyle \textbf{Question 13: } \text{In how many years will Rs. }7000\text{ amount to Rs. }9317$
$\displaystyle \text{at }10\%\text{ per annum compound interest?}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }7000,\ A=\text{Rs. }9317,\ r=10\%,\ n=n\text{ years}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle 9317=7000\left(1+\frac{10}{100}\right)^n$
$\displaystyle \left(1.1\right)^n=\frac{9317}{7000}=1.331=(1.1)^3$
$\displaystyle \therefore\ n=3\text{ years}$
$\\$

$\displaystyle \textbf{Question 14: } \text{Find the time, in years, in which Rs. }4000\text{ will produce}$
$\displaystyle \text{Rs. }630.50\text{ as compound interest at }5\%\text{ p.a.}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }4000,\ r=5\%,\ n=n\text{ years}$
$\displaystyle \text{Compound Interest}=\text{Rs. }630.50$
$\displaystyle \therefore\ A=4000+630.50=\text{Rs. }4630.50$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle 4630.50=4000\left(1+\frac{5}{100}\right)^n$
$\displaystyle \left(1.05\right)^n=\frac{4630.50}{4000}=1.157625=(1.05)^3$
$\displaystyle \therefore\ n=3\text{ years}$
$\\$

$\displaystyle \textbf{Question 15: } \text{Divide Rs. }28730\text{ between }A\text{ and }B\text{ so that when}$
$\displaystyle \text{their shares are lent at }10\%\text{ compound interest, A's amount in }3\text{ years}$
$\displaystyle \text{is the same as B's amount in }5\text{ years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let A's share be Rs. }x$
$\displaystyle \therefore\ \text{B's share}=\text{Rs. }(28730-x)$
$\displaystyle \text{For A: }P=\text{Rs. }x,\ r=10\%,\ n=3\text{ years}$
$\displaystyle \text{Amount received by A}=x\left(1+\frac{10}{100}\right)^3$
$\displaystyle \text{For B: }P=\text{Rs. }(28730-x),\ r=10\%,\ n=5\text{ years}$
$\displaystyle \text{Amount received by B}=(28730-x)\left(1+\frac{10}{100}\right)^5$
$\displaystyle \text{According to the question,}$
$\displaystyle x\left(1+\frac{10}{100}\right)^3=(28730-x)\left(1+\frac{10}{100}\right)^5$
$\displaystyle x=(28730-x)(1.1)^2$
$\displaystyle x=1.21(28730-x)$
$\displaystyle x=34763.30-1.21x$
$\displaystyle 2.21x=34763.30$
$\displaystyle x=\text{Rs. }15730$
$\displaystyle \therefore\ \text{A's share}=\text{Rs. }15730$
$\displaystyle \text{B's share}=28730-15730=\text{Rs. }13000$
$\\$

$\displaystyle \textbf{Question 16: } \text{A sum of Rs. }34522\text{ is divided between }A\text{ and }B,$
$\displaystyle 18\text{ years and }21\text{ years old respectively, so that at }5\%\text{ compound interest}$
$\displaystyle \text{both receive equal money at the age of }30\text{ years. Find their shares.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let A's share be Rs. }x$
$\displaystyle \therefore\ \text{B's share}=\text{Rs. }(34522-x)$
$\displaystyle \text{A is }18\text{ years old, so time for A}=30-18=12\text{ years}$
$\displaystyle \text{B is }21\text{ years old, so time for B}=30-21=9\text{ years}$
$\displaystyle \text{For A: }P=\text{Rs. }x,\ r=5\%,\ n=12\text{ years}$
$\displaystyle \text{Amount received by A}=x\left(1+\frac{5}{100}\right)^{12}$
$\displaystyle \text{For B: }P=\text{Rs. }(34522-x),\ r=5\%,\ n=9\text{ years}$
$\displaystyle \text{Amount received by B}=(34522-x)\left(1+\frac{5}{100}\right)^9$
$\displaystyle \text{According to the question,}$
$\displaystyle x\left(1+\frac{5}{100}\right)^{12}=(34522-x)\left(1+\frac{5}{100}\right)^9$
$\displaystyle x(1.05)^3=34522-x$
$\displaystyle 1.157625x=34522-x$
$\displaystyle 2.157625x=34522$
$\displaystyle x=\text{Rs. }16000$
$\displaystyle \therefore\ \text{A's share}=\text{Rs. }16000$
$\displaystyle \text{B's share}=34522-16000=\text{Rs. }18522$
$\\$

$\displaystyle \textbf{Question 17: } \text{A sum of Rs. }44200\text{ is divided between }A\text{ and }B,$
$\displaystyle 12\text{ years and }14\text{ years old respectively, so that at }10\%\text{ compound interest}$
$\displaystyle \text{they receive equal amounts on reaching }16\text{ years of age. Find their shares}$
$\displaystyle \text{and the amount each receives.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let A's share be Rs. }x$
$\displaystyle \therefore\ \text{B's share}=\text{Rs. }(44200-x)$
$\displaystyle \text{A is }12\text{ years old, so time for A}=16-12=4\text{ years}$
$\displaystyle \text{B is }14\text{ years old, so time for B}=16-14=2\text{ years}$
$\displaystyle \text{For A: }P=\text{Rs. }x,\ r=10\%,\ n=4\text{ years}$
$\displaystyle \text{Amount received by A}=x\left(1+\frac{10}{100}\right)^4$
$\displaystyle \text{For B: }P=\text{Rs. }(44200-x),\ r=10\%,\ n=2\text{ years}$
$\displaystyle \text{Amount received by B}=(44200-x)\left(1+\frac{10}{100}\right)^2$
$\displaystyle \text{According to the question,}$
$\displaystyle x\left(1+\frac{10}{100}\right)^4=(44200-x)\left(1+\frac{10}{100}\right)^2$
$\displaystyle x(1.1)^2=44200-x$
$\displaystyle 1.21x=44200-x$
$\displaystyle 2.21x=44200$
$\displaystyle x=\text{Rs. }20000$
$\displaystyle \therefore\ \text{A's share}=\text{Rs. }20000$
$\displaystyle \text{B's share}=44200-20000=\text{Rs. }24200$
$\displaystyle \text{Amount received by each}=20000(1.1)^4$
$\displaystyle =20000\times1.4641=\text{Rs. }29282$
$\\$

$\displaystyle \textbf{Question 18: } \text{At the beginning of the year }2011,\text{ a man had Rs. }22000$
$\displaystyle \text{in his bank account. He saved some money by the end of }2011\text{ and deposited it.}$
$\displaystyle \text{The bank pays }10\%\text{ p.a. compound interest. At the end of }2012,$
$\displaystyle \text{he had Rs. }39820.\text{ Find the amount he saved at the end of }2011.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Amount of Rs. }22000\text{ at the end of }2011$
$\displaystyle =22000\left(1+\frac{10}{100}\right)=\text{Rs. }24200$
$\displaystyle \text{Let the amount saved and deposited at the end of }2011\text{ be Rs. }x$
$\displaystyle \therefore\ \text{Principal for the year }2012=\text{Rs. }(24200+x)$
$\displaystyle 39820=(24200+x)\left(1+\frac{10}{100}\right)$
$\displaystyle 39820=(24200+x)\times1.1$
$\displaystyle 24200+x=\frac{39820}{1.1}=36200$
$\displaystyle x=36200-24200=\text{Rs. }12000$
$\displaystyle \therefore\ \text{Amount saved and deposited at the end of }2011=\text{Rs. }12000$
$\\$

$\displaystyle \textbf{Question 19: } \text{If the amounts of two consecutive years on a sum of money}$
$\displaystyle \text{are in the ratio }20:21,\text{ find the rate of interest.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum be Rs. }x\text{ and the rate be }r\%\text{ per annum.}$
$\displaystyle \text{Amount after one year}=x\left(1+\frac{r}{100}\right)$
$\displaystyle \text{Amount after two years}=x\left(1+\frac{r}{100}\right)^2$
$\displaystyle \text{Given,}$
$\displaystyle x\left(1+\frac{r}{100}\right):x\left(1+\frac{r}{100}\right)^2=20:21$
$\displaystyle \frac{x\left(1+\frac{r}{100}\right)}{x\left(1+\frac{r}{100}\right)^2}=\frac{20}{21}$
$\displaystyle \frac{1}{1+\frac{r}{100}}=\frac{20}{21}$
$\displaystyle 1+\frac{r}{100}=\frac{21}{20}$
$\displaystyle \frac{r}{100}=\frac{1}{20}$
$\displaystyle r=5\%$
$\displaystyle \therefore\ \text{Required rate}=5\%\text{ p.a.}$
$\\$

$\displaystyle \textbf{Question 20: } \text{On what sum of money will the difference between the compound}$
$\displaystyle \text{interest and simple interest for }3\text{ years be Rs. }930,\text{ if the rate is }10\%\text{ p.a.?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the principal be Rs. }x$
$\displaystyle \text{Simple Interest}=x\times\frac{10}{100}\times3=0.3x$
$\displaystyle \text{Amount under compound interest}=x\left(1+\frac{10}{100}\right)^3$
$\displaystyle =x(1.1)^3=1.331x$
$\displaystyle \therefore\ \text{Compound Interest}=1.331x-x=0.331x$
$\displaystyle \text{Given, Compound Interest}-\text{Simple Interest}=930$
$\displaystyle 0.331x-0.3x=930$
$\displaystyle 0.031x=930$
$\displaystyle x=\frac{930}{0.031}=\text{Rs. }30000$
$\displaystyle \therefore\ \text{Required sum}=\text{Rs. }30000$
$\\$