Class 10: Compound Interest

$\displaystyle \textbf{Question 1: }\text{Find the money invested at }10\%\text{ compounded annually, on which the sum}$
$\displaystyle \text{of interest for the first year and the third year is Rs. }1768.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the money invested be Rs. }x$
$\displaystyle \text{For 1st year: }P=\text{Rs. }x,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{x\times10\times1}{100}=\text{Rs. }0.1x$
$\displaystyle \text{Amount}=x+0.1x=\text{Rs. }1.1x$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }1.1x,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{1.1x\times10\times1}{100}=\text{Rs. }0.11x$
$\displaystyle \text{Amount}=1.1x+0.11x=\text{Rs. }1.21x$
$\displaystyle \text{For 3rd year: }P=\text{Rs. }1.21x,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{1.21x\times10\times1}{100}=\text{Rs. }0.121x$
$\displaystyle \text{Given, }0.1x+0.121x=1768$
$\displaystyle 0.221x=1768$
$\displaystyle x=\frac{1768}{0.221}=8000$
$\displaystyle \therefore\ \text{Money invested}=\text{Rs. }8000$
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$\displaystyle \textbf{Question 2: }\text{A sum of money is invested at compound interest payable annually.}$
$\displaystyle \text{The interest in the first two successive years is Rs. }1350\text{ and Rs. }1440\text{ respectively.}$
$\displaystyle \text{Find (i) the rate of interest, (ii) the original sum, and (iii) the interest earned in the third year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Difference between the compound interest of two successive years}$
$\displaystyle =1440-1350=\text{Rs. }90$
$\displaystyle \therefore\ \text{Rs. }90\text{ is the interest on Rs. }1350\text{ for }1\text{ year}$
$\displaystyle \therefore\ \text{Rate of Interest}=\frac{100\times I}{P\times T}\%$
$\displaystyle =\frac{100\times90}{1350\times1}\%=6\frac{2}{3}\%$
$\displaystyle \text{For 1st year: }P=\text{Rs. }P,\ R=6\frac{2}{3}\%,\ T=1\text{ year}$
$\displaystyle 1350=P\times\frac{6\frac{2}{3}}{100}\times1$
$\displaystyle P=1350\times\frac{100}{6\frac{2}{3}}=\text{Rs. }20250$
$\displaystyle \text{Amount at the end of the 1st year}=20250+1350=\text{Rs. }21600$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }21600,\ R=6\frac{2}{3}\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{21600\times6\frac{2}{3}\times1}{100}=\text{Rs. }1440$
$\displaystyle \text{Amount}=21600+1440=\text{Rs. }23040$
$\displaystyle \text{For 3rd year: }P=\text{Rs. }23040,\ R=6\frac{2}{3}\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{23040\times6\frac{2}{3}\times1}{100}=\text{Rs. }1536$
$\displaystyle \therefore\ \text{(i) Rate of interest}=6\frac{2}{3}\%\text{ per annum}$
$\displaystyle \therefore\ \text{(ii) Original sum}=\text{Rs. }20250$
$\displaystyle \therefore\ \text{(iii) Interest earned in the 3rd year}=\text{Rs. }1536$
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$\displaystyle \textbf{Question 3: }\text{A sum of money amounts to Rs. }46305\text{ in }1\text{ year and to}$
$\displaystyle \text{Rs. }48620.25\text{ in }1\frac{1}{2}\text{ years at compound interest, compounded semi-annually.}$
$\displaystyle \text{Find the sum and the rate of interest per annum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Interest earned in the 3rd half-year}=48620.25-46305=\text{Rs. }2315.25$
$\displaystyle \text{For 3rd half-year: }P=\text{Rs. }46305,\ I=\text{Rs. }2315.25,\ T=\frac{1}{2}\text{ year}$
$\displaystyle 2315.25=\frac{46305\times R\times\frac{1}{2}}{100}$
$\displaystyle R=\frac{2315.25\times100\times2}{46305}=10\%$
$\displaystyle \text{Let the original sum be Rs. }P$
$\displaystyle \text{For 1st half-year: }P=\text{Rs. }P,\ R=10\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=P\times\frac{10}{100}\times\frac{1}{2}=0.05P$
$\displaystyle \text{Amount}=P+0.05P=1.05P$
$\displaystyle \text{For 2nd half-year: }P=\text{Rs. }1.05P,\ R=10\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=1.05P\times\frac{10}{100}\times\frac{1}{2}=0.0525P$
$\displaystyle \text{Amount}=1.05P+0.0525P=1.1025P$
$\displaystyle \text{Given, }1.1025P=46305$
$\displaystyle P=\frac{46305}{1.1025}=42000$
$\displaystyle \therefore\ \text{Rate of interest}=10\%\text{ per annum}$
$\displaystyle \therefore\ \text{Original sum}=\text{Rs. }42000$
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$\displaystyle \textbf{Question 4: }\text{The cost of a machine is Rs. }32000.\text{ Its value depreciates at}$
$\displaystyle \text{the rate of }5\%\text{ every year. Find the total depreciation in its value by the end of }2\text{ years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }32000,\ R=5\%,\ T=1\text{ year}$
$\displaystyle \text{Depreciation}=32000\times\frac{5}{100}\times1=\text{Rs. }1600$
$\displaystyle \text{Value after 1st year}=32000-1600=\text{Rs. }30400$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }30400,\ R=5\%,\ T=1\text{ year}$
$\displaystyle \text{Depreciation}=30400\times\frac{5}{100}\times1=\text{Rs. }1520$
$\displaystyle \text{Value after 2nd year}=30400-1520=\text{Rs. }28880$
$\displaystyle \therefore\ \text{Total depreciation}=1600+1520=\text{Rs. }3120$
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$\displaystyle \textbf{Question 5: }\text{Find the sum, invested at }10\%\text{ compounded annually, on which}$
$\displaystyle \text{the interest for the third year exceeds the interest of the first year by Rs. }252.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum be Rs. }x$
$\displaystyle \text{For 1st year: }P=\text{Rs. }x,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=x\times\frac{10}{100}\times1=\text{Rs. }0.1x$
$\displaystyle \text{Amount}=x+0.1x=\text{Rs. }1.1x$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }1.1x,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=1.1x\times\frac{10}{100}\times1=\text{Rs. }0.11x$
$\displaystyle \text{Amount}=1.1x+0.11x=\text{Rs. }1.21x$
$\displaystyle \text{For 3rd year: }P=\text{Rs. }1.21x,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=1.21x\times\frac{10}{100}\times1=\text{Rs. }0.121x$
$\displaystyle \text{Given, }0.121x-0.1x=252$
$\displaystyle 0.021x=252$
$\displaystyle x=\frac{252}{0.021}=12000$
$\displaystyle \therefore\ \text{Required sum}=\text{Rs. }12000$
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$\displaystyle \textbf{Question 6: }\text{The compound interest, compounded annually, on a certain sum is}$
$\displaystyle \text{Rs. }9680\text{ in the second year and Rs. }10648\text{ in the third year. Calculate:}$
$\displaystyle \text{(i) the rate of interest, (ii) the sum borrowed, and (iii) the interest of the 1st year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Difference between the compound interest of two successive years}$
$\displaystyle =10648-9680=\text{Rs. }968$
$\displaystyle \therefore\ \text{Rs. }968\text{ is the interest on Rs. }9680\text{ for }1\text{ year}$
$\displaystyle \therefore\ \text{Rate of Interest}=\frac{100\times I}{P\times T}\%$
$\displaystyle =\frac{100\times968}{9680\times1}\%=10\%$
$\displaystyle \text{Let the sum be Rs. }100$
$\displaystyle \text{Interest for the 1st year}=10\%\text{ of Rs. }100=\text{Rs. }10$
$\displaystyle \text{Amount at the end of the 1st year}=100+10=\text{Rs. }110$
$\displaystyle \text{Interest for the 2nd year}=10\%\text{ of Rs. }110=\text{Rs. }11$
$\displaystyle \text{When the interest for the 2nd year is Rs. }11,\ \text{the sum is Rs. }100$
$\displaystyle \text{When the interest for the 2nd year is Rs. }9680,\ \text{the sum}=\frac{100}{11}\times9680$
$\displaystyle =\text{Rs. }88000$
$\displaystyle \text{Interest for the 1st year}=88000\times\frac{10}{100}=\text{Rs. }8800$
$\displaystyle \therefore\ \text{Rate of interest}=10\%\text{ per annum}$
$\displaystyle \therefore\ \text{Sum borrowed}=\text{Rs. }88000$
$\displaystyle \therefore\ \text{Interest of the 1st year}=\text{Rs. }8800$
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$\displaystyle \textbf{Question 7: }\text{A man borrows Rs. }10000\text{ at }10\%\text{ compound interest,}$
$\displaystyle \text{compounded yearly. At the end of each year, he pays back }30\%\text{ of the}$
$\displaystyle \text{sum borrowed. How much money is left unpaid just after the second year?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }10000,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=10000\times\frac{10}{100}=\text{Rs. }1000$
$\displaystyle \text{Amount}=10000+1000=\text{Rs. }11000$
$\displaystyle \text{Repayment}=30\%\text{ of Rs. }10000=\text{Rs. }3000$
$\displaystyle \text{Balance}=11000-3000=\text{Rs. }8000$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }8000,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=8000\times\frac{10}{100}=\text{Rs. }800$
$\displaystyle \text{Amount}=8000+800=\text{Rs. }8800$
$\displaystyle \text{Repayment}=30\%\text{ of Rs. }10000=\text{Rs. }3000$
$\displaystyle \text{Amount left unpaid}=8800-3000=\text{Rs. }5800$
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$\displaystyle \textbf{Question 8: }\text{A man borrows Rs. }10000\text{ at }10\%\text{ compound interest,}$
$\displaystyle \text{compounded yearly. At the end of each year, he pays back }20\%\text{ of the}$
$\displaystyle \text{amount due for that year. How much money is left unpaid just after the second year?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }10000,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=10000\times\frac{10}{100}=\text{Rs. }1000$
$\displaystyle \text{Amount}=10000+1000=\text{Rs. }11000$
$\displaystyle \text{Repayment}=20\%\text{ of Rs. }11000=\text{Rs. }2200$
$\displaystyle \text{Balance}=11000-2200=\text{Rs. }8800$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }8800,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=8800\times\frac{10}{100}=\text{Rs. }880$
$\displaystyle \text{Amount}=8800+880=\text{Rs. }9680$
$\displaystyle \text{Repayment}=20\%\text{ of Rs. }9680=\text{Rs. }1936$
$\displaystyle \text{Amount left unpaid}=9680-1936=\text{Rs. }7744$
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$\displaystyle \textbf{Question 9: }\text{The population of a town increases by }10\%\text{ every year.}$
$\displaystyle \text{If the present population is }72600,\text{ calculate (i) its population after }2\text{ years,}$
$\displaystyle \text{and (ii) its population }2\text{ years ago.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the population }2\text{ years ago be }x$
$\displaystyle \text{After 1 year, population}=x+\frac{10x}{100}=1.1x$
$\displaystyle \text{After 2 years, population}=1.1x+\frac{10}{100}(1.1x)=1.21x$
$\displaystyle \text{Given, }1.21x=72600$
$\displaystyle x=\frac{72600}{1.21}=60000$
$\displaystyle \therefore\ \text{Population }2\text{ years ago}=60000$
$\displaystyle \text{Population after 1 year}=72600+\frac{10}{100}\times72600=79860$
$\displaystyle \text{Population after 2 years}=79860+\frac{10}{100}\times79860=87846$
$\displaystyle \therefore\ \text{Population after }2\text{ years}=87846$
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$\displaystyle \textbf{Question 10: }\text{The compound interest, calculated yearly, on a certain sum of money}$
$\displaystyle \text{for the second year is Rs. }1320\text{ and for the third year is Rs. }1452.$
$\displaystyle \text{Calculate the rate of interest and the original sum of money. [ICSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Difference between the compound interest of two successive years}$
$\displaystyle =1452-1320=\text{Rs. }132$
$\displaystyle \therefore\ \text{Rs. }132\text{ is the interest on Rs. }1320\text{ for }1\text{ year}$
$\displaystyle \therefore\ \text{Rate of Interest}=\frac{100\times132}{1320\times1}\%=10\%$
$\displaystyle \text{Let the original sum be Rs. }100$
$\displaystyle \text{Interest for the 1st year}=10\%\text{ of Rs. }100=\text{Rs. }10$
$\displaystyle \text{Amount at the end of the 1st year}=100+10=\text{Rs. }110$
$\displaystyle \text{Interest for the 2nd year}=10\%\text{ of Rs. }110=\text{Rs. }11$
$\displaystyle \text{When the interest for the 2nd year is Rs. }11,\ \text{the sum is Rs. }100$
$\displaystyle \text{When the interest for the 2nd year is Rs. }1320,\ \text{the sum}=\frac{100}{11}\times1320$
$\displaystyle =\text{Rs. }12000$
$\displaystyle \therefore\ \text{Rate of interest}=10\%\text{ per annum}$
$\displaystyle \therefore\ \text{Original sum}=\text{Rs. }12000$
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