Class 10: Compound Interest

$\displaystyle \textbf{Question 1: }\text{Find the effective rate per cent per annum equivalent to a}$
$\displaystyle \text{nominal rate of }10\%\text{ per annum, interest payable half-yearly.}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }100,\ r=10\%\text{ p.a.}$
$\displaystyle \text{Rate per half-year}=5\%,\ \text{Number of half-years}=2$
$\displaystyle A=P\left(1+\frac{10}{2\times100}\right)^2$
$\displaystyle =100(1.05)^2=\text{Rs. }110.25$
$\displaystyle \text{Effective rate}=\frac{110.25-100}{100}\times100\%=10.25\%$
$\displaystyle \therefore\ \text{Effective rate}=10.25\%\text{ p.a.}$
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$\displaystyle \textbf{Question 2: }\text{A property decreases in value every year at the rate of }6\%.$
$\displaystyle \text{If its value at the end of }2\text{ years is Rs. }225000,\text{ find its value at the}$
$\displaystyle \text{beginning of these }2\text{ years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the value at the beginning of the first year be Rs. }x$
$\displaystyle \text{After the first year, value}=x\left(1-\frac{6}{100}\right)=0.94x$
$\displaystyle \text{After the second year, value}=0.94x\left(1-\frac{6}{100}\right)$
$\displaystyle =0.94x\times0.94=0.8836x$
$\displaystyle \text{Given, }0.8836x=225000$
$\displaystyle x=\frac{225000}{0.8836}=\text{Rs. }254640.11$
$\displaystyle \therefore\ \text{Value at the beginning of the }2\text{ years}=\text{Rs. }254640.11$
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$\displaystyle \textbf{Question 3: }\text{A man wishes to accumulate Rs. }50440\text{ at the beginning of }3\text{ years}$
$\displaystyle \text{from now. If his savings earn }5\%\text{ p.a. compound interest, what equal sum}$
$\displaystyle \text{must be invested at the end of each year?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let Rs. }x\text{ be invested at the end of each year.}$
$\displaystyle \text{The first investment earns interest for }2\text{ years.}$
$\displaystyle \therefore\ \text{Value of first investment}=x(1.05)^2=1.1025x$
$\displaystyle \text{The second investment earns interest for }1\text{ year.}$
$\displaystyle \therefore\ \text{Value of second investment}=x(1.05)=1.05x$
$\displaystyle \text{The third investment earns no interest.}$
$\displaystyle \therefore\ \text{Value of third investment}=x$
$\displaystyle 1.1025x+1.05x+x=50440$
$\displaystyle 3.1525x=50440$
$\displaystyle x=\frac{50440}{3.1525}=\text{Rs. }16000$
$\displaystyle \therefore\ \text{Equal sum to be invested each year}=\text{Rs. }16000$
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$\displaystyle \textbf{Question 4: }\text{Simple interest on a sum for }2\text{ years at }4\%\text{ is Rs. }450.$
$\displaystyle \text{Find the compound interest on the same sum at the same rate for }1\text{ year,}$
$\displaystyle \text{if the interest is reckoned half-yearly.}\hfill\text{[ICSE 1997]}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{Finding the principal}$
$\displaystyle P=\text{Rs. }x,\ T=2\text{ years},\ r=4\%$
$\displaystyle \text{Simple Interest}=x\times\frac{4}{100}\times2=\frac{2}{25}x$
$\displaystyle \frac{2}{25}x=450$
$\displaystyle x=\frac{450\times25}{2}=\text{Rs. }5625$
$\displaystyle \textbf{Compound Interest for }1\text{ year, compounded half-yearly}$
$\displaystyle \text{Rate per half-year}=2\%,\ \text{Number of half-years}=2$
$\displaystyle A=5625\left(1+\frac{4}{2\times100}\right)^2$
$\displaystyle =5625(1.02)^2=\text{Rs. }5852.25$
$\displaystyle \text{Compound Interest}=5852.25-5625=\text{Rs. }227.25$
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$\displaystyle \textbf{Question 5: }\text{Find the compound interest to the nearest rupee on Rs. }10800$
$\displaystyle \text{for }2\frac{1}{2}\text{ years at }10\%\text{ per annum.}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }10800,\ r=10\%\text{ p.a.},\ n=2\text{ years}$
$\displaystyle \text{Amount after }2\text{ years}=10800\left(1+\frac{10}{100}\right)^2$
$\displaystyle =10800(1.1)^2=\text{Rs. }13068$
$\displaystyle \text{Interest for the remaining }\frac{1}{2}\text{ year}$
$\displaystyle =13068\times\frac{10}{100}\times\frac{1}{2}=\text{Rs. }653.40$
$\displaystyle \therefore\ \text{Amount after }2\frac{1}{2}\text{ years}=13068+653.40=\text{Rs. }13721.40$
$\displaystyle \text{Compound Interest}=13721.40-10800=\text{Rs. }2921.40$
$\displaystyle \therefore\ \text{Compound Interest}\approx\text{Rs. }2921$
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$\displaystyle \textbf{Question 6: }\text{A bought a plot of land for Rs. }70000\text{ and a car for Rs. }32000.$
$\displaystyle \text{The plot appreciates at }10\%\text{ p.a. while the car depreciates by }20\%\text{ in the}$
$\displaystyle \text{first year and }10\%\text{ in the second year. Find the profit or loss after }2\text{ years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{Value of the plot after }2\text{ years}$
$\displaystyle =70000\left(1+\frac{10}{100}\right)^2$
$\displaystyle =70000(1.1)^2=\text{Rs. }84700$
$\displaystyle \textbf{Value of the car after the first year}$
$\displaystyle =32000\left(1-\frac{20}{100}\right)=\text{Rs. }25600$
$\displaystyle \textbf{Value of the car after the second year}$
$\displaystyle =25600\left(1-\frac{10}{100}\right)=\text{Rs. }23040$
$\displaystyle \text{Total investment}=70000+32000=\text{Rs. }102000$
$\displaystyle \text{Total selling price}=84700+23040=\text{Rs. }107740$
$\displaystyle \text{Profit}=107740-102000=\text{Rs. }5740$
$\displaystyle \therefore\ \text{A makes a profit of Rs. }5740$
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$\displaystyle \textbf{Question 7: }\text{The value of a machine, purchased }2\text{ years ago, depreciates}$
$\displaystyle \text{at }10\%\text{ per annum. If its present value is Rs. }97200,\text{ find:}$
$\displaystyle \text{(i) its value after }2\text{ years \qquad (ii) its value when it was purchased.}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(i) Value after }2\text{ years}$
$\displaystyle =97200\left(1-\frac{10}{100}\right)^2$
$\displaystyle =97200(0.9)^2$
$\displaystyle =\text{Rs. }78732$
$\displaystyle \textbf{(ii) Value when purchased}$
$\displaystyle \text{Let the value when purchased be Rs. }x$
$\displaystyle \text{Present value after }2\text{ years}=x\left(1-\frac{10}{100}\right)^2$
$\displaystyle 97200=x(0.9)^2$
$\displaystyle 97200=0.81x$
$\displaystyle x=\frac{97200}{0.81}=\text{Rs. }120000$
$\displaystyle \therefore\ \text{Value after }2\text{ years}=\text{Rs. }78732$
$\displaystyle \therefore\ \text{Value when purchased}=\text{Rs. }120000$
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$\displaystyle \textbf{Question 8: }\text{A man borrowed a sum and agrees to pay Rs. }9450\text{ at the end}$
$\displaystyle \text{of the first year and Rs. }13230\text{ at the end of the second year.}$
$\displaystyle \text{If the rate of compound interest is }5\%\text{ p.a., find the sum borrowed.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum borrowed be Rs. }x$
$\displaystyle \text{Amount due at the end of the first year}=x\left(1+\frac{5}{100}\right)=1.05x$
$\displaystyle \text{After paying Rs. }9450,\text{ balance}=1.05x-9450$
$\displaystyle \text{Amount due at the end of the second year}=(1.05x-9450)(1.05)$
$\displaystyle \text{Given, }(1.05x-9450)(1.05)=13230$
$\displaystyle 1.05x-9450=\frac{13230}{1.05}=12600$
$\displaystyle 1.05x=22050$
$\displaystyle x=\frac{22050}{1.05}=\text{Rs. }21000$
$\displaystyle \therefore\ \text{Sum borrowed}=\text{Rs. }21000$
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$\displaystyle \textbf{Question 9: }\text{A and B each lent the same sum for }2\text{ years at }8\%$
$\displaystyle \text{simple interest and compound interest respectively. B received Rs. }64\text{ more}$
$\displaystyle \text{than A. Find the money lent by each and the interest received.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let each lend Rs. }x$
$\displaystyle \text{Simple Interest received by A}=x\times\frac{8}{100}\times2=0.16x$
$\displaystyle \text{Amount under compound interest for B}=x\left(1+\frac{8}{100}\right)^2$
$\displaystyle =x(1.08)^2=1.1664x$
$\displaystyle \text{Compound Interest received by B}=1.1664x-x=0.1664x$
$\displaystyle \text{Given, }0.1664x-0.16x=64$
$\displaystyle 0.0064x=64$
$\displaystyle x=\frac{64}{0.0064}=\text{Rs. }10000$
$\displaystyle \therefore\ \text{Money lent by each}=\text{Rs. }10000$
$\displaystyle \text{Interest received by A}=0.16\times10000=\text{Rs. }1600$
$\displaystyle \text{Interest received by B}=0.1664\times10000=\text{Rs. }1664$
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$\displaystyle \textbf{Question 10: }\text{Calculate the sum on which the compound interest payable annually}$
$\displaystyle \text{for }2\text{ years is four times the simple interest on Rs. }4715\text{ for }5\text{ years,}$
$\displaystyle \text{both at the rate of }5\%\text{ per annum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Simple Interest on Rs. }4715\text{ for }5\text{ years at }5\%\text{ p.a.}$
$\displaystyle =4715\times\frac{5}{100}\times5=\text{Rs. }1178.75$
$\displaystyle \therefore\ \text{Required Compound Interest}=4\times1178.75=\text{Rs. }4715$
$\displaystyle \text{Let the required sum be Rs. }x$
$\displaystyle \text{Compound Interest for }2\text{ years at }5\%\text{ p.a.}$
$\displaystyle =x\left(1+\frac{5}{100}\right)^2-x$
$\displaystyle =x(1.1025)-x=0.1025x$
$\displaystyle 0.1025x=4715$
$\displaystyle x=\frac{4715}{0.1025}=\text{Rs. }46000$
$\displaystyle \therefore\ \text{Required sum}=\text{Rs. }46000$
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$\displaystyle \textbf{Question 11: }\text{A sum of money was invested for }3\text{ years. Interest was}$
$\displaystyle \text{compounded annually at }10\%,\ 15\%\text{ and }18\%\text{ for successive years.}$
$\displaystyle \text{If the compound interest for the second year was Rs. }4950,\text{ find the sum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum invested be Rs. }x$
$\displaystyle \text{Amount after the first year}=x\left(1+\frac{10}{100}\right)=1.1x$
$\displaystyle \text{Amount after the second year}=1.1x\left(1+\frac{15}{100}\right)=1.265x$
$\displaystyle \text{Compound interest for the second year}=1.265x-1.1x$
$\displaystyle \therefore\ 1.265x-1.1x=4950$
$\displaystyle 0.165x=4950$
$\displaystyle x=\frac{4950}{0.165}=\text{Rs. }30000$
$\displaystyle \therefore\ \text{Sum invested}=\text{Rs. }30000$
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$\displaystyle \textbf{Question 12: }\text{A sum of money is invested at }10\%\text{ per annum, compounded}$
$\displaystyle \text{half-yearly. If the difference of amounts at the end of }6\text{ months and }12\text{ months}$
$\displaystyle \text{is Rs. }189,\text{ find the sum of money invested.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum invested be Rs. }x$
$\displaystyle \text{Rate per half-year}=\frac{10}{2}\%=5\%$
$\displaystyle \text{Amount after }6\text{ months}=x\left(1+\frac{5}{100}\right)=1.05x$
$\displaystyle \text{Amount after }12\text{ months}=x\left(1+\frac{5}{100}\right)^2=1.1025x$
$\displaystyle \text{Given, }1.1025x-1.05x=189$
$\displaystyle 0.0525x=189$
$\displaystyle x=\frac{189}{0.0525}=\text{Rs. }3600$
$\displaystyle \therefore\ \text{Sum of money invested}=\text{Rs. }3600$
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$\displaystyle \textbf{Question 13: }\text{Rohit borrows Rs. }86000\text{ from Arun for }2\text{ years at }5\%$
$\displaystyle \text{simple interest. He lends it to Akshay at }5\%\text{ compound interest annually.}$
$\displaystyle \text{Calculate Rohit's profit at the end of }2\text{ years.}\hfill\text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{Simple Interest paid to Arun}$
$\displaystyle S.I.=86000\times\frac{5}{100}\times2=\text{Rs. }8600$
$\displaystyle \textbf{Compound Interest received from Akshay}$
$\displaystyle A=86000\left(1+\frac{5}{100}\right)^2$
$\displaystyle =86000(1.05)^2=\text{Rs. }94815$
$\displaystyle C.I.=94815-86000=\text{Rs. }8815$
$\displaystyle \text{Profit}=8815-8600=\text{Rs. }215$
$\displaystyle \therefore\ \text{Rohit's profit}=\text{Rs. }215$
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$\displaystyle \textbf{Question 14: }\text{A person borrowed Rs. }60000\text{ at }5\%\text{ simple interest for }2\text{ years.}$
$\displaystyle \text{The interest was deducted in advance and the balance was deposited in a bank}$
$\displaystyle \text{paying }5\%\text{ compound interest per annum. How much must he add to repay the loan?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Simple Interest deducted}=60000\times\frac{5}{100}\times2=\text{Rs. }6000$
$\displaystyle \text{Amount received}=60000-6000=\text{Rs. }54000$
$\displaystyle \text{Amount in bank after }2\text{ years}=54000\left(1+\frac{5}{100}\right)^2$
$\displaystyle =54000(1.05)^2=\text{Rs. }59535$
$\displaystyle \text{Amount to be repaid to the moneylender}=\text{Rs. }60000$
$\displaystyle \text{Amount to be added}=60000-59535=\text{Rs. }465$
$\displaystyle \therefore\ \text{He must add Rs. }465$
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$\displaystyle \textbf{Question 15: }\text{The simple interest on a certain sum for }3\text{ years at }5\%$
$\displaystyle \text{per annum is Rs. }1200.\text{ Find the amount and the compound interest on the same}$
$\displaystyle \text{sum for }2\text{ years at the same rate, interest compounded annually.}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{Finding the principal}$
$\displaystyle \text{Let the principal be Rs. }x$
$\displaystyle 1200=x\times\frac{5}{100}\times3$
$\displaystyle 1200=\frac{3}{20}x$
$\displaystyle x=\frac{1200\times20}{3}=\text{Rs. }8000$
$\displaystyle \textbf{Compound amount after }2\text{ years}$
$\displaystyle A=8000\left(1+\frac{5}{100}\right)^2$
$\displaystyle =8000(1.05)^2=\text{Rs. }8820$
$\displaystyle \text{Compound Interest}=8820-8000=\text{Rs. }820$
$\displaystyle \therefore\ \text{Amount}=\text{Rs. }8820,\ \text{Compound Interest}=\text{Rs. }820$
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$\displaystyle \textbf{Question 16: }\text{A person invests Rs. }6000\text{ for }2\text{ years at compound}$
$\displaystyle \text{interest. At the end of the first year it amounts to Rs. }6720.\text{ Calculate:}$
$\displaystyle \text{(i) the rate of interest \qquad (ii) the amount at the end of the second year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(i) Finding the rate of interest}$
$\displaystyle 6720=6000\left(1+\frac{x}{100}\right)$
$\displaystyle 1+\frac{x}{100}=\frac{6720}{6000}=1.12$
$\displaystyle x=12\%$
$\displaystyle \textbf{(ii) Amount at the end of the second year}$
$\displaystyle A=6000\left(1+\frac{12}{100}\right)^2$
$\displaystyle =6000(1.12)^2$
$\displaystyle =\text{Rs. }7526.40$
$\displaystyle \therefore\ \text{Rate}=12\%\text{ p.a. and Amount}=\text{Rs. }7526.40$
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