Class 10: Compound Interest

$\displaystyle \textbf{Question 1: } \text{Calculate the amount and the compound interest on:}$
$\displaystyle \text{(i) Rs. }12000\text{ for }2\text{ years at }5\%\text{ per annum compounded annually.}$
$\displaystyle \text{(ii) Rs. }8000\text{ for }1\frac{1}{2}\text{ years at }10\%\text{ per annum compounded yearly.}$
$\displaystyle \text{(iii) Rs. }8000\text{ for }1\frac{1}{2}\text{ years at }10\%\text{ per annum compounded half-yearly.}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(i)}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }12000,\ R=5\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{12000\times5\times1}{100}=\text{Rs. }600$
$\displaystyle \text{Amount}=12000+600=\text{Rs. }12600$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }12600,\ R=5\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{12600\times5\times1}{100}=\text{Rs. }630$
$\displaystyle \text{Amount}=12600+630=\text{Rs. }13230$
$\displaystyle \therefore\ \text{Compound Interest}=600+630=\text{Rs. }1230$
$\displaystyle \therefore\ \text{Amount}=\text{Rs. }13230$
$\\$

$\displaystyle \textbf{(ii)}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }8000,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{8000\times10\times1}{100}=\text{Rs. }800$
$\displaystyle \text{Amount}=8000+800=\text{Rs. }8800$
$\displaystyle \text{For next }\frac{1}{2}\text{ year: }P=\text{Rs. }8800,\ R=10\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=\frac{8800\times10\times1}{100\times2}=\text{Rs. }440$
$\displaystyle \text{Amount}=8800+440=\text{Rs. }9240$
$\displaystyle \therefore\ \text{Compound Interest}=800+440=\text{Rs. }1240$
$\displaystyle \therefore\ \text{Amount}=\text{Rs. }9240$
$\\$

$\displaystyle \textbf{(iii)}$
$\displaystyle \text{For 1st half-year: }P=\text{Rs. }8000,\ R=10\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=\frac{8000\times10\times1}{100\times2}=\text{Rs. }400$
$\displaystyle \text{Amount}=8000+400=\text{Rs. }8400$
$\displaystyle \text{For 2nd half-year: }P=\text{Rs. }8400,\ R=10\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=\frac{8400\times10\times1}{100\times2}=\text{Rs. }420$
$\displaystyle \text{Amount}=8400+420=\text{Rs. }8820$
$\displaystyle \text{For 3rd half-year: }P=\text{Rs. }8820,\ R=10\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=\frac{8820\times10\times1}{100\times2}=\text{Rs. }441$
$\displaystyle \text{Amount}=8820+441=\text{Rs. }9261$
$\displaystyle \therefore\ \text{Compound Interest}=400+420+441=\text{Rs. }1261$
$\displaystyle \therefore\ \text{Amount}=\text{Rs. }9261$
$\\$

$\displaystyle \textbf{Question 2: } \text{Calculate the amount and the compound interest on Rs. }12500\text{ in } \\ 3\text{ years when the rates of interest for successive years are }8\%,\ 10\%\text{ and }10\%\text{ respectively:}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }12500,\ R=8\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{12500\times8\times1}{100}=\text{Rs. }1000$
$\displaystyle \text{Amount}=12500+1000=\text{Rs. }13500$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }13500,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{13500\times10\times1}{100}=\text{Rs. }1350$
$\displaystyle \text{Amount}=13500+1350=\text{Rs. }14850$
$\displaystyle \text{For 3rd year: }P=\text{Rs. }14850,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{14850\times10\times1}{100}=\text{Rs. }1485$
$\displaystyle \text{Amount}=14850+1485=\text{Rs. }16335$
$\displaystyle \therefore\ \text{Compound Interest}=1000+1350+1485=\text{Rs. }3835$
$\displaystyle \therefore\ \text{Amount}=\text{Rs. }16335$
$\\$

$\displaystyle \textbf{Question 3: } \text{A man lends Rs. }5500\text{ at the rate of }8\%\text{ per annum. Find the amount} \\ \text{if the interest is compounded half-yearly and the duration is one year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st half-year: }P=\text{Rs. }5500,\ R=8\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=\frac{5500\times8\times1}{100\times2}=\text{Rs. }220$
$\displaystyle \text{Amount}=5500+220=\text{Rs. }5720$
$\displaystyle \text{For 2nd half-year: }P=\text{Rs. }5720,\ R=8\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=\frac{5720\times8\times1}{100\times2}=\text{Rs. }228.80$
$\displaystyle \text{Amount}=5720+228.80=\text{Rs. }5948.80$
$\displaystyle \therefore\ \text{Compound Interest}=220+228.80=\text{Rs. }448.80$
$\displaystyle \therefore\ \text{Amount}=\text{Rs. }5948.80$
$\\$

$\displaystyle \textbf{Question 4: } \text{A man borrows Rs. }8500\text{ at }10\%\text{ compound interest. If he repays} \\ \text{Rs. }2700\text{ at the end of each year, find the amount of the loan outstanding at the beginning of the third year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }8500,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{8500\times10\times1}{100}=\text{Rs. }850$
$\displaystyle \text{Amount}=8500+850=\text{Rs. }9350$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }(9350-2700)=\text{Rs. }6650,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{6650\times10\times1}{100}=\text{Rs. }665$
$\displaystyle \text{Amount}=6650+665=\text{Rs. }7315$
$\displaystyle \therefore\ \text{Amount outstanding at the beginning of the 3rd year}=7315-2700=\text{Rs. }4615$
$\\$

$\displaystyle \textbf{Question 5: } \text{A man borrows Rs. }10000\text{ at }5\%\text{ per annum compound interest. He} \\ \text{repays }35\%\text{ of the sum borrowed at the end of the first year and }42\%\text{ of the sum} \\ \text{borrowed at the end of the second year. How much must he pay at the end of} \\ \text{the third year in order to clear the debt?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }10000,\ R=5\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{10000\times5\times1}{100}=\text{Rs. }500$
$\displaystyle \text{Amount}=10000+500=\text{Rs. }10500$
$\displaystyle \text{Amount repaid at the end of the 1st year}=35\%\text{ of Rs. }10000=\text{Rs. }3500$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }(10500-3500)=\text{Rs. }7000,\ R=5\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{7000\times5\times1}{100}=\text{Rs. }350$
$\displaystyle \text{Amount}=7000+350=\text{Rs. }7350$
$\displaystyle \text{Amount repaid at the end of the 2nd year}=42\%\text{ of Rs. }10000=\text{Rs. }4200$
$\displaystyle \text{Amount outstanding at the beginning of the 3rd year}=7350-4200=\text{Rs. }3150$
$\displaystyle \text{For 3rd year: }P=\text{Rs. }3150,\ R=5\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{3150\times5\times1}{100}=\text{Rs. }157.50$
$\displaystyle \text{Amount}=3150+157.50=\text{Rs. }3307.50$
$\displaystyle \therefore\ \text{Amount to be paid at the end of the 3rd year}=\text{Rs. }3307.50$
$\\$

$\displaystyle \textbf{Question 6: } \text{Rachana borrows Rs. }12000\text{ at }10\%\text{ per annum interest compounded} \\ \text{half-yearly. She repays Rs. }4000\text{ at the end of every six months. Calculate the third} \\ \text{payment she has to make at the end of }18\text{ months in order to clear the} \\ \text{entire loan.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st half-year: }P=\text{Rs. }12000,\ R=10\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=\frac{12000\times10\times1}{100\times2}=\text{Rs. }600$
$\displaystyle \text{Amount}=12000+600=\text{Rs. }12600$
$\displaystyle \text{Amount repaid at the end of the 1st half-year}=\text{Rs. }4000$
$\displaystyle \text{For 2nd half-year: }P=\text{Rs. }(12600-4000)=\text{Rs. }8600,\ R=10\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=\frac{8600\times10\times1}{100\times2}=\text{Rs. }430$
$\displaystyle \text{Amount}=8600+430=\text{Rs. }9030$
$\displaystyle \text{Amount repaid at the end of the 2nd half-year}=\text{Rs. }4000$
$\displaystyle \text{For 3rd half-year: }P=\text{Rs. }(9030-4000)=\text{Rs. }5030,\ R=10\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=\frac{5030\times10\times1}{100\times2}=\text{Rs. }251.50$
$\displaystyle \text{Amount}=5030+251.50=\text{Rs. }5281.50$
$\displaystyle \therefore\ \text{Third payment she has to make}=\text{Rs. }5281.50$
$\\$

$\displaystyle \textbf{Question 7: } \text{On a certain sum of money, invested at the rate of }10\%\text{ per annum} \\ \text{compounded annually, the interest for the first year plus the interest for the third year is} \\ \text{Rs. }2652\text{. Find the sum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum be Rs. }x$
$\displaystyle \text{For 1st year: }P=\text{Rs. }x,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{x\times10\times1}{100}=\text{Rs. }0.1x$
$\displaystyle \text{Amount}=x+0.1x=\text{Rs. }1.1x$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }1.1x,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{1.1x\times10\times1}{100}=\text{Rs. }0.11x$
$\displaystyle \text{Amount}=1.1x+0.11x=\text{Rs. }1.21x$
$\displaystyle \text{For 3rd year: }P=\text{Rs. }1.21x,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{1.21x\times10\times1}{100}=\text{Rs. }0.121x$
$\displaystyle \text{Given, interest for the 1st year}+\text{interest for the 3rd year}=\text{Rs. }2652$
$\displaystyle 0.1x+0.121x=2652$
$\displaystyle 0.221x=2652$
$\displaystyle x=\frac{2652}{0.221}=12000$
$\displaystyle \therefore\ \text{Required sum}=\text{Rs. }12000$
$\\$

$\displaystyle \textbf{Question 8: } \text{A sum of money is lent at }8\%\text{ per annum compound interest. If the} \\ \text{interest for the second year exceeds that for the first year by Rs. }96\text{, find the sum} \\ \text{of money.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum of money be Rs. }x$
$\displaystyle \text{For 1st year: }P=\text{Rs. }x,\ R=8\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{x\times8\times1}{100}=\text{Rs. }0.08x$
$\displaystyle \text{Amount}=x+0.08x=\text{Rs. }1.08x$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }1.08x,\ R=8\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{1.08x\times8\times1}{100}=\text{Rs. }0.0864x$
$\displaystyle \text{Amount}=1.08x+0.0864x=\text{Rs. }1.1664x$
$\displaystyle \text{Given, interest for the 2nd year exceeds that for the 1st year by Rs. }96$
$\displaystyle 0.0864x-0.08x=96$
$\displaystyle 0.0064x=96$
$\displaystyle x=\frac{96}{0.0064}=15000$
$\displaystyle \therefore\ \text{Required sum of money}=\text{Rs. }15000$
$\\$

$\displaystyle \textbf{Question 9: } \text{A person invested Rs. }8000\text{ every year at the beginning of the year,} \\ \text{at }10\%\text{ per annum compound interest. Calculate his total savings at the} \\ \text{beginning of the third year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }8000,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{8000\times10\times1}{100}=\text{Rs. }800$
$\displaystyle \text{Amount}=8000+800=\text{Rs. }8800$
$\displaystyle \text{At the beginning of the 2nd year, he invests Rs. }8000\text{ again.}$
$\displaystyle \text{Principal for 2nd year}=8800+8000=\text{Rs. }16800$
$\displaystyle \text{Interest}=\frac{16800\times10\times1}{100}=\text{Rs. }1680$
$\displaystyle \text{Amount at the end of the 2nd year}=16800+1680=\text{Rs. }18480$
$\displaystyle \text{At the beginning of the 3rd year, he invests Rs. }8000\text{ again.}$
$\displaystyle \therefore\ \text{Total savings at the beginning of the 3rd year}=18480+8000=\text{Rs. }26480$
$\\$

$\displaystyle \textbf{Question 10: } \text{A person saves Rs. }8000\text{ every year and invests it at the end of the year}$
$\displaystyle \text{at }10\%\text{ per annum compound interest. Calculate her total savings at the end of the third year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }8000,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{8000\times10\times1}{100}=\text{Rs. }800$
$\displaystyle \text{Amount}=8000+800=\text{Rs. }8800$
$\displaystyle \text{At the end of the 2nd year, she invests another Rs. }8000$
$\displaystyle \text{Principal for 2nd year}=8800+8000=\text{Rs. }16800$
$\displaystyle \text{Interest}=\frac{16800\times10\times1}{100}=\text{Rs. }1680$
$\displaystyle \text{Amount}=16800+1680=\text{Rs. }18480$
$\displaystyle \text{At the end of the 3rd year, she invests another Rs. }8000$
$\displaystyle \text{Principal for 3rd year}=18480+8000=\text{Rs. }26480$
$\displaystyle \text{Interest}=\frac{26480\times10\times1}{100}=\text{Rs. }2648$
$\displaystyle \therefore\ \text{Total savings at the end of the 3rd year}=26480+2648=\text{Rs. }29128$
$\\$

$\displaystyle \textbf{Question 11: } \text{During every financial year, the value of the machine depreciates by }12\%.$
$\displaystyle \text{Find the original cost of the machine which depreciates by Rs. }2640\text{ during the}$
$\displaystyle \text{second financial year of its purchase.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original cost of the machine be Rs. }x$
$\displaystyle \text{For 1st year: }P=\text{Rs. }x,\ R=12\%,\ T=1\text{ year}$
$\displaystyle \text{Depreciation}=\frac{x\times12\times1}{100}=\text{Rs. }0.12x$
$\displaystyle \text{Value after 1st year}=x-0.12x=\text{Rs. }0.88x$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }0.88x,\ R=12\%,\ T=1\text{ year}$
$\displaystyle \text{Depreciation}=\frac{0.88x\times12\times1}{100}=\text{Rs. }0.1056x$
$\displaystyle \text{Given, depreciation during the 2nd year}=\text{Rs. }2640$
$\displaystyle 0.1056x=2640$
$\displaystyle x=\frac{2640}{0.1056}=25000$
$\displaystyle \therefore\ \text{Original cost of the machine}=\text{Rs. }25000$
$\\$

$\displaystyle \textbf{Question 12: } \text{Find the sum on which the difference between the simple interest and}$
$\displaystyle \text{the compound interest at }8\%\text{ per annum compounded annually is Rs. }64\text{ in }2\text{ years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum be Rs. }x$
$\displaystyle \text{Simple Interest}=\frac{x\times8\times2}{100}=\text{Rs. }0.16x$
$\displaystyle \text{For 1st year: }P=\text{Rs. }x,\ R=8\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{x\times8\times1}{100}=\text{Rs. }0.08x$
$\displaystyle \text{Amount}=x+0.08x=\text{Rs. }1.08x$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }1.08x,\ R=8\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{1.08x\times8\times1}{100}=\text{Rs. }0.0864x$
$\displaystyle \text{Total Compound Interest}=0.08x+0.0864x=\text{Rs. }0.1664x$
$\displaystyle \text{Given, }0.1664x-0.16x=64$
$\displaystyle 0.0064x=64$
$\displaystyle x=\frac{64}{0.0064}=10000$
$\displaystyle \therefore\ \text{Required sum}=\text{Rs. }10000$
$\\$

$\displaystyle \textbf{Question 13: } \text{A person borrows Rs. }18000\text{ at }10\%\text{ simple interest. He immediately}$
$\displaystyle \text{invests the money at }10\%\text{ compound interest compounded half-yearly.}$
$\displaystyle \text{How much money does he gain in one year?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Sum borrowed}=\text{Rs. }18000$
$\displaystyle \text{Simple Interest}=\frac{18000\times10\times1}{100}=\text{Rs. }1800$
$\displaystyle \text{For 1st half-year: }P=\text{Rs. }18000,\ R=10\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=\frac{18000\times10\times1}{100\times2}=\text{Rs. }900$
$\displaystyle \text{Amount}=18000+900=\text{Rs. }18900$
$\displaystyle \text{For 2nd half-year: }P=\text{Rs. }18900,\ R=10\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=\frac{18900\times10\times1}{100\times2}=\text{Rs. }945$
$\displaystyle \text{Total Compound Interest}=900+945=\text{Rs. }1845$
$\displaystyle \therefore\ \text{Gain}=1845-1800=\text{Rs. }45$
$\\$

$\displaystyle \textbf{Question 14: } \text{A sum of Rs. }13500\text{ is invested at }16\%\text{ per annum compound interest}$
$\displaystyle \text{for }5\text{ years. Calculate (i) interest for the first year, (ii) amount at the end of the}$
$\displaystyle \text{first year, (iii) interest for the second year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }13500,\ R=16\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{13500\times16\times1}{100}=\text{Rs. }2160$
$\displaystyle \text{Amount}=13500+2160=\text{Rs. }15660$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }15660,\ R=16\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{15660\times16\times1}{100}=\text{Rs. }2505.60$
$\displaystyle \text{Amount}=15660+2505.60=\text{Rs. }18165.60$
$\displaystyle \therefore\ \text{(i) Interest for the 1st year}=\text{Rs. }2160$
$\displaystyle \therefore\ \text{(ii) Amount at the end of the 1st year}=\text{Rs. }15660$
$\displaystyle \therefore\ \text{(iii) Interest for the 2nd year}=\text{Rs. }2505.60$
$\\$

$\displaystyle \textbf{Question 15: } \text{A person invests Rs. }48000\text{ for }7\text{ years at }10\%\text{ per annum}$
$\displaystyle \text{compound interest. Calculate (i) interest for the first year, (ii) amount at the end of}$
$\displaystyle \text{the second year, (iii) interest for the third year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }48000,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{48000\times10\times1}{100}=\text{Rs. }4800$
$\displaystyle \text{Amount}=48000+4800=\text{Rs. }52800$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }52800,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{52800\times10\times1}{100}=\text{Rs. }5280$
$\displaystyle \text{Amount}=52800+5280=\text{Rs. }58080$
$\displaystyle \text{For 3rd year: }P=\text{Rs. }58080,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{58080\times10\times1}{100}=\text{Rs. }5808$
$\displaystyle \text{Amount}=58080+5808=\text{Rs. }63888$
$\displaystyle \therefore\ \text{(i) Interest for the 1st year}=\text{Rs. }4800$
$\displaystyle \therefore\ \text{(ii) Amount at the end of the 2nd year}=\text{Rs. }58080$
$\displaystyle \therefore\ \text{(iii) Interest for the 3rd year}=\text{Rs. }5808$
$\\$

$\displaystyle \textbf{Question 16: } \text{A person borrowed Rs. }7500\text{ at }8\%\text{ per annum compound interest.}$
$\displaystyle \text{After }2\text{ years he paid Rs. }6248\text{ and a TV set to clear the debt.}$
$\displaystyle \text{Find the value of the TV set.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }7500,\ R=8\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{7500\times8\times1}{100}=\text{Rs. }600$
$\displaystyle \text{Amount}=7500+600=\text{Rs. }8100$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }8100,\ R=8\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{8100\times8\times1}{100}=\text{Rs. }648$
$\displaystyle \text{Amount}=8100+648=\text{Rs. }8748$
$\displaystyle \text{Amount paid in cash}=\text{Rs. }6248$
$\displaystyle \therefore\ \text{Value of the TV set}=8748-6248=\text{Rs. }2500$
$\\$

$\displaystyle \textbf{Question 17: } \text{It is estimated that every year, the value of an asset depreciates by }20\%$
$\displaystyle \text{of its value at the beginning of the year. Calculate the original value of the asset if}$
$\displaystyle \text{its value after two years is Rs. }10240.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original value of the asset be Rs. }x$
$\displaystyle \text{For 1st year: }P=\text{Rs. }x,\ R=20\%,\ T=1\text{ year}$
$\displaystyle \text{Depreciation}=\frac{x\times20\times1}{100}=\text{Rs. }0.2x$
$\displaystyle \text{Value after 1st year}=x-0.2x=\text{Rs. }0.8x$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }0.8x,\ R=20\%,\ T=1\text{ year}$
$\displaystyle \text{Depreciation}=\frac{0.8x\times20\times1}{100}=\text{Rs. }0.16x$
$\displaystyle \text{Value after 2nd year}=0.8x-0.16x=\text{Rs. }0.64x$
$\displaystyle \text{Given, value after 2 years}=\text{Rs. }10240$
$\displaystyle 0.64x=10240$
$\displaystyle x=\frac{10240}{0.64}=16000$
$\displaystyle \therefore\ \text{Original value of the asset}=\text{Rs. }16000$
$\\$

$\displaystyle \textbf{Question 18: } \text{Find the sum that will amount to Rs. }4928\text{ in }2\text{ years at compound}$
$\displaystyle \text{interest, if the rates for the successive years are }10\%\text{ and }12\%\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum be Rs. }x$
$\displaystyle \text{For 1st year: }P=\text{Rs. }x,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{x\times10\times1}{100}=\text{Rs. }0.1x$
$\displaystyle \text{Amount}=x+0.1x=\text{Rs. }1.1x$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }1.1x,\ R=12\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{1.1x\times12\times1}{100}=\text{Rs. }0.132x$
$\displaystyle \text{Amount}=1.1x+0.132x=\text{Rs. }1.232x$
$\displaystyle \text{Given, }1.232x=4928$
$\displaystyle x=\frac{4928}{1.232}=4000$
$\displaystyle \therefore\ \text{Required sum}=\text{Rs. }4000$
$\\$

$\displaystyle \textbf{Question 19: } \text{A person opens a bank account on 1st Jan 2010 with Rs. }24000.$
$\displaystyle \text{If the bank pays }10\%\text{ per annum and he deposits Rs. }4000\text{ at the end of each year,}$
$\displaystyle \text{find the sum in the account on 1st Jan 2012.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the year 2010: }P=\text{Rs. }24000,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{24000\times10\times1}{100}=\text{Rs. }2400$
$\displaystyle \text{Amount at the end of 2010}=24000+2400=\text{Rs. }26400$
$\displaystyle \text{After depositing Rs. }4000,\ \text{balance on 1st Jan 2011}=\text{Rs. }30400$
$\displaystyle \text{For the year 2011: }P=\text{Rs. }30400,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{30400\times10\times1}{100}=\text{Rs. }3040$
$\displaystyle \text{Amount at the end of 2011}=30400+3040=\text{Rs. }33440$
$\displaystyle \text{After depositing Rs. }4000,\ \text{balance on 1st Jan 2012}=33440+4000=\text{Rs. }37440$
$\displaystyle \therefore\ \text{Required sum in the account on 1st Jan 2012}=\text{Rs. }37440$
$\\$

$\displaystyle \textbf{Question 20: } \text{A person borrows Rs. }12000\text{ at some rate per cent compound interest.}$
$\displaystyle \text{After one year, the person repays Rs. }4000\text{. If the compound interest for the}$
$\displaystyle \text{second year is Rs. }920,\text{ find (i) the rate of interest, (ii) the debt at the end of the second year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the rate of interest be }x\%\text{ per annum.}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }12000,\ R=x\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{12000\times x\times1}{100}=\text{Rs. }120x$
$\displaystyle \text{Amount at the end of the 1st year}=\text{Rs. }(12000+120x)$
$\displaystyle \text{After repaying Rs. }4000,\ \text{principal for the 2nd year}=\text{Rs. }(8000+120x)$
$\displaystyle \text{Interest for the 2nd year}=\frac{(8000+120x)\times x}{100}$
$\displaystyle \text{Given, }\frac{(8000+120x)x}{100}=920$
$\displaystyle 8000x+120x^2=92000$
$\displaystyle 3x^2+200x-2300=0$
$\displaystyle (x-10)(3x+230)=0$
$\displaystyle x=10\%\quad(\because\ x>0)$
$\displaystyle \text{Debt at the beginning of the 2nd year}=\text{Rs. }(8000+120\times10)=\text{Rs. }9200$
$\displaystyle \text{Debt at the end of the 2nd year}=9200+920=\text{Rs. }10120$
$\displaystyle \therefore\ \text{(i) Rate of interest}=10\%\text{ per annum}$
$\displaystyle \therefore\ \text{(ii) Amount of debt at the end of the 2nd year}=\text{Rs. }10120$
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