Class 10: Compound Interest

$\displaystyle \textbf{Question 1: }\text{A sum is invested at compound interest compounded yearly. If the}$
$\displaystyle \text{interest for two successive years is Rs. }5700\text{ and Rs. }7410,\text{ calculate the rate}$
$\displaystyle \text{of interest.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Difference between the compound interest of two successive years}$
$\displaystyle =7410-5700=\text{Rs. }1710$
$\displaystyle \therefore\ \text{Rs. }1710\text{ is the interest on Rs. }5700\text{ for }1\text{ year}$
$\displaystyle \therefore\ \text{Rate of Interest}=\frac{100\times I}{P\times T}\%$
$\displaystyle =\frac{100\times1710}{5700\times1}\%=30\%$
$\displaystyle \text{Alternatively,}$
$\displaystyle \text{Rate of Interest}=\frac{\text{Difference in interest of two consecutive periods}\times100}{\text{Compound interest of preceding year}\times\text{Time}}\%$
$\displaystyle =\frac{(7410-5700)\times100}{5700\times1}\%=30\%$
$\\$

$\displaystyle \textbf{Question 2: }\text{A certain sum of money is invested at compound interest compounded}$
$\displaystyle \text{half-yearly. If the interests for two successive half-years are Rs. }650\text{ and}$
$\displaystyle \text{Rs. }760.50,\text{ find the rate of interest.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Difference between the compound interest of two successive half-years}$
$\displaystyle =760.50-650=\text{Rs. }110.50$
$\displaystyle \therefore\ \text{Rs. }110.50\text{ is the interest on Rs. }650\text{ for }\frac{1}{2}\text{ year}$
$\displaystyle \therefore\ \text{Rate of Interest}=\frac{100\times I}{P\times\frac{1}{2}}\%$
$\displaystyle =\frac{100\times110.50}{650\times\frac{1}{2}}\%=34\%$
$\displaystyle \text{Alternatively,}$
$\displaystyle \text{Rate of Interest}=\frac{\text{Difference in interest of two consecutive periods}\times100}{\text{Compound interest of preceding period}\times\text{Time}}\%$
$\displaystyle =\frac{(760.50-650)\times100}{650\times\frac{1}{2}}\%=34\%$
$\\$

$\displaystyle \textbf{Question 3: }\text{A certain sum amounts to Rs. }5292\text{ in two years and Rs. }5556.60$
$\displaystyle \text{in three years, interest being compounded annually. Find (i) the rate of interest}$
$\displaystyle \text{and (ii) the original sum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Difference between the compound interest of two successive years}$
$\displaystyle =5556.60-5292=\text{Rs. }264.60$
$\displaystyle \therefore\ \text{Rs. }264.60\text{ is the interest on Rs. }5292\text{ for }1\text{ year}$
$\displaystyle \therefore\ \text{Rate of Interest}=\frac{100\times I}{P\times T}\%$
$\displaystyle =\frac{100\times264.60}{5292\times1}\%=5\%$
$\displaystyle \text{Alternatively,}$
$\displaystyle \text{Rate of Interest}=\frac{\text{Difference in interest of two consecutive periods}\times100}{\text{Compound interest of preceding year}\times\text{Time}}\%$
$\displaystyle =\frac{(5556.60-5292)\times100}{5292\times1}\%=5\%$
$\displaystyle \text{Let the original sum be Rs. }100$
$\displaystyle \text{Interest for the 1st year}=5\%\text{ of Rs. }100=\text{Rs. }5$
$\displaystyle \text{Amount at the end of the 1st year}=100+5=\text{Rs. }105$
$\displaystyle \text{Interest for the 2nd year}=5\%\text{ of Rs. }105=\text{Rs. }5.25$
$\displaystyle \text{Amount at the end of the 2nd year}=105+5.25=\text{Rs. }110.25$
$\displaystyle \text{When the amount after 2 years is Rs. }110.25,\ \text{the principal is Rs. }100$
$\displaystyle \text{When the amount after 2 years is Rs. }5292,\ \text{the principal}=\frac{100}{110.25}\times5292$
$\displaystyle =\text{Rs. }4800$
$\displaystyle \therefore\ \text{Rate of interest}=5\%\text{ and original sum}=\text{Rs. }4800$
$\\$

$\displaystyle \textbf{Question 4: }\text{The compound interest, calculated yearly, on a certain sum of money}$
$\displaystyle \text{for the second year is Rs. }1089\text{ and for the third year is Rs. }1197.90.$
$\displaystyle \text{Calculate the rate of interest and the sum of money.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Difference between the compound interest of two successive years}$
$\displaystyle =1197.90-1089=\text{Rs. }108.90$
$\displaystyle \therefore\ \text{Rs. }108.90\text{ is the interest on Rs. }1089\text{ for }1\text{ year}$
$\displaystyle \therefore\ \text{Rate of Interest}=\frac{100\times I}{P\times T}\%$
$\displaystyle =\frac{100\times108.90}{1089\times1}\%=10\%$
$\displaystyle \text{Alternatively,}$
$\displaystyle \text{Rate of Interest}=\frac{\text{Difference in interest of two consecutive periods}\times100}{\text{Compound interest of preceding year}\times\text{Time}}\%$
$\displaystyle =\frac{(1197.90-1089)\times100}{1089\times1}\%=10\%$
$\displaystyle \text{Let the original sum be Rs. }100$
$\displaystyle \text{Interest for the 1st year}=10\%\text{ of Rs. }100=\text{Rs. }10$
$\displaystyle \text{Amount at the end of the 1st year}=100+10=\text{Rs. }110$
$\displaystyle \text{Interest for the 2nd year}=10\%\text{ of Rs. }110=\text{Rs. }11$
$\displaystyle \text{When the interest for the 2nd year is Rs. }11,\ \text{the principal is Rs. }100$
$\displaystyle \text{When the interest for the 2nd year is Rs. }1089,\ \text{the principal}=\frac{100}{11}\times1089$
$\displaystyle =\text{Rs. }9900$
$\displaystyle \therefore\ \text{Rate of interest}=10\%\text{ and sum of money}=\text{Rs. }9900$
$\\$

$\displaystyle \textbf{Question 5: }\text{A person invests Rs. }8000\text{ for }3\text{ years at a certain rate of interest,}$
$\displaystyle \text{compounded annually. At the end of one year it amounts to Rs. }9440.\text{ Calculate:}$
$\displaystyle \text{(i) the rate of interest per annum, (ii) the amount at the end of the second year,}$
$\displaystyle \text{and (iii) the interest accrued in the third year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the rate of interest be }x\%\text{ per annum.}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }8000,\ R=x\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{8000\times x\times1}{100}=\text{Rs. }80x$
$\displaystyle \text{Amount}=8000+80x$
$\displaystyle \text{Given, }8000+80x=9440$
$\displaystyle 80x=1440$
$\displaystyle x=18$
$\displaystyle \therefore\ \text{Rate of interest}=18\%\text{ per annum}$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }9440,\ R=18\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{9440\times18\times1}{100}=\text{Rs. }1699.20$
$\displaystyle \text{Amount at the end of the 2nd year}=9440+1699.20=\text{Rs. }11139.20$
$\displaystyle \text{For 3rd year: }P=\text{Rs. }11139.20,\ R=18\%,\ T=1\text{ year}$
$\displaystyle \text{Interest accrued in the 3rd year}=\frac{11139.20\times18\times1}{100}=\text{Rs. }2005.06$
$\displaystyle \therefore\ \text{(i) Rate of interest}=18\%\text{ per annum}$
$\displaystyle \therefore\ \text{(ii) Amount at the end of the 2nd year}=\text{Rs. }11139.20$
$\displaystyle \therefore\ \text{(iii) Interest accrued in the 3rd year}=\text{Rs. }2005.06$
$\\$

$\displaystyle \textbf{Question 6: }\text{A person borrowed Rs. }15000\text{ for }18\text{ months at a certain rate of interest}$
$\displaystyle \text{compounded semi-annually. If at the end of six months it amounted to Rs. }15600,$
$\displaystyle \text{calculate (i) the rate of interest per annum and (ii) the total amount to be paid}$
$\displaystyle \text{at the end of }18\text{ months in order to clear the account.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the rate of interest be }x\%\text{ per annum.}$
$\displaystyle \text{For 1st half-year: }P=\text{Rs. }15000,\ R=x\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=15600-15000=\text{Rs. }600$
$\displaystyle 600=\frac{15000\times x\times\frac{1}{2}}{100}$
$\displaystyle x=8$
$\displaystyle \therefore\ \text{Rate of interest}=8\%\text{ per annum}$
$\displaystyle \text{For 2nd half-year: }P=\text{Rs. }15600,\ R=8\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=\frac{15600\times8\times\frac{1}{2}}{100}=\text{Rs. }624$
$\displaystyle \text{Amount}=15600+624=\text{Rs. }16224$
$\displaystyle \text{For 3rd half-year: }P=\text{Rs. }16224,\ R=8\%,\ T=\frac{1}{2}\text{ year}$
$\displaystyle \text{Interest}=\frac{16224\times8\times\frac{1}{2}}{100}=\text{Rs. }648.96$
$\displaystyle \text{Amount}=16224+648.96=\text{Rs. }16872.96$
$\displaystyle \therefore\ \text{Total amount to be paid at the end of }18\text{ months}=\text{Rs. }16872.96$
$\\$

$\displaystyle \textbf{Question 7: }\text{Ramesh invests Rs. }12800\text{ for three years at }10\%\text{ per annum}$
$\displaystyle \text{compound interest. Find (i) the sum due at the end of the first year,}$
$\displaystyle \text{(ii) the interest earned during the second year, and (iii) the amount due at the}$
$\displaystyle \text{end of the third year. [ICSE 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }12800,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{12800\times10\times1}{100}=\text{Rs. }1280$
$\displaystyle \text{Amount}=12800+1280=\text{Rs. }14080$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }14080,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{14080\times10\times1}{100}=\text{Rs. }1408$
$\displaystyle \text{Amount}=14080+1408=\text{Rs. }15488$
$\displaystyle \text{For 3rd year: }P=\text{Rs. }15488,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{15488\times10\times1}{100}=\text{Rs. }1548.80$
$\displaystyle \text{Amount}=15488+1548.80=\text{Rs. }17036.80$
$\displaystyle \therefore\ \text{(i) Sum due at the end of the 1st year}=\text{Rs. }14080$
$\displaystyle \therefore\ \text{(ii) Interest for the 2nd year}=\text{Rs. }1408$
$\displaystyle \therefore\ \text{(iii) Amount due at the end of the 3rd year}=\text{Rs. }17036.80$
$\\$

$\displaystyle \textbf{Question 8: }\text{The simple interest on a certain sum is Rs. }256\text{ in }2\text{ years,}$
$\displaystyle \text{whereas the compound interest on the same sum at the same rate and for the same}$
$\displaystyle \text{time is Rs. }276.48.\text{ Find the rate per cent and the sum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the principal be Rs. }P\text{ and the rate be }x\%\text{ per annum.}$
$\displaystyle \text{Simple Interest}=P\times\frac{x}{100}\times2=256$
$\displaystyle \therefore\ \frac{Px}{100}=128$
$\displaystyle \text{Difference between compound and simple interest}=276.48-256=20.48$
$\displaystyle \text{For }2\text{ years, }\text{C.I.}-\text{S.I.}=\frac{Pr^2}{10000}$
$\displaystyle \therefore\ \frac{128x}{100}=20.48$
$\displaystyle 1.28x=20.48$
$\displaystyle x=16$
$\displaystyle \therefore\ \frac{P\times16}{100}=128$
$\displaystyle P=\frac{128\times100}{16}=800$
$\displaystyle \therefore\ \text{Rate of interest}=16\%\text{ per annum}$
$\displaystyle \therefore\ \text{Required sum}=\text{Rs. }800$
$\\$

$\displaystyle \textbf{Question 9: }\text{On a certain sum and at a certain rate per cent, the simple interest}$
$\displaystyle \text{for the first year is Rs. }270\text{ and the compound interest for the first two years}$
$\displaystyle \text{is Rs. }580.50.\text{ Find the sum and the rate per cent.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the principal be Rs. }P\text{ and the rate be }x\%\text{ per annum.}$
$\displaystyle \text{Simple Interest for 1 year}=P\times\frac{x}{100}=270$
$\displaystyle \therefore\ \frac{Px}{100}=270$
$\displaystyle \text{Compound Interest for 2 years}=580.50$
$\displaystyle \frac{Px}{100}+P\left(1+\frac{x}{100}\right)\times\frac{x}{100}=580.50$
$\displaystyle 270+270\left(1+\frac{x}{100}\right)=580.50$
$\displaystyle 540+2.7x=580.50$
$\displaystyle 2.7x=40.50$
$\displaystyle x=15$
$\displaystyle \therefore\ \frac{P\times15}{100}=270$
$\displaystyle P=\frac{270\times100}{15}=1800$
$\displaystyle \therefore\ \text{Rate of interest}=15\%\text{ per annum}$
$\displaystyle \therefore\ \text{Required sum}=\text{Rs. }1800$
$\\$

$\displaystyle \textbf{Question 10: }\text{The interest charged on a certain sum is Rs. }720\text{ for one year}$
$\displaystyle \text{and Rs. }1497.60\text{ for two years. Find whether the interest is simple or}$
$\displaystyle \text{compound. Also calculate the rate per cent and the sum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Interest for the 1st year}=\text{Rs. }720$
$\displaystyle \text{Interest for 2 years}=\text{Rs. }1497.60$
$\displaystyle \text{Interest for the 2nd year}=1497.60-720=\text{Rs. }777.60$
$\displaystyle \text{Since the interest for the 2nd year is greater than that for the 1st year,}$
$\displaystyle \therefore\ \text{the interest is compound interest.}$
$\displaystyle \text{Difference between the compound interest of two successive years}$
$\displaystyle =777.60-720=\text{Rs. }57.60$
$\displaystyle \therefore\ \text{Rs. }57.60\text{ is the interest on Rs. }720\text{ for }1\text{ year}$
$\displaystyle \therefore\ \text{Rate of Interest}=\frac{100\times I}{P\times T}\%$
$\displaystyle =\frac{100\times57.60}{720\times1}\%=8\%$
$\displaystyle \text{Alternatively,}$
$\displaystyle \text{Rate of Interest}=\frac{\text{Difference in interest of two consecutive periods}\times100}{\text{Compound interest of preceding year}\times\text{Time}}\%$
$\displaystyle =\frac{(777.60-720)\times100}{720\times1}\%=8\%$
$\displaystyle \text{Principal}=\frac{720\times100}{8}=\text{Rs. }9000$
$\displaystyle \therefore\ \text{The interest is compound, the rate is }8\%\text{ per annum}$
$\displaystyle \therefore\ \text{and the required sum is Rs. }9000$
$\\$

$\displaystyle \textbf{Question 11: }\text{The compound interest, calculated yearly, on a certain sum of money}$
$\displaystyle \text{for the second year is Rs. }864\text{ and for the third year is Rs. }933.12.$
$\displaystyle \text{Calculate the rate of interest and the compound interest for the fourth year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Difference between the compound interest of two successive years}$
$\displaystyle =933.12-864=\text{Rs. }69.12$
$\displaystyle \therefore\ \text{Rs. }69.12\text{ is the interest on Rs. }864\text{ for }1\text{ year}$
$\displaystyle \therefore\ \text{Rate of Interest}=\frac{100\times I}{P\times T}\%$
$\displaystyle =\frac{100\times69.12}{864\times1}\%=8\%$
$\displaystyle \text{Compound interest for the 4th year}=933.12+\frac{933.12\times8}{100}$
$\displaystyle =933.12+74.6496=\text{Rs. }1007.7696$
$\displaystyle \therefore\ \text{Rate of interest}=8\%\text{ per annum}$
$\displaystyle \therefore\ \text{Compound interest for the 4th year}=\text{Rs. }1007.77$
$\\$

$\displaystyle \textbf{Question 12: }\text{A sum of money amounts to Rs. }20160\text{ in }3\text{ years and to}$
$\displaystyle \text{Rs. }24192\text{ in }4\text{ years. Calculate (i) the rate of interest, (ii) the amount}$
$\displaystyle \text{in }2\text{ years, and (iii) the amount in }5\text{ years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Difference between the compound interest of two successive years}$
$\displaystyle =24192-20160=\text{Rs. }4032$
$\displaystyle \therefore\ \text{Rs. }4032\text{ is the interest on Rs. }20160\text{ for }1\text{ year}$
$\displaystyle \therefore\ \text{Rate of Interest}=\frac{100\times I}{P\times T}\%$
$\displaystyle =\frac{100\times4032}{20160\times1}\%=20\%$
$\displaystyle \text{Amount in }3\text{ years}=\text{Rs. }20160$
$\displaystyle \therefore\ \text{Amount in }2\text{ years}=\frac{20160}{1+\frac{20}{100}}$
$\displaystyle =\frac{20160}{1.2}=\text{Rs. }16800$
$\displaystyle \text{Amount in }4\text{ years}=\text{Rs. }24192$
$\displaystyle \therefore\ \text{Amount in }5\text{ years}=24192\left(1+\frac{20}{100}\right)$
$\displaystyle =24192\times1.2=\text{Rs. }29030.40$
$\displaystyle \therefore\ \text{Rate of interest}=20\%\text{ per annum}$
$\displaystyle \therefore\ \text{Amount in }2\text{ years}=\text{Rs. }16800$
$\displaystyle \therefore\ \text{Amount in }5\text{ years}=\text{Rs. }29030.40$
$\\$

$\displaystyle \textbf{Question 13: }\text{Rs. }8000\text{ is lent out at }7\%\text{ compound interest for }2\text{ years.}$
$\displaystyle \text{At the end of the first year, Rs. }3560\text{ are returned. Calculate:}$
$\displaystyle \text{(i) the interest paid for the second year, (ii) the total interest paid in two years,}$
$\displaystyle \text{and (iii) the total amount paid in two years to clear the debt.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }8000,\ R=7\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{8000\times7\times1}{100}=\text{Rs. }560$
$\displaystyle \text{Amount}=8000+560=\text{Rs. }8560$
$\displaystyle \text{After paying Rs. }3560,\ \text{balance for the 2nd year}=8560-3560=\text{Rs. }5000$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }5000,\ R=7\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{5000\times7\times1}{100}=\text{Rs. }350$
$\displaystyle \text{Amount}=5000+350=\text{Rs. }5350$
$\displaystyle \therefore\ \text{(i) Interest for the 2nd year}=\text{Rs. }350$
$\displaystyle \therefore\ \text{(ii) Total interest paid}=560+350=\text{Rs. }910$
$\displaystyle \therefore\ \text{(iii) Total amount paid}=3560+5350=\text{Rs. }8910$
$\\$

$\displaystyle \textbf{Question 14: }\text{A sum of Rs. }24000\text{ is lent for }2\text{ years at }10\%\text{ compound}$
$\displaystyle \text{interest per annum. The borrower returns some money at the end of the first year}$
$\displaystyle \text{and pays Rs. }12540\text{ at the end of the second year to clear the debt.}$
$\displaystyle \text{Calculate the amount returned at the end of the first year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the amount returned at the end of the first year be Rs. }x$
$\displaystyle \text{For 1st year: }P=\text{Rs. }24000,\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{24000\times10\times1}{100}=\text{Rs. }2400$
$\displaystyle \text{Amount}=24000+2400=\text{Rs. }26400$
$\displaystyle \text{Balance after repayment}=\text{Rs. }(26400-x)$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }(26400-x),\ R=10\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{(26400-x)\times10\times1}{100}=\text{Rs. }\left(2640-\frac{x}{10}\right)$
$\displaystyle \text{Amount}=(26400-x)+\left(2640-\frac{x}{10}\right)=12540$
$\displaystyle 29040-\frac{11x}{10}=12540$
$\displaystyle \frac{11x}{10}=16500$
$\displaystyle x=\frac{16500\times10}{11}=15000$
$\displaystyle \therefore\ \text{Amount returned at the end of the first year}=\text{Rs. }15000$
$\\$

$\displaystyle \textbf{Question 15: }\text{A man invests Rs. }1200\text{ for two years at compound interest.}$
$\displaystyle \text{After one year his money amounts to Rs. }1275.\text{ Find the interest for the}$
$\displaystyle \text{second year, correct to the nearest rupee.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the rate of interest be }x\%\text{ per annum.}$
$\displaystyle \text{For 1st year: }P=\text{Rs. }1200,\ R=x\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{1200\times x\times1}{100}=\text{Rs. }12x$
$\displaystyle \text{Amount}=1200+12x=1275$
$\displaystyle 12x=75$
$\displaystyle x=\frac{75}{12}=6.25\%$
$\displaystyle \text{For 2nd year: }P=\text{Rs. }1275,\ R=6.25\%,\ T=1\text{ year}$
$\displaystyle \text{Interest}=\frac{1275\times6.25\times1}{100}=\text{Rs. }79.6875$
$\displaystyle \therefore\ \text{Interest for the 2nd year}\approx\text{Rs. }80$
$\\$

$\displaystyle \textbf{Question 16: }\text{The compound interest, calculated yearly, on a certain sum of money}$
$\displaystyle \text{for the second year is Rs. }880\text{ and for the third year is Rs. }968.$
$\displaystyle \text{Calculate the rate of interest and the sum of money. [ICSE 1995]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Difference between the compound interest of two successive years}$
$\displaystyle =968-880=\text{Rs. }88$
$\displaystyle \therefore\ \text{Rs. }88\text{ is the interest on Rs. }880\text{ for }1\text{ year}$
$\displaystyle \therefore\ \text{Rate of Interest}=\frac{100\times I}{P\times T}\%$
$\displaystyle =\frac{100\times88}{880\times1}\%=10\%$
$\displaystyle \text{Alternatively,}$
$\displaystyle \text{Rate of Interest}=\frac{\text{Difference in interest of two consecutive periods}\times100}{\text{Compound interest of preceding year}\times\text{Time}}\%$
$\displaystyle =\frac{(968-880)\times100}{880\times1}\%=10\%$
$\displaystyle \text{Let the principal be Rs. }x$
$\displaystyle \text{For the 2nd year, principal}=1.1x$
$\displaystyle \text{Interest for the 2nd year}=\frac{1.1x\times10}{100}=0.11x$
$\displaystyle \text{Given, }0.11x=880$
$\displaystyle x=\frac{880}{0.11}=8000$
$\displaystyle \therefore\ \text{Rate of interest}=10\%\text{ per annum}$
$\displaystyle \therefore\ \text{Required sum}=\text{Rs. }8000$
$\\$

$\displaystyle \textbf{Question 17: }\text{The cost of a machine depreciated by Rs. }4000\text{ during the first}$
$\displaystyle \text{year and by Rs. }3600\text{ during the second year. Calculate:}$
$\displaystyle \text{(i) the rate of depreciation, (ii) the original cost of the machine, and}$
$\displaystyle \text{(iii) its cost at the end of the third year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original cost of the machine be Rs. }x\text{ and the rate of depreciation be }r\%\text{.}$
$\displaystyle \text{For the 1st year: }P=\text{Rs. }x,\ R=r\%,\ T=1\text{ year}$
$\displaystyle \text{Depreciation}=\frac{xr}{100}=\text{Rs. }4000\qquad\ldots\text{(i)}$
$\displaystyle \text{Value after the 1st year}=x\left(1-\frac{r}{100}\right)$
$\displaystyle \text{For the 2nd year: }P=x\left(1-\frac{r}{100}\right),\ R=r\%,\ T=1\text{ year}$
$\displaystyle x\left(1-\frac{r}{100}\right)\times\frac{r}{100}=3600\qquad\ldots\text{(ii)}$
$\displaystyle \text{Dividing (ii) by (i),}$
$\displaystyle 1-\frac{r}{100}=\frac{3600}{4000}=\frac{9}{10}$
$\displaystyle \frac{r}{100}=\frac{1}{10}$
$\displaystyle r=10\%$
$\displaystyle \text{From (i), }\frac{10x}{100}=4000$
$\displaystyle x=\text{Rs. }40000$
$\displaystyle \text{Value at the end of the 2nd year}=40000-4000-3600=\text{Rs. }32400$
$\displaystyle \text{Depreciation during the 3rd year}=\frac{32400\times10}{100}=\text{Rs. }3240$
$\displaystyle \text{Value at the end of the 3rd year}=32400-3240=\text{Rs. }29160$
$\displaystyle \therefore\ \text{(i) Rate of depreciation}=10\%$
$\displaystyle \therefore\ \text{(ii) Original cost of the machine}=\text{Rs. }40000$
$\displaystyle \therefore\ \text{(iii) Cost at the end of the 3rd year}=\text{Rs. }29160$
$\\$