Class 8-12 Mathematics Practice Questions, Solutions | ICSE ISC CBSE Math Mathematics Made Easy for School Students

$\displaystyle \textbf{Queston 1. }\text{Find the amount and the compound interest on }$
$\displaystyle \text{Rs }10000\text{ at }8\%\text{ per annum in }2\text{ years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year : }P=\text{Rs }10000,\ R=8\%\text{ and }T=1\text{ year.}$
$\displaystyle \therefore I=\text{Rs }\frac{10000\times8\times1}{100}=\text{Rs }800$
$\displaystyle \text{And, amount }(A)=P+I$
$\displaystyle =\text{Rs }10000+\text{Rs }800=\text{Rs }10800$
$\displaystyle \text{For 2nd year : }P=\text{Rs }10800,\ R=8\%\text{ and }T=1\text{ year.}$
$\displaystyle \therefore I=\text{Rs }\frac{10800\times8\times1}{100}=\text{Rs }864$
$\displaystyle \text{And, amount }(A)=\text{Rs }10800+\text{Rs }864=\text{Rs }11664$
$\displaystyle \therefore \text{Required amount }=\text{Rs }11664$
$\displaystyle \text{And, compound interest}=A-P$
$\displaystyle =\text{Rs }11664-\text{Rs }10000=\text{Rs }1664$
$\displaystyle \\$

$\displaystyle \textbf{Question 2. }\text{Find the amount and the compound interest on }$
$\displaystyle \text{Rs }10000\text{ at }8\%\text{ per annum in }1\text{ year; interest being}$
$\displaystyle \text{compounded half-yearly.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st }\frac{1}{2}\text{ year : }P=\text{Rs }10000,\ R=8\%\text{ and }T=\frac{1}{2}\text{ year.}$
$\displaystyle \therefore I=\text{Rs }\frac{10000\times8\times1}{100\times2}=\text{Rs }400$
$\displaystyle \text{And, amount }(A)=P+I$
$\displaystyle =\text{Rs }10000+\text{Rs }400=\text{Rs }10400$
$\displaystyle \text{For 2nd }\frac{1}{2}\text{ year : }P=\text{Rs }10400,\ R=8\%\text{ and }T=\frac{1}{2}\text{ year.}$
$\displaystyle \therefore I=\text{Rs }\frac{10400\times8\times1}{100\times2}=\text{Rs }416$
$\displaystyle \text{And, amount }(A)=P+I$
$\displaystyle =\text{Rs }10400+\text{Rs }416=\text{Rs }10816$
$\displaystyle \therefore \text{Required amount }=\text{Rs }10816$
$\displaystyle \text{And, compound interest}=A-P$
$\displaystyle =\text{Rs }10816-\text{Rs }10000=\text{Rs }816$
$\displaystyle \\$

$\displaystyle \textbf{Question 3. }\text{Calculate the compound interest accrued on }$
$\displaystyle \text{Rs }16000\text{ in }3\text{ years, when the rates of interest for}$
$\displaystyle \text{successive years are }10\%,\ 12\%\text{ and }15\%\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st year : }P=\text{Rs }16000,\ R=10\%\text{ and }T=1\text{ year.}$
$\displaystyle \therefore I=\text{Rs }\frac{16000\times10\times1}{100}=\text{Rs }1600$
$\displaystyle \text{And, amount }(A)=P+I$
$\displaystyle =\text{Rs }16000+\text{Rs }1600=\text{Rs }17600$
$\displaystyle \text{For 2nd year : }P=\text{Rs }17600,\ R=12\%\text{ and }T=1\text{ year.}$
$\displaystyle \therefore I=\text{Rs }\frac{17600\times12\times1}{100}=\text{Rs }2112$
$\displaystyle \text{And, amount }(A)=P+I$
$\displaystyle =\text{Rs }17600+\text{Rs }2112=\text{Rs }19712$
$\displaystyle \text{For 3rd year : }P=\text{Rs }19712,\ R=15\%\text{ and }T=1\text{ year.}$
$\displaystyle \therefore I=\text{Rs }\frac{19712\times15\times1}{100}=\text{Rs }2956.80$
$\displaystyle \text{And, amount }(A)=P+I$
$\displaystyle =\text{Rs }19712+\text{Rs }2956.80=\text{Rs }22668.80$
$\displaystyle \therefore \text{C.I. accrued}=\text{Final amount}-\text{Initial principal}$
$\displaystyle =\text{Rs }22668.80-\text{Rs }16000=\text{Rs }6668.80$
$\displaystyle \\$

$\displaystyle \textbf{Question 4. }\text{A man borrows Rs }8000\text{ at }10\%\text{ compound interest}$
$\displaystyle \text{payable every six months. He repays Rs }2500\text{ at the end of every}$
$\displaystyle \text{six months. Calculate the third payment he has to make at the end}$
$\displaystyle \text{of }18\text{ months in order to clear the entire loan.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For 1st six months : }P=\text{Rs }8000,\ R=10\%\text{ and }T=\frac{1}{2}\text{ year.}$
$\displaystyle \therefore I=\text{Rs }\frac{8000\times10\times1}{100\times2}=\text{Rs }400$
$\displaystyle \text{And, amount }(A)=P+I$
$\displaystyle =\text{Rs }8000+\text{Rs }400=\text{Rs }8400$
$\displaystyle \text{Money repaid}=\text{Rs }2500$
$\displaystyle \therefore \text{Balance}=\text{Rs }8400-\text{Rs }2500=\text{Rs }5900$
$\displaystyle \text{For the 2nd six months : }P=\text{Rs }5900,\ R=10\%\text{ and }T=\frac{1}{2}\text{ year.}$
$\displaystyle \therefore I=\text{Rs }\frac{5900\times10\times1}{100\times2}=\text{Rs }295$
$\displaystyle \text{And, amount }(A)=P+I$
$\displaystyle =\text{Rs }5900+\text{Rs }295=\text{Rs }6195$
$\displaystyle \text{Again, money repaid}=\text{Rs }2500$
$\displaystyle \therefore \text{Balance}=\text{Rs }6195-\text{Rs }2500=\text{Rs }3695$
$\displaystyle \text{For the 3rd six months : }P=\text{Rs }3695,\ R=10\%\text{ and }T=\frac{1}{2}\text{ year.}$
$\displaystyle \therefore I=\text{Rs }\frac{3695\times10\times1}{100\times2}=\text{Rs }184.75$
$\displaystyle \text{And, amount }(A)=P+I$
$\displaystyle =\text{Rs }3695+\text{Rs }184.75=\text{Rs }3879.75$
$\displaystyle \therefore \text{The 3rd instalment to be made to clear the entire loan}$
$\displaystyle =\text{Rs }3879.75$
$\displaystyle \\$

$\displaystyle \textbf{Question 5. }\text{On a certain sum of money, invested at the rate of }5\%$
$\displaystyle \text{per annum compounded annually, the difference between the interest}$
$\displaystyle \text{of the first year and the interest of the third year is Rs }61.50.$
$\displaystyle \text{Find the sum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum (principal)}=\text{Rs }100$
$\displaystyle \text{C.I. of 1st year}=\text{Rs }\frac{100\times5\times1}{100}=\text{Rs }5$
$\displaystyle \text{And, amount of 1st year}=\text{Rs }100+\text{Rs }5=\text{Rs }105$
$\displaystyle \therefore \text{The principal for 2nd year}=\text{Rs }105$
$\displaystyle \text{C.I. of 2nd year}=\text{Rs }\frac{105\times5\times1}{100}=\text{Rs }5.25$
$\displaystyle \text{And, amount of 2nd year}=\text{Rs }105+\text{Rs }5.25=\text{Rs }110.25$
$\displaystyle \therefore \text{The principal for 3rd year}=\text{Rs }110.25$
$\displaystyle \text{C.I. of 3rd year}=\text{Rs }\frac{110.25\times5\times1}{100}=\text{Rs }5.5125$
$\displaystyle \text{Difference between C.I. of 1st year and C.I. of 3rd year}$
$\displaystyle =\text{Rs }5.5125-\text{Rs }5=\text{Rs }0.5125$
$\displaystyle \text{Now, when the difference of interest}=\text{Rs }0.5125,\ \text{sum}=\text{Rs }100$
$\displaystyle \text{And, when the difference of interest}=\text{Rs }61.50,\ \text{sum}$
$\displaystyle =\text{Rs }\frac{100}{0.5125}\times61.50=\text{Rs }12000$
$\displaystyle \\$

$\displaystyle \textbf{Question 6. }\text{Mrs. Kapoor invested Rs }6000\text{ every year at the}$
$\displaystyle \text{beginning of the year, at }10\%\text{ per annum compound interest.}$
$\displaystyle \text{Calculate the amount of her total savings :}$
$\displaystyle \text{(i) upto the end of the second year \qquad (ii) at the beginning of}$
$\displaystyle \text{the third year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) For 1st year :}$
$\displaystyle \text{Since, money invested at the beginning of the year}=\text{Rs }6000$
$\displaystyle \therefore \text{Principal for 1st year}=\text{Rs }6000$
$\displaystyle \therefore \text{Interest}=\text{Rs }\frac{6000\times10\times1}{100}=\text{Rs }600$
$\displaystyle \text{And, amount}=\text{Rs }6000+\text{Rs }600=\text{Rs }6600$
$\displaystyle \text{For 2nd year :}$
$\displaystyle \text{Since, Rs }6000\text{ is invested again at the beginning of the second}$
$\displaystyle \text{year, therefore principal for the second year}$
$\displaystyle =\text{Rs }6600+\text{Rs }6000=\text{Rs }12600$
$\displaystyle \therefore \text{Interest}=\text{Rs }\frac{12600\times10\times1}{100}=\text{Rs }1260$
$\displaystyle \text{And, amount}=\text{Rs }12600+\text{Rs }1260=\text{Rs }13860$
$\displaystyle \therefore \text{Amount of her total savings upto the end of the second year}$
$\displaystyle =\text{Rs }13860$
$\displaystyle \text{(ii) Since, Rs }6000\text{ is invested again at the beginning of the}$
$\displaystyle \text{third year}$
$\displaystyle \therefore \text{Amount of her total savings at the beginning of the third year}$
$\displaystyle =\text{Rs }13860+\text{Rs }6000=\text{Rs }19860$
$\displaystyle \\$

$\displaystyle \textbf{Question 7. }\text{During every financial year, the value of a machine}$
$\displaystyle \text{depreciates by }10\%.\text{ Find the original value (cost) of a machine}$
$\displaystyle \text{which depreciates by Rs }2250\text{ during the second year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original cost of the machine}=\text{Rs }100$
$\displaystyle \therefore \text{Depreciation during 1st year}=10\%\text{ of Rs }100=\text{Rs }10$
$\displaystyle \text{Value of the machine at the beginning of 2nd year}$
$\displaystyle =\text{Rs }100-\text{Rs }10=\text{Rs }90$
$\displaystyle \therefore \text{Depreciation during 2nd year}=10\%\text{ of Rs }90=\text{Rs }9$
$\displaystyle \text{Now, when depreciation during 2nd year}=\text{Rs }9,\ \text{original cost}$
$\displaystyle =\text{Rs }100$
$\displaystyle \therefore \text{When depreciation during 2nd year}=\text{Rs }2250$
$\displaystyle \text{Original cost}=\text{Rs }\frac{100}{9}\times2250=\text{Rs }25000$
$\displaystyle \\$

$\displaystyle \textbf{Question 8. }\text{Calculate the difference between the compound interest}$
$\displaystyle \text{and the simple interest on Rs }4000\text{ at }8\%\text{ per annum and}$
$\displaystyle \text{in }2\text{ years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For S.I. : }P=\text{Rs }4000,\ R=8\%\text{ and }T=2\text{ years}$
$\displaystyle \therefore \text{Simple interest}=\text{Rs }\frac{4000\times8\times2}{100}$
$\displaystyle =\text{Rs }640$
$\displaystyle \text{For C.I. :}$
$\displaystyle \text{Principal for 1st year}=\text{Rs }4000$
$\displaystyle \therefore \text{Interest on it}=\text{Rs }\frac{4000\times8\times1}{100}$
$\displaystyle =\text{Rs }320$
$\displaystyle \text{Amount}=\text{Rs }4000+\text{Rs }320=\text{Rs }4320$
$\displaystyle \therefore \text{Principal for 2nd year}=\text{Rs }4320$
$\displaystyle \therefore \text{Interest on it}=\text{Rs }\frac{4320\times8\times1}{100}$
$\displaystyle =\text{Rs }345.60$
$\displaystyle \therefore \text{C.I. of 2 years}=\text{Rs }320+\text{Rs }345.60$
$\displaystyle =\text{Rs }665.60$
$\displaystyle \therefore \text{Required difference between C.I. and S.I.}$
$\displaystyle =\text{C.I.}-\text{S.I.}$
$\displaystyle =\text{Rs }665.60-\text{Rs }640$
$\displaystyle =\text{Rs }25.60$
$\displaystyle \\$

$\displaystyle \textbf{Question 9. }\text{Ashok borrowed Rs }16000\text{ at }10\%\text{ simple interest.}$
$\displaystyle \text{He immediately invested this money at }10\%\text{ compound interest}$
$\displaystyle \text{compounded half-yearly. Calculate Ashok's gain in }18\text{ months.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{To calculate the S.I. paid by Ashok :}$
$\displaystyle P=\text{Rs }16000,\ R=10\%\text{ and }T=18\text{ months}=\frac{3}{2}\text{ years}$
$\displaystyle \therefore \text{S.I.}=\text{Rs }\frac{16000\times10\times3}{100\times2}=\text{Rs }2400$
$\displaystyle \text{To calculate the C.I. earned by Ashok :}$
$\displaystyle \text{For 1st half-year :}$
$\displaystyle P=\text{Rs }16000,\ R=10\%\text{ and }T=\frac{1}{2}\text{ year}$
$\displaystyle \therefore \text{Interest}=\text{Rs }\frac{16000\times10\times1}{100\times2}=\text{Rs }800$
$\displaystyle \text{And, amount}=\text{Rs }16000+\text{Rs }800=\text{Rs }16800$
$\displaystyle \text{For 2nd half-year :}$
$\displaystyle P=\text{Rs }16800,\ R=10\%\text{ and }T=\frac{1}{2}\text{ year}$
$\displaystyle \therefore \text{Interest}=\text{Rs }\frac{16800\times10\times1}{100\times2}=\text{Rs }840$
$\displaystyle \text{And, amount}=\text{Rs }16800+\text{Rs }840=\text{Rs }17640$
$\displaystyle \text{For 3rd half-year :}$
$\displaystyle P=\text{Rs }17640,\ R=10\%\text{ and }T=\frac{1}{2}\text{ year}$
$\displaystyle \therefore \text{Interest}=\text{Rs }\frac{17640\times10\times1}{100\times2}=\text{Rs }882$
$\displaystyle \text{And, amount}=\text{Rs }17640+\text{Rs }882=\text{Rs }18522$
$\displaystyle \therefore \text{Total C.I. earned}=\text{Rs }18522-\text{Rs }16000=\text{Rs }2522$
$\displaystyle \therefore \text{Ashok's gain in }18\text{ months}=\text{C.I. earned}-\text{S.I. paid}$
$\displaystyle =\text{Rs }2522-\text{Rs }2400=\text{Rs }122$
$\displaystyle \\$

$\displaystyle \textbf{Question 10. }\text{A sum of money is invested at C.I. payable annually.}$
$\displaystyle \text{The amounts of interest in two successive years are Rs }2700\text{ and}$
$\displaystyle \text{Rs }2880.\text{ Find the rate of interest.}$
$\displaystyle \text{Answer:}$
$\displaystyle \therefore \text{Difference between the C.I. of two successive years}$
$\displaystyle =\text{Rs }2880-\text{Rs }2700=\text{Rs }180$
$\displaystyle \therefore \text{Rs }180\text{ is the interest of one year on Rs }2700.$
$\displaystyle \therefore \text{Rate of interest}=\frac{100\times I}{P\times T}\%$
$\displaystyle =\frac{100\times180}{2700\times1}\%=6\frac{2}{3}\%$
$\displaystyle \text{Directly :}$
$\displaystyle \text{Rate of interest}$
$\displaystyle =\frac{\text{Difference in interest of two consecutive periods}\times100}{\text{C.I. of preceding year}\times\text{Time}}\%$
$\displaystyle =\frac{(2880-2700)\times100}{2700\times1}\%=6\frac{2}{3}\%$
$\displaystyle \\$

$\displaystyle \textbf{Question 11. }\text{A certain sum of money, placed out at compound interest,}$
$\displaystyle \text{amounts to Rs }6272\text{ in }2\text{ years and to Rs }7024.64\text{ in }3\text{ years.}$
$\displaystyle \text{Find the rate of interest and the sum of money.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Difference between the amounts of two consecutive years}$
$\displaystyle =\text{Rs }7024.64-\text{Rs }6272=\text{Rs }752.64$
$\displaystyle \therefore \text{Interest for one year on Rs }6272=\text{Rs }752.64$
$\displaystyle \therefore \text{Rate of interest}=\frac{752.64}{6272}\times100\%=12\%$
$\displaystyle \text{Directly :}$
$\displaystyle \text{Rate of interest}$
$\displaystyle =\frac{\text{Difference between the amounts of two consecutive periods}\times100}{\text{Preceding amount}\times\text{Time}}\%$
$\displaystyle =\frac{(7024.64-6272)\times100}{6272\times1}\%=12\%$
$\displaystyle \text{Let the sum of money}=\text{Rs }100$
$\displaystyle \therefore \text{Interest on it for 1st year}=12\%\text{ of Rs }100=\text{Rs }12$
$\displaystyle \therefore \text{Amount in one year}=\text{Rs }100+\text{Rs }12=\text{Rs }112$
$\displaystyle \text{Similarly, amount in two years}=\text{Rs }112+12\%\text{ of Rs }112=\text{Rs }125.44$
$\displaystyle \text{When amount in two years}=\text{Rs }125.44,\ \text{sum}=\text{Rs }100$
$\displaystyle \therefore \text{When amount in two years}=\text{Rs }6272,\ \text{sum}$
$\displaystyle =\text{Rs }\frac{100}{125.44}\times6272$
$\displaystyle =\text{Rs }5000$
$\displaystyle \\$

$\displaystyle \textbf{Question 12. }\text{The simple interest on a certain sum computes to}$
$\displaystyle \text{Rs }600\text{ in }3\text{ years and the compound interest on the same sum,}$
$\displaystyle \text{at the same rate and for }2\text{ years computes to Rs }410.$
$\displaystyle \text{Find the rate per cent.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since, S.I. of 3 years}=\text{Rs }600$
$\displaystyle \therefore \text{S.I. of 1 year}=\text{Rs }\frac{600}{3}=\text{Rs }200$
$\displaystyle \therefore \text{C.I. for first year}=\text{Rs }200$
$\displaystyle \text{Given, C.I. for two years}=\text{Rs }410$
$\displaystyle \therefore \text{C.I. for 2nd year}=\text{Rs }410-\text{Rs }200=\text{Rs }210$
$\displaystyle \text{Difference between the C.I. of two successive years}$
$\displaystyle =\text{Rs }210-\text{Rs }200=\text{Rs }10$
$\displaystyle \therefore \text{Rs }10\text{ is the interest for one year on the interest of}$
$\displaystyle \text{1st year i.e. on Rs }200$
$\displaystyle \therefore \text{Rate }\%=\frac{100\times I}{P\times T}\%$
$\displaystyle =\frac{100\times10}{200\times1}\%=5\%$
$\displaystyle \\$

$\displaystyle \textbf{Question 13. }\text{The compound interest calculated yearly at }10\%$
$\displaystyle \text{on a certain sum of money amounts to Rs }665.50\text{ in the fifth year.}$
$\displaystyle \text{Calculate :}$
$\displaystyle \text{(i) C.I. for the sixth year at the same rate and on the same sum.}$
$\displaystyle \text{(ii) C.I. for the fourth year on the same sum and at the same rate.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) C.I. for 6th year}=\text{C.I. of 5th year}+\text{Interest on it for 1 year}$
$\displaystyle =\text{Rs }665.50+10\%\text{ of Rs }665.50=\text{Rs }732.05$
$\displaystyle \text{(ii) Let C.I. for 4th year}=\text{Rs }x$
$\displaystyle \text{Since, C.I. for 5th year}=\text{C.I. of 4th year}+\text{Interest on it for 1 year}$
$\displaystyle \therefore \text{Rs }665.50=\text{Rs }x+10\%\text{ of Rs }x$
$\displaystyle \therefore 665.50=1.1x$
$\displaystyle \therefore x=605$
$\displaystyle \therefore \text{C.I. for 4th year}=\text{Rs }605$
$\displaystyle \\$

$\displaystyle \textbf{Question 14. }\text{A sum of money, at compound interest, amounts to}$
$\displaystyle \text{Rs }8100\text{ in }5\text{ years and to Rs }8748\text{ in }6\text{ years. Find :}$
$\displaystyle \text{(i) the rate per cent \qquad (ii) amount in 7 years and}$
$\displaystyle \text{(iii) amount in 4 years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Amount in 5 years}=\text{Rs }8100\text{ and amount in 6 years}=\text{Rs }8748$
$\displaystyle \therefore \text{Rs }8748-\text{Rs }8100=\text{Rs }648\text{ is the interest of 1 year}$
$\displaystyle \text{on Rs }8100$
$\displaystyle \therefore \text{Rate }\%=\frac{648\times100}{8100\times1}\%=8\%$
$\displaystyle \text{(ii) Amount in 7 years}=\text{Amount in 6 years}+\text{Interest on it for 1 year}$
$\displaystyle =\text{Rs }8748+8\%\text{ of Rs }8748=\text{Rs }9447.84$
$\displaystyle \text{(iii) Let amount in 4 years}=\text{Rs }x$
$\displaystyle \therefore \text{Amount in 5 years}=\text{Amount in 4 years}+\text{Interest on it for 1 year}$
$\displaystyle \therefore \text{Rs }8100=\text{Rs }x+8\%\text{ of Rs }x$
$\displaystyle \text{On solving, we get }x=7500$
$\displaystyle \therefore \text{Amount in 4 years}=\text{Rs }7500$
$\displaystyle \\$

$\displaystyle \textbf{Question 15. }\text{The cost of a machine depreciated by Rs }4752$
$\displaystyle \text{during the second year and by Rs }4181.76\text{ during the third year.}$
$\displaystyle \text{Calculate :}$
$\displaystyle \text{(i) the rate of depreciation;}$
$\displaystyle \text{(ii) the original cost;}$
$\displaystyle \text{(iii) the cost at the end of the third year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Difference between depreciations of 2nd year and 3rd year}$
$\displaystyle =\text{Rs }4752-\text{Rs }4181.76=\text{Rs }570.24$
$\displaystyle \therefore \text{Depreciation of one year on Rs }4752=\text{Rs }570.24$
$\displaystyle \therefore \text{Rate of depreciation}=\frac{\text{Rs }570.24}{\text{Rs }4752}\times100\%=12\%$
$\displaystyle \text{(ii) Let original cost of the machine be Rs }100$
$\displaystyle \therefore \text{Its value after one year}=\text{Rs }100-12\%\text{ of Rs }100=\text{Rs }88$
$\displaystyle \text{Depreciation during 2nd year}=12\%\text{ of Rs }88=\text{Rs }10.56$
$\displaystyle \text{When depreciation during 2nd year}=\text{Rs }10.56,\ \text{original cost}=\text{Rs }100$
$\displaystyle \therefore \text{When depreciation during 2nd year}=\text{Rs }4752$
$\displaystyle \text{Original cost}=\text{Rs }\frac{100}{10.56}\times4752$
$\displaystyle =\text{Rs }45000$
$\displaystyle \text{(iii) Since, total depreciation during all the three years}$
$\displaystyle =\text{Depreciation during }(1\text{st year}+2\text{nd year}+3\text{rd year})$
$\displaystyle =12\%\text{ of Rs }45000+\text{Rs }4752+\text{Rs }4181.76$
$\displaystyle =\text{Rs }14333.76$
$\displaystyle \therefore \text{The cost of machine at the end of the third year}$
$\displaystyle =\text{Rs }45000-\text{Rs }14333.76=\text{Rs }30666.24$
$\displaystyle \\$

$\displaystyle \textbf{Question 16. }\text{Calculate the amount and the compound interest on}$
$\displaystyle \text{Rs }7500\text{ in }2\text{ years and at }6\%\text{ compounded annually.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : }P=\text{Rs }7500,\ n=2\text{ years and }r=6\%$
$\displaystyle \therefore A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle =\text{Rs }7500\left(1+\frac{6}{100}\right)^2$
$\displaystyle =\text{Rs }7500\left(\frac{106}{100}\right)^2=\text{Rs }8427$
$\displaystyle \therefore \text{Required amount}=\text{Rs }8427$
$\displaystyle \text{And, C.I.}=\text{Rs }8427-\text{Rs }7500=\text{Rs }927$
$\displaystyle \\$

$\displaystyle \textbf{Question 17. }\text{Calculate the amount and the compound interest on}$
$\displaystyle \text{Rs }12000\text{ in }3\text{ years when the rates of interest for}$
$\displaystyle \text{successive years are }8\%,\ 10\%\text{ and }15\%\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Required amount, }A=P\left(1+\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right)\left(1+\frac{r_3}{100}\right)$
$\displaystyle \therefore A=\text{Rs }12000\left(1+\frac{8}{100}\right)\left(1+\frac{10}{100}\right)\left(1+\frac{15}{100}\right)$
$\displaystyle =\text{Rs }16394.40$
$\displaystyle \text{And, C.I.}=\text{Rs }16394.40-\text{Rs }12000=\text{Rs }4394.40$
$\displaystyle \\$

$\displaystyle \textbf{Question 18. }\text{Calculate the compound interest on Rs }18000$
$\displaystyle \text{in }2\text{ years at }15\%\text{ per annum.}$
$\displaystyle \text{Answer:}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle \therefore A=\text{Rs }18000\left(1+\frac{15}{100}\right)^2$
$\displaystyle =\text{Rs }23805$
$\displaystyle \therefore \text{Compound Interest}=A-P$
$\displaystyle =\text{Rs }23805-\text{Rs }18000=\text{Rs }5805$
$\displaystyle \text{Direct method :}$
$\displaystyle \text{Compound Interest}=P\left[\left(1+\frac{r}{100}\right)^n-1\right]$
$\displaystyle =\text{Rs }18000\left[\left(1+\frac{15}{100}\right)^2-1\right]$
$\displaystyle =\text{Rs }18000(1.3225-1)$
$\displaystyle =\text{Rs }18000\times0.3225=\text{Rs }5805$
$\displaystyle \\$

$\displaystyle \textbf{Question 19. }\text{On a certain sum, the compound interest in }3\text{ years}$
$\displaystyle \text{and at }10\%\text{ per cent per annum amounts to Rs }2317.$
$\displaystyle \text{Find the sum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : }n=3\text{ years, }r=10\%\text{ and C.I.}=\text{Rs }2317$
$\displaystyle \therefore 2317=P\left[\left(1+\frac{10}{100}\right)^3-1\right]$
$\displaystyle \therefore 2317=P(1.331-1)$
$\displaystyle \therefore P=\frac{2317}{0.331}=7000$
$\displaystyle \therefore \text{Required sum}=\text{Rs }7000$
$\displaystyle \\$

$\displaystyle \textbf{Question 20. }\text{On a certain sum, the compound interest in two years}$
$\displaystyle \text{amounts to Rs }2256.\text{ If the rates of interest for successive years}$
$\displaystyle \text{are }8\%\text{ and }10\%\text{ respectively, find the sum.}$
$\displaystyle \text{Answer:}$
$\displaystyle A=P\left(1+\frac{8}{100}\right)\left(1+\frac{10}{100}\right)$
$\displaystyle =\frac{297}{250}P$
$\displaystyle \text{Given : C.I.}=\text{Rs }2256$
$\displaystyle \therefore \frac{297}{250}P-P=2256$
$\displaystyle \therefore \frac{297P-250P}{250}=2256$
$\displaystyle \therefore P=\frac{2256\times250}{47}=12000$
$\displaystyle \therefore \text{Required sum}=\text{Rs }12000$
$\displaystyle \\$

$\displaystyle \textbf{Question 21. }\text{Ramesh lends Rs }15000\text{ for }2\text{ years at a certain}$
$\displaystyle \text{rate of compound interest. If after }2\text{ years, it amounts to}$
$\displaystyle \text{Rs }16224,\text{ find the rate of interest.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : }P=\text{Rs }15000,\ n=2\text{ years and }A=\text{Rs }16224$
$\displaystyle \therefore 16224=15000\left(1+\frac{r}{100}\right)^2$
$\displaystyle \therefore \left(1+\frac{r}{100}\right)^2=\frac{16224}{15000}$
$\displaystyle =\frac{2\times2\times2\times2\times3\times13\times13}{2\times2\times3\times5\times5\times5\times5}$
$\displaystyle =\frac{2\times2\times13\times13}{5\times5\times5\times5}=\left(\frac{26}{25}\right)^2$
$\displaystyle \therefore 1+\frac{r}{100}=\frac{26}{25}$
$\displaystyle \therefore \frac{r}{100}=\frac{26}{25}-1=\frac{1}{25}$
$\displaystyle \therefore r=4\%$
$\displaystyle \\$

$\displaystyle \textbf{Question 22. }\text{Divide Rs }36465\text{ between A and B so that when}$
$\displaystyle \text{their shares are lent out at }10\%\text{ compound interest per year, the}$
$\displaystyle \text{amount that A receives in }7\text{ years is the same as what B receives}$
$\displaystyle \text{in }5\text{ years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let A's share}=\text{Rs }x$
$\displaystyle \therefore \text{B's share}=\text{Rs }(36465-x)$
$\displaystyle \text{Given : Amount of A in }7\text{ years}=\text{Amount of B in }5\text{ years}$
$\displaystyle \therefore x\left(1+\frac{10}{100}\right)^7=(36465-x)\left(1+\frac{10}{100}\right)^5$
$\displaystyle \therefore x\left(1+\frac{10}{100}\right)^2=36465-x$
$\displaystyle \therefore \frac{121}{100}x=36465-x$
$\displaystyle \text{On solving, we get }x=16500$
$\displaystyle \therefore 36465-x=36465-16500=19965$
$\displaystyle \therefore \text{A's share}=\text{Rs }16500\text{ and B's share}=\text{Rs }19965$
$\displaystyle \\$

$\displaystyle \textbf{Question 23. }\text{Calculate the compound interest on Rs }4000\text{ in}$
$\displaystyle 1\frac{1}{2}\text{ years at }10\%\text{ per annum compounded half-yearly.}$
$\displaystyle \text{Answer:}$
$\displaystyle A=P\left(1+\frac{r}{2\times100}\right)^{n\times2}$
$\displaystyle \therefore A=\text{Rs }4000\left(1+\frac{10}{2\times100}\right)^{\frac{3}{2}\times2}$
$\displaystyle =\text{Rs }4630.50$
$\displaystyle \therefore \text{C.I.}=A-P$
$\displaystyle =\text{Rs }4630.50-\text{Rs }4000=\text{Rs }630.50$
$\displaystyle \\$

$\displaystyle \textbf{Question 24. }\text{Find the amount when Rs }10000\text{ is invested for}$
$\displaystyle 2\frac{1}{2}\text{ years at }10\%\text{ interest compounded yearly.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{First of all find the amount in }2\text{ years.}$
$\displaystyle \text{Amount in 2 years}=\text{Rs }10000\left(1+\frac{10}{100}\right)^2$
$\displaystyle =\text{Rs }12100$
$\displaystyle \text{After two years, Rs }12100\text{ is the principal for the remaining}$
$\displaystyle \frac{1}{2}\text{ year and so :}$
$\displaystyle A=P\left(1+\frac{r}{2\times100}\right)^{n\times2}$
$\displaystyle \therefore A=\text{Rs }12100\left(1+\frac{10}{2\times100}\right)^{\frac{1}{2}\times2}$
$\displaystyle =\text{Rs }12705$
$\displaystyle \therefore \text{Amount in }2\frac{1}{2}\text{ years}=\text{Rs }12705$
$\displaystyle \text{Note : For the example given above, we can directly use the formula :}$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^2\left(1+\frac{r}{2\times100}\right)^{\frac{1}{2}\times2}$
$\displaystyle \therefore \text{Amount in }2\frac{1}{2}\text{ years}$
$\displaystyle =\text{Rs }10000\left(1+\frac{10}{100}\right)^2\left(1+\frac{10}{2\times100}\right)$
$\displaystyle =\text{Rs }12705$
$\displaystyle \\$

$\displaystyle \textbf{Question 25. }\text{John borrowed Rs }20000\text{ for }4\text{ years under the}$
$\displaystyle \text{following conditions :}$
$\displaystyle 10\%\text{ simple interest for the first }2\frac{1}{2}\text{ years.}$
$\displaystyle 10\%\text{ C.I. for the remaining }1\frac{1}{2}\text{ years on the amount due}$
$\displaystyle \text{after }2\frac{1}{2}\text{ years, the interest being compounded half-yearly.}$
$\displaystyle \text{Find the total amount to be paid at the end of the four years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For first }2\frac{1}{2}\text{ years : }P=\text{Rs }20000,\ R=10\%\text{ and }T=\frac{5}{2}\text{ years}$
$\displaystyle \therefore \text{Interest}=\text{Rs }\frac{20000\times10\times5}{100\times2}$
$\displaystyle =\text{Rs }5000$
$\displaystyle \therefore \text{Amount due after }2\frac{1}{2}\text{ years}$
$\displaystyle =\text{Rs }20000+\text{Rs }5000=\text{Rs }25000$
$\displaystyle \text{For remaining }1\frac{1}{2}\text{ years : }P=\text{Rs }25000,\ n=\frac{3}{2}\text{ years}$
$\displaystyle \text{and }r=10\%\text{ per annum compounded half-yearly.}$
$\displaystyle \therefore A=P\left(1+\frac{r}{2\times100}\right)^{n\times2}$
$\displaystyle =\text{Rs }25000\left(1+\frac{10}{2\times100}\right)^{\frac{3}{2}\times2}$
$\displaystyle =\text{Rs }28940.63$
$\displaystyle \therefore \text{The total amount to be paid by John at the end of }4\text{ years}$
$\displaystyle =\text{Rs }28940.63$
$\displaystyle \\$

$\displaystyle \textbf{Question 26. }\text{A sum of money is lent out at compound interest for}$
$\displaystyle 2\text{ years at }20\%\text{ per annum, compound interest being reckoned yearly.}$
$\displaystyle \text{If the same sum of money was lent out at compound interest at the}$
$\displaystyle \text{same rate per cent per annum, compound interest being reckoned}$
$\displaystyle \text{half-yearly, it would have fetched Rs }482\text{ more by way of interest}$
$\displaystyle \text{in two years. Calculate the sum of money lent out.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the sum of money lent out be Rs }x$
$\displaystyle \text{In the 1st case :}$
$\displaystyle A_1=\text{Rs }x\left(1+\frac{20}{100}\right)^2=\frac{36x}{25}$
$\displaystyle \text{In the 2nd case :}$
$\displaystyle A_2=\text{Rs }x\left(1+\frac{20}{2\times100}\right)^{2\times2}$
$\displaystyle =\text{Rs }\frac{14641x}{10000}$
$\displaystyle \text{Given, C.I. in the 2nd case is Rs }482\text{ more than the C.I. in the}$
$\displaystyle \text{1st case.}$
$\displaystyle \therefore \text{For the same principal, amount in the 2nd case is Rs }482$
$\displaystyle \text{more than the amount in the 1st case.}$
$\displaystyle \therefore \frac{14641x}{10000}-\frac{36x}{25}=482$
$\displaystyle \therefore x=20000$
$\displaystyle \therefore \text{The sum of money lent out}=\text{Rs }20000$
$\displaystyle \\$

$\displaystyle \textbf{Question 27. }\text{A sum of Rs }6400\text{ earns a compound interest of}$
$\displaystyle \text{Rs }1008.80\text{ in }18\text{ months, when the interest is reckoned}$
$\displaystyle \text{half-yearly. Find the rate of interest.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : }P=\text{Rs }6400,\ \text{C.I.}=\text{Rs }1008.80\text{ and time}$
$\displaystyle =18\text{ months}=\frac{3}{2}\text{ years}$
$\displaystyle \therefore \text{Amount }(A)=P+\text{C.I.}=\text{Rs }6400+\text{Rs }1008.80$
$\displaystyle =\text{Rs }7408.80$
$\displaystyle \therefore 7408.80=6400\left(1+\frac{r}{2\times100}\right)^{\frac{3}{2}\times2}$
$\displaystyle \therefore \left(1+\frac{r}{200}\right)^3=\frac{7408.80}{6400}=\left(\frac{21}{20}\right)^3$
$\displaystyle \text{On simplifying, we get }r=10\%$
$\displaystyle \\$

$\displaystyle \textbf{Question 28. }\text{Calculate the amount when a sum of Rs }4800\text{ is}$
$\displaystyle \text{invested at }8\%\text{ per annum for }4\text{ years, the C.I. being}$
$\displaystyle \text{compounded half-yearly.}$
$\displaystyle \text{Do not use mathematical tables, use the necessary information from}$
$\displaystyle \text{the following :}$
$\displaystyle (1.08)^4=1.3605\qquad (1.04)^8=1.3686$
$\displaystyle (1.08)^8=1.8509\qquad (1.04)^4=1.1699$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }P=\text{Rs }4800,\ r=8\%\text{ compounded half-yearly and}$
$\displaystyle n=4\text{ years}$
$\displaystyle \therefore A=P\left(1+\frac{r}{2\times100}\right)^{n\times2}$
$\displaystyle =\text{Rs }4800\left(1+\frac{8}{2\times100}\right)^{4\times2}$
$\displaystyle =\text{Rs }4800(1.04)^8$
$\displaystyle =\text{Rs }4800\times1.3686$
$\displaystyle =\text{Rs }6569.28$
$\displaystyle \\$

$\displaystyle \textbf{Question 29. }\text{The total number of industries in a particular portion}$
$\displaystyle \text{of the country is approximately }1600.\text{ If the government has decided}$
$\displaystyle \text{to increase the number of industries in the area by }20\%\text{ every year;}$
$\displaystyle \text{find the approximate number of industries after }2\text{ years.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Number of industries after }2\text{ years}$
$\displaystyle =\text{Original number of industries}\left(1+\frac{r}{100}\right)^n$
$\displaystyle =1600\left(1+\frac{20}{100}\right)^2$
$\displaystyle =2304$
$\displaystyle \\$

$\displaystyle \textbf{Question 30. }\text{The cost of a machine depreciates by }10\%\text{ every year.}$
$\displaystyle \text{If its present worth is Rs }18000,\text{ what will be its value after}$
$\displaystyle \text{three years?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Applying the formula, we get :}$
$\displaystyle \text{Value after }3\text{ years}=\text{Rs }18000\left(1-\frac{10}{100}\right)^3$
$\displaystyle =\text{Rs }13122$
$\displaystyle \\$

$\displaystyle \textbf{Question 31. }\text{A machine depreciates every year at the rate of }20\%$
$\displaystyle \text{of its value at the beginning of the year. The machine was purchased}$
$\displaystyle \text{for Rs }250000\text{ when new, and the scrap value realised when sold was}$
$\displaystyle \text{Rs }128000.\text{ Find the number of years that the machine was used.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required number of years}=n\text{ years}$
$\displaystyle \therefore \text{Value of machine after }n\text{ years}$
$\displaystyle =\text{Its value when new}\left(1-\frac{r}{100}\right)^n$
$\displaystyle \therefore \text{Rs }128000=\text{Rs }250000\left(1-\frac{20}{100}\right)^n$
$\displaystyle \therefore \frac{128000}{250000}=\left(\frac{80}{100}\right)^n=\left(\frac{4}{5}\right)^n$
$\displaystyle \text{On simplifying, we get :}$
$\displaystyle \left(\frac{4}{5}\right)^3=\left(\frac{4}{5}\right)^n$
$\displaystyle \therefore n=3\text{ years}$
$\displaystyle \\$

$\displaystyle \textbf{Question 32. }\text{The population of a town in China increases by }20\%$
$\displaystyle \text{every year. If its present population is }216000,\text{ find :}$
$\displaystyle \text{(i) its population after }2\text{ years,}$
$\displaystyle \text{(ii) its population }2\text{ years ago.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Population after }n\text{ years}=\text{Present population}\left(1+\frac{r}{100}\right)^n$
$\displaystyle \therefore \text{Population after }2\text{ years}$
$\displaystyle =216000\left(1+\frac{20}{100}\right)^2=311040$
$\displaystyle \text{(ii) Present population}$
$\displaystyle =\text{Population }n\text{ years ago}\left(1+\frac{r}{100}\right)^n$
$\displaystyle \therefore 216000=\text{Population }2\text{ years ago}\left(1+\frac{20}{100}\right)^2$
$\displaystyle \therefore \text{Population }2\text{ years ago}=150000$
$\displaystyle \\$

$\displaystyle \textbf{Question 33. }\text{A certain sum of money, lent out at compound interest,}$
$\displaystyle \text{amounts to Rs }14520\text{ in }2\text{ years and to Rs }17569.20\text{ in }4\text{ years.}$
$\displaystyle \text{Find the rate of interest per annum and the sum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Amount in 2 years}=\text{Rs }14520$
$\displaystyle \therefore P\left(1+\frac{r}{100}\right)^2=14520 \qquad \text{...(I)}$
$\displaystyle \text{Amount in 4 years}=\text{Rs }17569.20$
$\displaystyle \therefore P\left(1+\frac{r}{100}\right)^4=17569.20 \qquad \text{...(II)}$
$\displaystyle \text{On dividing (II) by (I), we get :}$
$\displaystyle \frac{P\left(1+\frac{r}{100}\right)^4}{P\left(1+\frac{r}{100}\right)^2}=\frac{17569.20}{14520}$
$\displaystyle \therefore \left(1+\frac{r}{100}\right)^2=\frac{121}{100}$
$\displaystyle \therefore 1+\frac{r}{100}=\frac{11}{10}$
$\displaystyle \therefore r=10$
$\displaystyle \text{Now, }P\left(1+\frac{r}{100}\right)^2=14520$
$\displaystyle \therefore P\left(1+\frac{10}{100}\right)^2=14520$
$\displaystyle \therefore P\times\frac{121}{100}=14520$
$\displaystyle \therefore P=14520\times\frac{100}{121}=12000$
$\displaystyle \therefore \text{Rate of interest}=10\%\text{ and sum}=\text{Rs }12000$
$\displaystyle \text{Alternative method :}$
$\displaystyle \text{For the last 2 years :}$
$\displaystyle P=\text{Rs }14520,\ A=\text{Rs }17569.20\text{ and }n=2\text{ years}$
$\displaystyle \therefore 17569.20=14520\left(1+\frac{r}{100}\right)^2$
$\displaystyle \therefore \frac{17569.20}{14520}=\left(1+\frac{r}{100}\right)^2$
$\displaystyle \text{On solving, we get : }r=10\%$
$\displaystyle \text{For the 1st 2 years :}$
$\displaystyle A=\text{Rs }14520,\ P=?\text{ and }r=10\%\text{ and }n=2\text{ years}$
$\displaystyle \therefore 14520=P\left(1+\frac{10}{100}\right)^2$
$\displaystyle \therefore P=\text{Rs }12000$
$\displaystyle \therefore \text{Rate of interest}=10\%\text{ and sum}=\text{Rs }12000$
$\displaystyle \\$

$\displaystyle \textbf{Question 34. }\text{The difference between the compound interest and the}$
$\displaystyle \text{simple interest on Rs }9500\text{ for }2\text{ years is Rs }95\text{ at the same}$
$\displaystyle \text{rate of interest per year. Find the rate of interest.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the rate of interest be }r\%$
$\displaystyle \therefore \text{S.I. in 2 years}=\frac{9500\times r\times2}{100}=190r$
$\displaystyle \text{And, C.I. in 2 years}=A-P$
$\displaystyle =9500\left(1+\frac{r}{100}\right)^2-9500$
$\displaystyle \text{Given : C.I.}-\text{S.I.}=\text{Rs }95$
$\displaystyle \therefore \left[9500\left(1+\frac{r}{100}\right)^2-9500\right]-190r=95$
$\displaystyle \therefore 9500\left(1+\frac{r^2}{10000}+\frac{2r}{100}\right)-9500-190r=95$
$\displaystyle \therefore 9500+\frac{95r^2}{100}+190r-9500-190r=95$
$\displaystyle \therefore \frac{95r^2}{100}=95$
$\displaystyle \therefore r^2=95\times\frac{100}{95}=100$
$\displaystyle \therefore r=10$
$\displaystyle \therefore \text{Rate of interest}=10\%$
$\displaystyle \\$

$\displaystyle \textbf{Question 35. }\text{A sum of money lent out at C.I. at a certain rate}$
$\displaystyle \text{per annum doubles itself in }5\text{ years. Find in how many years}$
$\displaystyle \text{will the money become eight times of itself at the same rate of}$
$\displaystyle \text{interest p.a.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let Principal}=\text{Rs }x$
$\displaystyle \therefore \text{Amount in }5\text{ years}=\text{Rs }2x$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle \therefore 2x=x\left(1+\frac{r}{100}\right)^5$
$\displaystyle \therefore 2=\left(1+\frac{r}{100}\right)^5 \qquad \text{...(I)}$
$\displaystyle \text{For the second part : }P=\text{Rs }x\text{ and }A=\text{Rs }8x$
$\displaystyle \therefore A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle \therefore 8x=x\left(1+\frac{r}{100}\right)^n$
$\displaystyle \therefore 8=\left(1+\frac{r}{100}\right)^n$
$\displaystyle \textit{i.e.}\ (2)^3=\left(1+\frac{r}{100}\right)^n$
$\displaystyle \therefore \left[\left(1+\frac{r}{100}\right)^5\right]^3=\left(1+\frac{r}{100}\right)^n$
$\displaystyle \therefore \left(1+\frac{r}{100}\right)^{15}=\left(1+\frac{r}{100}\right)^n$
$\displaystyle \therefore n=15$
$\displaystyle \therefore \text{Time required}=15\text{ years}$
$\displaystyle \\$

$\displaystyle \textbf{Question 36. }\text{A man borrowed a sum of money and agrees to pay it}$
$\displaystyle \text{off by paying Rs }43200\text{ at the end of the first year and Rs }34992$
$\displaystyle \text{at the end of the second year. If the rate of compound interest is}$
$\displaystyle 8\%\text{ per annum, find the sum borrowed.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the payment of Rs }43200\text{ at the end of the first year :}$
$\displaystyle A=\text{Rs }43200,\ n=1\text{ year and }r=8\%.\text{ To find }P.$
$\displaystyle A=P\left(1+\frac{r}{100}\right)^n$
$\displaystyle \therefore 43200=P\left(1+\frac{8}{100}\right)^1$
$\displaystyle \therefore P=43200\times\frac{100}{108}=\text{Rs }40000$
$\displaystyle \text{For the payment of Rs }34992\text{ at the end of the second year :}$
$\displaystyle 34992=P\left(1+\frac{8}{100}\right)^2$
$\displaystyle \therefore P=34992\times\left(\frac{100}{108}\right)^2=\text{Rs }30000$
$\displaystyle \therefore \text{Sum borrowed}=\text{Rs }40000+\text{Rs }30000=\text{Rs }70000$
$\displaystyle \\$

$\displaystyle \textbf{Question 37. }\text{A, B and C are three persons with ages }26\text{ years,}$
$\displaystyle 27\text{ years and }28\text{ years respectively. In what ratio must they invest}$
$\displaystyle \text{money at }10\%\text{ p.a. compounded yearly so that each gets the same sum}$
$\displaystyle \text{at the age of his retirement }(\textit{i.e.}\text{ at the age of }58\text{ years).}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the investments made be Rs }x,\text{ Rs }y\text{ and Rs }z\text{ respectively. Then :}$
$\displaystyle x\left(1+\frac{10}{100}\right)^{58-26}=y\left(1+\frac{10}{100}\right)^{58-27}$
$\displaystyle =z\left(1+\frac{10}{100}\right)^{58-28}$
$\displaystyle \therefore x\left(\frac{11}{10}\right)^{32}=y\left(\frac{11}{10}\right)^{31}=z\left(\frac{11}{10}\right)^{30}$
$\displaystyle \text{Dividing each term by }\left(\frac{11}{10}\right)^{30}$
$\displaystyle \therefore x\left(\frac{11}{10}\right)^2=y\left(\frac{11}{10}\right)=z$
$\displaystyle \therefore \frac{121}{100}x=\frac{11}{10}y=z$
$\displaystyle \therefore 121x=110y=100z=k\text{ (let)}$
$\displaystyle \therefore x=\frac{k}{121},\ y=\frac{k}{110}\text{ and }z=\frac{k}{100}$
$\displaystyle \therefore x:y:z=\frac{k}{121}:\frac{k}{110}:\frac{k}{100}$
$\displaystyle =\frac{k}{121}\times12100:\frac{k}{110}\times12100:\frac{k}{100}\times12100$
$\displaystyle \text{[L.C.M. of }121,\ 110\text{ and }100=12100\text{]}$
$\displaystyle =100k:110k:121k$
$\displaystyle =100:110:121$
$\displaystyle \\$