\displaystyle \textbf{Question 1: }\text{Find the simple interest and amount on:}
\displaystyle \text{(i) Rs. }4500\text{ for }2\frac{1}{2}\text{ years at }7\frac{2}{3}\%\text{ per annum.}
\displaystyle \text{(ii) Rs. }6360\text{ for }6\text{ years }3\text{ months at }8\%\text{ per annum.}
\displaystyle \text{(iii) Rs. }19200\text{ for }11\text{ months at }9\frac{3}{4}\%\text{ per annum.}
\displaystyle \text{(iv) Rs. }58400\text{ for }75\text{ days at }6\frac{1}{2}\%\text{ per annum.}
\displaystyle \text{Answer:}
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.

\displaystyle \text{(i) }P=\text{Rs. }4500,\quad R=7\frac{2}{3}\%=\frac{23}{3}\%.
\displaystyle T=2\frac{1}{2}\text{ years}=\frac{5}{2}\text{ years}.
\displaystyle \text{S.I.} =\frac{4500\times\frac{23}{3}\times\frac{5}{2}}{100}.
\displaystyle =\text{Rs. }862.50.
\displaystyle \text{Amount}=P+\text{S.I.}=4500+862.50=\text{Rs. }5362.50.

\displaystyle \text{(ii) }P=\text{Rs. }6360,\quad R=8\%.
\displaystyle T=6\text{ years }3\text{ months} =6+\frac{3}{12}=6\frac{1}{4}=\frac{25}{4}\text{ years}.
\displaystyle \text{S.I.} =\frac{6360\times8\times\frac{25}{4}}{100}.
\displaystyle =\text{Rs. }3180.
\displaystyle \text{Amount}=P+\text{S.I.}=6360+3180=\text{Rs. }9540.

\displaystyle \text{(iii) }P=\text{Rs. }19200,\quad R=9\frac{3}{4}\%=\frac{39}{4}\%.
\displaystyle T=11\text{ months}=\frac{11}{12}\text{ years}.
\displaystyle \text{S.I.} =\frac{19200\times\frac{39}{4}\times\frac{11}{12}}{100}.
\displaystyle =\text{Rs. }1716.
\displaystyle \text{Amount}=P+\text{S.I.}=19200+1716=\text{Rs. }20916.

\displaystyle \text{(iv) }P=\text{Rs. }58400,\quad R=6\frac{1}{2}\%=\frac{13}{2}\%.
\displaystyle T=75\text{ days}=\frac{75}{365}\text{ years}.
\displaystyle \text{S.I.} =\frac{58400\times\frac{13}{2}\times\frac{75}{365}}{100}.
\displaystyle =\text{Rs. }780.
\displaystyle \text{Amount}=P+\text{S.I.}=58400+780=\text{Rs. }59180.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the simple interest on Rs. }8600\text{ from}
\displaystyle \text{18th October, 2006 to 13th March, 2007 at }8\%\text{ per annum. Also find}
\displaystyle \text{the amount.}
\displaystyle \text{Answer:}
\displaystyle P=\text{Rs. }8600,\qquad R=8\%.
\displaystyle \text{The day on which the money is borrowed is not counted, while the day}
\displaystyle \text{on which it is returned is counted.}
\displaystyle \text{Number of days} =13+30+31+31+28+13=146\text{ days}.
\displaystyle T=\frac{146}{365}\text{ years}=\frac{2}{5}\text{ year}.
\displaystyle \text{S.I.} =\frac{P\times R\times T}{100}.
\displaystyle =\frac{8600\times8\times\frac{2}{5}}{100}.
\displaystyle =\text{Rs. }275.20.
\displaystyle \text{Amount}=P+\text{S.I.}=8600+275.20=\text{Rs. }8875.20.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Ashish lent Rs. }10500\text{ to Mark at }7\%\text{ per annum}
\displaystyle \text{simple interest. After }5\text{ years, Mark discharged the debt by giving a}
\displaystyle \text{watch and Rs. }3000\text{ in cash. What is the value of the watch?}
\displaystyle \text{Answer:}
\displaystyle P=\text{Rs. }10500,\qquad R=7\%,\qquad T=5\text{ years}.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle =\frac{10500\times7\times5}{100}=\text{Rs. }3675.
\displaystyle \text{Amount}=P+\text{S.I.}=10500+3675=\text{Rs. }14175.
\displaystyle \text{Value of the watch} =\text{Amount}-\text{Cash paid}.
\displaystyle =14175-3000=\text{Rs. }11175.
\displaystyle \therefore \text{the value of the watch is Rs. }11175.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In what time will the simple interest on Rs. }7560\text{ be}
\displaystyle \text{Rs. }1102.50\text{ at }6\frac{1}{4}\%\text{ per annum?}
\displaystyle \text{Answer:}
\displaystyle P=\text{Rs. }7560,\qquad \text{S.I.}=\text{Rs. }1102.50.
\displaystyle R=6\frac{1}{4}\%=\frac{25}{4}\%.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle T=\frac{\text{S.I.}\times100}{P\times R}.
\displaystyle =\frac{1102.50\times100}{7560\times\frac{25}{4}} =\frac{7}{3}\text{ years}.
\displaystyle \frac{7}{3}\text{ years} =2\frac{1}{3}\text{ years}=2\text{ years }4\text{ months}.
\displaystyle \therefore \text{the required time is }2\text{ years }4\text{ months}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In how much time will Rs. }25600\text{ amount to Rs. }35664
\displaystyle \text{when money is worth }9\frac{1}{4}\%\text{ per annum simple interest?}
\displaystyle \text{Answer:}
\displaystyle P=\text{Rs. }25600,\qquad A=\text{Rs. }35664.
\displaystyle \text{S.I.}=A-P=35664-25600=\text{Rs. }10064.
\displaystyle R=9\frac{1}{4}\%=\frac{37}{4}\%.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle T=\frac{\text{S.I.}\times100}{P\times R}.
\displaystyle =\frac{10064\times100}{25600\times\frac{37}{4}} =\frac{17}{4}\text{ years}.
\displaystyle \frac{17}{4}\text{ years} =4\frac{1}{4}\text{ years}=4\text{ years }3\text{ months}.
\displaystyle \therefore \text{the required time is }4\text{ years }3\text{ months}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{At what rate per cent per annum will Rs. }1625\text{ amount}
\displaystyle \text{to Rs. }2080\text{ in }3\frac{1}{2}\text{ years at simple interest?}
\displaystyle \text{Answer:}
\displaystyle P=\text{Rs. }1625,\qquad A=\text{Rs. }2080.
\displaystyle T=3\frac{1}{2}\text{ years}=\frac{7}{2}\text{ years}.
\displaystyle \text{S.I.}=A-P=2080-1625=\text{Rs. }455.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle R=\frac{\text{S.I.}\times100}{P\times T}.
\displaystyle =\frac{455\times100}{1625\times\frac{7}{2}}=8\%.
\displaystyle \therefore \text{the required rate of interest is }8\%\text{ per annum}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{At what rate per cent per annum will the simple interest}
\displaystyle \text{on Rs. }6720\text{ be Rs. }1911\text{ in }3\text{ years }3\text{ months?}
\displaystyle \text{Answer:}
\displaystyle P=\text{Rs. }6720,\qquad \text{S.I.}=\text{Rs. }1911.
\displaystyle T=3\text{ years }3\text{ months} =3\frac{1}{4}=\frac{13}{4}\text{ years}.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle R=\frac{\text{S.I.}\times100}{P\times T}.
\displaystyle =\frac{1911\times100}{6720\times\frac{13}{4}} =8.75\%=8\frac{3}{4}\%.
\displaystyle \therefore \text{the required rate is }8\frac{3}{4}\%\text{ per annum}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{At what rate per cent per annum simple interest will a}
\displaystyle \text{sum of money double itself in }12\text{ years?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the Principal be Rs. }x.
\displaystyle \text{Since the sum doubles itself, Amount}=2x.
\displaystyle \therefore \text{S.I.}=2x-x=x.
\displaystyle T=12\text{ years}.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle R=\frac{\text{S.I.}\times100}{P\times T}.
\displaystyle =\frac{x\times100}{x\times12} =\frac{25}{3}\%=8\frac{1}{3}\%.
\displaystyle \therefore \text{the required rate is }8\frac{1}{3}\%\text{ per annum}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The simple interest on a sum is }\frac{9}{16}\text{ of the sum.}
\displaystyle \text{Find the rate per cent per annum and the time in years, if both are}
\displaystyle \text{numerically equal.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the Principal be Rs. }x.
\displaystyle \therefore \text{S.I.}=\frac{9}{16}x.
\displaystyle \text{Since the rate and time are numerically equal, let }R=T=y.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle \frac{9}{16}x=\frac{x\times y\times y}{100}.
\displaystyle \frac{9}{16}=\frac{y^2}{100}.
\displaystyle y^2=\frac{9}{16}\times100=\frac{225}{4}.
\displaystyle y=\sqrt{\frac{225}{4}}=\frac{15}{2}=7\frac{1}{2}.
\displaystyle \therefore R=7\frac{1}{2}\%\text{ per annum and }T=7\frac{1}{2}\text{ years}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{What sum will yield Rs. }406\text{ as simple interest in}
\displaystyle 1\text{ year }2\text{ months at }6\frac{1}{4}\%\text{ per annum?}
\displaystyle \text{Answer:}
\displaystyle \text{S.I.}=\text{Rs. }406.
\displaystyle T=1\text{ year }2\text{ months} =1+\frac{2}{12}=\frac{7}{6}\text{ years}.
\displaystyle R=6\frac{1}{4}\%=\frac{25}{4}\%.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle P=\frac{\text{S.I.}\times100}{R\times T}.
\displaystyle =\frac{406\times100}{\frac{25}{4}\times\frac{7}{6}} =\text{Rs. }5568.
\displaystyle \therefore \text{the required sum is Rs. }5568.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{What sum will amount to Rs. }1748\text{ in}
\displaystyle 2\frac{1}{2}\text{ years at }7\frac{1}{2}\%\text{ per annum simple interest?}
\displaystyle \text{Answer:}
\displaystyle A=\text{Rs. }1748,\qquad R=7\frac{1}{2}\%=\frac{15}{2}\%.
\displaystyle T=2\frac{1}{2}\text{ years}=\frac{5}{2}\text{ years}.
\displaystyle \text{Let the Principal be Rs. }P.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle =\frac{P\times\frac{15}{2}\times\frac{5}{2}}{100} =\frac{3P}{16}.
\displaystyle A=P+\text{S.I.}
\displaystyle 1748=P+\frac{3P}{16}=\frac{19P}{16}.
\displaystyle P=1748\times\frac{16}{19}=\text{Rs. }1472.
\displaystyle \therefore \text{the required sum is Rs. }1472.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{A sum of money becomes }\frac{8}{5}\text{ of itself in}
\displaystyle 5\text{ years at a certain rate of simple interest. Find the rate of interest.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the Principal be Rs. }x.
\displaystyle \text{Amount}=\text{Rs. }\frac{8}{5}x,\qquad T=5\text{ years}.
\displaystyle \text{S.I.}=\text{Amount}-\text{Principal}.
\displaystyle =\frac{8}{5}x-x=\frac{3}{5}x.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle R=\frac{\text{S.I.}\times100}{P\times T}.
\displaystyle =\frac{\frac{3}{5}x\times100}{x\times5}=12\%.
\displaystyle \therefore \text{the required rate of interest is }12\%\text{ per annum}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{What sum of money lent at }12\frac{1}{2}\%\text{ per annum}
\displaystyle \text{will produce the same interest in }4\text{ years as Rs. }8560\text{ produces}
\displaystyle \text{in }5\text{ years at }12\%\text{ per annum simple interest?}
\displaystyle \text{Answer:}
\displaystyle \text{For the first sum, }P=\text{Rs. }8560,\quad R=12\%,\quad T=5\text{ years}.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle =\frac{8560\times12\times5}{100}=\text{Rs. }5136.

\displaystyle \text{For the second sum, S.I.}=\text{Rs. }5136,\quad T=4\text{ years}.
\displaystyle R=12\frac{1}{2}\%=\frac{25}{2}\%.
\displaystyle P=\frac{\text{S.I.}\times100}{R\times T}.
\displaystyle =\frac{5136\times100}{\frac{25}{2}\times4} =\text{Rs. }10272.
\displaystyle \therefore \text{the required sum is Rs. }10272.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If Rs. }1250\text{ amounts to Rs. }1550\text{ in }3\text{ years}
\displaystyle \text{at simple interest, what will Rs. }3100\text{ amount to in }4\text{ years}
\displaystyle \text{at the same rate of interest?}
\displaystyle \text{Answer:}
\displaystyle \text{For the first sum, }P=\text{Rs. }1250,\quad A=\text{Rs. }1550.
\displaystyle T=3\text{ years}.
\displaystyle \text{S.I.}=A-P=1550-1250=\text{Rs. }300.
\displaystyle R=\frac{\text{S.I.}\times100}{P\times T}.
\displaystyle =\frac{300\times100}{1250\times3}=8\%\text{ per annum}.

\displaystyle \text{For the second sum, }P=\text{Rs. }3100,\quad R=8\%,\quad T=4\text{ years}.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle =\frac{3100\times8\times4}{100}=\text{Rs. }992.
\displaystyle \text{Amount}=P+\text{S.I.}=3100+992=\text{Rs. }4092.
\displaystyle \therefore \text{the required amount is Rs. }4092.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A sum of money lent at simple interest amounts to}
\displaystyle \text{Rs. }3224\text{ in }2\text{ years and Rs. }4160\text{ in }5\text{ years. Find the}
\displaystyle \text{principal and the rate of interest.}
\displaystyle \text{Answer:}
\displaystyle \text{Amount after }2\text{ years}=\text{Rs. }3224.
\displaystyle \text{Amount after }5\text{ years}=\text{Rs. }4160.
\displaystyle \text{Interest for }5-2=3\text{ years}=4160-3224=\text{Rs. }936.
\displaystyle \text{Interest for }1\text{ year}=\frac{936}{3}=\text{Rs. }312.
\displaystyle \text{Interest for }2\text{ years}=312\times2=\text{Rs. }624.
\displaystyle \text{Principal}=3224-624=\text{Rs. }2600.
\displaystyle \text{Now, S.I.}=\text{Rs. }624,\quad P=\text{Rs. }2600,\quad T=2\text{ years}.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle R=\frac{\text{S.I.}\times100}{P\times T}.
\displaystyle =\frac{624\times100}{2600\times2}=12\%.
\displaystyle \therefore \text{the principal is Rs. }2600\text{ and the rate is }12\%\text{ p.a.}
\displaystyle \\

\displaystyle \textbf{Question 16: }A\text{ lends Rs. }2500\text{ to }B\text{ and a certain sum to }C
\displaystyle \text{at the same time at }7\%\text{ per annum simple interest. If after }4\text{ years,}
\displaystyle A\text{ receives altogether Rs. }1120\text{ as interest from }B\text{ and }C,\text{ find}
\displaystyle \text{the sum lent to }C.
\displaystyle \text{Answer:}
\displaystyle \text{For the sum lent to }B,\quad P=\text{Rs. }2500,\quad R=7\%,\quad T=4\text{ years}.
\displaystyle \text{S.I. received from }B =\frac{2500\times7\times4}{100}=\text{Rs. }700.
\displaystyle \text{Total interest received from }B\text{ and }C=\text{Rs. }1120.
\displaystyle \text{Interest received from }C=1120-700=\text{Rs. }420.
\displaystyle \text{Let the sum lent to }C\text{ be Rs. }P.
\displaystyle \text{For }C,\quad \text{S.I.}=\text{Rs. }420,\quad R=7\%,\quad T=4\text{ years}.
\displaystyle P=\frac{\text{S.I.}\times100}{R\times T}.
\displaystyle =\frac{420\times100}{7\times4}=\text{Rs. }1500.
\displaystyle \therefore A\text{ lent Rs. }1500\text{ to }C.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The simple interest on a certain sum for }3\text{ years}
\displaystyle \text{at }8\%\text{ per annum is Rs. }96\text{ more than the simple interest on the}
\displaystyle \text{same sum for }2\text{ years at }9\%\text{ per annum. Find the sum.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the Principal be Rs. }P.
\displaystyle \text{For }R=8\%\text{ and }T=3\text{ years,}
\displaystyle \text{S.I.}_1=\frac{P\times8\times3}{100} =\frac{24P}{100}.
\displaystyle \text{For }R=9\%\text{ and }T=2\text{ years,}
\displaystyle \text{S.I.}_2=\frac{P\times9\times2}{100} =\frac{18P}{100}.
\displaystyle \text{Given, S.I.}_1-\text{S.I.}_2=\text{Rs. }96.
\displaystyle \frac{24P}{100}-\frac{18P}{100}=96.
\displaystyle \frac{6P}{100}=96.
\displaystyle P=\frac{96\times100}{6}=\text{Rs. }1600.
\displaystyle \therefore \text{the required sum is Rs. }1600.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Two equal sums of money were lent at simple interest at}
\displaystyle 11\%\text{ per annum for }3\frac{1}{2}\text{ years and }4\frac{1}{2}\text{ years, respectively.}
\displaystyle \text{If the difference between the interests is Rs. }412.50,\text{ find each sum.}
\displaystyle \text{Answer:}
\displaystyle \text{Let each sum be Rs. }P.
\displaystyle \text{For the first sum, }R=11\%,\quad T=3\frac{1}{2}\text{ years}.
\displaystyle \text{S.I.}_1 =\frac{P\times11\times3\frac{1}{2}}{100}.
\displaystyle \text{For the second sum, }R=11\%,\quad T=4\frac{1}{2}\text{ years}.
\displaystyle \text{S.I.}_2 =\frac{P\times11\times4\frac{1}{2}}{100}.
\displaystyle \text{Given, S.I.}_2-\text{S.I.}_1=\text{Rs. }412.50.
\displaystyle \frac{P\times11}{100} \left(4\frac{1}{2}-3\frac{1}{2}\right)=412.50.
\displaystyle \frac{11P}{100}=412.50.
\displaystyle P=\frac{412.50\times100}{11}=\text{Rs. }3750.
\displaystyle \therefore \text{each sum is Rs. }3750.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Divide Rs. }6000\text{ into two parts so that the simple}
\displaystyle \text{interest on the first part for }9\text{ months at }12\%\text{ per annum is}
\displaystyle \text{equal to the simple interest on the second part for }1\frac{1}{2}\text{ years}
\displaystyle \text{at }10\%\text{ per annum.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first part be Rs. }x.
\displaystyle \text{Then the second part is Rs. }(6000-x).
\displaystyle \text{For the first part, }T=9\text{ months}=\frac{3}{4}\text{ year},\quad R=12\%.
\displaystyle \text{S.I.}_1 =\frac{x\times12\times\frac{3}{4}}{100} =\frac{9x}{100}.
\displaystyle \text{For the second part, }T=1\frac{1}{2}\text{ years}=\frac{3}{2}\text{ years}.
\displaystyle R=10\%.
\displaystyle \text{S.I.}_2 =\frac{(6000-x)\times10\times\frac{3}{2}}{100}.
\displaystyle =\frac{15(6000-x)}{100}.
\displaystyle \text{Since the two simple interests are equal, S.I.}_1=\text{S.I.}_2.
\displaystyle \frac{9x}{100}=\frac{15(6000-x)}{100}.
\displaystyle 9x=15(6000-x).
\displaystyle 9x=90000-15x.
\displaystyle 24x=90000.
\displaystyle x=\frac{90000}{24}=\text{Rs. }3750.
\displaystyle \text{Second part}=6000-3750=\text{Rs. }2250.
\displaystyle \therefore \text{the two parts are Rs. }3750\text{ and Rs. }2250.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Divide Rs. }13500\text{ into two parts such that if one}
\displaystyle \text{part is lent at }8\frac{1}{3}\%\text{ per annum for }2\text{ years }9\text{ months and}
\displaystyle \text{the other at }7\frac{1}{2}\%\text{ per annum for }1\text{ year }8\text{ months, the}
\displaystyle \text{total simple interest received is Rs. }2375.\text{ Find the two parts.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first part be Rs. }x.
\displaystyle \text{Then the second part is Rs. }(13500-x).
\displaystyle \text{For the first part, }R=8\frac{1}{3}\%=\frac{25}{3}\%.
\displaystyle T=2\text{ years }9\text{ months} =2\frac{3}{4}=\frac{11}{4}\text{ years}.
\displaystyle \text{S.I.}_1 =\frac{x\times\frac{25}{3}\times\frac{11}{4}}{100} =\frac{275x}{1200}.
\displaystyle \text{For the second part, }R=7\frac{1}{2}\%=\frac{15}{2}\%.
\displaystyle T=1\text{ year }8\text{ months} =1\frac{2}{3}=\frac{5}{3}\text{ years}.
\displaystyle \text{S.I.}_2 =\frac{(13500-x)\times\frac{15}{2}\times\frac{5}{3}}{100}.
\displaystyle =\frac{(13500-x)}{8}.
\displaystyle \text{Given, S.I.}_1+\text{S.I.}_2=\text{Rs. }2375.
\displaystyle \frac{275x}{1200}+\frac{13500-x}{8}=2375.
\displaystyle 275x+150(13500-x)=2375\times1200.
\displaystyle 275x+2025000-150x=2850000.
\displaystyle 125x=825000.
\displaystyle x=\text{Rs. }6600.
\displaystyle \text{Second part}=13500-6600=\text{Rs. }6900.
\displaystyle \therefore \text{the two parts are Rs. }6600\text{ and Rs. }6900.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{In what time will a sum of money lent at }8\frac{1}{3}\%
\displaystyle \text{per annum simple interest become four times itself?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the Principal be Rs. }P.
\displaystyle \text{Amount}=4P,\qquad R=8\frac{1}{3}\%=\frac{25}{3}\%.
\displaystyle \text{S.I.}=\text{Amount}-\text{Principal}=4P-P=3P.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle 3P=\frac{P\times\frac{25}{3}\times T}{100}.
\displaystyle 3=\frac{25T}{300}.
\displaystyle T=\frac{3\times300}{25}=36\text{ years}.
\displaystyle \therefore \text{the required time is }36\text{ years}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{A certain sum of money lent at }6\frac{2}{3}\%\text{ per annum}
\displaystyle \text{produced the same simple interest in }6\text{ years as Rs. }3200\text{ lent at}
\displaystyle 8\frac{2}{5}\%\text{ per annum for }7\text{ years. Find the sum.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required sum be Rs. }P.
\displaystyle R=6\frac{2}{3}\%=\frac{20}{3}\%,\qquad T=6\text{ years}.
\displaystyle \text{S.I.}_1 =\frac{P\times\frac{20}{3}\times6}{100} =\frac{2P}{5}.
\displaystyle \text{For the second sum, }P=\text{Rs. }3200,\quad T=7\text{ years}.
\displaystyle R=8\frac{2}{5}\%=\frac{42}{5}\%.
\displaystyle \text{S.I.}_2 =\frac{3200\times\frac{42}{5}\times7}{100} =\text{Rs. }1881.60.
\displaystyle \text{Since the simple interests are equal, S.I.}_1=\text{S.I.}_2.
\displaystyle \frac{2P}{5}=1881.60.
\displaystyle P=1881.60\times\frac{5}{2}=\text{Rs. }4704.
\displaystyle \therefore \text{the required sum is Rs. }4704.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Naveen and Praveen borrowed Rs. }42000\text{ and Rs. }55000
\displaystyle \text{respectively for }3\frac{1}{2}\text{ years at the same rate of simple interest.}
\displaystyle \text{If Praveen has to pay Rs. }3640\text{ more interest than Naveen, find the rate.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }R\%\text{ per annum be the rate of interest}.
\displaystyle T=3\frac{1}{2}\text{ years}=\frac{7}{2}\text{ years}.
\displaystyle \text{S.I.}_1 =\frac{42000\times R\times\frac{7}{2}}{100}=1470R.
\displaystyle \text{S.I.}_2 =\frac{55000\times R\times\frac{7}{2}}{100}=1925R.
\displaystyle \text{Given, S.I.}_2-\text{S.I.}_1=\text{Rs. }3640.
\displaystyle 1925R-1470R=3640.
\displaystyle 455R=3640.
\displaystyle R=\frac{3640}{455}=8\%.
\displaystyle \therefore \text{the required rate of interest is }8\%\text{ per annum}.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{A sum of money was invested at simple interest at a}
\displaystyle \text{certain rate for }2\text{ years. If it had been invested at a }3\%\text{ higher rate,}
\displaystyle \text{it would have earned Rs. }720\text{ more as interest. Find the sum.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the Principal be Rs. }P\text{ and the original rate be }R\%\text{ p.a.}
\displaystyle \text{At the original rate, S.I.}_1=\frac{P\times R\times2}{100}.
\displaystyle \text{At the higher rate, S.I.}_2 =\frac{P\times(R+3)\times2}{100}.
\displaystyle \text{Given, S.I.}_2-\text{S.I.}_1=\text{Rs. }720.
\displaystyle \frac{P\times(R+3)\times2}{100} -\frac{P\times R\times2}{100}=720.
\displaystyle \frac{6P}{100}=720.
\displaystyle P=\frac{720\times100}{6}=\text{Rs. }12000.
\displaystyle \therefore \text{the required sum is Rs. }12000.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{A sum of money invested at }6\%\text{ per annum simple}
\displaystyle \text{interest for a certain period yields Rs. }960\text{ as interest. If the sum}
\displaystyle \text{had been invested for }5\text{ years more, it would have yielded Rs. }2160
\displaystyle \text{as interest. Find the sum.}
\displaystyle \text{Answer:}
\displaystyle \text{The additional interest for }5\text{ years}=2160-960=\text{Rs. }1200.
\displaystyle \text{Let the Principal be Rs. }P.
\displaystyle R=6\%,\qquad T=5\text{ years for the additional interest}.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle 1200=\frac{P\times6\times5}{100}.
\displaystyle P=\frac{1200\times100}{6\times5}=\text{Rs. }4000.
\displaystyle \therefore \text{the required sum is Rs. }4000.
\displaystyle \\

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