\displaystyle \textbf{Question 1: }\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 2: }\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 3: }\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 \\

\displaystyle \textbf{Question 4: }\text{Find the amount and compound interest on Rs. }2500
\displaystyle \text{for }3\text{ years at }6\%\text{ per annum, compounded annually.}
\displaystyle \text{Answer:}
\displaystyle \text{Principal for the first year}=\text{Rs. }2500.
\displaystyle \text{Interest for the first year} =\frac{6}{100}\times2500=\text{Rs. }150.
\displaystyle \text{Amount at the end of the first year} =2500+150=\text{Rs. }2650.
\displaystyle \text{This amount becomes the principal for the second year}.
\displaystyle \text{Interest for the second year} =\frac{6}{100}\times2650=\text{Rs. }159.
\displaystyle \text{Amount at the end of the second year} =2650+159=\text{Rs. }2809.
\displaystyle \text{This amount becomes the principal for the third year}.
\displaystyle \text{Interest for the third year} =\frac{6}{100}\times2809=\text{Rs. }168.54.
\displaystyle \text{Amount at the end of the third year} =2809+168.54=\text{Rs. }2977.54.
\displaystyle \text{Compound interest} =2977.54-2500=\text{Rs. }477.54.
\displaystyle \therefore \text{Amount}=\text{Rs. }2977.54,\qquad \text{C.I.}=\text{Rs. }477.54.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the amount and compound interest on Rs. }10000
\displaystyle \text{for }3\text{ years at }10\%\text{ per annum, compounded annually.}
\displaystyle \text{Answer:}
\displaystyle \text{Principal for the first year}=\text{Rs. }10000.
\displaystyle \text{Interest for the first year} =\frac{10}{100}\times10000=\text{Rs. }1000.
\displaystyle \text{Amount at the end of the first year} =10000+1000=\text{Rs. }11000.
\displaystyle \text{This amount becomes the principal for the second year}.
\displaystyle \text{Interest for the second year} =\frac{10}{100}\times11000=\text{Rs. }1100.
\displaystyle \text{Amount at the end of the second year} =11000+1100=\text{Rs. }12100.
\displaystyle \text{This amount becomes the principal for the third year}.
\displaystyle \text{Interest for the third year} =\frac{10}{100}\times12100=\text{Rs. }1210.
\displaystyle \text{Amount at the end of the third year} =12100+1210=\text{Rs. }13310.
\displaystyle \text{Compound interest} =13310-10000=\text{Rs. }3310.
\displaystyle \therefore \text{Amount}=\text{Rs. }13310,\qquad \text{C.I.}=\text{Rs. }3310.
\displaystyle \\

\displaystyle \textbf{Question 6: }A\text{ took a loan of Rs. }25000\text{ from a bank at }12\%
\displaystyle \text{per annum, compounded annually. How much will he have to pay at the end}
\displaystyle \text{of }3\text{ years?}
\displaystyle \text{Answer:}
\displaystyle \text{Principal for the first year}=\text{Rs. }25000.
\displaystyle \text{Interest for the first year} =\frac{12}{100}\times25000=\text{Rs. }3000.
\displaystyle \text{Amount at the end of the first year} =25000+3000=\text{Rs. }28000.
\displaystyle \text{This amount becomes the principal for the second year}.
\displaystyle \text{Interest for the second year} =\frac{12}{100}\times28000=\text{Rs. }3360.
\displaystyle \text{Amount at the end of the second year} =28000+3360=\text{Rs. }31360.
\displaystyle \text{This amount becomes the principal for the third year}.
\displaystyle \text{Interest for the third year} =\frac{12}{100}\times31360=\text{Rs. }3763.20.
\displaystyle \text{Amount at the end of the third year} =31360+3763.20=\text{Rs. }35123.20.
\displaystyle \text{Compound interest} =35123.20-25000=\text{Rs. }10123.20.
\displaystyle \therefore A\text{ will have to pay Rs. }35123.20\text{ after }3\text{ years}.
\displaystyle \\

\displaystyle \textbf{Question 7: }A\text{ deposited Rs. }15625\text{ in a bank at }8\%\text{ per annum,}
\displaystyle \text{compounded annually. What amount will he get after }3\text{ years?}
\displaystyle \text{Answer:}
\displaystyle \text{Principal for the first year}=\text{Rs. }15625.
\displaystyle \text{Interest for the first year} =\frac{8}{100}\times15625=\text{Rs. }1250.
\displaystyle \text{Amount at the end of the first year} =15625+1250=\text{Rs. }16875.
\displaystyle \text{This amount becomes the principal for the second year}.
\displaystyle \text{Interest for the second year} =\frac{8}{100}\times16875=\text{Rs. }1350.
\displaystyle \text{Amount at the end of the second year} =16875+1350=\text{Rs. }18225.
\displaystyle \text{This amount becomes the principal for the third year}.
\displaystyle \text{Interest for the third year} =\frac{8}{100}\times18225=\text{Rs. }1458.
\displaystyle \text{Amount at the end of the third year} =18225+1458=\text{Rs. }19683.
\displaystyle \therefore A\text{ will receive Rs. }19683\text{ after }3\text{ years}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A person lent Rs. }16000\text{ at simple interest and the}
\displaystyle \text{same sum at compound interest for }2\text{ years at }12.5\%\text{ per annum.}
\displaystyle \text{Find the ratio of the interests received by him after }2\text{ years.}
\displaystyle \text{Answer:}
\displaystyle \textbf{Simple Interest}
\displaystyle P=\text{Rs. }16000,\qquad R=12.5\%,\qquad T=2\text{ years}.
\displaystyle \text{S.I.}=\frac{P\times R\times T}{100}.
\displaystyle =\frac{16000\times12.5\times2}{100}=\text{Rs. }4000.

\displaystyle \textbf{Compound Interest}
\displaystyle \text{Principal for the first year}=\text{Rs. }16000.
\displaystyle \text{Interest for the first year} =\frac{12.5}{100}\times16000=\text{Rs. }2000.
\displaystyle \text{Amount at the end of the first year} =16000+2000=\text{Rs. }18000.
\displaystyle \text{Interest for the second year} =\frac{12.5}{100}\times18000=\text{Rs. }2250.
\displaystyle \text{C.I.}=2000+2250=\text{Rs. }4250.
\displaystyle \text{Required ratio}=\text{S.I.}:\text{C.I.}=4000:4250=16:17.
\displaystyle \therefore \text{the required ratio is }16:17.
\displaystyle \\

\displaystyle \textbf{Direct Formula for Compound Interest}
\displaystyle \text{When the rate of interest remains the same and interest is compounded}
\displaystyle \text{annually, the amount after }n\text{ years is given by}
\displaystyle A=P\left(1+\frac{R}{100}\right)^n,
\displaystyle \text{where }P=\text{Principal},\quad R=\text{Rate per annum},
\displaystyle n=\text{Number of years},\quad A=\text{Amount}.
\displaystyle \text{Compound interest is the difference between the amount and principal}.
\displaystyle \therefore \text{C.I.}=A-P.
\displaystyle \text{If the rates are different in successive years, the corresponding}
\displaystyle \text{growth factors are applied separately for each year}.
\displaystyle \text{For rates }R_1,R_2,\ldots,R_n,\text{ in successive years,}
\displaystyle A=P\left(1+\frac{R_1}{100}\right) \left(1+\frac{R_2}{100}\right)\cdots \left(1+\frac{R_n}{100}\right).

\displaystyle \text{For Question 1:}
\displaystyle A=5000\left(1+\frac{8}{100}\right)^2 =\text{Rs. }5832.

\displaystyle \text{For Question 2:}
\displaystyle A=8000\left(1+\frac{6}{100}\right)^2 =\text{Rs. }8988.80.

\displaystyle \text{For Question 3, the rates are different in the two years}.
\displaystyle A=2500\left(1+\frac{6}{100}\right) \left(1+\frac{8}{100}\right)=\text{Rs. }2862.

\displaystyle \text{For Question 4:}
\displaystyle A=2500\left(1+\frac{6}{100}\right)^3 =\text{Rs. }2977.54.

\displaystyle \text{For Question 5:}
\displaystyle A=10000\left(1+\frac{10}{100}\right)^3 =\text{Rs. }13310.

\displaystyle \text{For Question 6:}
\displaystyle A=25000\left(1+\frac{12}{100}\right)^3 =\text{Rs. }35123.20.

\displaystyle \text{For Question 7:}
\displaystyle A=15625\left(1+\frac{8}{100}\right)^3 =\text{Rs. }19683.

\displaystyle \text{For Question 8:}
\displaystyle A=16000\left(1+\frac{12.5}{100}\right)^2 =\text{Rs. }20250.
\displaystyle \text{C.I.}=20250-16000=\text{Rs. }4250.
\displaystyle \\

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