Question 15 Marks
The difference between the compound interest and the simple interest on a certain sum for $3$ years at $10\%$ per annum is $Rs. 93.$ Find the sum.
Answer$\therefore$ Simple interest $(S.I) =\frac{\text{prt}}{100}$
$=\frac{100\times10\times3}{100}=\text{Rs. }30$ Amount $(A)=\text{P}\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$=\text{Rs. }100\Big(1+\frac{10}{100}\Big)^3$
$=\text{Rs. }100\times\Big(\frac{11}{10}\Big)^3$
$=\text{Rs. }100\times\frac{11}{10}\times\frac{11}{10}\times\frac{11}{10}$
$=\text{Rs. }\frac{1331}{10}$
$\therefore C.I = A - P =\text{Rs. }\frac{1331}{10}-\text{Rs. }100$
$=\text{Rs. }\frac{331}{10}$
Difference between $C.I$ and $S.I =\text{Rs. }\frac{1331}{10}-\text{Rs. }30=\text{Rs. }\frac{31}{10}$
If difference is $=\text{Rs. }\frac{31}{10},$
then sum and if difference is $Rs. 93$
then sum $=\text{Rs. }\frac{100\times93\times10}{31}$
$=\text{Rs. }3000$
View full question & answer→Question 25 Marks
Neeraj lent $Rs. 65536$ for $2$ years at $12\frac{1}{2}\%$ per annum, compounded annually. How much more could he earn if the interest were compounded half-yearly$?$
AnswerAmount of loan $= Rs. 65536$
Rate of interest $(R) =12\frac{1}{2}\%=\frac{25}{2}\%\text{ p.a. or }\frac{25}{4}\%$
half yearly Period $(n) = 2$ year or $4$ half years
$\therefore$ Amount $(A) =\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }65536\Big(1+\frac{25}{4\times100}\Big)^4$
$=\text{Rs. }65536\times\Big(\frac{17}{16}\Big)^4$
$=\text{Rs. }65536\times\frac{17}{16}\times\frac{17}{16}\times\frac{17}{16}\times\frac{17}{16}$
$=\text{Rs. }83521$ If the interest in compounded annually, then Amount $=\text{Rs. }65536\Big(1+\frac{25}{2\times100}\Big)^2$
$=\text{Rs. }65536\times\Big(\frac{9}{8}\Big)^2$
$=\text{Rs. }65536\times\frac{9}{8}\times\frac{9}{8}=\text{Rs. }82944$
$\therefore$ More interest recieved $= Rs. 183521 - 82944 = Rs. 577$
View full question & answer→Question 35 Marks
Find the amount and the compound interest on $Rs. 3000$ for $2$ years at $10\%$ per annum.
AnswerHere, $\text{A}=\text{P}\times\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs}.\ 3000\times\Big(1+\frac{10}{100}\Big)^2$
$=\text{Rs}.\ 3000\times\Big(\frac{110}{100}\Big)^2$
$=\text{Rs}.\ 3000\times\Big(\frac{11}{10}\Big)\times\Big(\frac{11}{10}\Big)$
$=\text{Rs}.(30\times11\times11)$
$=\text{Rs}. 3630$
$\therefore CI = A = P = Rs. (3630 - 3000) = Rs. 630$
Hence, the amount is $Rs. 3630$ and the $CI$ is $Rs. 630.$
View full question & answer→Question 45 Marks
Find the amount and the compound interest on $Rs. 31250$ for $1\frac{1}{2}$ years at $8\%$ per annum, compounded half-yearly.
AnswerPrincipal, $(P) = Rs. 31250$
Annual rate of interest, $\text{R}=\frac{3}{2}\%$
Rate of interest for a half year $=\frac{1}{2}\Big(\frac{3}{2}\Big)\%=\frac{3}{4}\%$
Time, $\text{n}=1\frac{1}{2}$
years $= 3$ half years Then the amount with the compound interest is given by $\text{A}=\text{P }\times \Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$= 31250 \times\Big(1+\frac{\frac{3}{4}}{100}\Big)^3$
$=31250\times\Big(1+\frac{3}{100\times4}\Big)^3$
$=31250\times\Big(\frac{400+3}{400}\Big)^3$
$=31250\times\Big(\frac{403}{400}\Big)^3$
$=31250\times\Big(\frac{403}{400}\Big)\times\Big(\frac{403}{400}\Big)\times\Big(\frac{403}{400}\Big)$
$= Rs. 31250 \times (1.0075 \times 1.0075 \times 1.0075)$
$ = Rs. 31958.41$
Therefore, compound interest $=$ amount $-$ principal $= Rs. (31958.41 - 31250) = Rs. 708.41.$
View full question & answer→Question 55 Marks
Manoj deposited a sum of $Rs. 64000$ in a post office for $3$ years, compounded annually at $7\frac{1}{2}\%$ per annum. What amount will he get on maturity$?$
AnswerPrincipal $(P) = Rs. 64000$
Rate $(r) =7\frac{1}{2}\%=\frac{15}{2}\%$
Period $(t) = 3$ years
Interest for the first year $=\frac{\text{prt}}{100}$
$=\text{Rs.}\ \frac{61000\times15\times1}{100\times2}$
$ =\text{Rs.}\ 4800$
Amount at the end of first year $= Rs. 64000 + Rs. 4800 = Rs. 68800$
Principal to the second year $= Rs. 68800$
Interest for the second year $=\text{Rs.}\ \frac{6880\times15\times1}{100\times2}$
$ =\text{Rs.}\ 5160$
Amount at the end of $2nd$ year $= Rs. 68800 + 5160 = Rs. 73960$
Principal for the third year $=\frac{73960\times15\times1}{100\times2}$
$ =\text{Rs.}\ 5547$
$\therefore$ Amount at the end of $3rd$ year $= Rs. 73960 + Rs. 5547 = Rs. 79507$
View full question & answer→Question 65 Marks
Find the amount and the compound interest on $Rs. 12800$ for 1 year at $7\frac{1}{2}\%$ per annum, compounded semi-annually.
AnswerPrincipal $(P) = Rs. 12800$
Rate $(R) =7\frac{1}{2}\%\text{ p.a }\frac{15}{2}\%$
half yearly Period $(n) = 1$ year or $2$ half years
$\therefore$ Amount $(A) =\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }12800\Big(1+\frac{15}{4\times100}\Big)^2$
$=\text{Rs. }12800\times\frac{83}{80}\times\frac{83}{80} =\text{Rs. } 13778$
$\therefore C.I. = A - P= Rs. 13778 - Rs. 12800$
$= Rs. 978.$
View full question & answer→Question 75 Marks
Swati borrowed $Rs. 40960$ from a bank to buy a piece of land. If the bank charges $12\frac{1}{2}\%$ per annum, compounded half-yearly, what amount will she have to pay after $1\frac{1}{2}$ years? Also, find the interest paid by her.
AnswerPrincipal $(P) = Rs. 40960$
Rate $(R) =12\frac{1}{2}=\frac{25}{2}\%\text{ p.a or }\frac{25}{4}\%$
half yearly Period $(n) =1\frac{1}{2}$ year or $3$ half years
$\therefore$ Amount $(A) =\text{P}\Big(1+\frac{R}{100}\Big)^{\text{n}}$
$=\text{Rs. }40960\Big(1+\frac{25}{4\times100}\Big)^3$
$=\text{Rs. }40960\times\Big(\frac{17}{16}\Big)^3$
$=\text{Rs. }40960\times\frac{17}{16}\times\frac{17}{16}\times\frac{17}{16}$
$=\text{Rs. }49130$
$\therefore C.I$. paid $= A - P = Rs. 49130 - Rs. 40960 = Rs. 8170.$
View full question & answer→Question 85 Marks
Find the difference between the simple interest and the compound interest on $Rs. 5000$ for $2$ years at $9\%$ per annum.
AnswerPrincipal $(p) = Rs. 5000$
Rate $(R) = 9\%$ p.a.
Time $(n) = 2$ years
$\therefore\text{S.I.}=\frac{\text{PRT}}{100}$
$=\frac{5000\times9\times2}{100}$
$=\text{Rs.}\ 900$
New amount at $C.I. =\text{P}\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs.}\ 5000\Big(1+\frac{9}{100}\Big)^\text{2}$
$=\text{Rs.}\ 5000\times\frac{109}{100}\times\frac{109}{100}$
$=\text{Rs.}\ \frac{11881}{2}$
$=\text{Rs.}\ 5940.50$
$\therefore C.I = A - P = Rs. 5940.50 - Rs. 5000 = Rs. 940.50$
$\therefore$ Difference between $C.I.$ and $S.I. = Rs. 940.50 - Rs. 900 = Rs. 40.50$
View full question & answer→Question 95 Marks
Abhay borrowed $Rs. 16000$ at $7\frac{1}{2}\%$ per annum simple interest. On the same day, he lent it to Gurmeet at the same rate but compounded annually. What does he gain at the end of $2$ years$?$
AnswerIn case of Abhay, Principal $(p) = Rs. 16000$
Rate $(r) 7\frac{1}{2}\%=\frac{15}{2}\%\text{p.a}$
Period $(t) = 2$ years
$\therefore$ S.Interest $=\frac{\text{prt}}{100}$
$=\text{Rs. }\frac{16000\times15\times2}{100\times2}=\text{Rs. }2400$
In case of Gurmeet, Principal $(p) = Rs. 16000$
Rate $(r)$ $7\frac{1}{2}\%=\frac{15}{2}\%\text{p.a}$
Period $(t) = 2$ years Amount $(A)$
$=\text{P}\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$=\text{Rs. }16000\times\Big(1+\frac{15}{2\times100}\Big)^2$
$=\text{Rs. }16000\Big(\frac{43}{40}\Big)^2$
$=\text{Rs. }16000\times\frac{43}{40}\times\frac{43}{40}$
$=\text{Rs. }18490$
$C.I = A - P = Rs. 18490 - Rs. 16000 = Rs. 2490$
gain to Abhay $= Rs. 2490 - Rs. 2400 = Rs. 90$
View full question & answer→Question 105 Marks
A scooter is bought for $Rs. 32000.$ Its value depreciates at $10\%$ per annum. What will be its value after $2$ years$?$
AnswerLet the principal amount be $P = Rs. 32000.$
Rate of interest $(R) = 10\%$
Time $(n) = 2$ years
Now, $\text{A}=\text{Rs}. \text{P}\times\Big(1-\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs}.\ 32000\times\Big(1-\frac{10}{100}\Big)^2$
$=\text{Rs}.\ 32000\times\Big(\frac{90}{100}\Big)^2$
$=\text{Rs}.\ 32000\times\Big(\frac{9}{10}\Big)\times\Big(\frac{9}{10}\Big)$
$=\text{Rs}.\ (320\times9\times9)$
$= \text{Rs}. 25920$
$\therefore$ The value of the scooter after $2$ years is $Rs. 25920$
View full question & answer→Question 115 Marks
Michael borrowed $Rs. 16000$ from a finance company at $10\%$ per annum, compounded half-yearly. What amount of money will discharge his debt after $1\frac{1}{2}$ years$?$
AnswerPrincipal $(p) = Rs. 16000$
Rate $(r) = 10\%$ p.a. or $5\%$ half yearly
Period $(t) = 1\frac{1}{2}$ year or $3$ half years
Interest for the first year $=\frac{\text{prt}}{100}$
$=\text{Rs.}\ \frac{16000\times5\times1}{100}$
$=\text{Rs.}\ 800$
Amount at the end of first half year $= Rs. 16000 + Rs. 800 = Rs. 16800$
Principal for the second half year $= Rs. 16800$
Interest for the second half year $=\text{Rs.}\ \frac{16800\times5\times1}{100}$
$=\text{Rs.}\ 840$
Amount at the end of second half year $= Rs. 16800 + Rs. 840 = Rs. 17640$
Interest for the third half year $=\text{Rs.}\ \frac{17640\times5\times1}{100}$
$=\text{Rs.}\ 882$
Amount at the end of $3rd$ half year $= Rs. 17640 + 882 = Rs. 18522$
View full question & answer→Question 125 Marks
Find the amount of $Rs. 10000$ after $2$ years compounded annually; the rate of interest being $10\%$ per annum during the first year and $12\%$ per annum during the second year. Also, find the compound interest.
AnswerHere, $\text{A}=\text{P}\times\Big(1+\frac{\text{P}}{100}\Big)\times\Big(1+\frac{\text{r}}{100}\Big)$
$=\text{Rs.}\ 10000\times\Big(1+\frac{10}{100}\Big)\times\Big(1+\frac{12}{100}\Big)$
$=\text{Rs}.\ 10000\times\Big(\frac{110}{100}\Big)\times\Big(\frac{112}{100}\Big)$
$=\text{Rs}.\ 10000\times\Big(\frac{11}{10}\Big)\times\Big(\frac{28}{25}\Big)$
$=\text{Rs}.(40\times11\times28)$
$=\text{Rs}.\ 12320$
$\therefore CI = A - P = Rs. (12320 - 10000) = Rs. 2320$
Hence, the amount is $Rs. 12320$ and the $CI$ is $Rs. 2320$
View full question & answer→Question 135 Marks
Ratna obtained a loan of $Rs. 25000$ from the Syndicate Bank to renovate her house. If the rate of interest is 8% per annum, what amount will she have to pay to the bank after $2$ years to discharge her debt$?$
AnswerAmount of loan $(p) = Rs. 25000$
Rate of interest $(r) = 8\%$ p.a.
Period $(t) = 2$ years
Interest for the first year $=\frac{\text{prt}}{100}$
$=\text{Rs.}\ \frac{25000\times8\times1}{100}$
$=\text{Rs.}\ 2000$
$\therefore$ Amount at the end of first year $= Rs. 25000 + Rs. 2000 = Rs. 27000$
Principal for the second year $= Rs. 27000$
Interest for the second year $=\text{Rs.}\ \frac{27000\times8\times1}{100}$
$=\text{Rs.}\ 2000$
$\therefore$ Amount at the end of second year $= Rs. 27000 + Rs. 2160 = Rs. 29160$
$\therefore$ She would repay her debt $= Rs. 29160$
View full question & answer→Question 145 Marks
Find the amount and the compound interest on $Rs. 160000$ for $2$ years at $10\%$ per annum, compounded half-yearly.
AnswerPrincipal $(P) = Rs. 160000$
Rate $(R)\ 10\%$ p.a. or $5\%$
half yearly Period $(n) = 2$ year or $4$ half years
$\therefore$ Amount $(A)=\text{P}\Big(1+\frac{R}{100}\Big)^{\text{n}}$
$=\text{Rs. }160000\Big(1+\frac{5}{100}\Big)^4$
$=\text{Rs. }160000\times\Big(\frac{21}{20}\Big)^4$
$=\text{Rs. }160000\times\frac{21}{20}\times\frac{21}{20}\times\frac{21}{20}\times\frac{21}{20}$
$=\text{Rs. }194481$
$\therefore C.I. = A - P = Rs. 194481 - Rs. 160000 = Rs. 34481.$
View full question & answer→Question 155 Marks
The difference between the compound interest and the simple interest on a certain sum for $2$ years at $6\%$ per annum is $Rs. 90.$ Find the sum.
AnswerDifference between $C.I$ and $S.I. = Rs. 90$
Rate $(R) = 6\%\ p.a$
Period $(n) = 2$ years
Let principal $(P) = Rs. 100$
$\therefore$ Amount on $C.I =\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=100\Big(1+\frac{6}{100}\Big)^2$
$=\text{Rs. }100\times\Big(\frac{53}{50}\Big)^2$
$=\text{Rs. }100\times\frac{53}{50}\times\frac{53}{50}=\text{Rs. }\frac{2809}{25}$
$\therefore\text{C.I}=\text{A}-\text{P}=\text{Rs. }\frac{2809}{25}-100$
$=\text{Rs. }\frac{2809-2500}{25}=\frac{309}{25}$ and
$S.I =\frac{\text{PRT}}{100}=\frac{100\times6\times2}{100}=\text{Rs. }12$
Difference between $C.I$ and $S.I. =\text{Rs. }\frac{309}{25}-12$
$\text{Rs. }\frac{309-300}{25}=\frac{9}{25}$
Now if difference is $\text{Rs. }\frac{9}{25}$
then Principal $= Rs. 100$ and if difference is $Rs. 90,$
then principal$=\frac{100\times25\times90}{9}=\text{Rs. }25000$
View full question & answer→Question 165 Marks
Find the amount and the compound interest on $Rs. 8000$ for $1$ year at $10\%$ per annum, compounded half-yearly.
AnswerPrincipal $(P) = Rs. 12800$
Rate $(R) = 10\%$ p.a $5\%$ half yearly
Period $(n) = 1$ year or $2$ half years
$\therefore$ Amount after $1$ year $(A)$
$\text{P}= \Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }8000\times\Big(1+\frac{5}{100}\Big)^2$
$=\text{Rs. }8000\times\Big(\frac{21}{20}\Big)^2$
$= \text{Rs. }8000\times\frac{21}{20}\times\frac{21}{20}=\text{Rs. 8820}$ and
$C.I. = A - P = Rs. 8820 - Rs. 8000 = Rs. 820.$
View full question & answer→Question 175 Marks
Arun took a loan of $Rs. 390625$ from Kuber Finance. If the company charges interest at $16\%$ per annum, compounded quarterly, what amount will discharge his debt after one year$?$
AnswerAmount taken from finance company $(P) = Rs. 390625$
Rate of interest $(R) = 16\%$ p.a. or $4\%$ quarterly
Period $(n) = 1$ year or $4$ quarterly
$\therefore$ Amount $(A)=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=390625\Big(1+\frac{4}{100}\Big)^4$
$=\text{Rs. }390625\times\Big(\frac{26}{25}\Big)^4$
$=\text{Rs. }390625\times\frac{26}{25}\times\frac{26}{25}\times\frac{26}{25}\times\frac{26}{25}$
$=\text{Rs. }456976$
View full question & answer→Question 185 Marks
Harpreet borrowed $Rs. 20000$ from her friend at $12\%$ per annum simple interest. She lent it to Alam at the same rate but compounded annually. Find her gain after $2$ years.
AnswerIn case of Harpreet:
Amount borrowed by Harpreet $(P) = Rs. 20000$
Rate $(r) = 12\%$
Period $(t) = 2$ Years
$\therefore\text{S.I.}=\frac{\text{prt}}{100}$
$=\frac{20000\times12\times2}{100}$
$=\text{Rs.}\ 4800$
In case of Alam:
$\therefore$ Interest for the $1st$ year $=\frac{20000\times12\times1}{100}$
$=\text{Rs.}\ 2400$
Amount at the end of first year $= Rs. 20000 + Rs. 2400 = Rs. 22400$
Interest for the second year $=\frac{22400\times12\times1}{100}$
$=\text{Rs.}\ 2688$
$\therefore$ $C.I$. for $3$ years $= Rs. 2400 + 2688 = Rs. 5088$
$\therefore$ Amount of gain $= Rs. 5088 - Rs. 4800 = Rs. 288$
View full question & answer→Question 195 Marks
Mohd. Aslam purchased a house from Avas Vikas Parishad on credit. If the cost of the house is $Rs. 125000$ and the Parishad charges interest at $12\%$ per annum compounded half-yearly, find the interest paid by Aslam after a year and a half.
AnswerLoan recieved for the cost of the house $(P) = Rs. 125000$
Rate of interest $(R) = 12\%$ p.a. or $6\%$ half yearly
Period $(n) =1\frac{1}{2}$ year or $3$ half years
$\therefore$ Amount $(A) =\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }125000\Big(1+\frac{6}{100}\Big)^3$
$=\text{Rs. }125000\times\Big(\frac{53}{50}\Big)^3$
$=\text{Rs. }125000\times\frac{53}{50}\times\frac{53}{50}\times\frac{53}{50}$
$=\text{Rs. }148877$
$\therefore C.I.$ paid $= A - P$
$= Rs. 148877- Rs. 125000$
$= Rs. 23877.$
View full question & answer→Question 205 Marks
Find the amount and the compound interest on $Rs. 6000$ for $1$ year at $10\%$ per annum compounded half-yearly.
AnswerLet the principal amount be $P = Rs. 6000.$
Rate of interest $= 10\%$
$p.a.= 5\%$ for half yearly.
Time $(n) = 1$ year $= 2$ half years Now, $\text{A}=\text{P}\times\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs}.\ 6000\times\Big(1+\frac{5}{100}\Big)^2$
$=\text{Rs}.\ 6000\times\Big(\frac{105}{100}\Big)^2$
$=\text{Rs}.\ 6000\times\Big(\frac{21}{20}\Big)\times\Big(\frac{21}{20}\Big)$
$=\text{Rs}.(15\times21\times21)$
$=\text{Rs}. 6615$
$\therefore CI = A - P = Rs. (6615 - 6000) = Rs. 615$
Hence, the amount is $Rs. 6615$ and the $CI$ is $Rs. 615$
View full question & answer→Question 215 Marks
Sheela deposited $Rs. 20000$ in a bank, where the interest is credited half-yearly. If the rate of interest paid by the bank is $6\%$ per annum, what amount will she get after $1$ year$?$
AnswerAmount deposite in the bank $= Rs. 20000$
Rate of interest $(R) = 6\%$ p.a. or $3\%$ half yearly
Period $(n) = 1$ year or $2$ half years
Amount recieved after $1$ year
$=\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }20000\Big(1+\frac{3}{100}\Big)^2$
$=20000\Big(\frac{103}{100}\Big)^2$
$=\text{Rs. }20000\times\frac{103}{100}\times\frac{103}{100}\times= \text{Rs. }21218$
View full question & answer→Question 225 Marks
Divakaran deposited a sum of $Rs. 6250$ in the Allahabad Bank for $1$ year, compounded half-yearly at $8\%$ per annum. Find the compound interest he gets.
AnswerPrincipal $(p) = Rs. 6250$
Rate $(r) 8\%$ pa or $4\%$ half yearly
Period $(t) = 1$ year $= 2$ half years
Interest for the first half year $=\frac{\text{prt}}{100}$
$=\frac{6250\times4\times1}{100}$
$=\text{Rs.}\ 250$
Amount at the end of first half years $= Rs. 6250 + 250 = Rs. 6500$
Principal for the second half year $= Rs. 6500$
Interest for the second half year $=\text{Rs.}\ \frac{6500\times4\times1}{100}$
$=\text{Rs.}\ 260$
Compound Interest for one year $= Rs. 250 + 260 = Rs. 510$
View full question & answer→Question 235 Marks
Sudershan deposited $Rs. 32000$ in a bank, where the interest is credited quarterly. If the rate of interest be $5\%$ per annum, what amount will he receive after $6$ months$?$
AnswerAmount deposit in the bank $(P) = Rs. 32000$
Rate of interest $(R) = 5\%$ p.a. or $\frac{5}{4}\%$
quarterly Period $(n) = 6$ months or $2$ quarters
$\therefore$ Amount $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }32000\Big(1+\frac{5}{4\times100}\Big)^2$
$=\text{Rs. }32000\times\Big(\frac{81}{80}\Big)^2$
$=\text{Rs. }32000\times\frac{81}{80}\times\frac{81}{80}=\text{Rs. }32805$
View full question & answer→Question 245 Marks
A sum amounts to $Rs. 23762$ in $2$ years at $9\%$ per annum, compounded annually. Find the sum.
AnswerAmount $(A) = Rs. 23762$
Rate of interest $(R) = 9\%$
Time $(n) = 2$ years
Now, $\text{A}=\text{P}\times\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$\Rightarrow\text{Rs}.\ 23762=\text{P}\times\Big(1+\frac{9}{100}\Big)^2$
$\Rightarrow\text{Rs}.\ 23762=\text{P}\times\Big(\frac{109}{100}\Big)\times\Big(\frac{109}{100}\Big)$
$\Rightarrow\text{P}=\text{Rs}. \frac{23762\times100\times100}{100\times100}$
$\Rightarrow\text{P}=\text{Rs.}\ 20000$
$\therefore$ The principal amount is $Rs. 20000.$
View full question & answer→