Compound Interest Questions in English

(C) Let the principal sum be $P$. The rate of interest $R = 5 \%$ per annum and time $T = 3$ years.
Simple Interest $(SI)$ is given by: $SI = \frac{P \times R \times T}{100} = \frac{P \times 5 \times 3}{100} = \frac{15P}{100} = \frac{3P}{20}$.
Compound Interest $(CI)$ is given by: $CI = P \left[ (1 + \frac{R}{100})^T - 1 \right] = P \left[ (1 + \frac{5}{100})^3 - 1 \right] = P \left[ (\frac{21}{20})^3 - 1 \right] = P \left[ \frac{9261}{8000} - 1 \right] = P \left[ \frac{1261}{8000} \right]$.
According to the problem,$CI - SI = 183$.
Substituting the values: $\frac{1261P}{8000} - \frac{3P}{20} = 183$.
To subtract,make the denominators equal: $\frac{1261P}{8000} - \frac{1200P}{8000} = 183$.
$\frac{61P}{8000} = 183$.
$P = \frac{183 \times 8000}{61} = 3 \times 8000 = 24000$.
Thus,the sum is $Rs. 24000$.

(D) The formula for compound interest is $CI = P \left[\left(1+\frac{R}{100}\right)^{n}-1\right]$.
Given $CI = 504.40$,$R = 5\%$,and $n = 3$ years.
$504.40 = P \left[\left(1+\frac{5}{100}\right)^{3}-1\right]$
$504.40 = P \left[\left(\frac{21}{20}\right)^{3}-1\right]$
$504.40 = P \left[\frac{9261}{8000}-1\right]$
$504.40 = P \left[\frac{1261}{8000}\right]$
$P = \frac{504.40 \times 8000}{1261} = 400 \times 0.8 = 3200$.
So,the principal $P = Rs. 3200$.
Now,calculate simple interest: $SI = \frac{P \times R \times T}{100}$.
$SI = \frac{3200 \times 5 \times 3}{100} = 32 \times 15 = 480$.
Thus,the simple interest is $Rs. 480$.

(A) Given: Principal $(P) = Rs. 1500$,Rate $(R) = 25 \% \text{ per annum}$.
We need to find the minimum number of years $(n)$ such that the amount $(A)$ becomes more than double the principal,i.e.,$A > 2P$.
The formula for compound interest is $A = P(1 + R/100)^n$.
Substituting the values: $P(1 + 25/100)^n > 2P$.
Dividing both sides by $P$: $(1 + 1/4)^n > 2$,which simplifies to $(5/4)^n > 2$.
Calculating for different values of $n$:
For $n = 1: (1.25)^1 = 1.25 < 2$.
For $n = 2: (1.25)^2 = 1.5625 < 2$.
For $n = 3: (1.25)^3 = 1.953125 < 2$.
For $n = 4: (1.25)^4 = 2.44140625 > 2$.
Thus,the least number of complete years required is $4$.

(C) Let the principal be $P$ and the rate of interest be $R\%$.
Simple Interest $(SI)$ for $3$ years $= Rs. 240$.
Therefore,$SI$ for $1$ year $= \frac{240}{3} = Rs. 80$.
$SI$ for $2$ years $= 80 \times 2 = Rs. 160$.
We know that $SI = \frac{P \times R \times T}{100}$,so for $2$ years: $160 = \frac{P \times R \times 2}{100} \Rightarrow PR = 8000$.
Compound Interest $(CI)$ for $2$ years $= Rs. 164$.
The difference between $CI$ and $SI$ for $2$ years is given by $CI - SI = P \left( \frac{R}{100} \right)^2$.
$164 - 160 = P \left( \frac{R}{100} \right)^2 \Rightarrow 4 = P \left( \frac{R}{100} \right)^2$.
Since $PR = 8000$,we have $R = \frac{8000}{P}$.
Substituting $R$ in the equation: $4 = P \left( \frac{8000/P}{100} \right)^2 = P \left( \frac{80}{P} \right)^2 = P \times \frac{6400}{P^2} = \frac{6400}{P}$.
$P = \frac{6400}{4} = 1600$.
Thus,the sum is $Rs. 1600$.

(A) The formula for compound interest is $CI = P \left[\left(1+\frac{R}{100}\right)^{n}-1\right]$.
Given: Principal $(P)$ = $Rs. 7850$,Rate $(R)$ = $14\%$,Time $(n)$ = $2$ years.
Substituting the values into the formula:
$CI = 7850 \times \left[\left(1+\frac{14}{100}\right)^{2}-1\right]$
$CI = 7850 \times \left[(1.14)^{2}-1\right]$
$CI = 7850 \times [1.2996 - 1]$
$CI = 7850 \times 0.2996$
$CI = 2351.86$
Thus,the compound interest is $Rs. 2351.86$.

(A) The formula for compound interest is $CI = P \left[ \left( 1 + \frac{R}{100} \right)^n - 1 \right]$.
Given $CI = 6500.52$,$R = 15$,and $n = 3$.
$6500.52 = P \left[ \left( 1 + \frac{15}{100} \right)^3 - 1 \right]$
$6500.52 = P \left[ \left( 1.15 \right)^3 - 1 \right]$
$6500.52 = P \left[ 1.520875 - 1 \right]$
$6500.52 = P \times 0.520875$
$P = \frac{6500.52}{0.520875} = 12480$.
Thus,the principal amount is $Rs. 12480$.

(D) The formula for compound interest is $CI = P \left[ \left( 1 + \frac{R}{100} \right)^n - 1 \right]$.
Given: Principal $(P)$ = $Rs. 7400$,Rate $(R)$ = $13.5\%$,Time $(n)$ = $2$ years.
$CI = 7400 \left[ \left( 1 + \frac{13.5}{100} \right)^2 - 1 \right]$
$CI = 7400 \left[ (1.135)^2 - 1 \right]$
$CI = 7400 \left[ 1.288225 - 1 \right]$
$CI = 7400 \times 0.288225$
$CI = 2132.865$
Rounding off to two decimal places,we get $Rs. 2132.87$.

(B) The formula for compound interest is $CI = P \left[\left(1 + \frac{R}{100}\right)^n - 1\right]$.
Given $CI = 1414.4$,$R = 8$,and $n = 2$.
$1414.4 = P \left[\left(1 + \frac{8}{100}\right)^2 - 1\right]$
$1414.4 = P \left[(1.08)^2 - 1\right]$
$1414.4 = P \left[1.1664 - 1\right]$
$1414.4 = P \times 0.1664$
$P = \frac{1414.4}{0.1664} = 8500$.
The total amount received is $Principal + CI = 8500 + 1414.4 = 9914.4$.

(B) The formula for compound interest with varying rates is given by:
$A = P \left(1 + \frac{R_1}{100}\right) \left(1 + \frac{R_2}{100}\right) \left(1 + \frac{R_3}{100}\right)$
Given:
$A = 5305.53$
$R_1 = 1\%$,$R_2 = 2\%$,$R_3 = 3\%$
Substituting the values:
$5305.53 = P \left(1 + \frac{1}{100}\right) \left(1 + \frac{2}{100}\right) \left(1 + \frac{3}{100}\right)$
$5305.53 = P \times (1.01) \times (1.02) \times (1.03)$
$5305.53 = P \times 1.061106$
Solving for $P$:
$P = \frac{5305.53}{1.061106}$
$P = 5000$
Thus,the sum of money is $Rs. 5000$.

(A) The formula for the difference between compound interest $(CI)$ and simple interest $(SI)$ for $3$ years is given by:
Difference $= \frac{P \times r^2}{100^2} \times \left(\frac{r}{100} + 3\right)$
Here,$r = 15$ and the difference $= 595.35$.
Substituting the values:
$595.35 = \frac{P \times 15^2}{100^2} \times \left(\frac{15}{100} + 3\right)$
$595.35 = \frac{P \times 225}{10000} \times (0.15 + 3)$
$595.35 = \frac{P \times 225}{10000} \times 3.15$
$595.35 = \frac{P \times 708.75}{10000}$
$P = \frac{595.35 \times 10000}{708.75}$
$P = \frac{5953500}{708.75} = 8400$
Thus,the principal amount is $Rs. 8400$.

(C) Step $1$: Calculate the original principal $(P)$.
Simple Interest $(SI)$ $= \frac{P \times R \times T}{100}$
$1000 = \frac{P \times 5 \times 4}{100}$
$1000 = \frac{P \times 20}{100}$
$P = \frac{1000 \times 100}{20} = Rs. 5000$
Step $2$: Calculate the new principal $(P')$ and Compound Interest $(CI)$.
New Principal $(P')$ $= 2 \times P = 2 \times 5000 = Rs. 10000$
Rate $(R)$ $= 5\%$,Time $(T)$ $= 2$ years.
$CI = P' \left[ \left( 1 + \frac{R}{100} \right)^T - 1 \right]$
$CI = 10000 \left[ \left( 1 + \frac{5}{100} \right)^2 - 1 \right]$
$CI = 10000 \left[ \left( \frac{105}{100} \right)^2 - 1 \right]$
$CI = 10000 \left[ \frac{11025}{10000} - 1 \right]$
$CI = 10000 \left[ \frac{11025 - 10000}{10000} \right]$
$CI = 10000 \times \frac{1025}{10000} = Rs. 1025$

(B) Step $1$: Calculate the rate of interest $(R)$ using the simple interest formula.
$SI = \frac{P \times R \times T}{100}$
$7200 = \frac{20000 \times R \times 3}{100}$
$7200 = 600 \times R$
$R = \frac{7200}{600} = 12\% \text{ per annum}$.
Step $2$: Calculate the compound interest $(CI)$ for the same principal,rate,and time.
$CI = P \left[ \left( 1 + \frac{R}{100} \right)^T - 1 \right]$
$CI = 20000 \left[ \left( 1 + \frac{12}{100} \right)^3 - 1 \right]$
$CI = 20000 \left[ (1.12)^3 - 1 \right]$
$CI = 20000 \left[ 1.404928 - 1 \right]$
$CI = 20000 \times 0.404928 = 8098.56$.
Thus,the compound interest is $Rs. 8098.56$.

(D) The population growth follows the compound interest formula: $A = P(1 + \frac{r}{100})^n$.
Here,the initial population $P = 15$ lakhs.
The rate of increase $r = 10 \%$ per year.
The time period $n = 2005 - 2003 = 2$ years.
Substituting the values into the formula:
$A = 15 \times (1 + \frac{10}{100})^2$
$A = 15 \times (1 + 0.1)^2$
$A = 15 \times (1.1)^2$
$A = 15 \times 1.21$
$A = 18.15$ lakhs.
Therefore,the population in the year $2005$ was $18.15$ lakhs.

(B) The present value of the machine is $P = Rs. 29644.032$.
The rate of depreciation is $r = 12\%$ per annum.
The time period is $n = 3$ years.
The formula for the depreciated value is given by $V = P_0 \times (1 - \frac{r}{100})^n$,where $P_0$ is the initial purchase price.
To find the purchase price $P_0$,we rearrange the formula:
$P_0 = \frac{V}{(1 - \frac{r}{100})^n}$
Substituting the given values:
$P_0 = \frac{29644.032}{(1 - \frac{12}{100})^3}$
$P_0 = \frac{29644.032}{(0.88)^3}$
$P_0 = \frac{29644.032}{0.681472}$
$P_0 = 43500$
Therefore,the purchase price of the machine was $Rs. 43500$.

(A) The initial number of students admitted in $2008$ $(P)$ is $5000$.
The rate of increase $(r)$ is $24\%$ per year.
The time period $(n)$ from $2008$ to $2010$ is $2010 - 2008 = 2$ years.
Since the number of students increases at a constant percentage rate,we use the compound growth formula: $A = P \left(1 + \frac{r}{100}\right)^n$.
Substituting the values: $A = 5000 \left(1 + \frac{24}{100}\right)^2$.
$A = 5000 \times (1.24)^2$.
$A = 5000 \times 1.5376$.
$A = 7688$.
Therefore,the number of students admitted in the year $2010$ will be $7688$.

(B) Given: Principal $(P) = Rs. 19800$,Simple Interest $(SI) = Rs. 7128$,Time $(T) = 3$ years.
First,calculate the rate of interest $(R)$ using the formula: $SI = \frac{P \times R \times T}{100}$.
$7128 = \frac{19800 \times R \times 3}{100} \Rightarrow 7128 = 198 \times R \times 3$.
$7128 = 594 \times R \Rightarrow R = \frac{7128}{594} = 12 \%$.
Now,calculate the compound interest $(CI)$ using the formula: $CI = P \left[ (1 + \frac{R}{100})^T - 1 \right]$.
$CI = 19800 \left[ (1 + \frac{12}{100})^3 - 1 \right] = 19800 \left[ (1.12)^3 - 1 \right]$.
$(1.12)^3 = 1.404928$.
$CI = 19800 \times (1.404928 - 1) = 19800 \times 0.404928$.
$CI = Rs. 8017.5744$.

(B) For a period of $2$ $years$,the difference between Compound Interest $(CI)$ and Simple Interest $(SI)$ is given by the formula:
$\text{Difference} = P \times \left( \frac{R}{100} \right)^2$
Where:
$P$ = Principal
$R$ = Rate of interest $(5\%)$
$\text{Difference} = 16$
Substituting the values:
$16 = P \times \left( \frac{5}{100} \right)^2$
$16 = P \times \left( \frac{1}{20} \right)^2$
$16 = P \times \frac{1}{400}$
$P = 16 \times 400$
$P = 6400$
Thus,the principal amount is $Rs. 6400$.

(B) The production of the factory in $2008$ is given as $P = 70$ $lakh$ tonnes.
The growth rate is $R = 8 \%$ $p.a.$
The time period from $2008$ to $2010$ is $n = 2$ years.
Since the production grows at a compound rate,we use the formula for compound growth:
$A = P \left(1 + \frac{R}{100}\right)^n$
Substituting the values:
$A = 70 \left(1 + \frac{8}{100}\right)^2$
$A = 70 \left(1 + 0.08\right)^2$
$A = 70 \times (1.08)^2$
$A = 70 \times 1.1664$
$A = 81.648$ $lakh$ tonnes.
Therefore,the production in $2010$ will be $81.648$ $lakh$ tonnes.

(B) For $A$ (Simple Interest):
$SI = \frac{P \times R \times T}{100} = \frac{6000 \times 5 \times 2}{100} = Rs. 600$.
For $B$ (Compound Interest):
Amount $= P \left(1 + \frac{R}{100}\right)^n = 5000 \left(1 + \frac{8}{100}\right)^2 = 5000 \left(\frac{108}{100}\right)^2 = 5000 \times 1.1664 = Rs. 5832$.
$CI = \text{Amount} - P = 5832 - 5000 = Rs. 832$.
Difference between their interests $= 832 - 600 = Rs. 232$.

(A) The formula for the amount $A$ with varying rates of interest $R_1, R_2, R_3$ is given by $A = P \times (1 + \frac{R_1}{100}) \times (1 + \frac{R_2}{100}) \times (1 + \frac{R_3}{100})$.
Given $P = 10000$,$R_1 = 4$,$R_2 = 5$,$R_3 = 6$.
$A = 10000 \times (1 + \frac{4}{100}) \times (1 + \frac{5}{100}) \times (1 + \frac{6}{100})$
$A = 10000 \times \frac{104}{100} \times \frac{105}{100} \times \frac{106}{100}$
$A = 10000 \times \frac{26}{25} \times \frac{21}{20} \times \frac{53}{50}$
$A = 10000 \times \frac{28938}{25000} = 4 \times 2893.8 = 11575.20$
Compound Interest $(CI) = A - P = 11575.20 - 10000 = Rs. 1575.20$.

(C) Principal $(P) = Rs. 5000$,Rate $(R) = 10\%$ per annum,Time $(T) = 2$ years.
Simple Interest $(SI) = \frac{P \times R \times T}{100} = \frac{5000 \times 10 \times 2}{100} = Rs. 1000$.
For compound interest compounded semi-annually:
Rate $(r) = \frac{10}{2} = 5\%$ per half-year.
Time $(n) = 2 \times 2 = 4$ half-years.
Amount $(A) = P \left(1 + \frac{r}{100}\right)^n = 5000 \left(1 + \frac{5}{100}\right)^4 = 5000 \left(1.05\right)^4$.
$A = 5000 \times 1.21550625 = Rs. 6077.53125$.
Compound Interest $(CI) = A - P = 6077.53125 - 5000 = Rs. 1077.53$.
Difference $= CI - SI = 1077.53 - 1000 = Rs. 77.53$.
Rounding to the nearest option,the difference is $Rs. 77.50$.

(C) Let the principal sum be $P$.
Given: Compound Interest $(CI)$ = $Rs. 101.50$,Rate $(r)$ = $3 \%$,Time $(t)$ = $2$ $years$.
The formula for compound interest is $CI = P \left[ \left( 1 + \frac{r}{100} \right)^t - 1 \right]$.
Substituting the values: $101.50 = P \left[ \left( 1 + \frac{3}{100} \right)^2 - 1 \right]$.
$101.50 = P \left[ (1.03)^2 - 1 \right] = P [1.0609 - 1] = P(0.0609)$.
$P = \frac{101.50}{0.0609} = \frac{1015000}{609} = Rs. \frac{5000}{3}$.
Now,calculate Simple Interest $(SI)$ using $SI = \frac{P \times r \times t}{100}$.
$SI = \frac{(5000/3) \times 3 \times 2}{100} = \frac{5000 \times 2}{100} = Rs. 100$.

(D) Given: Compound Interest $(CI)$ = $Rs. 1272$,Time $(t)$ = $2$ years,Rate $(r)$ = $12 \%$ per annum.
The formula for compound interest is: $CI = P \left[\left(1 + \frac{r}{100}\right)^t - 1\right]$
Substituting the values: $1272 = P \left[\left(1 + \frac{12}{100}\right)^2 - 1\right]$
$1272 = P \left[(1.12)^2 - 1\right]$
$1272 = P [1.2544 - 1]$
$1272 = P \times 0.2544$
$P = \frac{1272}{0.2544} = 5000$
Now,calculate the simple interest $(SI)$ using the formula: $SI = \frac{P \times r \times t}{100}$
$SI = \frac{5000 \times 12 \times 2}{100}$
$SI = 50 \times 24 = 1200$
Therefore,the simple interest is $Rs. 1200$.

(A) The formula for compound interest is $A = P(1 + \frac{r}{100})^t$,where $A$ is the amount,$P$ is the principal,$r$ is the rate of interest,and $t$ is the time in years.
For $t = 4$ years,$3840 = P(1 + \frac{r}{100})^4$ --- $(i)$
For $t = 5$ years,$3936 = P(1 + \frac{r}{100})^5$ --- (ii)
Dividing equation (ii) by equation $(i)$:
$\frac{3936}{3840} = \frac{P(1 + \frac{r}{100})^5}{P(1 + \frac{r}{100})^4}$
$\frac{3936}{3840} = 1 + \frac{r}{100}$
$\frac{3936}{3840} - 1 = \frac{r}{100}$
$\frac{3936 - 3840}{3840} = \frac{r}{100}$
$\frac{96}{3840} = \frac{r}{100}$
$r = \frac{96 \times 100}{3840} = \frac{9600}{3840} = 2.5$
Therefore,the rate of interest is $2.5 \%$ per annum.

(A) The formula for the difference between Compound Interest $(CI)$ and Simple Interest $(SI)$ for $2$ years is given by:
Difference $= \frac{P \times r^2}{100^2}$
Where $P$ is the principal sum and $r$ is the rate of interest per annum.
Given: Difference $= 4$,$r = 4 \%$,time $= 2$ years.
Substituting the values into the formula:
$4 = \frac{P \times 4^2}{100^2}$
$4 = \frac{P \times 16}{10000}$
$P = \frac{4 \times 10000}{16}$
$P = \frac{10000}{4}$
$P = 2500$
Thus,the sum is $Rs. 2500$.

(B) Given: Principal $(P) = 64000$,Rate $(R) = 5 \%$ per annum,Time $(n) = 18$ months.
Since the interest is compounded half-yearly,the rate for half-year $(r) = \frac{5}{2} = 2.5 \%$.
The number of half-yearly periods $(n) = \frac{18}{6} = 3$.
The formula for the amount $(A)$ is $A = P(1 + \frac{r}{100})^n$.
Substituting the values: $A = 64000(1 + \frac{2.5}{100})^3$.
$A = 64000(1 + 0.025)^3 = 64000(1.025)^3$.
$A = 64000 \times (\frac{41}{40})^3 = 64000 \times \frac{68921}{64000}$.
$A = 68921$.

(B) The formula for the difference between compound interest and simple interest for $2$ years is given by: $\text{Difference} = P \times \left(\frac{r}{100}\right)^2$.
Here,the principal $P = ₹ 12500$ and the rate $r = 4 \%$.
Substituting the values:
$\text{Difference} = 12500 \times \left(\frac{4}{100}\right)^2$
$\text{Difference} = 12500 \times \left(\frac{1}{25}\right)^2$
$\text{Difference} = 12500 \times \frac{1}{625}$
$\text{Difference} = 20$.
Thus,the difference between the simple interest and the compound interest is ₹ $20$.

(C) The formula for the difference between compound interest and simple interest for $3$ years is given by: $\text{Difference} = \frac{P \times r^2(300 + r)}{100^3}$.
Here,$P = 8000$,$r = 7.5$,and $n = 3$ years.
Substituting the values into the formula:
$\text{Difference} = \frac{8000 \times (7.5)^2 \times (300 + 7.5)}{100^3}$
$\text{Difference} = \frac{8000 \times 56.25 \times 307.5}{1000000}$
$\text{Difference} = \frac{450000 \times 307.5}{1000000}$
$\text{Difference} = \frac{138375000}{1000000} = ₹ 138.375$.

(C) Let the principal be $P$ and the rate of interest be $R \%$.
The amount after $n$ years is given by $A = P(1 + \frac{R}{100})^n$.
For $2$ years: $2809 = P(1 + \frac{R}{100})^2$ --- $(i)$
For $3$ years: $2977.54 = P(1 + \frac{R}{100})^3$ --- $(ii)$
Dividing $(ii)$ by $(i)$:
$\frac{2977.54}{2809} = 1 + \frac{R}{100}$
$1.06 = 1 + \frac{R}{100}$
$\frac{R}{100} = 0.06 \implies R = 6 \%$.
Now,substitute $R = 6$ in equation $(i)$:
$2809 = P(1 + \frac{6}{100})^2$
$2809 = P(1.06)^2$
$2809 = P(1.1236)$
$P = \frac{2809}{1.1236} = 2500$.
Thus,the rate of interest is $6 \%$ and the original sum is ₹ $2500$.

(B) For a period of $3$ years,the difference $(D)$ between compound interest $(CI)$ and simple interest $(SI)$ is given by the formula: $D = P \times \left(\frac{r}{100}\right)^2 \times \left(\frac{300+r}{100}\right)$.
Given: $D = ₹ 122$,$r = 5 \%$,$t = 3$ years.
Substituting the values: $122 = P \times \left(\frac{5}{100}\right)^2 \times \left(\frac{300+5}{100}\right)$.
$122 = P \times \left(\frac{1}{20}\right)^2 \times \left(\frac{305}{100}\right)$.
$122 = P \times \frac{1}{400} \times \frac{305}{100}$.
$122 = P \times \frac{305}{40000}$.
$P = \frac{122 \times 40000}{305}$.
$P = \frac{122 \times 8000}{61}$.
$P = 2 \times 8000 = ₹ 16000$.

(C) Let $P$ be the principal and $R$ be the rate of interest per annum.
The amount after $n$ years at compound interest is given by $A = P(1 + R/100)^n$.
For $3$ years: $P(1 + R/100)^3 = 800$ --- (Equation $1$)
For $4$ years: $P(1 + R/100)^4 = 840$ --- (Equation $2$)
Dividing Equation $2$ by Equation $1$:
$\frac{P(1 + R/100)^4}{P(1 + R/100)^3} = \frac{840}{800}$
$(1 + R/100) = 1.05$
$R/100 = 1.05 - 1 = 0.05$
$R = 0.05 \times 100 = 5 \%$
Thus,the rate of interest is $5 \%$ per annum.

(B) Let the principal be $P$.
Simple Interest $(SI)$ for $2$ years at $10 \%$ per annum:
$SI = P \times \frac{10 \times 2}{100} = 0.2P$
Compound Interest $(CI)$ for $2$ years compounded half-yearly:
Rate $R = 10 \% / 2 = 5 \%$ per half-year.
Time $n = 2 \times 2 = 4$ half-years.
$CI = P \left(1 + \frac{5}{100}\right)^4 - P = P \left(1.05^4 - 1\right)$
$CI = P (1.21550625 - 1) = 0.21550625P$
Difference $= CI - SI = 0.21550625P - 0.2P = 0.01550625P$
Given,$0.01550625P = 124.05$
$P = \frac{124.05}{0.01550625} = 8000$
Thus,the principal is ₹ $8000$.

(A) Let the principal sum be $P$.
For $2$ years,the simple interest $(SI)$ is given by $SI = \frac{P \times R \times T}{100} = \frac{P \times 4 \times 2}{100} = \frac{8P}{100} = 0.08P$.
The compound interest $(CI)$ compounded annually is given by $CI = P \left(1 + \frac{R}{100}\right)^T - P = P \left(1 + \frac{4}{100}\right)^2 - P = P \left(1.04^2 - 1\right) = P(1.0816 - 1) = 0.0816P$.
The difference between $CI$ and $SI$ is given as ₹ $1$.
$0.0816P - 0.08P = 1$
$0.0016P = 1$
$P = \frac{1}{0.0016} = \frac{10000}{16} = 625$.
Thus,the sum is ₹ $625$.

(C) Let the principal sum be $P$.
The Simple Interest ($S$.$I$.) for $3$ years at $8 \%$ per annum is given by:
$S.I. = \frac{P \times 3 \times 8}{100} = \frac{24P}{100} = 0.24P$ (Equation $1$)
The Compound Interest ($C$.$I$.) on ₹ $4000$ for $2$ years at $10 \%$ per annum is:
$C.I. = 4000 \left(1 + \frac{10}{100}\right)^2 - 4000$
$C.I. = 4000 \left(\frac{11}{10}\right)^2 - 4000 = 4000 \times 1.21 - 4000 = 4840 - 4000 = 840$ (Equation $2$)
According to the problem,the simple interest is half of the compound interest:
$S.I. = \frac{1}{2} \times C.I.$
$0.24P = \frac{1}{2} \times 840$
$0.24P = 420$
$P = \frac{420}{0.24} = \frac{42000}{24} = 1750$
Thus,the sum placed on simple interest is ₹ $1750$.

(D) Let the principal amount be $P$.
Given rate $R = 12 \frac{1}{2} \% = \frac{25}{2} \% = 12.5 \% = \frac{1}{8}$ per annum.
Time $T = 2$ years.
Compound Interest ($C$.$I$.) formula: $C.I. = P \left[ \left(1 + \frac{R}{100} \right)^T - 1 \right]$.
$510 = P \left[ \left(1 + \frac{1}{8} \right)^2 - 1 \right]$.
$510 = P \left[ \left( \frac{9}{8} \right)^2 - 1 \right] = P \left[ \frac{81}{64} - 1 \right] = P \left( \frac{17}{64} \right)$.
$P = \frac{510 \times 64}{17} = 30 \times 64 = ₹ 1920$.
Now,Simple Interest ($S$.$I$.) = $\frac{P \times R \times T}{100} = \frac{1920 \times 12.5 \times 2}{100} = \frac{1920 \times 25}{100} = 1920 \times 0.25 = ₹ 480$.

(D) Let the principal be $P$.
The formula for compound interest is $A = P(1 + \frac{R}{100})^n$.
Given: $A = 4913$,$R = 6 \frac{1}{4} \% = \frac{25}{4} \%$,$n = 3$ years.
Substituting the values:
$4913 = P(1 + \frac{25/4}{100})^3$
$4913 = P(1 + \frac{25}{400})^3$
$4913 = P(1 + \frac{1}{16})^3$
$4913 = P(\frac{17}{16})^3$
$P = 4913 \times (\frac{16}{17})^3$
Since $17^3 = 4913$,we have:
$P = 4913 \times \frac{4096}{4913} = 4096$.
Thus,the principal is ₹ $4096$.

(A) Let the period of deposit be $N$ years.
Given,Principal $(P)$ = ₹ $30000$,Rate $(R)$ = $7 \%$,Compound Interest $(CI)$ = ₹ $4347$.
The formula for compound interest is $CI = P \left[ \left(1 + \frac{R}{100} \right)^N - 1 \right]$.
Substituting the values: $4347 = 30000 \left[ \left(1 + \frac{7}{100} \right)^N - 1 \right]$.
$4347 = 30000 \left[ \left(1.07 \right)^N - 1 \right]$.
Divide both sides by $30000$: $\frac{4347}{30000} = (1.07)^N - 1$.
$0.1449 = (1.07)^N - 1$.
$(1.07)^N = 1.1449$.
Since $(1.07)^2 = 1.1449$,we have $(1.07)^N = (1.07)^2$.
Therefore,$N = 2$ years.

(D) Simple Interest ($S$.$I$.) on ₹ $1000$ at $10 \%$ per annum for $4$ years is calculated as:
$S.I. = \frac{P \times R \times T}{100} = \frac{1000 \times 10 \times 4}{100} = ₹ 400$.
Compound Interest ($C$.$I$.) on ₹ $1000$ at $10 \%$ per annum for $4$ years is calculated as:
$C.I. = P \left[ \left( 1 + \frac{R}{100} \right)^T - 1 \right] = 1000 \left[ \left( 1 + \frac{10}{100} \right)^4 - 1 \right]$
$= 1000 \left[ (1.1)^4 - 1 \right] = 1000 [1.4641 - 1] = 1000 \times 0.4641 = ₹ 464.10$.
The difference between Compound Interest and Simple Interest is:
$C.I. - S.I. = 464.10 - 400 = ₹ 64.10$.

(C) Given: Principal $(P) = ₹ 15625$,Rate $(R) = 16 \%$ per annum,Time $(n) = 9$ months.
Since the interest is compounded quarterly,the rate per quarter is $R' = 16 / 4 = 4 \%$ and the number of quarters is $n' = 9 / 3 = 3$.
The formula for compound interest is $CI = P \left[ (1 + R'/100)^{n'} - 1 \right]$.
Substituting the values: $CI = 15625 \left[ (1 + 4/100)^3 - 1 \right]$.
$CI = 15625 \left[ (1 + 1/25)^3 - 1 \right] = 15625 \left[ (26/25)^3 - 1 \right]$.
$CI = 15625 \times \left( \frac{17576 - 15625}{15625} \right)$.
$CI = 17576 - 15625 = ₹ 1951$.

(B) The interest is calculated half-yearly at a rate of $5 \%$ per annum,so the half-yearly rate is $r = 2.5 \% = 0.025$.
For the first deposit of ₹ $1600$ made on $1^{st}$ January,it earns interest for two half-yearly periods (one year).
Amount after one year for the first deposit $= 1600(1 + 0.025)^2 = 1600(1.025)^2 = 1600(1.050625) = 1681$.
For the second deposit of ₹ $1600$ made on $1^{st}$ July,it earns interest for one half-yearly period (six months).
Amount after six months for the second deposit $= 1600(1 + 0.025)^1 = 1600(1.025) = 1640$.
Total amount at the end of the year $= 1681 + 1640 = 3321$.
Total principal deposited $= 1600 + 1600 = 3200$.
Total interest earned $= 3321 - 3200 = ₹ 121$.

(C) The formula for compound interest is $CI = P \left[ \left( 1 + \frac{R}{100} \right)^n - 1 \right]$.
Given: Principal $(P) = ₹ 25000$,Rate $(R) = 12 \%$,Time $(n) = 3 \text{ years}$.
$CI = 25000 \left[ \left( 1 + \frac{12}{100} \right)^3 - 1 \right]$
$CI = 25000 \left[ \left( 1.12 \right)^3 - 1 \right]$
$CI = 25000 \left[ 1.404928 - 1 \right]$
$CI = 25000 \times 0.404928$
$CI = ₹ 10123.20$

(D) The formula for the amount $A$ under compound interest is given by $A = P(1 + \frac{R}{100})^N$,where $P$ is the principal,$R$ is the rate of interest,and $N$ is the time period in years.
Given: $P = ₹ 8000$,$R = 5 \%$,$N = 2$ years.
Substituting the values into the formula:
$A = 8000(1 + \frac{5}{100})^2$
$A = 8000(1 + 0.05)^2$
$A = 8000(1.05)^2$
$A = 8000 \times 1.1025$
$A = 8820$
Therefore,Albert will get ₹ $8820$ on maturity.

(C) The formula for compound interest is $A = P(1 + \frac{R}{n \times 100})^{n \times T}$,where $P$ is the principal,$R$ is the annual rate,$n$ is the number of times interest is compounded per year,and $T$ is the time in years.
Given: $P = 15000$,$R = 10 \%$,$T = 1$ year,and interest is compounded half-yearly $(n = 2)$.
Effective rate per half-year = $\frac{10 \%}{2} = 5 \%$.
Number of half-years $(N)$ = $1 \times 2 = 2$.
Amount $A = 15000(1 + \frac{5}{100})^2$.
$A = 15000(1 + 0.05)^2 = 15000(1.05)^2$.
$A = 15000 \times 1.1025 = 16537.50$.
Thus,the amount received at the end of the year is ₹ $16537.50$.

(A) Principal $(P)$ = ₹ $5000$,Rate $(R)$ = $4 \%$ per annum,Time $(n)$ = $1 \frac{1}{2}$ years = $3$ half-years.
Case $1$: Compounded yearly.
For the first year,interest is calculated yearly. For the remaining half-year,simple interest is applied.
Amount $= 5000 \times (1 + \frac{4}{100})^1 \times (1 + \frac{2}{100})^1 = 5000 \times \frac{26}{25} \times \frac{51}{50} = 200 \times 26 \times \frac{51}{50} = 4 \times 26 \times 51 = ₹ 5304$.
Compound Interest $(CI_1)$ $= 5304 - 5000 = ₹ 304$.
Case $2$: Compounded half-yearly.
Rate $(r)$ = $4/2 = 2 \%$ per half-year,Time $(n)$ = $3$ half-years.
Amount $= 5000 \times (1 + \frac{2}{100})^3 = 5000 \times (1.02)^3 = 5000 \times 1.061208 = ₹ 5306.04$.
Compound Interest $(CI_2)$ $= 5306.04 - 5000 = ₹ 306.04$.
Difference $= CI_2 - CI_1 = 306.04 - 304 = ₹ 2.04$.

(B) Let each annual instalment be ₹ $x$.
The present value of the instalments must equal the total debt:
$\frac{x}{(1 + \frac{5}{100})^1} + \frac{x}{(1 + \frac{5}{100})^2} = 1025$
Simplify the fractions:
$\frac{x}{1.05} + \frac{x}{(1.05)^2} = 1025$
$\frac{x}{21/20} + \frac{x}{(21/20)^2} = 1025$
$\frac{20x}{21} + \frac{400x}{441} = 1025$
Find a common denominator $(441)$:
$\frac{420x + 400x}{441} = 1025$
$\frac{820x}{441} = 1025$
$x = \frac{1025 \times 441}{820}$
$x = 1.25 \times 441 = 551.25$
Thus,the annual payment is ₹ $551.25$.

(C) The formula for compound interest is $A = P(1 + \frac{R}{100})^n$, where $A$ is the amount, $P$ is the principal, $R$ is the rate, and $n$ is the time in years.
Given that the sum doubles in $5 \, \text{years}$, we have $2P = P(1 + \frac{R}{100})^5$, which simplifies to $(1 + \frac{R}{100})^5 = 2$.
We want to find the time $N$ such that the sum becomes $8$ times itself: $8P = P(1 + \frac{R}{100})^N$.
This simplifies to $(1 + \frac{R}{100})^N = 8$.
Since $8 = 2^3$, we can substitute $2 = (1 + \frac{R}{100})^5$ into the equation:
$(1 + \frac{R}{100})^N = [(1 + \frac{R}{100})^5]^3$.
$(1 + \frac{R}{100})^N = (1 + \frac{R}{100})^{15}$.
Comparing the exponents, we get $N = 15 \, \text{years}$.

(A) Let the amount invested in scheme $A$ be $₹ x$.
Then,the amount invested in scheme $B$ is $₹ (27000 - x)$.
Compound interest for $2$ years is given by the formula $P[(1 + r/100)^n - 1]$.
For scheme $A$: $CI_A = x[(1 + 8/100)^2 - 1] = x[(1.08)^2 - 1] = x[1.1664 - 1] = 0.1664x$.
For scheme $B$: $CI_B = (27000 - x)[(1 + 9/100)^2 - 1] = (27000 - x)[(1.09)^2 - 1] = (27000 - x)[1.1881 - 1] = 0.1881(27000 - x)$.
The total interest is $4818.30$.
So,$0.1664x + 0.1881(27000 - x) = 4818.30$.
$0.1664x + 5078.7 - 0.1881x = 4818.30$.
$-0.0217x = 4818.30 - 5078.7$.
$-0.0217x = -260.4$.
$x = 260.4 / 0.0217 = 12000$.
Thus,the amount invested in scheme $A$ is ₹ $12000$.

(D) Let the principal be $P$ and the rate of interest be $R \%$.
Simple Interest $(S.I.)$ for $2$ years is given by:
$S.I. = \frac{P \times R \times 2}{100} = 660$ --- $(1)$
Compound Interest $(C.I.)$ for $2$ years is given by:
$C.I. = P \left[ \left(1 + \frac{R}{100}\right)^2 - 1 \right] = 696.30$ --- $(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{P \left[ \left(1 + \frac{R}{100}\right)^2 - 1 \right]}{\frac{P \times 2R}{100}} = \frac{696.30}{660}$
$\frac{\frac{R^2}{10000} + \frac{2R}{100}}{\frac{2R}{100}} = 1.055$
$\frac{R}{200} + 1 = 1.055$
$\frac{R}{200} = 0.055$
$R = 0.055 \times 200 = 11$
Thus,the rate of interest is $11 \%$.

(C) Simple Interest ($S$.$I$.) for $6$ years $= 0.6P$.
Since $S.I. = \frac{P \times R \times T}{100}$,we have $0.6P = \frac{P \times R \times 6}{100}$.
Solving for $R$,we get $R = \frac{0.6 \times 100}{6} = 10 \% \text{ per annum}$.
Now,we calculate the Compound Interest ($C$.$I$.) for $P = ₹ 12000$,$R = 10 \%$,and $T = 3$ years.
$C.I. = P \left[ \left( 1 + \frac{R}{100} \right)^T - 1 \right]$.
$C.I. = 12000 \left[ \left( 1 + \frac{10}{100} \right)^3 - 1 \right] = 12000 \left[ (1.1)^3 - 1 \right]$.
$C.I. = 12000 \times (1.331 - 1) = 12000 \times 0.331 = ₹ 3972$.

(B) Let the principal sum be $P$.
Given compound interest ($C$.$I$.) for $n = 2$ years at $r = 10 \%$ per annum is ₹ $525$.
Formula for $C$.$I$. $= P \left[ \left( 1 + \frac{r}{100} \right)^n - 1 \right] = 525$.
$P \left[ \left( 1 + \frac{10}{100} \right)^2 - 1 \right] = 525$
$P \left[ (1.1)^2 - 1 \right] = 525$
$P [1.21 - 1] = 525$
$P (0.21) = 525$
$P = \frac{525}{0.21} = \frac{52500}{21} = ₹ 2500$.
Now,we need to find simple interest ($S$.$I$.) for double the time ($n' = 2 \times 2 = 4$ years) at half the rate ($r' = \frac{10}{2} = 5 \%$ per annum).
$S$.$I$. $= \frac{P \times r' \times n'}{100} = \frac{2500 \times 5 \times 4}{100} = 25 \times 20 = ₹ 500$.