Compound Interest Questions in English

(B) The formula for Compound Interest $(CI)$ is given by:
$CI = P \left[\left(1+\frac{R}{100}\right)^{n}-1\right]$
Where:
$P = 17500$
$R = 12\%$
$n = 2$
Substituting the values:
$CI = 17500 \left[\left(1+\frac{12}{100}\right)^{2}-1\right]$
$CI = 17500 \left[(1.12)^{2}-1\right]$
$CI = 17500 \left[1.2544-1\right]$
$CI = 17500 \times 0.2544$
$CI = 4452$
Thus,the compound interest is $Rs. 4452$.

(C) The formula for Compound Interest $(C.I.)$ is given by: $C.I. = P \left[\left(1 + \frac{R}{100}\right)^{n} - 1\right]$
Given:
Principal $(P)$ = $Rs. 12000$
Rate $(R)$ = $9\% \, p.c.p.a$
Time $(n)$ = $3 \, \text{years}$
Substituting the values in the formula:
$C.I. = 12000 \left[\left(1 + \frac{9}{100}\right)^{3} - 1\right]$
$C.I. = 12000 \left[(1.09)^{3} - 1\right]$
$C.I. = 12000 \left[1.295029 - 1\right]$
$C.I. = 12000 \times 0.295029$
$C.I. = 3540.348$
Rounding to the nearest integer,the value is approximately $Rs. 3540$.

(D) The formula for Compound Interest $(C.I.)$ is given by:
$C.I. = P \left[ \left( 1 + \frac{R}{100} \right)^n - 1 \right]$
Where:
$P = 4800$
$R = 5$
$n = 3$
Substituting the values:
$C.I. = 4800 \left[ \left( 1 + \frac{5}{100} \right)^3 - 1 \right]$
$C.I. = 4800 \left[ (1.05)^3 - 1 \right]$
$C.I. = 4800 \left[ 1.157625 - 1 \right]$
$C.I. = 4800 \times 0.157625$
$C.I. = 756.6$
Thus,the compound interest is $Rs. 756.6$.

(A) The formula for Compound Interest $(C.I.)$ is given by:
$C.I. = P \left[ \left( 1 + \frac{R}{100} \right)^n - 1 \right]$
Given:
Principal $(P)$ = $12500$
Rate $(R)$ = $12 \, p.c.p.a.$
Time $(n)$ = $2 \, \text{years}$
Substituting the values:
$C.I. = 12500 \left[ \left( 1 + \frac{12}{100} \right)^2 - 1 \right]$
$C.I. = 12500 \left[ (1.12)^2 - 1 \right]$
$C.I. = 12500 \left[ 1.2544 - 1 \right]$
$C.I. = 12500 \times 0.2544$
$C.I. = 3180$
Thus, the compound interest is $Rs. 3180$.

(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 amount and $R$ is the rate of interest.
Given: $P = 10000$,Difference $= 64$.
Substituting the values:
$64 = \frac{10000 \times R^{2}}{100 \times 100}$
$64 = \frac{10000 \times R^{2}}{10000}$
$64 = R^{2}$
$R = \sqrt{64} = 8 \%$

(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 $= 1$,$R = 4$,and time $= 2$ years.
Substituting the values into the formula:
$1 = \frac{P \times 4^2}{100^2}$
$1 = \frac{P \times 16}{10000}$
$P = \frac{10000}{16}$
$P = 625$
Thus,the sum is $Rs. 625$.

(D) Given: Principal $(P) = 12500$,Rate $(R) = 8 \%$ per annum,Time $(T) = 9$ months.
Since the interest is compounded quarterly,we adjust the rate and time period:
Number of quarters in $9$ months $= 9 / 3 = 3$ quarters.
Quarterly rate $= 8 \% / 4 = 2 \%$.
Using the compound interest formula: $CI = P \times [(1 + R/100)^n - 1]$,where $n$ is the number of quarters.
$CI = 12500 \times [(1 + 2/100)^3 - 1]$
$CI = 12500 \times [(1.02)^3 - 1]$
$CI = 12500 \times [1.061208 - 1]$
$CI = 12500 \times 0.061208 = 765.1$
Rounding to the nearest integer,the compound interest is $Rs. 765$.

(B) Given: Principal $(P) = Rs. 32000$,Rate $(R) = 20 \%$ per annum,Time $(T) = 1$ year.
Since the interest is compounded half-yearly,the rate becomes $R' = \frac{20}{2} = 10 \%$ per half-year,and the time becomes $n = 1 \times 2 = 2$ half-years.
The formula for compound interest is $CI = P \left[ \left( 1 + \frac{R'}{100} \right)^n - 1 \right]$.
Substituting the values: $CI = 32000 \left[ \left( 1 + \frac{10}{100} \right)^2 - 1 \right]$.
$CI = 32000 \left[ (1.1)^2 - 1 \right] = 32000 [1.21 - 1] = 32000 \times 0.21$.
$CI = 6720$.
Therefore,the compound interest is $Rs. 6720$.

(D) The formula for the difference between Compound Interest $(C.I.)$ and Simple Interest $(S.I.)$ for $2$ years is given by: $\text{Difference} = \frac{P \times R^2}{100^2}$
Given:
Principal $(P)$ = $Rs. 700$
Rate $(R)$ = $5 \%$
Time $(n)$ = $2$ years
Substituting the values into the formula:
$\text{Difference} = \frac{700 \times 5^2}{100^2}$
$\text{Difference} = \frac{700 \times 25}{10000}$
$\text{Difference} = \frac{17500}{10000} = Rs. 1.75$

(C) The formula for the difference between compound interest $(CI)$ and simple interest $(SI)$ for $2$ years is given by: $\text{Difference} = \frac{PR^2}{100^2}$.
Given:
$\text{Difference} = Rs. 10$
$R = 6 \frac{1}{4} \% = \frac{25}{4} \%$
Substituting the values into the formula:
$10 = P \times \left[ \frac{25/4}{100} \right]^2$
$10 = P \times \left[ \frac{25}{400} \right]^2$
$10 = P \times \left[ \frac{1}{16} \right]^2$
$10 = P \times \frac{1}{256}$
$P = 10 \times 256 = 2560$.
Therefore,the sum is $Rs. 2560$.

(C) Let the principal sum be $P$.
Given time $T = 1 \text{ year}$ and rate $R = 10 \% \text{ per annum}$.
For simple interest $(SI)$: $SI = \frac{P \times R \times T}{100} = \frac{P \times 10 \times 1}{100} = 0.1P$.
For compound interest $(CI)$ compounded half-yearly: The rate becomes $R' = \frac{10}{2} = 5 \% \text{ per half-year}$ and time $n = 2 \text{ half-years}$.
$CI = P(1 + \frac{R'}{100})^n - P = P(1 + \frac{5}{100})^2 - P = P(1.05)^2 - P = P(1.1025) - P = 0.1025P$.
The difference between $CI$ and $SI$ is $0.1025P - 0.1P = 0.0025P$.
Given $0.0025P = 25$.
$P = \frac{25}{0.0025} = \frac{250000}{25} = 10000$.
Thus,the sum is $Rs. 10000$.

(B) 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 amount and $R$ is the rate of interest per annum.
Given: $P = 2000$,Difference $= 12.8$.
Substituting the values in the formula:
$12.8 = \frac{2000 \times R^2}{10000}$
$12.8 = \frac{2 \times R^2}{10}$
$12.8 = 0.2 \times R^2$
$R^2 = \frac{12.8}{0.2} = 64$
$R = \sqrt{64} = 8$
Therefore,the rate of interest is $8 \%$ per annum.

(D) Let the principal amount be $P$.
According to the compound interest formula,$A = P(1 + r/100)^n$.
Given that the sum doubles in $3$ years: $2P = P(1 + r/100)^3$,which implies $(1 + r/100)^3 = 2$.
We want to find the time $t$ in which the sum becomes $4$ times: $4P = P(1 + r/100)^t$.
This simplifies to $4 = (1 + r/100)^t$.
Since $4 = 2^2$,we can substitute $2 = (1 + r/100)^3$ into the equation:
$4 = ((1 + r/100)^3)^2 = (1 + r/100)^6$.
Comparing the exponents,we get $t = 6$ years.

(D) Let the principal be $P = 100$.
The amount $A$ after $2$ years is $6 \frac{1}{4}$ times the principal,so $A = 100 \times \frac{25}{4} = 625$.
The formula for compound interest is $A = P(1 + \frac{r}{100})^n$,where $n = 2$.
Substituting the values: $625 = 100(1 + \frac{r}{100})^2$.
Dividing both sides by $100$: $6.25 = (1 + \frac{r}{100})^2$.
Taking the square root of both sides: $\sqrt{6.25} = 1 + \frac{r}{100}$.
$2.5 = 1 + \frac{r}{100}$.
$1.5 = \frac{r}{100}$.
$r = 1.5 \times 100 = 150 \%$.

(D) The formula for the amount $A$ under compound interest is given by $A = P \left(1 + \frac{R}{100}\right)^n$,where $P$ is the principal,$R$ is the rate of interest,and $n$ is the time in years.
Given: $P = 25000$,$R = 8 \%$,$n = 2$.
Substituting the values into the formula:
$A = 25000 \left(1 + \frac{8}{100}\right)^2$
$A = 25000 \left(1 + 0.08\right)^2$
$A = 25000 \times (1.08)^2$
$A = 25000 \times 1.1664$
$A = 29160$.
Therefore,Amit will receive $Rs. 29160$.

(B) Let the principal amount be $P = 100$.
Rate $R = 5 \%$ per annum and time $T = 2$ years.
Simple Interest $(S.I.)$ $= \frac{P \times R \times T}{100} = \frac{100 \times 5 \times 2}{100} = 10$.
Compound Interest $(C.I.)$ $= P \left[ \left( 1 + \frac{R}{100} \right)^T - 1 \right] = 100 \left[ \left( 1 + \frac{5}{100} \right)^2 - 1 \right]$.
$C.I. = 100 \left[ \left( \frac{105}{100} \right)^2 - 1 \right] = 100 \left[ \left( \frac{21}{20} \right)^2 - 1 \right] = 100 \left[ \frac{441}{400} - 1 \right] = 100 \times \frac{41}{400} = \frac{41}{4} = 10.25$.
Ratio of $S.I. : C.I. = 10 : 10.25 = 1000 : 1025$.
Dividing both by $25$,we get $40 : 41$.

(C) The formula for the difference between Compound Interest $(C.I.)$ and Simple Interest $(S.I.)$ for $2$ years is given by:
Difference $= \frac{P \times R^2}{100^2}$
Where $P$ is the principal amount,and $R$ is the rate of interest per annum.
Given $P = 12000$ and $R = 5 \%$.
Difference $= \frac{12000 \times 5^2}{100^2}$
Difference $= \frac{12000 \times 25}{10000}$
Difference $= \frac{12000}{400} = 30$
Thus,the difference is $Rs. 30$.

(A) The formula for the difference between Compound Interest $(C.I.)$ and Simple Interest $(S.I.)$ for $3$ years is given by:
Difference $= \frac{P \times R^2 \times (300 + R)}{100^3}$
Given:
Principal $(P)$ $= Rs. 15000$
Rate $(R)$ $= 3 \%$
Time $(n)$ $= 3$ years
Substituting the values into the formula:
Difference $= \frac{15000 \times 3^2 \times (300 + 3)}{100^3}$
Difference $= \frac{15000 \times 9 \times 303}{1000000}$
Difference $= \frac{135000 \times 303}{1000000}$
Difference $= \frac{40905000}{1000000} = 40.905$
Rounding to two decimal places,the difference is $Rs. 40.91$.

(A) The formula for the difference between Compound Interest $(C.I.)$ and Simple Interest $(S.I.)$ for $3$ years is given by:
Difference $= \frac{P \times R^2 \times (300 + R)}{100^3}$
Given:
Principal $(P)$ $= Rs. 13000$
Rate $(R)$ $= 4 \%$
Time $(T)$ $= 3$ years
Substituting the values into the formula:
Difference $= \frac{13000 \times 4^2 \times (300 + 4)}{100^3}$
Difference $= \frac{13000 \times 16 \times 304}{1000000}$
Difference $= \frac{13000 \times 4864}{1000000}$
Difference $= \frac{63232000}{1000000} = Rs. 63.232$
Rounding to two decimal places,the difference is $Rs. 63.23$.

(D) Step $1$: Calculate the Principal $(P)$.
Given $S.I. = Rs. 500$,$R = 5 \%$,$T = 1$ year.
$P = \frac{S.I. \times 100}{R \times T} = \frac{500 \times 100}{5 \times 1} = Rs. 10,000$.
Step $2$: Calculate the Compound Interest $(C.I.)$ for $2$ years.
$C.I. = P \left[ \left( 1 + \frac{R}{100} \right)^n - 1 \right]$
$C.I. = 10,000 \left[ \left( 1 + \frac{5}{100} \right)^2 - 1 \right]$
$C.I. = 10,000 \left[ (1.05)^2 - 1 \right]$
$C.I. = 10,000 \left[ 1.1025 - 1 \right]$
$C.I. = 10,000 \times 0.1025 = Rs. 1,025$.

(C) The formula for the difference between Compound Interest $(C.I.)$ and Simple Interest $(S.I.)$ for $2$ years is given by: $\text{Difference} = \frac{P \times R^2}{100^2}$
Given:
$\text{Difference} = 35$
$R = 5\%$
$T = 2 \text{ years}$
Substituting the values into the formula:
$35 = \frac{P \times 5^2}{100^2}$
$35 = \frac{P \times 25}{10000}$
$35 = \frac{P}{400}$
$P = 35 \times 400$
$P = 14000$
Therefore,the principal amount is $Rs. 14000$.

(C) Given: Principal $(P) = Rs. 10000$,Annual Rate $= 20 \%$,Time $= 1$ year $6$ months $= 1.5$ years.
Since the interest is calculated half-yearly:
Rate $(R) = \frac{20}{2} = 10 \%$ per half-year.
Time $(n) = 1.5 \times 2 = 3$ half-years.
Compound Interest $(C.I.) = P \left[ \left( 1 + \frac{R}{100} \right)^n - 1 \right]$.
$C.I. = 10000 \left[ \left( 1 + \frac{10}{100} \right)^3 - 1 \right]$.
$C.I. = 10000 \left[ \left( \frac{11}{10} \right)^3 - 1 \right]$.
$C.I. = 10000 \left[ \frac{1331}{1000} - 1 \right]$.
$C.I. = 10000 \left[ \frac{331}{1000} \right] = Rs. 3310$.

(D) Step $1$: Calculate the principal sum $(P)$ using the Simple Interest formula: $S.I. = (P \times R \times T) / 100$.
Given $S.I. = 7200$,$R = 12$,$T = 6$.
$7200 = (P \times 12 \times 6) / 100$
$7200 = (P \times 72) / 100$
$P = (7200 \times 100) / 72 = 10000$.
Step $2$: Calculate the Compound Interest $(C.I.)$ on the principal $P = 10000$ at $R = 5$ $p.c.p.a.$ for $T = 2$ years.
$C.I. = P \times [(1 + R/100)^T - 1]$
$C.I. = 10000 \times [(1 + 5/100)^2 - 1]$
$C.I. = 10000 \times [(1.05)^2 - 1]$
$C.I. = 10000 \times [1.1025 - 1]$
$C.I. = 10000 \times 0.1025 = 1025$.
Thus,the $C.I.$ is $Rs. 1025$.

(C) The formula for the difference between $Compound Interest$ $(C.I.)$ and $Simple Interest$ $(S.I.)$ for $2$ years is given by:
$Difference = \frac{P \times R^2}{100^2}$
Given:
$Principal (P) = 10000$
$Difference = 25$
$Time (T) = 2$ years
Substituting the values into the formula:
$25 = \frac{10000 \times R^2}{10000}$
$25 = R^2$
$R = \sqrt{25} = 5 \%$
Therefore,the rate of interest is $5 \%$ per annum.

(D) For $2$ years,the Compound Interest $(C.I.)$ is given by $C.I. = P \left[ \left( 1 + \frac{R}{100} \right)^2 - 1 \right] = P \left[ \frac{2R}{100} + \frac{R^2}{10000} \right]$.
The Simple Interest $(S.I.)$ for $2$ years is $S.I. = \frac{P \times R \times 2}{100} = \frac{2PR}{100}$.
The ratio is given as $\frac{C.I.}{S.I.} = 1.2$.
Substituting the expressions: $\frac{P \left( \frac{2R}{100} + \frac{R^2}{10000} \right)}{\frac{2PR}{100}} = 1.2$.
Simplifying the expression: $\frac{\frac{2R}{100} (1 + \frac{R}{200})}{\frac{2R}{100}} = 1 + \frac{R}{200} = 1.2$.
Therefore,$\frac{R}{200} = 0.2$,which gives $R = 0.2 \times 200 = 40 \%.$

(B) The formula for the amount $A$ with varying rates of interest is $A = P \times (1 + \frac{R_1}{100})^{n_1} \times (1 + \frac{R_2}{100})^{n_2}$.
Given: Principal $P = 7500$,$R_1 = 10 \%$,$n_1 = 2$ years,$R_2 = 20 \%$,$n_2 = 2$ years.
First,calculate the total amount $A$:
$A = 7500 \times (1 + \frac{10}{100})^2 \times (1 + \frac{20}{100})^2$
$A = 7500 \times (1.1)^2 \times (1.2)^2$
$A = 7500 \times 1.21 \times 1.44$
$A = 7500 \times 1.7424 = 13068$.
Now,calculate the Compound Interest $(C.I.)$:
$C.I. = A - P$
$C.I. = 13068 - 7500 = 5568$.
Thus,the compound interest is $Rs. 5568$.

(C) Given: Principal $(P) = Rs. 10000$,Rate $(R) = 20 \% \text{ per annum}$,Time $(T) = 2 \text{ years}$.
Since the interest is compounded half-yearly:
New Rate $(r) = \frac{20}{2} = 10 \% \text{ per half-year}$.
New Time $(n) = 2 \times 2 = 4 \text{ half-years}$.
The formula for Compound Interest $(C.I.)$ is $C.I. = P \left[ \left( 1 + \frac{r}{100} \right)^n - 1 \right]$.
Substituting the values: $C.I. = 10000 \left[ \left( 1 + \frac{10}{100} \right)^4 - 1 \right]$.
$C.I. = 10000 \left[ (1.1)^4 - 1 \right]$.
$C.I. = 10000 \left[ 1.4641 - 1 \right]$.
$C.I. = 10000 \times 0.4641 = Rs. 4641$.

(C) The formula for the difference between Simple Interest $(S.I.)$ and Compound Interest $(C.I.)$ for $2$ years is given by:
Difference $= \frac{P \times R^{2}}{100^{2}}$
Where $P$ is the principal amount and $R$ is the rate of interest.
Given: $P = 10000$,$R = 12 \%$,and $T = 2$ years.
Substituting the values in the formula:
Difference $= \frac{10000 \times (12)^{2}}{100^{2}}$
Difference $= \frac{10000 \times 144}{10000}$
Difference $= 144$
Thus,the difference between $S.I.$ and $C.I.$ is $Rs. 144$.

(D) The formula for the difference between Simple Interest $(S.I.)$ and Compound Interest $(C.I.)$ for $2$ years is given by: $\text{Difference} = \frac{P \times r^2}{100^2}$.
Here,the principal $P = 10000$ and the rate $r = 14 \%$.
Substituting the values into the formula:
$\text{Difference} = \frac{10000 \times 14^2}{100^2}$
$\text{Difference} = \frac{10000 \times 196}{10000}$
$\text{Difference} = 196$.
Thus,the difference between $S.I.$ and $C.I.$ is $Rs. 196$.

(C) Step $1$: Calculate the amount after $2$ years of compound interest $(C.I.)$.
Amount $A = P(1 + R/100)^n$
$A = 50000(1 + 10/100)^2 = 50000(1.1)^2 = 50000 \times 1.21 = Rs. 60500$.
Step $2$: Calculate the simple interest $(S.I.)$ on this amount for $3$ years at $12 \%$ per annum.
$S.I. = (P \times R \times T) / 100$
$S.I. = (60500 \times 12 \times 3) / 100 = 605 \times 36 = Rs. 21780$.
Step $3$: Calculate the final total amount.
Final Amount = Principal for $S.I. + S.I. = 60500 + 21780 = Rs. 82280$.

(C) Let the principal $P = 100$. The amount increases by $60 \%$,so the simple interest $I = 60$.
Using the simple interest formula $I = \frac{P \times R \times T}{100}$,we have $60 = \frac{100 \times R \times 6}{100}$,which gives $R = 10 \%$.
Now,for compound interest on $P = 12,000$ at $R = 10 \%$ for $T = 3$ years:
The amount $A = P(1 + \frac{R}{100})^T = 12,000(1 + \frac{10}{100})^3 = 12,000(1.1)^3$.
$A = 12,000 \times 1.331 = 15,972$.
The compound interest $CI = A - P = 15,972 - 12,000 = 3,972$.

(B) The interest is calculated on a half-yearly basis,so the rate per half-year is $r = \frac{5}{2} \% = 2.5 \% = 0.025$.
The first deposit of $Rs. 1600$ is made on $1^{st}$ January. It earns interest for two half-yearly periods (one year).
Amount of first deposit at the end of the year: $A_1 = 1600 \times (1 + 0.025)^2 = 1600 \times (1.025)^2 = 1600 \times 1.050625 = Rs. 1681$.
The second deposit of $Rs. 1600$ is made on $1^{st}$ July. It earns interest for one half-yearly period (six months).
Amount of second deposit at the end of the year: $A_2 = 1600 \times (1 + 0.025)^1 = 1600 \times 1.025 = Rs. 1640$.
Total amount at the end of the year: $A = A_1 + A_2 = 1681 + 1640 = Rs. 3321$.
Total principal deposited: $P = 1600 + 1600 = Rs. 3200$.
Interest gained: $CI = A - P = 3321 - 3200 = Rs. 121$.

(B) Let the amount invested in scheme $A$ be $P_A$ and in scheme $B$ be $P_B$.
Given,total investment $P_A + P_B = 35000$.
For scheme $A$,$S.I. = 3600$,$R = 12\%$,$T = 2$ years.
Using the formula $S.I. = (P_A \times R \times T) / 100$:
$3600 = (P_A \times 12 \times 2) / 100$
$3600 = P_A \times 0.24$
$P_A = 3600 / 0.24 = 15000$.
Now,$P_B = 35000 - 15000 = 20000$.
For scheme $B$,$P_B = 20000$,$R = 10\%$,$T = 2$ years,compounded annually.
$C.I. = P_B \times [(1 + R/100)^T - 1]$
$C.I. = 20000 \times [(1 + 10/100)^2 - 1]$
$C.I. = 20000 \times [(1.1)^2 - 1]$
$C.I. = 20000 \times [1.21 - 1]$
$C.I. = 20000 \times 0.21 = 4200$.
Thus,the interest accrued in scheme $B$ is $Rs. 4200$.

(D) Step $1$: Calculate the Principal $(P)$.
Given $SI = Rs. 3800$,$R = 8\%$,and $T = 5$ years.
Formula: $SI = \frac{P \times R \times T}{100}$
$3800 = \frac{P \times 8 \times 5}{100}$
$3800 = \frac{40P}{100} = 0.4P$
$P = \frac{3800}{0.4} = Rs. 9500$.
Step $2$: Calculate the Compound Interest $(CI)$ for $2$ years.
Formula: $CI = P \left[ \left( 1 + \frac{R}{100} \right)^n - 1 \right]$
$CI = 9500 \left[ \left( 1 + \frac{8}{100} \right)^2 - 1 \right]$
$CI = 9500 \left[ (1.08)^2 - 1 \right]$
$CI = 9500 \left[ 1.1664 - 1 \right]$
$CI = 9500 \times 0.1664 = Rs. 1580.8$.

(C) To find the rate of interest $R$,we analyze the statements:
From statement $(I)$: The difference between $C.I.$ and $S.I.$ for $2$ years is given by the formula $D = \frac{P R^2}{100^2}$. Here,$P = 15000$ and $D = 170$. Since $P$ and $D$ are known,we can solve for $R$.
From statement $(II)$: The $S.I.$ for $2$ years is given by $S.I. = \frac{P \times R \times T}{100}$. Here,$P = 15000$,$T = 2$,and $S.I. = 2500$. Since $P$,$T$,and $S.I.$ are known,we can solve for $R$.
Therefore,either statement $(I)$ or statement $(II)$ is sufficient to answer the question.

(C) To find the rate of interest $(R)$,we can use either statement.
From statement $(I)$: Simple Interest $(S.I.)$ = $19$,Principal $(P)$ = $800$,Time $(T)$ = $3$ years. Using the formula $S.I. = (P \times R \times T) / 100$,we get $19 = (800 \times R \times 3) / 100$,which allows us to solve for $R$.
From statement $(II)$: The difference between Compound Interest $(C.I.)$ and Simple Interest $(S.I.)$ for $2$ years is given by the formula $Difference = P \times (R/100)^2$. Given $Difference = 15.76$ and $P = 800$,we can substitute these values to solve for $R$.
Since both statements independently provide enough information to calculate the rate of interest,the correct answer is 'Either $I$ or $II$'.

(A) The formula for the difference between $C.I.$ and $S.I.$ for $2$ years is given by: $Difference = P \times (R/100)^2$.
From statement $(I)$,we can calculate the rate of interest $(R)$:
$S.I. = (P \times R \times T) / 100$
$120 = (1000 \times R \times 2) / 100$
$120 = 20R \implies R = 6\%$.
From statement $(II)$,we are given the principal $(P = 2000)$.
To find the principal amount when the difference is $18$,we need both the rate $(R)$ and the difference. Since statement $(I)$ provides the rate and statement $(II)$ provides the principal,we need to check if they are consistent or if both are required to solve for a missing variable. Actually,if we use the formula $18 = P \times (R/100)^2$,we have two variables ($P$ and $R$). Statement $(I)$ gives $R=6\%$. If we use this in the formula,we can find $P$. Thus,statement $(I)$ alone is sufficient. However,if the question asks to verify the principal,statement $(I)$ is sufficient. Given the options,$(I)$ alone is sufficient.

(B) Let the principal sum be $P$ and the rate of interest be $R \%$.
Given: Simple Interest $(SI) = Rs. 100$,Compound Interest $(CI) = Rs. 110$,Time $(t) = 2$ years.
For $2$ years,the formula for Simple Interest is $SI = \frac{P \times R \times t}{100} = \frac{P \times R \times 2}{100} = 100$,which implies $P \times R = 5000$.
The difference between Compound Interest and Simple Interest for $2$ years is given by $CI - SI = \frac{P \times R^2}{100^2}$.
Substituting the values: $110 - 100 = \frac{P \times R^2}{10000} \Rightarrow 10 = \frac{(P \times R) \times R}{10000}$.
Since $P \times R = 5000$,we have $10 = \frac{5000 \times R}{10000} \Rightarrow 10 = \frac{R}{2} \Rightarrow R = 20 \%$.
Now,substitute $R = 20$ into $P \times R = 5000$: $P \times 20 = 5000 \Rightarrow P = 250$.
Therefore,the sum is $Rs. 250$ and the rate of interest is $20 \%$ p.a.

(B) Let the principal sum be $P$.
The formula for the amount $A$ under compound interest is $A = P(1 + \frac{R}{100})^n$,where $R$ is the rate of interest and $n$ is the number of years.
According to the question,we want the amount to be more than double the principal,so $A > 2P$.
Substituting the values: $P(1 + \frac{20}{100})^n > 2P$.
Dividing both sides by $P$: $(1.2)^n > 2$.
Now,test values for $n$:
For $n = 3$: $(1.2)^3 = 1.728 < 2$.
For $n = 4$: $(1.2)^4 = 2.0736 > 2$.
Thus,the least number of complete years required is $4$.

(C) Given,initial population $P = 10$ crore.
Population after $n = 3$ years $A = 13.31$ crore.
The formula for population growth is $A = P(1 + \frac{R}{100})^n$.
Substituting the values: $13.31 = 10(1 + \frac{R}{100})^3$.
Dividing both sides by $10$: $\frac{13.31}{10} = (1 + \frac{R}{100})^3$.
$1.331 = (1 + \frac{R}{100})^3$.
Since $1.331 = (1.1)^3$ or $(\frac{11}{10})^3$,we have $(1.1)^3 = (1 + \frac{R}{100})^3$.
Taking the cube root on both sides: $1.1 = 1 + \frac{R}{100}$.
Subtracting $1$ from both sides: $0.1 = \frac{R}{100}$.
Therefore,$R = 0.1 \times 100 = 10 \%$.
The annual growth rate is $10 \%$.

(A) Let the value of each equal annual instalment be $P$.
The formula for the present value of instalments paid at the end of each year under compound interest is:
$P \times (1 + r/100)^{-1} + P \times (1 + r/100)^{-2} = \text{Principal}$
Given,Principal = $5100$,Rate $(r)$ = $4 \%$,Time = $2$ years.
$\Rightarrow \frac{P}{(1 + 4/100)} + \frac{P}{(1 + 4/100)^2} = 5100$
$\Rightarrow \frac{P}{(26/25)} + \frac{P}{(26/25)^2} = 5100$
$\Rightarrow \frac{25P}{26} + \frac{625P}{676} = 5100$
Taking $676$ as the common denominator:
$\Rightarrow \frac{25 \times 26 \times P + 625P}{676} = 5100$
$\Rightarrow \frac{650P + 625P}{676} = 5100$
$\Rightarrow \frac{1275P}{676} = 5100$
$P = \frac{5100 \times 676}{1275}$
$P = 4 \times 676 = 2704$
Thus,each instalment is $Rs. 2704$.

(A) Let the first part given to $X$ be $Rs. a$ and the second part given to $Y$ be $Rs. (2602 - a)$.
According to the question,the amount of $X$ after $7 \,yr$ is equal to the amount of $Y$ after $9 \,yr$ with interest compounded at $4 \%$ per annum.
Formula for amount: $A = P(1 + \frac{r}{100})^n$.
So,$a(1 + \frac{4}{100})^7 = (2602 - a)(1 + \frac{4}{100})^9$.
Dividing both sides by $(1 + \frac{4}{100})^7$,we get:
$a = (2602 - a)(1 + \frac{4}{100})^2$.
$a = (2602 - a)(1 + \frac{1}{25})^2 = (2602 - a)(\frac{26}{25})^2$.
$a = (2602 - a) \times \frac{676}{625}$.
$625a = 676(2602 - a)$.
$625a = 1758952 - 676a$.
$1301a = 1758952$.
$a = \frac{1758952}{1301} = 1352$.
Thus,the first part is $Rs. 1352$ and the second part is $2602 - 1352 = Rs. 1250$.

(B) Let the rate of interest be $R\%$ per annum and the time be $n$ years.
Given that the amount $A = P(1 + R/100)^n$.
Substituting the given values: $4320 = 3000(1 + R/100)^n$.
Therefore,$(1 + R/100)^n = 4320 / 3000 = 1.44$.
We need to find the amount for half the time,i.e.,for $n/2$ years.
The required amount is $A' = 3000(1 + R/100)^{n/2}$.
Since $(1 + R/100)^n = 1.44$,taking the square root on both sides gives $(1 + R/100)^{n/2} = \sqrt{1.44} = 1.2$.
Thus,the required amount $A' = 3000 \times 1.2 = Rs. 3600$.

(A) Let the principal be $P$.
When interest is compounded half-yearly,the rate $R = 20/2 = 10 \% \, p.a.$ and time $T = 2 \times 2 = 4$ half-years.
$CI_{half-yearly} = P \left[ (1 + 10/100)^4 - 1 \right] = P \left[ (1.1)^4 - 1 \right] = P (1.4641 - 1) = 0.4641 P$.
When interest is compounded annually,the rate $R = 20 \% \, p.a.$ and time $T = 2$ years.
$CI_{annually} = P \left[ (1 + 20/100)^2 - 1 \right] = P \left[ (1.2)^2 - 1 \right] = P (1.44 - 1) = 0.44 P$.
According to the question,the difference is $Rs. 964$:
$0.4641 P - 0.44 P = 964$
$0.0241 P = 964$
$P = 964 / 0.0241 = 40000$.
Thus,the sum is $Rs. 40000$.

(C) Let the shares of $X$ and $Y$ be $Rs. x$ and $Rs. (8448 - x)$ respectively.
$X$ is $18$ years old,so the time remaining until age $21$ is $21 - 18 = 3$ years.
$Y$ is $19$ years old,so the time remaining until age $21$ is $21 - 19 = 2$ years.
The rate of interest $R = 6.25 \% = \frac{6.25}{100} = \frac{1}{16}$.
According to the problem,the amounts received by both at age $21$ are equal:
$x(1 + \frac{1}{16})^3 = (8448 - x)(1 + \frac{1}{16})^2$
Dividing both sides by $(1 + \frac{1}{16})^2$:
$x(1 + \frac{1}{16}) = 8448 - x$
$x(\frac{17}{16}) = 8448 - x$
$17x = 16(8448 - x)$
$17x = 135168 - 16x$
$33x = 135168$
$x = \frac{135168}{33} = 4096$.
Thus,the present share of $X$ is $Rs. 4096$.

(B) Let the amount invested in the first scheme be $Rs. x$ and in the second scheme be $Rs. (10500 - x)$.
The rate of interest is $10 \% \, p.a.$ compounded annually.
The amount after $n$ years is given by $A = P(1 + r/100)^n$.
For the first scheme,the amount after $2 \, years$ is $A_1 = x(1 + 10/100)^2 = x(1.1)^2 = 1.21x$.
For the second scheme,the amount after $3 \, years$ is $A_2 = (10500 - x)(1 + 10/100)^3 = (10500 - x)(1.1)^3 = 1.331(10500 - x)$.
Since the amounts are equal,$1.21x = 1.331(10500 - x)$.
$1.21x = 13975.5 - 1.331x$.
$1.21x + 1.331x = 13975.5$.
$2.541x = 13975.5$.
$x = 13975.5 / 2.541 = 5500$.
The amount deposited for $3 \, years$ is $(10500 - x) = 10500 - 5500 = Rs. 5000$.

(A) The amount of $Rs. 1000$ at $10 \% \, p.a.$ after $3 \, \text{years}$ becomes:
$\Rightarrow A = P(1 + \frac{R}{100})^n = 1000(1 + \frac{10}{100})^3 = 1000(1.1)^3 = 1000 \times 1.331 = Rs. 1331$.
Now,$Rs. 1728$ depreciates at rate $R_d \%$ per annum for $3 \, \text{years}$ to become $Rs. 1331$:
$\Rightarrow A = P(1 - \frac{R_d}{100})^n$
$\Rightarrow 1331 = 1728(1 - \frac{R_d}{100})^3$
$\Rightarrow (1 - \frac{R_d}{100})^3 = \frac{1331}{1728} = (\frac{11}{12})^3$
$\Rightarrow 1 - \frac{R_d}{100} = \frac{11}{12}$
$\Rightarrow \frac{R_d}{100} = 1 - \frac{11}{12} = \frac{1}{12}$
$\Rightarrow R_d = \frac{100}{12} = 8.33 \%$.
The difference between the rates is $10 \% - 8.33 \% = 1.67 \%$,which is approximately $1.7 \%$.

(D) Let the principal (sum of money) be $P = 100$.
Given that the amount after $2$ years at compound interest is $1.44$ times the principal, so $A = 144$.
Using the compound interest formula: $A = P(1 + \frac{R}{100})^n$
$144 = 100(1 + \frac{R}{100})^2$
$(1 + \frac{R}{100})^2 = \frac{144}{100} = 1.44$
$1 + \frac{R}{100} = \sqrt{1.44} = 1.2$
$\frac{R}{100} = 0.2 \Rightarrow R = 20\%$.
Now, we consider twice the original sum, so the new principal $P' = 2 \times 100 = 200$.
We want this sum to double itself, so the new amount $A' = 2 \times 200 = 400$.
The simple interest required is $SI = A' - P' = 400 - 200 = 200$.
Using the simple interest formula: $SI = \frac{P' \times R \times T}{100}$
$200 = \frac{200 \times 20 \times T}{100}$
$200 = 40 \times T$
$T = \frac{200}{40} = 5$ years.

(D) The difference in the amount of money over one year represents the compound interest for that year.
Hence,the interest for the third year is Rs. $[2977.54 - 2809] = Rs. 168.54$.
This interest is earned on the principal amount at the end of the second year,which is Rs. $2809$.
Using the formula for simple interest for one year (since $CI$ for one year equals $SI$):
$R = \frac{Interest \times 100}{Principal \times Time} = \frac{168.54 \times 100}{2809 \times 1} = \frac{16854}{2809} = 6 \%$.
Now,to find the original sum $P$,we use the amount formula for $2$ years:
$A = P(1 + \frac{R}{100})^n$
$2809 = P(1 + \frac{6}{100})^2$
$2809 = P(1.06)^2$
$2809 = P(1.1236)$
$P = \frac{2809}{1.1236} = 2500$.
Thus,the rate of interest is $6 \%$ and the original sum is Rs. $2500$.

(B) The formula for the amount under compound interest is $A = P(1 + \frac{R}{100})^n$.
Given $P = 1200$,$A = 1323$,and $n = 2$ years.
$1323 = 1200(1 + \frac{R}{100})^2$
$\frac{1323}{1200} = (1 + \frac{R}{100})^2$
$\frac{441}{400} = (1 + \frac{R}{100})^2$
$(\frac{21}{20})^2 = (1 + \frac{R}{100})^2$
Taking the square root on both sides: $1 + \frac{R}{100} = \frac{21}{20} = 1.05$.
Thus,$\frac{R}{100} = 0.05$,which means $R = 5\%$.
Now,for $P = 1600$,$n = 3$ years,and $R = 5\%$:
$A = 1600(1 + \frac{5}{100})^3$
$A = 1600(1.05)^3$
$A = 1600 \times 1.157625 = 1852.50$.
Therefore,the amount is $Rs. 1852.50$.