Compound Interest Questions in English

(A) Given,Simple Interest ($S$.$I$.) = ₹ $50$,Time $(T)$ = $2$ years,Rate $(R)$ = $5 \%$.
Using the formula for Simple Interest: $S.I. = \frac{P \times R \times T}{100}$.
$50 = \frac{P \times 5 \times 2}{100} \implies 50 = \frac{10P}{100} \implies 50 = \frac{P}{10}$.
Therefore,Principal $(P)$ = ₹ $500$.
Now,calculating Compound Interest ($C$.$I$.) for the same principal,rate,and time:
$C.I. = P \left[ \left( 1 + \frac{R}{100} \right)^T - 1 \right]$.
$C.I. = 500 \left[ \left( 1 + \frac{5}{100} \right)^2 - 1 \right] = 500 \left[ \left( \frac{21}{20} \right)^2 - 1 \right]$.
$C.I. = 500 \left[ \frac{441}{400} - 1 \right] = 500 \times \frac{41}{400}$.
$C.I. = \frac{5 \times 41}{4} = \frac{205}{4} = ₹ 51.25$.

(B) Let the principal amount be $P$ and the number of years be $N$.
Given that the rate of interest $R = 20 \%$.
The formula for the amount $A$ in compound interest is $A = P(1 + R/100)^N$.
We want the amount to be more than doubled,so $A > 2P$.
Substituting the values: $P(1 + 20/100)^N > 2P$.
$(1.2)^N > 2$.
For $N = 3$: $(1.2)^3 = 1.728$ (which is $< 2$).
For $N = 4$: $(1.2)^4 = 2.0736$ (which is $> 2$).
Therefore,the least number of complete years is $4$.

(D) Let the rate of interest be $R \%$ per annum.
The formula for the difference between compound interest $(C.I.)$ and simple interest $(S.I.)$ for $2$ years is given by:
Difference $= P \times \left(\frac{R}{100}\right)^2$
Given:
Principal $(P)$ $= ₹ 18000$
Difference $= ₹ 405$
Time $(T)$ $= 2$ years
Substituting the values into the formula:
$405 = 18000 \times \left(\frac{R}{100}\right)^2$
$\left(\frac{R}{100}\right)^2 = \frac{405}{18000}$
$\left(\frac{R}{100}\right)^2 = \frac{81}{3600}$
Taking the square root on both sides:
$\frac{R}{100} = \sqrt{\frac{81}{3600}}$
$\frac{R}{100} = \frac{9}{60}$
$R = \frac{9}{60} \times 100$
$R = 15$
Therefore,the rate of interest is $15 \%$ per annum.

(B) Let the required number of years be $N$.
Since the interest is compounded semi-annually,the rate per period is $r = 10 / 2 = 5 \%$ and the number of periods is $2N$.
The formula for compound interest is $A = P(1 + r/100)^n$.
Substituting the given values: $926.10 = 800(1 + 5/100)^{2N}$.
$926.10 = 800(1.05)^{2N}$.
$(1.05)^{2N} = 926.10 / 800$.
$(1.05)^{2N} = 1.157625$.
We know that $(1.05)^3 = 1.157625$.
Therefore,$2N = 3$.
$N = 3 / 2 = 1.5$ years.

(C) Let each instalment be ₹ $x$.
The present value of the instalments must equal the borrowed amount.
$\frac{x}{(1 + \frac{4}{100})^1} + \frac{x}{(1 + \frac{4}{100})^2} = 3825$
$\frac{x}{(26/25)} + \frac{x}{(26/25)^2} = 3825$
$\frac{25x}{26} + \frac{625x}{676} = 3825$
Multiply by $676$ to clear the denominators:
$25x \times 26 + 625x = 3825 \times 676$
$650x + 625x = 3825 \times 676$
$1275x = 3825 \times 676$
$x = \frac{3825 \times 676}{1275}$
$x = 3 \times 676 = ₹ 2028$
Thus,each instalment is ₹ $2028$.

(D) The formula for compound interest is $A = P(1 + \frac{R}{100})^n$.
Given that the sum becomes double in $5$ years:
$15000(1 + \frac{R}{100})^5 = 30000$
$(1 + \frac{R}{100})^5 = 2$
We need to find the amount after $20$ years:
$A = 15000(1 + \frac{R}{100})^{20}$
$A = 15000[(1 + \frac{R}{100})^5]^4$
Substituting the value $(1 + \frac{R}{100})^5 = 2$:
$A = 15000 \times 2^4$
$A = 15000 \times 16 = 240000$
Thus,the amount after $20$ years is ₹ $240000$.

(B) Step $1$: Find the principal $P$ using the annual compounding condition.
For $2$ years at $10 \%$ p.a.:
$C.I. = P[(1 + 0.1)^2 - 1] = P[1.21 - 1] = 0.21P$
$S.I. = (P \times 2 \times 10) / 100 = 0.2P$
Given $C.I. - S.I. = 20$,so $0.21P - 0.2P = 20 \Rightarrow 0.01P = 20 \Rightarrow P = 2000$.
Step $2$: Calculate the difference if interest is compounded half-yearly.
For half-yearly compounding,rate $r = 5 \%$ per half-year,and time $n = 4$ half-years.
$C.I. = P[(1 + 0.05)^4 - 1] = 2000[(1.05)^4 - 1]$
$(1.05)^4 = 1.21550625$
$C.I. = 2000[0.21550625] = 431.0125$
$S.I. = (P \times 2 \times 10) / 100 = 2000 \times 0.2 = 400$
Difference $= 431.0125 - 400 = 31.0125 \approx ₹ 31.01$.

(A) Let the principal be $P = 25000$,rate $R = 20 \%$,and annual repayment $A = 5000$.
After the $1^{st}$ year,the amount becomes $25000 \times 1.2 = 30000$. After paying $5000$,the balance is $30000 - 5000 = 25000$.
After the $2^{nd}$ year,the amount becomes $25000 \times 1.2 = 30000$. After paying $5000$,the balance is $30000 - 5000 = 25000$.
After the $3^{rd}$ year,the amount becomes $25000 \times 1.2 = 30000$. After paying $5000$,the balance is $30000 - 5000 = 25000$.
Thus,after three instalments,the remaining amount is ₹ $25000$.

(B) Let the sum borrowed be $P$. The formula for the annual instalment $I$ for a loan repaid in $n$ equal annual instalments at a rate of $R \%$ per annum compound interest is given by $P = \sum_{k=1}^{n} \frac{I}{(1 + R/100)^k}$.
Given $I = ₹ 882$,$R = 5 \%$,and $n = 2$.
$P = \frac{882}{(1 + 5/100)^1} + \frac{882}{(1 + 5/100)^2}$
$P = \frac{882}{21/20} + \frac{882}{(21/20)^2}$
$P = 882 \times \frac{20}{21} + 882 \times \frac{400}{441}$
$P = 42 \times 20 + 2 \times 400$
$P = 840 + 800 = ₹ 1640$.

(D) The formula for compound interest is $CI = P \left[ \left(1 + \frac{R}{100} \right)^n - 1 \right]$.
Given: Principal $P = ₹ 50000$,Rate $R = 10 \%$,Time $n = 4$ years.
$CI = 50000 \left[ \left(1 + \frac{10}{100} \right)^4 - 1 \right]$
$CI = 50000 \left[ \left(1.1 \right)^4 - 1 \right]$
$CI = 50000 \left[ 1.4641 - 1 \right]$
$CI = 50000 \times 0.4641 = 23205$.
Therefore,the compound interest is ₹ $23205$.

(A) Given: Principal $(P)$ = ₹ $8000$,Time $(t)$ = $3$ years,Rate of interest $(R)$ = $5 \%$ per annum.
The formula for the amount $(A)$ under compound interest compounded annually is: $A = P(1 + \frac{R}{100})^t$.
Substituting the values:
$A = 8000(1 + \frac{5}{100})^3$
$A = 8000(1 + \frac{1}{20})^3$
$A = 8000(\frac{21}{20})^3$
$A = 8000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}$
$A = 8000 \times \frac{9261}{8000}$
$A = ₹ 9261$.
Therefore,Nikita will get ₹ $9261$ after $3$ years.

(C) Given: Principal $(P)$ = ₹ $2000$,Rate $(R)$ = $5 \%$ per annum,Time $(t)$ = $2$ years.
The formula for compound interest $(CI)$ is:
$CI = P \left[ \left( 1 + \frac{R}{100} \right)^t - 1 \right]$
Substituting the values:
$CI = 2000 \left[ \left( 1 + \frac{5}{100} \right)^2 - 1 \right]$
$CI = 2000 \left[ \left( \frac{105}{100} \right)^2 - 1 \right]$
$CI = 2000 \left[ \left( \frac{21}{20} \right)^2 - 1 \right]$
$CI = 2000 \left[ \frac{441}{400} - 1 \right]$
$CI = 2000 \left[ \frac{441 - 400}{400} \right]$
$CI = 2000 \times \frac{41}{400}$
$CI = 5 \times 41 = ₹ 205$
Therefore,the compound interest is ₹ $205$.

(A) Given: Principal $(P)$ = ₹ $1000$,Amount $(A)$ = ₹ $1331$,Time $(t)$ = $3$ years.
The formula for compound interest is $A = P(1 + \frac{r}{100})^t$.
Substituting the values: $1331 = 1000(1 + \frac{r}{100})^3$.
Divide both sides by $1000$: $\frac{1331}{1000} = (1 + \frac{r}{100})^3$.
Since $1331 = 11^3$ and $1000 = 10^3$,we have $(\frac{11}{10})^3 = (1 + \frac{r}{100})^3$.
Taking the cube root on both sides: $\frac{11}{10} = 1 + \frac{r}{100}$.
$1.1 = 1 + \frac{r}{100}$.
$0.1 = \frac{r}{100}$.
$r = 0.1 \times 100 = 10 \%$.
Therefore,the rate of interest is $10 \%$ per annum.

(B) Given: Amount $(A)$ = ₹ $9261$,Time $(t)$ = $3$ years,Rate $(R)$ = $5 \%$ per annum.
The formula for the present worth $(P)$ in compound interest is given by:
$P = \frac{A}{(1 + \frac{R}{100})^t}$
Substituting the values:
$P = \frac{9261}{(1 + \frac{5}{100})^3}$
$P = \frac{9261}{(1 + \frac{1}{20})^3}$
$P = \frac{9261}{(\frac{21}{20})^3}$
$P = \frac{9261 \times 20 \times 20 \times 20}{21 \times 21 \times 21}$
$P = \frac{9261 \times 8000}{9261}$
$P = ₹ 8000$
Thus,the present worth is ₹ $8000$.

(B) Given: Principal $(P)$ = ₹ $10000$,Rate $(R)$ = $20 \%$ per annum,Time $(t)$ = $1$ year $6$ months = $1.5$ years.
Since interest is calculated half-yearly,the rate per half-year = $\frac{20}{2} = 10 \%$.
The number of half-years $(n)$ = $1.5 \times 2 = 3$.
Using the compound interest formula: $CI = P \left[ \left( 1 + \frac{R}{100} \right)^n - 1 \right]$.
$CI = 10000 \left[ \left( 1 + \frac{10}{100} \right)^3 - 1 \right]$.
$CI = 10000 \left[ \left( \frac{11}{10} \right)^3 - 1 \right]$.
$CI = 10000 \left[ \frac{1331}{1000} - 1 \right]$.
$CI = 10000 \left[ \frac{331}{1000} \right]$.
$CI = 10 \times 331 = ₹ 3310$.

(B) Let the principal sum be $P = ₹ x$.
Given rate of interest $R = 4 \%$ per annum,compounded half-yearly.
Therefore,the half-yearly rate $r = \frac{4}{2} = 2 \% = 0.02$.
Time period $n = 1 \frac{1}{2}$ years $= 3$ half-years.
The formula for compound interest is $A = P(1 + r)^n$.
Substituting the values: $6632.55 = x(1 + 0.02)^3$.
$6632.55 = x(1.02)^3$.
$6632.55 = x(1.061208)$.
$x = \frac{6632.55}{1.061208} = 6250$.
Thus,the sum is ₹ $6250$.

(A) Given: Principal $(P)$ = ₹ $12000$,Rate $(R)$ = $20 \%$ per annum,Time $(t)$ = $9$ months = $\frac{9}{12} = 0.75$ years.
Since the interest is compounded quarterly,the rate per quarter = $\frac{20}{4} = 5 \%$ and the number of quarters $(n)$ = $9$ months / $3$ months = $3$.
Using the compound interest formula: $CI = P \left[ (1 + \frac{r}{100})^n - 1 \right]$,where $r$ is the quarterly rate.
$CI = 12000 \left[ (1 + \frac{5}{100})^3 - 1 \right]$
$CI = 12000 \left[ (1.05)^3 - 1 \right]$
$CI = 12000 \left[ 1.157625 - 1 \right]$
$CI = 12000 \times 0.157625 = ₹ 1891.50$.

(A) When compounded half-yearly:
Here,$P = 800, R = 20 \%, t = 1$ year.
Since it is compounded half-yearly,the rate $R' = 20/2 = 10 \%$ and time $n = 1 \times 2 = 2$ half-years.
$CI = P[(1 + R'/100)^n - 1] = 800[(1 + 10/100)^2 - 1] = 800[(1.1)^2 - 1] = 800[1.21 - 1] = 800 \times 0.21 = ₹ 168$.
When compounded quarterly:
Here,$P = 800, R = 20 \%, t = 1$ year.
Since it is compounded quarterly,the rate $R'' = 20/4 = 5 \%$ and time $n = 1 \times 4 = 4$ quarters.
$CI = P[(1 + R''/100)^n - 1] = 800[(1 + 5/100)^4 - 1] = 800[(1.05)^4 - 1]$.
$(1.05)^4 = 1.21550625$.
$CI = 800[1.21550625 - 1] = 800 \times 0.21550625 = ₹ 172.405$.
Difference $= ₹ 172.405 - ₹ 168 = ₹ 4.405 \approx ₹ 4.40$.

(B) Simple Interest $(SI)$ for $1$ year at $10 \%$ per annum on ₹ $600$ is calculated as:
$SI = \frac{P \times R \times T}{100} = \frac{600 \times 10 \times 1}{100} = ₹ 60$.
For Compound Interest $(CI)$ reckoned half-yearly,the rate $R$ becomes $10/2 = 5 \%$ per half-year and the time $T$ becomes $1 \times 2 = 2$ half-years.
$CI = P \left[ (1 + \frac{R}{100})^n - 1 \right] = 600 \left[ (1 + \frac{5}{100})^2 - 1 \right]$
$CI = 600 \left[ (1.05)^2 - 1 \right] = 600 \left[ 1.1025 - 1 \right] = 600 \times 0.1025 = ₹ 61.50$.
Difference $= CI - SI = ₹ 61.50 - ₹ 60 = ₹ 1.50$.

(B) Let the time be $t$ years.
Using the compound interest formula: $A = P(1 + \frac{r}{100})^t$
Given: $A = 882$,$P = 800$,$r = 5 \%$
$882 = 800(1 + \frac{5}{100})^t$
$882 = 800(1 + \frac{1}{20})^t$
$882 = 800(\frac{21}{20})^t$
$\frac{882}{800} = (\frac{21}{20})^t$
$\frac{441}{400} = (\frac{21}{20})^t$
$(\frac{21}{20})^2 = (\frac{21}{20})^t$
Comparing the exponents,we get $t = 2$ years.

(A) Given: Principal $(P) = ₹ 1875$,Rate for first year $(R_1) = 4 \%$,Rate for second year $(R_2) = 8 \%$,Time $(n) = 2 \text{ years}$.
The formula for the amount $(A)$ when rates are different for different years is:
$A = P \times (1 + \frac{R_1}{100}) \times (1 + \frac{R_2}{100})$
$A = 1875 \times (1 + \frac{4}{100}) \times (1 + \frac{8}{100})$
$A = 1875 \times (\frac{104}{100}) \times (\frac{108}{100})$
$A = 1875 \times (\frac{26}{25}) \times (\frac{27}{25})$
$A = 1875 \times \frac{702}{625}$
$A = 3 \times 702 = ₹ 2106$
Compound Interest $(CI) = A - P$
$CI = 2106 - 1875 = ₹ 231$.

(B) Given: Principal $(P) = ₹ 5000$,$R_1 = 2\%$,$R_2 = 3\%$,$R_3 = 4\%$,and time $(n) = 3$ years.
The formula for the amount $(A)$ with varying rates of interest is:
$A = P \times (1 + \frac{R_1}{100}) \times (1 + \frac{R_2}{100}) \times (1 + \frac{R_3}{100})$
Substituting the values:
$A = 5000 \times (1 + \frac{2}{100}) \times (1 + \frac{3}{100}) \times (1 + \frac{4}{100})$
$A = 5000 \times \frac{102}{100} \times \frac{103}{100} \times \frac{104}{100}$
$A = 5000 \times \frac{51}{50} \times \frac{103}{100} \times \frac{26}{25}$
$A = 5000 \times \frac{51 \times 103 \times 26}{50 \times 100 \times 25}$
$A = 5000 \times \frac{136578}{125000}$
$A = 5000 \times 1.092624 = ₹ 5463.12$
Therefore,the amount is ₹ $5463.12$.

(C) Let the principal sum be ₹ $P$.
The formula for the amount $A$ with varying rates of interest $R_1, R_2, R_3$ for $3$ years 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 $A = 15916.59$,$R_1 = 3$,$R_2 = 2$,and $R_3 = 1$:
$15916.59 = P \left(1 + \frac{3}{100}\right) \left(1 + \frac{2}{100}\right) \left(1 + \frac{1}{100}\right)$
$15916.59 = P \left(\frac{103}{100} \times \frac{102}{100} \times \frac{101}{100}\right)$
$15916.59 = P \left(\frac{1061106}{1000000}\right)$
$P = \frac{15916.59 \times 1000000}{1061106}$
$P = \frac{15916590000}{1061106} = 15000$
Thus,the required sum is ₹ $15000$.

(B) The formula for compound interest for fractional years is $CI = P \left[ \left(1 + \frac{R}{100} \right)^n \left(1 + \frac{\frac{a}{b} \times R}{100} \right) - 1 \right]$.
Here,$P = 800$,$R = 5 \%$,$n = 2$ years,and the remaining time is $\frac{1}{2}$ year.
$CI = 800 \left[ \left(1 + \frac{5}{100} \right)^2 \left(1 + \frac{\frac{1}{2} \times 5}{100} \right) - 1 \right]$
$CI = 800 \left[ \left(\frac{21}{20} \right)^2 \left(1 + \frac{5}{200} \right) - 1 \right]$
$CI = 800 \left[ \frac{441}{400} \times \frac{41}{40} - 1 \right]$
$CI = 800 \left[ \frac{18081}{16000} - 1 \right]$
$CI = 800 \times \frac{2081}{16000} = \frac{2081}{20} = ₹ 104.05$.

(C) The formula for the amount $A$ with compound interest for a fractional time period $n = a \frac{b}{c}$ is given by $A = P(1 + \frac{R}{100})^a (1 + \frac{\frac{b}{c} \times R}{100})$.
Here,$A = P + CI$. Since the interest is ₹ $6352.50$,the total amount $A = P + 6352.50$.
Substituting the values: $P + 6352.50 = P(1 + \frac{10}{100})^2 (1 + \frac{0.5 \times 10}{100})$.
$P + 6352.50 = P(1.1)^2 (1.05)$.
$P + 6352.50 = P(1.21)(1.05)$.
$P + 6352.50 = 1.2705P$.
$6352.50 = 1.2705P - P$.
$6352.50 = 0.2705P$.
$P = \frac{6352.50}{0.2705} = 23484.28$ (This suggests the question implies the total amount is ₹ $6352.50$ rather than the interest).
If we assume the total amount $A = 6352.50$:
$6352.50 = P(1 + \frac{10}{100})^2 (1 + \frac{5}{100})$.
$6352.50 = P(1.21)(1.05) = 1.2705P$.
$P = \frac{6352.50}{1.2705} = 5000$.
Thus,the principal sum is ₹ $5000$.

(C) Let the principal be $P$,rate $R = 5 \%$,and time $n = 3$ years.
Compound Interest $(CI)$ = $P \left[ (1 + \frac{R}{100})^n - 1 \right] = 1324.05$.
$P \left[ (1 + \frac{5}{100})^3 - 1 \right] = 1324.05$.
$P \left[ (1.05)^3 - 1 \right] = 1324.05$.
$P [1.157625 - 1] = 1324.05$.
$P \times 0.157625 = 1324.05$.
$P = \frac{1324.05}{0.157625} = 8400$.
Now,Simple Interest $(SI)$ = $\frac{P \times R \times n}{100} = \frac{8400 \times 5 \times 3}{100}$.
$SI = 84 \times 15 = 1260$.
Therefore,the simple interest is ₹ $1260$.

(B) Given: Simple Interest $(SI)$ = ₹ $80$,Rate $(R)$ = $4 \%$,Time $(T)$ = $2$ years.
First,find the principal $(P)$:
$SI = \frac{P \times R \times T}{100} \Rightarrow 80 = \frac{P \times 4 \times 2}{100} \Rightarrow 80 = \frac{8P}{100} \Rightarrow P = \frac{8000}{8} = ₹ 1000$.
Now,calculate the Compound Interest $(CI)$:
$CI = P \left(1 + \frac{R}{100}\right)^T - P = 1000 \left(1 + \frac{4}{100}\right)^2 - 1000$
$CI = 1000 \left(\frac{104}{100}\right)^2 - 1000 = 1000 \times (1.04)^2 - 1000$
$CI = 1000 \times 1.0816 - 1000 = 1081.6 - 1000 = ₹ 81.60$.
Alternatively,for $2$ years: $CI - SI = \frac{SI \times R}{200} = \frac{80 \times 4}{200} = \frac{320}{200} = ₹ 1.60$.
Therefore,$CI = SI + 1.60 = 80 + 1.60 = ₹ 81.60$.

(A) Given that the compound interest $(CI)$ for $2$ years is ₹ $60.60$ and the simple interest $(SI)$ for $2$ years is ₹ $60$.
For a period of $2$ years,the difference between compound interest and simple interest is given by the formula:
$CI - SI = \frac{P \times R^2}{100^2}$
Alternatively,using the relation $CI - SI = \frac{R \times SI}{200}$ for $2$ years:
$60.60 - 60 = \frac{R \times 60}{200}$
$0.60 = \frac{R \times 60}{200}$
$0.60 = \frac{R \times 3}{10}$
$R = \frac{0.60 \times 10}{3}$
$R = \frac{6}{3} = 2 \%$
Thus,the rate of interest per annum is $2 \%$.

(B) Let the principal be $P$,the rate of interest be $R$ per annum,and the time be $T = 2$ years.
Simple Interest $(SI)$ for $2$ years = $P \times R \times 2 / 100 = 100$.
Thus,$P \times R = 5000$.
Compound Interest $(CI)$ for $2$ years = $P[(1 + R/100)^2 - 1] = 105$.
$P[1 + R^2/10000 + 2R/100 - 1] = 105$.
$P[R^2/10000 + 2R/100] = 105$.
Since $P \times R = 5000$,we have $P \times R / 100 = 50$.
Substituting $P \times R = 5000$ into the equation: $(5000 \times R) / 10000 + (2 \times 5000) / 100 = 105$.
$0.5R + 100 = 105$.
$0.5R = 5 \Rightarrow R = 10\%$.
Now,using $P \times R = 5000$,we get $P \times 10 = 5000$.
Therefore,$P = ₹ 500$.

(C) The formula for the difference between compound interest $(CI)$ and simple interest $(SI)$ for $2$ years is given by:
$CI - SI = P \left( \frac{R}{100} \right)^2$
Given:
Principal $(P)$ = ₹ $1250$
Rate $(R)$ = $4 \%$
Time $(n)$ = $2$ years
Substituting the values into the formula:
$CI - SI = 1250 \times \left( \frac{4}{100} \right)^2$
$CI - SI = 1250 \times \left( \frac{1}{25} \right)^2$
$CI - SI = 1250 \times \frac{1}{625}$
$CI - SI = 2$
Therefore,the difference is ₹ $2$.

(A) Given:
Simple Interest $(SI)$ for $2$ years = ₹ $200$.
Rate of interest $(R)$ = $7 \%$ per annum.
Time $(T)$ = $2$ years.
Step $1$: Calculate the Principal $(P)$.
$SI = \frac{P \times R \times T}{100}$
$200 = \frac{P \times 7 \times 2}{100}$
$200 = \frac{14P}{100}$
$P = \frac{20000}{14} = ₹ \frac{10000}{7}$.
Step $2$: Calculate the difference between $CI$ and $SI$ for $2$ years.
The formula for the difference between $CI$ and $SI$ for $2$ years is:
$Difference = P \times (\frac{R}{100})^2$
$Difference = \frac{10000}{7} \times (\frac{7}{100})^2$
$Difference = \frac{10000}{7} \times \frac{49}{10000}$
$Difference = 7$.
Alternatively,for $2$ years,$CI - SI = \frac{SI \times R}{200} = \frac{200 \times 7}{200} = 7$.
Thus,the difference is ₹ $7$.

(B) The formula for the difference between compound interest $(CI)$ and simple interest $(SI)$ for $2$ years is given by:
$CI - SI = P \left( \frac{R}{100} \right)^2$
Given:
Difference $(CI - SI)$ = ₹ $1.50 = \frac{3}{2}$
Rate $(R)$ = $5 \%$
Time $(T)$ = $2$ years
Substituting the values in the formula:
$\frac{3}{2} = P \left( \frac{5}{100} \right)^2$
$\frac{3}{2} = P \left( \frac{1}{20} \right)^2$
$\frac{3}{2} = P \times \frac{1}{400}$
$P = \frac{3 \times 400}{2}$
$P = 3 \times 200 = 600$
Therefore,the sum is ₹ $600$.

(C) The formula for the difference between compound interest $(CI)$ and simple interest $(SI)$ for $3$ years is given by:
$CI - SI = P \left[ \left( \frac{R}{100} \right)^3 + 3 \left( \frac{R}{100} \right)^2 \right]$
Given $R = 3 \%$ and $CI - SI = 27.27$.
Substituting the values:
$27.27 = P \left[ \left( \frac{3}{100} \right)^3 + 3 \left( \frac{3}{100} \right)^2 \right]$
$27.27 = P \left[ \frac{27}{1000000} + \frac{27}{10000} \right]$
$27.27 = P \left[ \frac{27 + 2700}{1000000} \right]$
$27.27 = P \left[ \frac{2727}{1000000} \right]$
$P = \frac{27.27 \times 1000000}{2727}$
$P = \frac{2727 \times 10000}{2727} = 10000$
Thus,the sum is ₹ $10000$.

(A) The formula for the difference between compound interest $(CI)$ and simple interest $(SI)$ for $3$ years is given by:
$CI - SI = P \left[ \left( \frac{R}{100} \right)^3 + 3 \left( \frac{R}{100} \right)^2 \right]$
Given:
Principal $(P)$ = ₹ $8000$
Rate $(R)$ = $5 \%$
Time $(n)$ = $3$ years
Substituting the values:
$CI - SI = 8000 \left[ \left( \frac{5}{100} \right)^3 + 3 \left( \frac{5}{100} \right)^2 \right]$
$CI - SI = 8000 \left[ \left( \frac{1}{20} \right)^3 + 3 \left( \frac{1}{20} \right)^2 \right]$
$CI - SI = 8000 \left[ \frac{1}{8000} + \frac{3}{400} \right]$
$CI - SI = 8000 \left[ \frac{1 + 60}{8000} \right]$
$CI - SI = 8000 \times \frac{61}{8000} = ₹ 61$
Thus,the difference is ₹ $61$.

(B) Let the principal be $P$. The amount $A$ after $t$ years is given by $A = P(1 + r/100)^t$.
According to the problem,the sum becomes $3P$ in $3$ years,so $3P = P(1 + r/100)^3$,which implies $(1 + r/100)^3 = 3$.
We want to find the time $T$ such that the sum becomes $9P$,so $9P = P(1 + r/100)^T$,which implies $(1 + r/100)^T = 9$.
Since $9 = 3^2$,we can write $(1 + r/100)^T = (3^2)$.
Substituting $(1 + r/100)^3 = 3$,we get $(1 + r/100)^T = ((1 + r/100)^3)^2 = (1 + r/100)^6$.
Comparing the exponents,we get $T = 6$ years.

(B) Let the principal be $P$ and the rate of interest be $R \%$ per annum.
According to the compound interest formula,the amount $A$ after $t$ years is given by $A = P(1 + R/100)^t$.
Given that the sum becomes $16$ times in $4$ years,we have $A = 16P$ and $t = 4$.
Substituting these values into the formula:
$16P = P(1 + R/100)^4$
$16 = (1 + R/100)^4$
$(2)^4 = (1 + R/100)^4$
Taking the fourth root on both sides:
$2 = 1 + R/100$
$R/100 = 2 - 1 = 1$
$R = 100 \%$
Thus,the rate of interest is $100 \%$.

(C) Let the principal be $P$ and the rate of interest be $R \%$ per annum.
For compound interest,the amount $A$ after $n$ years is given by $A = P(1 + R/100)^n$.
Given:
Amount after $2$ years $(A_2)$ = $P(1 + R/100)^2 = 12960$ --- (Equation $1$)
Amount after $3$ years $(A_3)$ = $P(1 + R/100)^3 = 13176$ --- (Equation $2$)
Dividing Equation $2$ by Equation $1$:
$\frac{P(1 + R/100)^3}{P(1 + R/100)^2} = \frac{13176}{12960}$
$(1 + R/100) = \frac{13176}{12960}$
$R/100 = \frac{13176}{12960} - 1$
$R/100 = \frac{13176 - 12960}{12960} = \frac{216}{12960}$
$R = \frac{216}{12960} \times 100 = \frac{2160}{129.6} = \frac{21600}{1296} = \frac{50}{3} \% = 16.66 \%$
Wait,checking the calculation: $\frac{216}{12960} = \frac{1}{60}$.
$R = \frac{1}{60} \times 100 = \frac{10}{6} = \frac{5}{3} = 1 \frac{2}{3} \%$.
Therefore,the correct option is $C$.

(C) Let the principal be $P$ and the rate of interest be $R \%$.
According to the formula for compound interest,the amount $A$ after $n$ years is given by $A = P(1 + R/100)^n$.
For the first year $(n=1)$: $650 = P(1 + R/100) \quad ... (1)$
For the second year $(n=2)$: $676 = P(1 + R/100)^2 \quad ... (2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{676}{650} = \frac{P(1 + R/100)^2}{P(1 + R/100)}$
$1.04 = 1 + R/100$
$R/100 = 0.04 \Rightarrow R = 4 \%$
Substituting $R=4$ in equation $(1)$:
$650 = P(1 + 4/100)$
$650 = P(104/100)$
$P = \frac{650 \times 100}{104} = \frac{65000}{104} = ₹ 625$.

(A) Let the annual instalment be $x$.
Given: Principal $P = 1260$,Rate $R = 10 \%$,Time $n = 2$ years.
The formula for the present value of two equal annual instalments is:
$P = \frac{x}{(1 + \frac{R}{100})} + \frac{x}{(1 + \frac{R}{100})^2}$
Substituting the values:
$1260 = \frac{x}{(1 + \frac{10}{100})} + \frac{x}{(1 + \frac{10}{100})^2}$
$1260 = \frac{x}{(1.1)} + \frac{x}{(1.1)^2}$
$1260 = \frac{x}{1.1} + \frac{x}{1.21}$
$1260 = \frac{1.1x + x}{1.21} = \frac{2.1x}{1.21}$
$x = \frac{1260 \times 1.21}{2.1}$
$x = 600 \times 1.21 = 726$
Therefore,the annual instalment is ₹ $726$.

(D) The tree grows annually by a factor of $(1 + \frac{1}{8}) = \frac{9}{8}$ of its current height.
Initial height of the tree $= 64 \ cm$.
Height after $1$ year $= 64 \times \frac{9}{8} = 72 \ cm$.
Height after $2$ years $= 72 \times \frac{9}{8} = 9 \times 9 = 81 \ cm$.
Therefore,the height of the tree after $2$ years will be $81 \ cm$.

(B) Let the principal amount be $P$. We want the amount $A$ to be more than double the principal,i.e.,$A > 2P$.
Using the compound interest formula $A = P(1 + r/100)^n$,we have $P(1 + 20/100)^n > 2P$.
This simplifies to $(1.2)^n > 2$ or $(6/5)^n > 2$.
For $n = 3$,$(6/5)^3 = 216/125 = 1.728$,which is less than $2$.
For $n = 4$,$(6/5)^4 = 1296/625 = 2.0736$,which is greater than $2$.
Therefore,the least number of completed years required is $4$.

(A) The rate of interest $R = 7 \frac{1}{2} \% = 7.5 \% = \frac{15}{2} \% = 0.075$.
Let the principal be $P = 4000$ and the annual installment be $I = 1500$.
After the $1^{st}$ year,the amount becomes $4000 \times (1 + 0.075) = 4300$. After paying $1500$,the balance is $4300 - 1500 = 2800$.
After the $2^{nd}$ year,the amount becomes $2800 \times (1 + 0.075) = 3010$. After paying $1500$,the balance is $3010 - 1500 = 1510$.
After the $3^{rd}$ year,the amount becomes $1510 \times (1 + 0.075) = 1623.25$. After paying $1500$,the final balance is $1623.25 - 1500 = 123.25$.
Thus,the man still owes ₹ $123.25$ to the bank.

(B) Let $r \%$ be the rate of interest and $n$ be the number of years.
The formula for the amount in compound interest is $A = P(1 + \frac{r}{100})^n$.
Given that $P = 3000$ and $A = 4320$ after $n$ years:
$4320 = 3000(1 + \frac{r}{100})^n$
Dividing both sides by $3000$:
$(1 + \frac{r}{100})^n = \frac{4320}{3000} = 1.44$
We need to find the amount after half the time,i.e.,$n/2$ years:
$A' = 3000(1 + \frac{r}{100})^{n/2}$
Since $(1 + \frac{r}{100})^n = 1.44$,taking the square root of both sides gives:
$(1 + \frac{r}{100})^{n/2} = \sqrt{1.44} = 1.2$
Substituting this value into the expression for $A'$:
$A' = 3000 \times 1.2 = 3600$
Thus,the amount will be ₹ $3600$.

(A) Let $A$'s share be $₹ x$ and $B$'s share be $₹(3757 - x)$.
According to the problem,the amount of $A$ after $7$ years equals the amount of $B$ after $9$ years at $10 \%$ compound interest.
$x(1 + \frac{10}{100})^7 = (3757 - x)(1 + \frac{10}{100})^9$
Dividing both sides by $(1 + \frac{10}{100})^7$,we get:
$x = (3757 - x)(1 + \frac{10}{100})^2$
$x = (3757 - x)(\frac{11}{10})^2$
$x = (3757 - x)(\frac{121}{100})$
$100x = 121(3757 - x)$
$100x = 454597 - 121x$
$221x = 454597$
$x = \frac{454597}{221} = 2057$
So,$A$'s share is $₹ 2057$.
$B$'s share $= 3757 - 2057 = ₹ 1700$.

(A) Given:
Principal $(P)$ = ₹ $2,000$
Compound Interest $(C.I.)$ = ₹ $662$
Rate $(r)$ = $10 \%$ per annum
Time $(n)$ = ?
The formula for compound interest is:
$C.I. = P \left[ \left( 1 + \frac{r}{100} \right)^n - 1 \right]$
Substituting the given values:
$662 = 2000 \left[ \left( 1 + \frac{10}{100} \right)^n - 1 \right]$
$662 = 2000 \left[ (1.1)^n - 1 \right]$
$\frac{662}{2000} = (1.1)^n - 1$
$0.331 = (1.1)^n - 1$
$1.331 = (1.1)^n$
Since $(1.1)^3 = 1.331$,we have:
$(1.1)^3 = (1.1)^n$
Therefore,$n = 3$ years.

(A) Let the principal be $P$ and the rate of interest be $r \%$.
Simple Interest $(SI)$ for any year is constant. Given $SI$ for the $3^{rd}$ year $= ₹ 2,000$,therefore $SI$ for the $1^{st}$ year $= ₹ 2,000$ and $SI$ for the $2^{nd}$ year $= ₹ 2,000$.
Total $SI$ for $2$ years $= 2,000 + 2,000 = ₹ 4,000$.
Compound Interest $(CI)$ for $2$ years $= ₹ 4,160$.
The difference between $CI$ and $SI$ for $2$ years is the interest earned on the interest of the $1^{st}$ year.
Difference $= 4,160 - 4,000 = ₹ 160$.
This $₹ 160$ is the interest on the $1^{st}$ year's interest $(₹ 2,000)$.
Rate of interest $r = \frac{\text{Difference}}{SI \text{ of } 1^{st} \text{ year}} \times 100$.
$r = \frac{160}{2000} \times 100 = 8 \%$.

(A) Given: Principal $(P)$ = $₹ 25,000$,Rate $(r)$ = $20 \%$,Time $(n)$ = $4$ years.
Using the formula for compound interest: $\text{Amount} = P \left(1 + \frac{r}{100}\right)^n$.
Substituting the values: $\text{Amount} = 25000 \left(1 + \frac{20}{100}\right)^4$.
$\text{Amount} = 25000 \left(1 + \frac{1}{5}\right)^4 = 25000 \left(\frac{6}{5}\right)^4$.
$\text{Amount} = 25000 \times \frac{1296}{625}$.
$\text{Amount} = 40 \times 1296 = 51840$.
Thus,the total amount received is $₹ 51,840$.

(D) Let the principal amount be $P = ₹ 100$.
After $8$ years,the amount becomes $140 \%$ of $P$,which is $₹ 140$.
Therefore,Simple Interest $(S.I)$ $= 140 - 100 = ₹ 40$.
For $1$ year,$S.I = \frac{40}{8} = ₹ 5$.
Since $S.I$ on $₹ 100$ for $1$ year is $₹ 5$,the rate of interest $R = 5 \%$ per annum.
Now,calculate the Compound Interest $(C.I)$ on $P = ₹ 30000$ for $n = 2$ years at $r = 5 \%$ per annum.
$C.I = P \left(1 + \frac{r}{100}\right)^n - P$
$C.I = 30000 \left(1 + \frac{5}{100}\right)^2 - 30000$
$C.I = 30000 \left(\frac{105}{100}\right)^2 - 30000$
$C.I = 30000 \times 1.1025 - 30000$
$C.I = 33075 - 30000 = ₹ 3075$.

(C) Let the principal sum be $P$.
For $2$ years,the simple interest $(SI)$ is given by $SI = \frac{P \times R \times T}{100} = \frac{P \times 4 \times 2}{100} = 0.08P$.
The compound interest $(CI)$ compounded annually is given by $CI = P(1 + \frac{R}{100})^T - P = P(1 + \frac{4}{100})^2 - P = P(1.04)^2 - P = P(1.0816) - P = 0.0816P$.
The difference between $CI$ and $SI$ is $0.0816P - 0.08P = 0.0016P$.
Given that the difference is ₹ $8$,we have $0.0016P = 8$.
$P = \frac{8}{0.0016} = \frac{80000}{16} = 5000$.
Thus,the sum is ₹ $5000$.

(A) Simple Interest $(SI)$ for $2$ years $= ₹ 1400$.
Since simple interest is the same for every year,$SI$ for $1$ year $= ₹ 1400 / 2 = ₹ 700$.
The difference between Compound Interest $(CI)$ and Simple Interest $(SI)$ for $2$ years is the interest earned on the interest of the first year.
Difference $= CI - SI = ₹ 1449 - ₹ 1400 = ₹ 49$.
This $₹ 49$ is the interest on the $SI$ of the first year $(₹ 700)$.
Rate of interest $(R)$ $= (\text{Difference} / SI \text{ of } 1 \text{ year}) \times 100$.
$R = (49 / 700) \times 100 = 7 \%$.