Simple Interest Questions in English

(D) Let the principal be $P$ and the rate of interest be $R$ percent per annum.
In the first case,the time $T_1 = 4$ years. The simple interest $SI_1 = \frac{P \times R \times 4}{100}$.
In the second case,the time $T_2 = 6$ years. The simple interest $SI_2 = \frac{P \times R \times 6}{100}$.
According to the problem,$SI_2 = SI_1 + 50\% \text{ of } SI_1 = 1.5 \times SI_1$.
Substituting the formulas: $\frac{P \times R \times 6}{100} = 1.5 \times \frac{P \times R \times 4}{100}$.
Simplifying this: $\frac{6PR}{100} = 1.5 \times \frac{4PR}{100} \Rightarrow \frac{6PR}{100} = \frac{6PR}{100}$.
Since both sides result in the same expression,the variables $P$ and $R$ cancel out,meaning the relationship holds true for any rate of interest $R$. Therefore,the specific rate of interest cannot be determined from the given information.

(D) Let the principal amount be $P$.
The interest rate for the first year is $6\%$.
The interest rate for the second year is $6\% + 1.5\% = 7.5\%$.
The interest rate for the third year is $7.5\% + 1.5\% = 9\%$.
Since the interest is calculated on the principal $P$ for each year separately:
Total Interest = $\frac{P \times 6 \times 1}{100} + \frac{P \times 7.5 \times 1}{100} + \frac{P \times 9 \times 1}{100} = 8190$.
$\frac{P}{100} \times (6 + 7.5 + 9) = 8190$.
$\frac{P}{100} \times 22.5 = 8190$.
$P = \frac{8190 \times 100}{22.5}$.
$P = \frac{8190000}{225} = 36400$.
Therefore,the loan amount is $Rs. 36400$.

(A) Given:
Principal $(P) = Rs. 10250$
Amount $(A) = Rs. 12710$
Rate $(R) = 4\% \text{ p.c.p.a.}$
Simple Interest $(SI) = A - P = 12710 - 10250 = Rs. 2460$
We know the formula for Simple Interest: $SI = \frac{P \times R \times T}{100}$
Substituting the values:
$2460 = \frac{10250 \times 4 \times T}{100}$
$2460 = 102.5 \times 4 \times T$
$2460 = 410 \times T$
$T = \frac{2460}{410} = 6 \text{ years}$
Therefore,she invested the amount for $6$ years.

(D) Given: Principal $(P) = Rs. 600$,Amount $(A) = Rs. 720$,Time $(T) = 4$ $years$.
Simple Interest $(SI) = A - P = 720 - 600 = Rs. 120$.
Using the formula $SI = (P \times R \times T) / 100$,we find the initial rate $(R)$:
$120 = (600 \times R \times 4) / 100$
$120 = 24 \times R$
$R = 120 / 24 = 5 \%$.
Now,the new rate of interest is $R' = 5 \% + 2 \% = 7 \%$.
New Simple Interest $(SI') = (P \times R' \times T) / 100 = (600 \times 7 \times 4) / 100 = 6 \times 28 = Rs. 168$.
Total Amount $= P + SI' = 600 + 168 = Rs. 768$.

(C) Let the principal be $Rs. x$.
We know that the Amount $(A) = \text{Principal} (P) + \text{Simple Interest} (SI)$.
Given,$A = 19050$,$T = 3 \text{ years}$,and $R = 9\% \text{ p.c.p.a.}$
$SI = \frac{P \times R \times T}{100} = \frac{x \times 9 \times 3}{100} = \frac{27x}{100}$.
Substituting these into the formula $A = P + SI$:
$19050 = x + \frac{27x}{100}$
$19050 = \frac{100x + 27x}{100}$
$19050 = \frac{127x}{100}$
$127x = 1905000$
$x = \frac{1905000}{127} = 15000$.
Therefore,the principal amount invested is $Rs. 15000$.

(A) Step $1$: Find the rate of interest $(R)$.
Given that simple interest on $P = Rs. 1$ for $T = 4$ years is $SI = Rs. 0.40$.
Using the formula $SI = \frac{P \times R \times T}{100}$, we have:
$0.40 = \frac{1 \times R \times 4}{100}$
$R = \frac{0.40 \times 100}{4} = \frac{40}{4} = 10\% \text{ per annum}$.
Step $2$: Calculate the interest on $Rs. 450$ for $2$ years at the same rate.
Here, $P = Rs. 450$, $T = 2$ years, and $R = 10\%$.
$SI = \frac{P \times R \times T}{100} = \frac{450 \times 10 \times 2}{100}$
$SI = 450 \times 0.2 = Rs. 90$.

(C) Given:
Principal $(P) = Rs. 9535$
Amount $(A) = Rs. 11442$
Rate $(R) = 4\%$ per annum
Simple Interest $(SI) = A - P = 11442 - 9535 = Rs. 1907$
Using the formula for Simple Interest:
$SI = \frac{P \times R \times T}{100}$
$1907 = \frac{9535 \times 4 \times T}{100}$
Solving for $T$:
$T = \frac{1907 \times 100}{9535 \times 4}$
$T = \frac{190700}{38140}$
$T = 5 \text{ years}$
Therefore,she invested the amount for $5$ years.

(D) Let the amount lent at $5 \%$ p.a. be $P_2$.
The formula for Simple Interest is $SI = \frac{P \times R \times T}{100}$.
The total interest received is the sum of interest from both amounts:
$138 = \frac{3000 \times 4 \times 1}{100} + \frac{P_2 \times 5 \times 1}{100}$
Calculating the first part:
$\frac{3000 \times 4}{100} = 30 \times 4 = 120$.
Substituting this into the equation:
$138 = 120 + \frac{5 P_2}{100}$
Subtracting $120$ from both sides:
$138 - 120 = \frac{5 P_2}{100}$
$18 = \frac{P_2}{20}$
Solving for $P_2$:
$P_2 = 18 \times 20 = 360$.
Thus,the amount lent at $5 \%$ p.a. is $Rs. 360$.

(D) The formula for Simple Interest $(SI)$ is given by: $SI = \frac{P \times R \times T}{100}$.
For the first case:
Principal $(P_1) = Rs. 2,00,000$,Rate $(R_1) = 12 \%$,Time $(T) = 1$ year.
$SI_1 = \frac{2,00,000 \times 12 \times 1}{100} = Rs. 24,000$.
For the second case:
Principal $(P_2) = Rs. 2,00,000 + 500 = Rs. 2,00,500$,Rate $(R_2) = 13 \%$,Time $(T) = 1$ year.
$SI_2 = \frac{2,00,500 \times 13 \times 1}{100} = Rs. 26,065$.
The additional interest received is $SI_2 - SI_1 = 26,065 - 24,000 = Rs. 2,065$.

(C) The interest for $1.5$ $years$ is the difference between the amounts: $873 - 756 = Rs. 117$.
Since the interest for $1.5$ $years$ is $Rs. 117$,the interest for $1$ $year$ is $\frac{117}{1.5} = Rs. 78$.
The interest for $2$ $years$ is $78 \times 2 = Rs. 156$.
The principal amount is the amount after $2$ $years$ minus the interest for $2$ $years$: $756 - 156 = Rs. 600$.
Using the Simple Interest formula $SI = \frac{P \times R \times T}{100}$,where $SI = 78$,$P = 600$,and $T = 1$:
$78 = \frac{600 \times R \times 1}{100}$
$78 = 6 \times R$
$R = \frac{78}{6} = 13 \%$.
Therefore,the rate of interest is $13 \%$.

(D) The annual payment $P$ required to discharge a debt $A$ due in $t$ years at a simple interest rate of $r \%$ per annum is given by the formula:
$P = \frac{100 \times A}{100 \times t + \frac{r \times t \times (t - 1)}{2}}$
Given:
$A = 770$,$t = 5$,$r = 5$
Substituting the values into the formula:
$P = \frac{100 \times 770}{100 \times 5 + \frac{5 \times 5 \times (5 - 1)}{2}}$
$P = \frac{77000}{500 + \frac{5 \times 5 \times 4}{2}}$
$P = \frac{77000}{500 + 50}$
$P = \frac{77000}{550}$
$P = 140$
Thus,the annual payment is $Rs. 140$.

(D) Let the principal amount deposited be $P = Rs. 100$.
Interest for the first $2$ years at $3 \%$ per annum: $I_1 = \frac{100 \times 3 \times 2}{100} = Rs. 6$.
Interest for the next $3$ years at $8 \%$ per annum: $I_2 = \frac{100 \times 8 \times 3}{100} = Rs. 24$.
Interest for the remaining $1$ year (since total time is $6$ years,$6 - 2 - 3 = 1$ year) at $10 \%$ per annum: $I_3 = \frac{100 \times 10 \times 1}{100} = Rs. 10$.
Total interest for $6$ years $= 6 + 24 + 10 = Rs. 40$.
If the interest is $Rs. 40$,the principal is $Rs. 100$.
If the interest is $Rs. 1520$,the principal $= \frac{100}{40} \times 1520 = 2.5 \times 1520 = Rs. 3800$.

(B) Let the three parts be $P_1, P_2,$ and $P_3$. The amount $A$ after $t$ years at a simple interest rate $r = 5 \%$ is given by $A = P(1 + \frac{rt}{100})$.
For $t = 2, 3, 4$ years,the amounts are:
$A_1 = P_1(1 + \frac{5 \times 2}{100}) = P_1(1 + 0.1) = 1.1P_1 = \frac{110}{100}P_1$
$A_2 = P_2(1 + \frac{5 \times 3}{100}) = P_2(1 + 0.15) = 1.15P_2 = \frac{115}{100}P_2$
$A_3 = P_3(1 + \frac{5 \times 4}{100}) = P_3(1 + 0.2) = 1.2P_3 = \frac{120}{100}P_3$
Given $A_1 = A_2 = A_3$,we have $110P_1 = 115P_2 = 120P_3$.
Dividing by $5$,we get $22P_1 = 23P_2 = 24P_3$.
The ratio $P_1 : P_2 : P_3 = \frac{1}{22} : \frac{1}{23} : \frac{1}{24}$.
Multiplying by the $LCM$ of $22, 23, 24$ (which is $22 \times 23 \times 12 = 6072$):
$P_1 : P_2 : P_3 = 276 : 264 : 253$.
Sum of ratios $= 276 + 264 + 253 = 793$.
$P_1 = \frac{276}{793} \times 4758 = 276 \times 6 = 1656$.
Thus,the $1^{\text{st}}$ part is $Rs. 1656$.

(B) Let the principal be $P$ and the rate of interest be $R \%$ per annum.
For the first period $t$,the ratio of principal to amount is $P : (P + SI_t) = 4 : 5$.
This implies $P + SI_t = \frac{5}{4}P$,so $SI_t = \frac{1}{4}P$.
Using $SI = \frac{P \times R \times t}{100}$,we have $\frac{P \times R \times t}{100} = \frac{1}{4}P$,which gives $Rt = 25$.
After $3$ years,the time becomes $t + 3$. The ratio of principal to amount is $P : (P + SI_{t+3}) = 5 : 7$.
This implies $P + SI_{t+3} = \frac{7}{5}P$,so $SI_{t+3} = \frac{2}{5}P$.
Using the formula again,$\frac{P \times R \times (t+3)}{100} = \frac{2}{5}P$,which gives $R(t+3) = 40$.
Expanding this,$Rt + 3R = 40$.
Substituting $Rt = 25$,we get $25 + 3R = 40$.
$3R = 15$,so $R = 5$.
The rate of interest is $5 \%$ per annum.

(B) Initial population = $11200$.
Population after the first year (increase of $25\%$):
$= 11200 \times (1 + \frac{25}{100}) = 11200 \times 1.25 = 14000$.
Population after the second year (decrease of $15\%$ on the new population):
$= 14000 \times (1 - \frac{15}{100}) = 14000 \times 0.85 = 11900$.
Therefore,the population of the town at the end of $2$ years is $11900$.

(B) Step $1$: Calculate the Simple Interest for the first case.
Interest $= 832 - 640 = ₹ 192$.
Step $2$: Find the rate of interest $(R)$.
Using the formula $SI = \frac{P \times R \times T}{100}$,we have $192 = \frac{640 \times R \times 2}{100}$.
$R = \frac{192 \times 100}{640 \times 2} = 15 \%$.
Step $3$: Calculate the interest for the second case.
For $P = ₹ 860$,$T = 4$ years,and $R = 15 \%$,
$SI = \frac{860 \times 15 \times 4}{100} = ₹ 516$.
Step $4$: Calculate the final amount.
Amount $= P + SI = 860 + 516 = ₹ 1376$.

(C) Let the sum lent at $12 \%$ be $x$ and the sum lent at $12.5 \%$ be $(25400 - x)$.
The total annual interest is given by the equation:
$0.12x + 0.125(25400 - x) = 3124.2$
Expanding the equation:
$0.12x + 3175 - 0.125x = 3124.2$
Simplifying the terms:
$-0.005x = 3124.2 - 3175$
$-0.005x = -50.8$
Solving for $x$:
$x = \frac{50.8}{0.005} = 10160$
Therefore,the money lent at $12 \%$ is ₹ $10160$.

(B) Let the two parts be $P_1$ and $P_2$ such that $P_1 + P_2 = 26000$.
Given that the simple interest on both parts is equal:
$SI_1 = SI_2$
$\frac{P_1 \times 10 \times 5}{100} = \frac{P_2 \times 9 \times 6}{100}$
$50 P_1 = 54 P_2$
$\frac{P_1}{P_2} = \frac{54}{50} = \frac{27}{25}$
Thus,the ratio of the two parts is $27:25$.
The sum lent out at $10 \%$ $(P_1)$ is:
$P_1 = 26000 \times \frac{27}{27 + 25} = 26000 \times \frac{27}{52}$
$P_1 = 500 \times 27 = 13500$
Therefore,the sum lent out at $10 \%$ is ₹ $13500$.

(C) Let the principal amount be $P$.
Since the money doubles itself,the amount $A = 2P$.
The simple interest $SI = A - P = 2P - P = P$.
The time $T = 16$ years.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$.
Substituting the values: $P = \frac{P \times R \times 16}{100}$.
$1 = \frac{16R}{100}$.
$R = \frac{100}{16} = 6.25 \%$.
Thus,the rate is $6.25 \%$ or $6 \frac{1}{4} \%$.

(C) Let the principal amount be $P$.
Since the amount trebles itself,the final amount $A = 3P$.
The simple interest $SI = A - P = 3P - P = 2P$.
The time $T = 15$ years.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$.
Substituting the values: $2P = \frac{P \times R \times 15}{100}$.
$2 = \frac{15R}{100}$.
$R = \frac{200}{15} = \frac{40}{3} = 13 \frac{1}{3} \%$.
Therefore,the rate of interest is $13 \frac{1}{3} \%$.

(D) Given: Principal $(P) = ₹ 8000$,Amount $(A) = ₹ 9200$,Time $(T) = 3$ years.
Simple Interest $(SI) = A - P = 9200 - 8000 = ₹ 1200$.
Using the formula $SI = \frac{P \times R \times T}{100}$,we find the initial rate $(R)$:
$1200 = \frac{8000 \times R \times 3}{100} \implies 1200 = 240R \implies R = 5 \%$.
New interest rate $(R') = 5 \% + 3 \% = 8 \%$.
New Simple Interest $(SI') = \frac{8000 \times 8 \times 3}{100} = 80 \times 24 = ₹ 1920$.
New Amount $(A') = P + SI' = 8000 + 1920 = ₹ 9920$.

(C) Let the total capital be $x$.
The investments are:
$1$. $\frac{x}{3}$ at $7 \% = \frac{x}{3} \times \frac{7}{100} = \frac{7x}{300}$
$2$. $\frac{x}{4}$ at $8 \% = \frac{x}{4} \times \frac{8}{100} = \frac{8x}{400} = \frac{2x}{100} = \frac{6x}{300}$
$3$. The remainder is $x - (\frac{x}{3} + \frac{x}{4}) = x - \frac{7x}{12} = \frac{5x}{12}$.
Interest on the remainder at $10 \% = \frac{5x}{12} \times \frac{10}{100} = \frac{50x}{1200} = \frac{x}{24}$.
Total annual income is given as ₹ $561$:
$\frac{7x}{300} + \frac{6x}{300} + \frac{x}{24} = 561$
$\frac{13x}{300} + \frac{x}{24} = 561$
Finding the common denominator $(600)$:
$\frac{26x + 25x}{600} = 561$
$\frac{51x}{600} = 561$
$x = \frac{561 \times 600}{51}$
$x = 11 \times 600 = 6600$
Therefore,the total capital is ₹ $6600$.

(C) Let the rate of interest be $R \%$ per annum.
Simple Interest formula is $SI = \frac{P \times R \times T}{100}$.
According to the problem,the difference in interest is ₹ $500$ for a time period of $3$ $years$.
Difference in interest $= SI_1 - SI_2 = 500$.
$\frac{12000 \times R \times 3}{100} - \frac{10000 \times R \times 3}{100} = 500$.
$\frac{3R}{100} (12000 - 10000) = 500$.
$\frac{3R}{100} (2000) = 500$.
$3R \times 20 = 500$.
$60R = 500$.
$R = \frac{500}{60} = \frac{50}{6} = 8 \frac{1}{3} \%$.

(C) Let the principal sum be $P$ and the rate of interest be $r \%$ per annum.
Simple Interest $(SI)$ formula is given by $SI = \frac{P \times r \times t}{100}$.
According to the problem,the difference in interest for $2$ years at a rate of $(r+1) \%$ and $r \%$ is ₹ $24$.
$\frac{P \times (r+1) \times 2}{100} - \frac{P \times r \times 2}{100} = 24$
$\frac{2P}{100} (r + 1 - r) = 24$
$\frac{2P}{100} = 24$
$2P = 2400$
$P = 1200$
Therefore,the sum is ₹ $1200$.

(C) Let the principal sum be $P$.
According to the problem,the amount $A$ becomes $\frac{8}{5}P$ in $T = 5$ years.
The simple interest $SI$ is given by $SI = A - P = \frac{8}{5}P - P = \frac{3}{5}P$.
The formula for the rate of interest $R$ is $R = \frac{SI \times 100}{P \times T}$.
Substituting the values: $R = \frac{(\frac{3}{5}P) \times 100}{P \times 5}$.
$R = \frac{3 \times 100}{5 \times 5} = \frac{300}{25} = 12 \%$.
Therefore,the rate per cent per annum is $12 \%$.

(D) The formula for Simple Interest is $SI = \frac{P \times R \times T}{100}$.
Let the rates of the two banks be $R_1$ and $R_2$ respectively.
The difference in interest is given by: $\frac{P \times R_1 \times T}{100} - \frac{P \times R_2 \times T}{100} = 25$.
This can be written as: $\frac{P \times T}{100} \times (R_1 - R_2) = 25$.
Given $P = 5000$,$T = 2$,and the difference in interest $= 25$.
Substituting the values: $\frac{5000 \times 2}{100} \times (R_1 - R_2) = 25$.
$100 \times (R_1 - R_2) = 25$.
$(R_1 - R_2) = \frac{25}{100} = 0.25 \%$.
Thus,the difference between their rates is $0.25 \%$.

(C) Let the principal amount be $P$.
Simple Interest formula is $SI = \frac{P \times R \times T}{100}$.
Interest for the first bank ($T_1 = 3.5$ years): $SI_1 = \frac{P \times 15 \times 3.5}{100} = 0.525P$.
Interest for the second bank ($T_2 = 5$ years): $SI_2 = \frac{P \times 15 \times 5}{100} = 0.75P$.
The difference between the interests is $SI_2 - SI_1 = 1440$.
$0.75P - 0.525P = 1440$.
$0.225P = 1440$.
$P = \frac{1440}{0.225} = 6400$.
Thus,each sum is ₹ $6400$.

(B) Let the principal sum be $P$.
The simple interest formula is $SI = \frac{P \times R \times T}{100}$.
Given the interest rates for different periods:
For the first $2$ years $(T_1 = 2)$,$R_1 = 4 \%$.
For the next $4$ years $(T_2 = 4)$,$R_2 = 6 \%$.
For the remaining period ($T_3 = 9 - 2 - 4 = 3$ years),$R_3 = 8 \%$.
The total simple interest is given by $SI = P \times \left( \frac{R_1 T_1}{100} + \frac{R_2 T_2}{100} + \frac{R_3 T_3}{100} \right)$.
$1120 = P \times \left( \frac{4 \times 2}{100} + \frac{6 \times 4}{100} + \frac{8 \times 3}{100} \right)$.
$1120 = P \times \left( \frac{8 + 24 + 24}{100} \right)$.
$1120 = P \times \left( \frac{56}{100} \right)$.
$P = \frac{1120 \times 100}{56} = 20 \times 100 = 2000$.
Thus,the sum is ₹ $2000$.

(B) Let the principal be $P$ and the rate of interest be $r \% \text{ per annum}$.
According to the problem,the time $T = r \text{ years}$.
The simple interest $SI = \frac{1}{16} P$.
Using the formula $SI = \frac{P \times R \times T}{100}$,we have:
$\frac{P}{16} = \frac{P \times r \times r}{100}$
$\frac{1}{16} = \frac{r^2}{100}$
$r^2 = \frac{100}{16} = \frac{25}{4}$
$r = \sqrt{\frac{25}{4}} = \frac{5}{2} = 2.5 \% \text{ per annum}$.

(D) Let the rate of interest be $r \%$ per annum.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$.
For $B$: $P_1 = 6000$,$T_1 = 2$ years,$R = r$.
$SI_1 = \frac{6000 \times 2 \times r}{100} = 120r$.
For $C$: $P_2 = 1500$,$T_2 = 4$ years,$R = r$.
$SI_2 = \frac{1500 \times 4 \times r}{100} = 60r$.
The total interest received is $900$.
$120r + 60r = 900$.
$180r = 900$.
$r = \frac{900}{180} = 5$.
Thus,the rate of interest is $5 \%$.

(D) Let the three parts be $x, y$ and $z$ such that $x + y + z = 2379$.
The amount $A$ at simple interest is given by $A = P(1 + \frac{RT}{100})$.
Given that the amounts are equal after $2, 3$ and $4$ years respectively at $5 \%$ rate:
$x(1 + \frac{5 \times 2}{100}) = y(1 + \frac{5 \times 3}{100}) = z(1 + \frac{5 \times 4}{100}) = k$
$x(1 + \frac{1}{10}) = y(1 + \frac{3}{20}) = z(1 + \frac{1}{5}) = k$
$\frac{11x}{10} = \frac{23y}{20} = \frac{6z}{5} = k$
Thus,$x = \frac{10k}{11}, y = \frac{20k}{23}, z = \frac{5k}{6}$.
Substituting these into $x + y + z = 2379$:
$\frac{10k}{11} + \frac{20k}{23} + \frac{5k}{6} = 2379$
Taking the $LCM$ of $11, 23, 6$,which is $1518$:
$\frac{1380k + 1320k + 1265k}{1518} = 2379$
$\frac{3965k}{1518} = 2379$
$k = \frac{2379 \times 1518}{3965} = 910.8$
Now,$x = \frac{10 \times 910.8}{11} = 828$.
Therefore,the first part is ₹ $828$.

(A) Let the amount invested in scheme $B$ be $x$.
Given that the investment in scheme $C$ is $240 \%$ of the investment in scheme $B$, we have:
Investment in scheme $C = 2.4x$.
Given that the investment in scheme $C$ is $150 \%$ of the investment in scheme $A$, we have:
$1.5 \times (\text{Investment in } A) = 2.4x$
Investment in scheme $A = \frac{2.4x}{1.5} = 1.6x$.
The total interest accrued in one year is given by the sum of interests from all three schemes:
Interest from $A = 1.6x \times 0.10 = 0.16x$
Interest from $B = x \times 0.12 = 0.12x$
Interest from $C = 2.4x \times 0.15 = 0.36x$
Total interest $= 0.16x + 0.12x + 0.36x = 0.64x$
Given total interest $= 3200$, we have:
$0.64x = 3200$
$x = \frac{3200}{0.64} = 5000$
Thus, the amount invested in scheme $B$ is ₹ $5000$.

(C) Let the capital be $x$.
The difference in the annual rate of interest is $8 \% - 7 \frac{3}{4} \% = 8 \% - 7.75 \% = 0.25 \%$.
We are given that the decrease in yearly income is ₹ $61.50$. Therefore,$0.25 \%$ of the capital $x$ is equal to ₹ $61.50$.
Mathematically,this is represented as:
$x \times \frac{0.25}{100} = 61.50$
$x \times \frac{1}{400} = 61.50$
$x = 61.50 \times 400$
$x = 24600$
Thus,the capital is ₹ $24600$.

(B) Let the principal amount be $P$ (in ₹).
The initial rate of interest is $10 \%$ and the new rate of interest is $12.5 \%$.
The increase in the annual rate of interest is $12.5 \% - 10 \% = 2.5 \%$.
According to the problem,this increase in the rate of interest results in an increase of ₹ $1250$ in the yearly income.
Therefore,$2.5 \%$ of $P = 1250$.
$\frac{2.5}{100} \times P = 1250$.
$P = \frac{1250 \times 100}{2.5}$.
$P = \frac{125000}{2.5} = 50000$.
Thus,the principal amount is ₹ $50000$.

(A) Let the total amount invested be $x$.
The interest from the first part is $12000 \times 0.10 = 1200$.
The remaining amount is $(x - 12000)$,which earns interest at $20 \%$,so the interest is $(x - 12000) \times 0.20$.
The total interest earned is $14 \%$ of the total amount $x$,which is $0.14x$.
Setting up the equation:
$1200 + 0.20(x - 12000) = 0.14x$
$1200 + 0.20x - 2400 = 0.14x$
$0.20x - 0.14x = 2400 - 1200$
$0.06x = 1200$
$x = \frac{1200}{0.06} = 20000$
Thus,the total amount invested is ₹ $20000$.

(D) Let the rate of simple interest be $x \% $ per annum.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$.
Interest from $B = \frac{5000 \times x \times 2}{100} = 100x$.
Interest from $C = \frac{3000 \times x \times 4}{100} = 120x$.
Total interest received = $100x + 120x = 220x$.
Given that the total interest is ₹ $2,200$,we have:
$220x = 2200$.
$x = \frac{2200}{220} = 10$.
Therefore,the rate of interest is $10 \% $ per annum.

(B) Let the sum lent to $C$ be $x$.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$.
Interest from $B = 2500 \times \frac{7}{100} \times 4 = 700$.
Interest from $C = x \times \frac{7}{100} \times 4 = \frac{28x}{100} = \frac{7x}{25}$.
Given that the total interest received is ₹ $1120$,we have:
$700 + \frac{7x}{25} = 1120$.
Subtracting $700$ from both sides:
$\frac{7x}{25} = 1120 - 700 = 420$.
Solving for $x$:
$x = \frac{420 \times 25}{7} = 60 \times 25 = 1500$.
Therefore,the sum lent to $C$ is ₹ $1500$.

(A) Let the principal be $P$ and the rate of interest be $R \%$ per annum.
Given that the time $T = 10$ $years$ and the simple interest $SI = \frac{2}{5} P$.
Using the simple interest formula: $SI = \frac{P \times R \times T}{100}$.
Substituting the values: $\frac{2}{5} P = \frac{P \times R \times 10}{100}$.
Dividing both sides by $P$: $\frac{2}{5} = \frac{10 R}{100}$.
Simplifying the equation: $\frac{2}{5} = \frac{R}{10}$.
Solving for $R$: $R = \frac{2}{5} \times 10 = 4 \%$.
Therefore,the rate of interest is $4 \%$ per annum.

(B) Let the principal amount be $P$ and the rate of interest be $R \%$ per annum.
Since the sum doubles itself,the final amount $A = 2P$.
The simple interest $SI = A - P = 2P - P = P$.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$,where $T$ is the time in years.
Substituting the values,we get $P = \frac{P \times R \times 12}{100}$.
Dividing both sides by $P$,we get $1 = \frac{12R}{100}$.
$R = \frac{100}{12} = \frac{25}{3} = 8 \frac{1}{3} \%$.

(B) Let the principal sum be $P$ and the rate of interest be $R \%$.
Simple Interest $(SI)$ for $2$ years is given by: $SI_2 = P \times R \times 2 / 100$.
The amount after $2$ years is $A_2 = P + SI_2 = P(1 + 2R/100) = 720$.
The total time after a further period of $5$ years is $2 + 5 = 7$ years.
The amount after $7$ years is $A_7 = P(1 + 7R/100) = 1020$.
Subtracting the two equations:
$SI$ for $5$ years = $1020 - 720 = 300$.
$SI$ for $1$ year = $300 / 5 = 60$.
$SI$ for $2$ years = $60 \times 2 = 120$.
Principal $P = \text{Amount after } 2 \text{ years} - SI \text{ for } 2 \text{ years}$.
$P = 720 - 120 = 600$.
Thus,the sum is ₹ $600$.

(C) Let the principal be $P$ and the rate of interest be $R \%$ per annum.
Simple interest for $3$ $years$ ($8$ $years$ - $5$ $years$) = $₹ 12,005 - ₹ 9,800 = ₹ 2,205$.
Simple interest for $1$ $year$ = $₹ 2,205 / 3 = ₹ 735$.
Simple interest for $5$ $years$ = $₹ 735 \times 5 = ₹ 3,675$.
Principal $(P)$ = Amount after $5$ $years$ - Simple interest for $5$ $years$ = $₹ 9,800 - ₹ 3,675 = ₹ 6,125$.
Now,using the formula: $\text{Simple Interest} = (P \times R \times T) / 100$.
$735 = (6,125 \times R \times 1) / 100$.
$R = (735 \times 100) / 6,125 = 73,500 / 6,125 = 12$.
Therefore,the rate of interest is $12 \%$ per annum.

(A) Let the principal amount borrowed by Nitin be $₹ x$.
The total time period is $11 \text{ years}$.
Interest for the first $3 \text{ years}$ at $6 \% \text{ p.a.} = x \times 3 \times \frac{6}{100} = \frac{18x}{100}$.
Interest for the next $5 \text{ years}$ at $9 \% \text{ p.a.} = x \times 5 \times \frac{9}{100} = \frac{45x}{100}$.
Interest for the remaining period $(11 - 3 - 5 = 3 \text{ years})$ at $13 \% \text{ p.a.} = x \times 3 \times \frac{13}{100} = \frac{39x}{100}$.
Total interest = $\frac{18x}{100} + \frac{45x}{100} + \frac{39x}{100} = 8160$.
$\frac{102x}{100} = 8160$.
$x = \frac{8160 \times 100}{102} = 8000$.
Therefore,the amount borrowed by Nitin is $₹ 8000$.

(C) Let the principal sum be $P$.
The formula for the amount $A$ under simple interest is $A = P + \text{Simple Interest} = P + \frac{P \times R \times T}{100}$.
Given,$R = 5 \%$,$T = 4$ years,and $A = ₹ 504$.
$504 = P + \frac{P \times 5 \times 4}{100} = P + \frac{20P}{100} = P + \frac{P}{5} = \frac{6P}{5}$.
Therefore,$P = \frac{504 \times 5}{6} = 84 \times 5 = ₹ 420$.
Now,for the second case,$P = ₹ 420$,$R = 10 \%$,and $T = 2 \frac{1}{2} = 2.5$ years.
New Amount $A' = P + \frac{P \times R \times T}{100} = 420 + \frac{420 \times 10 \times 2.5}{100}$.
$A' = 420 + \frac{420 \times 25}{100} = 420 + 420 \times 0.25 = 420 + 105 = ₹ 525$.

(C) Let the required number of years be $N$.
The formula for Simple Interest is $SI = \frac{P \times R \times T}{100}$.
According to the problem,the interest on ₹ $150$ at $8 \%$ for $N$ years is equal to the interest on ₹ $800$ at $4.5 \%$ $(4 \frac{1}{2} \%)$ for $3$ years.
$\frac{150 \times 8 \times N}{100} = \frac{800 \times 4.5 \times 3}{100}$
$150 \times 8 \times N = 800 \times 4.5 \times 3$
$1200 \times N = 10800$
$N = \frac{10800}{1200} = 9$
Thus,the required time is $9$ years.

(C) Let the principal amount be $P$ and the rate of interest be $R$ per annum.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$.
For $T_1 = 6$ years,$SI_1 = \frac{P \times R \times 6}{100}$.
For $T_2 = 9$ years,$SI_2 = \frac{P \times R \times 9}{100}$.
The ratio of the simple interests is $\frac{SI_1}{SI_2} = \frac{\frac{P \times R \times 6}{100}}{\frac{P \times R \times 9}{100}}$.
Simplifying the expression,we get $\frac{SI_1}{SI_2} = \frac{6}{9} = \frac{2}{3}$.
Thus,the ratio is $2:3$.

(C) Given: Principal $(P) = ₹ 800$,Time $(T) = 3$ years,Amount $(A) = ₹ 956$.
Simple Interest $(SI) = A - P = 956 - 800 = ₹ 156$.
Using the formula $SI = \frac{P \times R \times T}{100}$,we have $156 = \frac{800 \times R \times 3}{100}$.
$156 = 24R \implies R = \frac{156}{24} = 6.5 \%$.
New rate of interest $(R') = 6.5 \% + 4 \% = 10.5 \%$.
New Simple Interest $(SI') = \frac{800 \times 10.5 \times 3}{100} = 8 \times 10.5 \times 3 = 252$.
New Amount $(A') = P + SI' = 800 + 252 = ₹ 1052$.

(D) Let the principal amount be $P$ and the rate of interest be $R \%$.
Simple Interest $(SI)$ is given by the formula: $SI = \frac{P \times R \times T}{100}$.
Given $SI = 1750$,$T = 7$ years.
So,$1750 = \frac{P \times R \times 7}{100} \implies P \times R = \frac{1750 \times 100}{7} = 25000$.
If the interest rate increases by $2 \%$,the new rate is $(R + 2) \%$.
The new interest earned would be $SI_{new} = \frac{P \times (R + 2) \times 7}{100}$.
The extra interest earned is $SI_{new} - SI = \frac{P \times (R + 2) \times 7}{100} - \frac{P \times R \times 7}{100}$.
Extra Interest $= \frac{P \times R \times 7 + P \times 2 \times 7 - P \times R \times 7}{100} = \frac{P \times 2 \times 7}{100} = \frac{14P}{100} = 0.14P$.
Since the principal $P$ is not provided,the exact value of the extra interest cannot be determined.

(C) Let the present worth be ₹ $x$.
The formula for the amount $A$ in terms of present worth $P$,rate $R$,and time $T$ is $A = P(1 + \frac{RT}{100})$.
Given $A = 132$,$R = 5\%$,and $T = 2$ years.
Substituting the values: $132 = x(1 + \frac{5 \times 2}{100})$.
$132 = x(1 + \frac{10}{100})$.
$132 = x(1 + 0.1) = 1.1x$.
$x = \frac{132}{1.1} = \frac{1320}{11} = 120$.
Therefore,the present worth is ₹ $120$.

(C) The formula for simple interest is $SI = \frac{P \times R \times T}{100}$,where $P$ is the principal,$R$ is the rate of interest,and $T$ is the time in years.
Given: $SI = ₹ 5400$,$R = 12 \%$,$T = 3 \text{ years}$.
Substituting the values into the formula:
$5400 = \frac{P \times 12 \times 3}{100}$
$5400 = \frac{P \times 36}{100}$
$P = \frac{5400 \times 100}{36}$
$P = 150 \times 100 = 15000$.
Therefore,the principal amount borrowed was ₹ $15000$.

(B) Let the rate of interest be $R \%$ per annum and the time period be $T$ years.
According to the problem,the time period is equal to the rate of interest,so $T = R$.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$.
Given: Principal $P = ₹ 1,200$,Simple Interest $SI = ₹ 432$,and $T = R$.
Substituting the values into the formula:
$432 = \frac{1200 \times R \times R}{100}$
$432 = 12 \times R^2$
$R^2 = \frac{432}{12}$
$R^2 = 36$
$R = \sqrt{36} = 6$.
Therefore,the rate of interest is $6 \%$.