Simple Interest Questions in English

(C) Given: Principal $(P)$ = ₹ $9500$,Amount $(A)$ = ₹ $17622.5$,Rate $(R)$ = $4.5 \%$ per annum.
Simple Interest $(S.I)$ = Amount - Principal = $17622.5 - 9500 = ₹ 8122.5$.
Using the formula for Simple Interest: $S.I = \frac{P \times N \times R}{100}$,where $N$ is the time in years.
$8122.5 = \frac{9500 \times N \times 4.5}{100}$.
$8122.5 = 95 \times N \times 4.5$.
$8122.5 = 427.5 \times N$.
$N = \frac{8122.5}{427.5} = 19$.
Therefore,the time required is $19$ years.

(C) Let the rate of interest be $r \%$ per annum.
Simple interest is calculated using the formula: $SI = \frac{P \times R \times T}{100}$.
For Jaabir: Principal $P_1 = ₹ 10800$,Time $T_1 = 3$ years,Rate $= r \%$.
$SI_1 = \frac{10800 \times r \times 3}{100} = 324r$.
For Kabir: Principal $P_2 = ₹ 7500$,Time $T_2 = 2$ years,Rate $= r \%$.
$SI_2 = \frac{7500 \times r \times 2}{100} = 150r$.
Total interest received $= SI_1 + SI_2 = 1422$.
$324r + 150r = 1422$.
$474r = 1422$.
$r = \frac{1422}{474} = 3$.
Therefore,the rate of interest is $3 \%$ per annum.

(D) Let the rate of interest be $r \%$ per annum.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$.
Interest from Jaichand $= \frac{8800 \times 13 \times r}{100} = 1144r$.
Interest from the second person $= \frac{5500 \times 12 \times r}{100} = 660r$.
Total interest received $= 1144r + 660r = 1804r$.
Given that the total interest is ₹ $14432$,we have:
$1804r = 14432$.
Solving for $r$:
$r = \frac{14432}{1804} = 8$.
Therefore,the rate of interest is $8 \%$ per annum.

(B) Let the principal sum be $P = x$.
Since the sum is doubled,the final amount becomes $2x$.
Therefore,the Simple Interest $(S.I.)$ earned is $Amount - Principal = 2x - x = x$.
The rate of interest $(R)$ is $12 \frac{1}{2} \% = 12.5 \% = \frac{25}{2} \%$.
The formula for Simple Interest is $S.I. = \frac{P \times N \times R}{100}$,where $N$ is the time in years.
Substituting the values: $x = \frac{x \times N \times 12.5}{100}$.
Dividing both sides by $x$: $1 = \frac{N \times 12.5}{100}$.
$N = \frac{100}{12.5} = 8$.
Thus,the time taken is $8$ years.

(D) The formula for simple interest is $S.I. = \frac{P \times N \times R}{100}$,where $P$ is the principal,$N$ is the time in years,and $R$ is the rate of interest per annum.
Given: $P = ₹ 5400$,$N = 5$ years,$R = 12 \%$.
Substituting the values into the formula:
$S.I. = \frac{5400 \times 5 \times 12}{100}$
$S.I. = 54 \times 5 \times 12$
$S.I. = 270 \times 12$
$S.I. = ₹ 3240$.

(D) Let the Principal be $P$ and the rate of interest be $R$ percent per annum.
Simple Interest $(SI)$ for $5$ years $= A_1 - P = 1020 - P$.
Simple Interest $(SI)$ for $8$ years $= A_2 - P = 1200 - P$.
Since simple interest is constant for every year,the interest for $(8 - 5) = 3$ years is $1200 - 1020 = ₹ 180$.
Interest for $1$ year $= 180 / 3 = ₹ 60$.
Interest for $5$ years $= 60 \times 5 = ₹ 300$.
Principal $P = \text{Amount after } 5 \text{ years} - \text{Interest for } 5 \text{ years}$.
$P = 1020 - 300 = ₹ 720$.

(C) Let the principal sum be $P$.
According to the simple interest formula,the amount $A = P + SI$,where $SI = (P \times R \times T) / 100$.
Given that the sum becomes $5$ times in $3$ years,the amount $A = 5P$.
So,$SI = 5P - P = 4P$.
Using $SI = (P \times R \times T) / 100$,we have $4P = (P \times R \times 3) / 100$,which implies $R = 400 / 3 \%$.
Now,we want the sum to become $13$ times,so $A = 13P$.
This means $SI = 13P - P = 12P$.
Using the formula $SI = (P \times R \times T) / 100$ again:
$12P = (P \times (400 / 3) \times T) / 100$
$12 = (4 / 3) \times T$
$T = 12 \times (3 / 4) = 9 \text{ years}$.
Therefore,the sum will become $13$ times in $9$ years.

(A) Simple Interest $(S.I.)$ for $1.5$ years $= 969 - 918 = ₹ 51$.
$S.I.$ for $1$ year $= \frac{51}{1.5} = ₹ 34$.
$S.I.$ for $2$ years $= 34 \times 2 = ₹ 68$.
Principal $(P)$ $= \text{Amount} - S.I. = 918 - 68 = ₹ 850$.
Using the formula $S.I. = \frac{P \times R \times T}{100}$:
$68 = \frac{850 \times R \times 2}{100}$.
$68 = 17 \times R$.
$R = \frac{68}{17} = 4 \% \text{ p.a.}$

(B) Given:
Principal $(P) = ₹ 9500$
Time $(T) = 10 \text{ years}$
Simple Interest $(S.I.) = 130\% \text{ of } P$
Calculation:
$S.I. = \frac{130}{100} \times 9500 = 130 \times 95 = ₹ 12350$
Using the formula for Simple Interest:
$S.I. = \frac{P \times R \times T}{100}$
$12350 = \frac{9500 \times R \times 10}{100}$
$12350 = 950 \times R$
$R = \frac{12350}{950}$
$R = 13\%$
Therefore,the rate of interest per annum is $13\%$.

(C) The formula for simple interest is given by $S.I = \frac{P \times R \times T}{100}$,where $P$ is the principal,$R$ is the rate of interest,and $T$ is the time period.
Given: $S.I = ₹ 3200$,$R = 6.25 \%$,and $T = 4 \text{ years}$.
Substituting the values into the formula:
$3200 = \frac{P \times 6.25 \times 4}{100}$
$3200 = \frac{P \times 25}{100}$
$3200 = \frac{P}{4}$
$P = 3200 \times 4 = ₹ 12800$.
Thus,the principal amount is ₹ $12800$.

(D) The nominal rate is $13 \%$ per annum,compounded half-yearly.
The rate per half-year is $r = \frac{13}{2} \% = 6.5 \%$.
The effective annual rate $(EAR)$ is calculated using the formula: $\text{EAR} = (1 + \frac{r}{100})^n - 1$,where $n$ is the number of compounding periods per year.
Here,$n = 2$ (since it is half-yearly).
$\text{EAR} = (1 + \frac{6.5}{100})^2 - 1$
$\text{EAR} = (1.065)^2 - 1$
$\text{EAR} = 1.134225 - 1 = 0.134225$
Converting to percentage: $0.134225 \times 100 = 13.4225 \% \approx 13.42 \%$.
Thus,the effective annual rate is $13.42 \%$.

(A) Let the rate of interest be $r \%$.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$.
Interest from $B = \frac{10000 \times r \times 3}{100} = 300r$.
Interest from $C = \frac{6000 \times r \times 4}{100} = 240r$.
Total interest received is $5400$.
Therefore,$300r + 240r = 5400$.
$540r = 5400$.
$r = \frac{5400}{540} = 10 \%$.
Thus,the rate of interest is $10 \%$.

(B) The amount after the first year is $x \times (1 + \frac{10}{100}) = x \times \frac{110}{100}$.
After the second year,the amount becomes $(x \times \frac{110}{100}) \times (1 + \frac{15}{100}) = x \times \frac{110}{100} \times \frac{115}{100}$.
Given that the final amount is ₹ $20,240$,we have:
$x \times \frac{110}{100} \times \frac{115}{100} = 20240$.
$x \times 1.1 \times 1.15 = 20240$.
$x \times 1.265 = 20240$.
$x = \frac{20240}{1.265} = 16000$.
Therefore,the principal amount $x$ is ₹ $16000$.

(C) Let the principal amount be $P = 10x$ and the simple interest be $S.I. = 3x$.
The rate of interest is $R = 6 \%$ per annum.
The formula for simple interest is $S.I. = \frac{P \times N \times R}{100}$,where $N$ is the number of years.
Substituting the given values into the formula:
$3x = \frac{10x \times N \times 6}{100}$
Dividing both sides by $x$ (assuming $x \neq 0$):
$3 = \frac{60N}{100}$
$3 = \frac{3N}{5}$
$N = 3 \times \frac{5}{3} = 5 \text{ years}$.
Therefore,the required time is $5$ years.

(B) The formula for simple interest is $S.I = \frac{P \times N \times R}{100}$,where $P$ is the principal,$N$ is the time in years,and $R$ is the rate of interest.
Given: $S.I = ₹ 840$,$N = 8$ years,$R = 5 \%$.
Substituting the values: $840 = \frac{P \times 8 \times 5}{100}$.
$840 = \frac{P \times 40}{100} \Rightarrow 840 = \frac{P}{2.5} \Rightarrow P = 840 \times 2.5 = ₹ 2100$.
Now,for the same principal $P = ₹ 2100$ and same interest $S.I = ₹ 840$,we need to find the new rate $R$ for $N = 5$ years.
$840 = \frac{2100 \times 5 \times R}{100}$.
$840 = 21 \times 5 \times R$.
$840 = 105 \times R$.
$R = \frac{840}{105} = 8 \%$.
Therefore,the required rate of interest is $8 \%$.

(B) Let the first part be $₹ x$. Then the second part is $₹ (2800 - x)$.
Since the simple interest on both parts is equal,we use the formula $SI = \frac{P \times R \times T}{100}$.
$\frac{x \times 9 \times 5}{100} = \frac{(2800 - x) \times 10 \times 6}{100}$
$45x = (2800 - x) \times 60$
Dividing both sides by $15$:
$3x = (2800 - x) \times 4$
$3x = 11200 - 4x$
$7x = 11200$
$x = 1600$
Therefore,the first part is $₹ 1600$ and the second part is $2800 - 1600 = ₹ 1200$.

(C) Given: Principal $(P) = ₹ 1$.
Simple Interest $(S.I.) = 1$ paisa $= ₹ \frac{1}{100}$.
Time $(T) = 1$ month $= \frac{1}{12}$ years.
We know the formula for Simple Interest: $S.I. = \frac{P \times R \times T}{100}$.
Substituting the values: $\frac{1}{100} = \frac{1 \times R \times (1/12)}{100}$.
By simplifying: $\frac{1}{100} = \frac{R}{1200}$.
$R = \frac{1200}{100} = 12$.
Therefore,the rate per cent per annum is $12 \%$.

(C) Let the principal be $P$ and the simple interest per year be $I$.
Given that the amount after $2$ years is $P + 2I = 5182$ and the amount after $3$ years is $P + 3I = 5832$.
Subtracting the first equation from the second: $(P + 3I) - (P + 2I) = 5832 - 5182$.
This gives the simple interest for $1$ year: $I = 650$.
Now,substitute $I$ into the equation for $2$ years: $P + 2(650) = 5182$.
$P + 1300 = 5182$.
$P = 5182 - 1300 = 3882$.
Therefore,the principal is ₹ $3882$.

(B) The formula for simple interest is given by $S.I. = \frac{P \times R \times T}{100}$,where $P$ is the principal sum,$R$ is the rate of interest per annum,and $T$ is the time in years.
Given that $S.I. = R$,$T = 2$ years,and the rate is $R \%$.
Substituting these values into the formula:
$R = \frac{P \times R \times 2}{100}$
Dividing both sides by $R$ (assuming $R \neq 0$):
$1 = \frac{P \times 2}{100}$
$1 = \frac{P}{50}$
$P = 50$
Therefore,the sum is $₹ 50$.

(C) Let the principal be $P$ and the rate of interest be $R\%$.
Given that the sum doubles in $7$ years,the simple interest $(S.I.)$ is $P$.
Using the formula $S.I. = \frac{P \times R \times T}{100}$,we have $P = \frac{P \times R \times 7}{100}$,which implies $R = \frac{100}{7}\%$.
Now,we want the sum to become $4$ times,meaning the amount $A = 4P$. Thus,the required simple interest is $S.I. = A - P = 4P - P = 3P$.
Using the formula $3P = \frac{P \times R \times T}{100}$,we substitute $R = \frac{100}{7}$:
$3P = \frac{P \times (100/7) \times T}{100}$
$3 = \frac{T}{7}$
$T = 21 \text{ years}$.
Therefore,the sum will become $4$ times in $21$ years.

(D) Let the principal be $P$ and the time period be $T$ years.
The formula for the amount $A$ is $A = P + \frac{P \times R \times T}{100}$.
For the first case: $2200 = P + \frac{P \times 5 \times T}{100} \implies 2200 = P(1 + 0.05T) \quad -(1)$
For the second case: $2320 = P + \frac{P \times 8 \times T}{100} \implies 2320 = P(1 + 0.08T) \quad -(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{2320}{2200} = \frac{1 + 0.08T}{1 + 0.05T}$
$\frac{232}{220} = \frac{1 + 0.08T}{1 + 0.05T} \implies \frac{58}{55} = \frac{1 + 0.08T}{1 + 0.05T}$
$58(1 + 0.05T) = 55(1 + 0.08T)$
$58 + 2.9T = 55 + 4.4T$
$58 - 55 = 4.4T - 2.9T$
$3 = 1.5T$
$T = \frac{3}{1.5} = 2$ years.

(B) Let the principal amount be $P$.
Since the sum doubles itself,the final amount $A = 2P$.
Therefore,the Simple Interest $(S.I.)$ is $A - P = 2P - P = P$.
We know the formula for Simple Interest is $S.I. = \frac{P \times R \times T}{100}$,where $R$ is the rate of interest and $T$ is the time in years.
Substituting the values: $P = \frac{P \times R \times 15}{100}$.
Dividing both sides by $P$,we get $1 = \frac{15R}{100}$.
$R = \frac{100}{15} = \frac{20}{3} = 6 \frac{2}{3} \%$.
Thus,the rate of interest is $6 \frac{2}{3} \%$.

(C) Let the principal amount be $P$.
Since the sum becomes $5$ times itself,the final amount $A = 5P$.
The simple interest $(S.I.)$ is given by $S.I. = A - P = 5P - P = 4P$.
The formula for simple interest is $S.I. = \frac{P \times R \times T}{100}$,where $R$ is the rate of interest and $T$ is the time in years.
Substituting the values: $4P = \frac{P \times R \times 8}{100}$.
Dividing both sides by $P$: $4 = \frac{8R}{100}$.
$400 = 8R$.
$R = \frac{400}{8} = 50 \%$.
Therefore,the rate of interest is $50 \%$.

(A) Let the time after which the amounts are equal be $N$ years.
Amount of debt for $A = P + \text{Simple Interest} = P + \frac{P \times x \times N}{100}$.
Amount of debt for $B = Q + \text{Simple Interest} = Q + \frac{Q \times y \times N}{100}$.
Since the amounts are equal,we set them equal to each other:
$P + \frac{PxN}{100} = Q + \frac{QyN}{100}$.
Rearranging the terms to solve for $N$:
$\frac{PxN}{100} - \frac{QyN}{100} = Q - P$.
$N \left(\frac{Px - Qy}{100}\right) = Q - P$.
$N = 100 \left(\frac{Q - P}{Px - Qy}\right)$ years.

(D) Let the principal amount be $P = 100$.
Given that the interest rate is $10 \%$ per annum.
The interest for one year is calculated as $I = \frac{P \times R \times T}{100} = \frac{100 \times 10 \times 1}{100} = 10$.
Since the money lender takes the interest in advance,the actual amount he gives to the borrower is $100 - 10 = 90$.
Therefore,the borrower effectively receives $90$ and pays $10$ as interest for one year.
The actual rate of interest is calculated as $\text{Rate} = \frac{\text{Interest}}{\text{Actual Principal}} \times 100$.
$\text{Rate} = \frac{10}{90} \times 100 = \frac{100}{9} = 11 \frac{1}{9} \%$.

(B) The selling price $(S.P.)$ is the present worth $(P.W.)$ of ₹ $2200$ due $1$ year hence.
The formula for present worth is $P.W. = \frac{\text{Amount} \times 100}{100 + (R \times T)}$.
Substituting the given values: $P.W. = \frac{2200 \times 100}{100 + (10 \times 1)} = \frac{220000}{110} = ₹ 2000$.
Since the cost price $(C.P.)$ is ₹ $1950$,the profit is $S.P. - C.P. = 2000 - 1950 = ₹ 50$.

(A) Let the rate of interest be $R\%$ per annum.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$.
For the first person: $SI_1 = \frac{400 \times R \times 3}{100} = 12R$.
For the second person: $SI_2 = \frac{500 \times R \times 4}{100} = 20R$.
According to the problem,the total interest received is ₹ $160$.
$12R + 20R = 160$
$32R = 160$
$R = \frac{160}{32} = 5$.
Thus,the rate of interest is $5\%$ per annum.

(D) Let the principal sum be $P$ and the rate of interest be $R \%$ per annum.
According to the problem,the simple interest $(S.I.)$ is $\frac{8}{25} P$.
The time period $(T)$ in years is numerically half the rate percent per annum,so $T = \frac{R}{2}$.
The formula for simple interest is $S.I. = \frac{P \times R \times T}{100}$.
Substituting the given values: $\frac{8}{25} P = \frac{P \times R \times (R/2)}{100}$.
Canceling $P$ from both sides: $\frac{8}{25} = \frac{R^2}{200}$.
Solving for $R^2$: $R^2 = \frac{8 \times 200}{25} = 8 \times 8 = 64$.
Taking the square root: $R = \sqrt{64} = 8$.
Thus,the rate percent per annum is $8 \%$.

(A) Let the principal be $P$ and the rate of interest be $R$ per annum.
Simple Interest $(SI)$ for $2$ years $= A_2 - P = 720 - P$.
Simple Interest $(SI)$ for $7$ years ($2 + 5$ years) $= A_7 - P = 1020 - P$.
Since simple interest is constant every year,the interest for $5$ years is $1020 - 720 = 300$.
Interest for $1$ year $= 300 / 5 = 60$.
Interest for $2$ years $= 60 \times 2 = 120$.
Principal $P = \text{Amount after } 2 \text{ years} - \text{Interest for } 2 \text{ years}$.
$P = 720 - 120 = 600$.
Therefore,the principal is $Rs. 600$.

(A) Let the principal amount deposited in each case be $P$.
For the company:
Principal = $P$,Rate $(r_1)$ = $12 \%$,Time $(t_1)$ = $4$ years.
Simple Interest $(SI_1)$ = $\frac{P \times r_1 \times t_1}{100} = \frac{P \times 12 \times 4}{100} = \frac{48P}{100}$.
For the bank:
Principal = $P$,Rate $(r_2)$ = $15 \%$,Time $(t_2)$ = $5$ years.
Simple Interest $(SI_2)$ = $\frac{P \times r_2 \times t_2}{100} = \frac{P \times 15 \times 5}{100} = \frac{75P}{100}$.
Given that the difference in interest is $Rs. 1350$:
$SI_2 - SI_1 = 1350$
$\frac{75P}{100} - \frac{48P}{100} = 1350$
$\frac{27P}{100} = 1350$
$P = \frac{1350 \times 100}{27}$
$P = 50 \times 100 = 5000$.
Thus,the sum deposited in each case is $Rs. 5000$.

(A) Let the amount lent at $8 \%$ rate of interest be $₹ x$.
Then,the amount lent at $\frac{4}{3} \%$ rate of interest is $₹(20,000 - x)$.
The formula for Simple Interest $(SI)$ is $SI = \frac{P \times R \times T}{100}$.
According to the problem,the total interest after $1$ year is $₹ 800$.
$\frac{x \times 8 \times 1}{100} + \frac{(20000 - x) \times (4/3) \times 1}{100} = 800$
$\frac{8x}{100} + \frac{4(20000 - x)}{300} = 800$
Multiply the entire equation by $300$ to clear the denominators:
$3(8x) + 4(20000 - x) = 800 \times 300$
$24x + 80000 - 4x = 240000$
$20x = 240000 - 80000$
$20x = 160000$
$x = \frac{160000}{20} = 8000$.
Thus,the sum lent at $8 \%$ interest is $₹ 8000$.

(C) Let the principal amount borrowed be $₹ x$.
The total interest is calculated based on the simple interest formula: $SI = \frac{P \times R \times T}{100}$.
For the first $3$ years at $6 \% \text{ p.a.}$,interest $= \frac{x \times 6 \times 3}{100} = \frac{18x}{100}$.
For the next $5$ years at $9 \% \text{ p.a.}$,interest $= \frac{x \times 9 \times 5}{100} = \frac{45x}{100}$.
For the remaining period of $11 - (3 + 5) = 3$ years at $13 \% \text{ p.a.}$,interest $= \frac{x \times 13 \times 3}{100} = \frac{39x}{100}$.
The total interest is given as $₹ 8,160$.
Therefore,$\frac{18x + 45x + 39x}{100} = 8160$.
$\frac{102x}{100} = 8160$.
$102x = 816000$.
$x = \frac{816000}{102} = 8000$.
Thus,the money borrowed by Nitin was ₹ $8,000$.

(A) Let the sum lent at $6 \%$ rate of interest be $₹ x$.
Then,the sum lent at $8 \%$ rate of interest is $₹(16800 - x)$.
The total simple interest earned is $SI = 19000 - 16800 = ₹ 2200$.
Using the simple interest formula $SI = \frac{P \times R \times T}{100}$,we have:
$\frac{x \times 6 \times 2}{100} + \frac{(16800 - x) \times 8 \times 2}{100} = 2200$
$12x + 16(16800 - x) = 220000$
$12x + 268800 - 16x = 220000$
$-4x = 220000 - 268800$
$-4x = -48800$
$x = \frac{48800}{4} = 12200$.
Therefore,the sum lent at $6 \%$ interest is ₹ $12200$.

(D) Let the principal be $P$ and the rate of interest be $R\%$ per annum.
Simple interest for $4$ years,$SI_1 = \frac{P \times R \times 4}{100}$.
Simple interest for $6$ years,$SI_2 = \frac{P \times R \times 6}{100}$.
According to the problem,$SI_2$ is $50\%$ more than $SI_1$.
So,$SI_2 = SI_1 + 0.50 \times SI_1 = 1.5 \times SI_1$.
Substituting the values: $\frac{P \times R \times 6}{100} = 1.5 \times \frac{P \times R \times 4}{100}$.
Simplifying both sides: $\frac{6PR}{100} = 1.5 \times \frac{4PR}{100}$.
$\frac{6PR}{100} = \frac{6PR}{100}$.
Since the equation holds true for any value of $R$,the rate of interest cannot be determined from the given information.

(D) Let the principal be $P$ and the rate of interest be $r \%$ per annum.
According to the problem,the time $T$ in years is equal to the rate $r$,so $T = r$.
The simple interest $SI$ is given as $\frac{1}{9}$ of the principal,so $SI = \frac{P}{9}$.
The formula for simple interest is $SI = \frac{P \times r \times T}{100}$.
Substituting the values: $\frac{P}{9} = \frac{P \times r \times r}{100}$.
Dividing both sides by $P$: $\frac{1}{9} = \frac{r^2}{100}$.
$r^2 = \frac{100}{9}$.
Taking the square root: $r = \sqrt{\frac{100}{9}} = \frac{10}{3} = 3 \frac{1}{3} \%$.

(D) Total principal amount = ₹ $20,000$.
Interest earned from the first friend on ₹ $12,000$ at $8 \%$ p.a. for $1$ year = $\frac{12000 \times 8 \times 1}{100} = ₹ 960$.
Total desired gain on ₹ $20,000$ at $10 \%$ p.a. = $\frac{20000 \times 10}{100} = ₹ 2,000$.
Remaining principal to be lent to the second friend = $20000 - 12000 = ₹ 8,000$.
Required interest from the second friend = $2000 - 960 = ₹ 1,040$.
Let the rate of interest for the second friend be $R \%$.
Using the formula,$SI = \frac{P \times R \times T}{100}$,we get $1040 = \frac{8000 \times R \times 1}{100}$.
$R = \frac{1040}{80} = 13 \%$.
Therefore,the required rate is $13 \%$ per annum.

(B) Let the principal amount be $P$.
Since the sum doubles itself,the final amount $A = 2P$.
The simple interest $SI$ is given by $SI = A - P = 2P - P = P$.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$,where $R = 20 \%$ and $T$ is the time in years.
Substituting the values: $P = \frac{P \times 20 \times T}{100}$.
Dividing both sides by $P$: $1 = \frac{20 \times T}{100}$.
$1 = \frac{T}{5}$.
Therefore,$T = 5$ years.

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

(B) Let the principal sum be $P$ and the rate of interest be $R = x \%$. Given that the time $T$ (in years) is equal to the rate $R$,so $T = x$.
Simple Interest $(SI)$ is given as $\frac{16}{25} P$.
The formula for Simple Interest is $SI = \frac{P \times R \times T}{100}$.
Substituting the values: $\frac{16}{25} P = \frac{P \times x \times x}{100}$.
Dividing both sides by $P$: $\frac{16}{25} = \frac{x^2}{100}$.
$x^2 = \frac{16}{25} \times 100$.
$x^2 = 16 \times 4 = 64$.
$x = \sqrt{64} = 8$.
Thus,the rate per cent is $8 \%$.

(B) The formula for simple interest $(SI)$ is given by $SI = \frac{P \times R \times T}{100}$,where $P$ is the principal,$R$ is the rate of interest,and $T$ is the time in years.
According to the problem,the simple interest on ₹ $x$ at $a \%$ for $m$ years is equal to the simple interest on ₹ $y$ at $a^2 \%$ for $m^2$ years.
Therefore,we can write:
$\frac{x \times a \times m}{100} = \frac{y \times a^2 \times m^2}{100}$
By canceling $100$ from both sides,we get:
$x \times a \times m = y \times a^2 \times m^2$
To find the ratio $x: y$,we divide both sides by $y$ and then by $(a \times m)$:
$\frac{x}{y} = \frac{a^2 \times m^2}{a \times m}$
$\frac{x}{y} = am$
Thus,$x: y = am: 1$.

(D) Let the amount borrowed by $A$ from $B$ be ₹ $x$.
Then,the amount borrowed from $C$ is ₹ $(1200 - x)$.
The simple interest formula is $SI = \frac{P \times R \times T}{100}$.
According to the question,the total interest paid in one year is ₹ $172$:
$\frac{x \times 14 \times 1}{100} + \frac{(1200 - x) \times 15 \times 1}{100} = 172$
Multiply the entire equation by $100$:
$14x + 15(1200 - x) = 17200$
$14x + 18000 - 15x = 17200$
$-x + 18000 = 17200$
$x = 18000 - 17200$
$x = 800$
Therefore,$A$ borrowed ₹ $800$ from $B$.

(B) Let the younger son's share be $₹ x$. Then the elder son's share is $₹ (120000 - x)$.
The younger son is $12$ years old,so he will receive the amount after $18 - 12 = 6$ years.
The elder son is $14$ years old,so he will receive the amount after $18 - 14 = 4$ years.
Since the amounts become equal at the age of $18$ with $5 \%$ simple interest per annum:
Amount for younger son $= x + \frac{x \times 5 \times 6}{100} = x(1 + 0.3) = 1.3x$
Amount for elder son $= (120000 - x) + \frac{(120000 - x) \times 5 \times 4}{100} = (120000 - x)(1 + 0.2) = 1.2(120000 - x)$
Equating both amounts:
$1.3x = 1.2(120000 - x)$
$1.3x = 144000 - 1.2x$
$2.5x = 144000$
$x = \frac{144000}{2.5} = 57600$
Therefore,the younger son's share is $₹ 57600$.

(B) Step $1$: Calculate the rate of interest $(r)$ using the simple interest formula: $SI = \frac{P \times r \times t}{100}$.
Given $SI = 10800$,$P = 22500$,and $t = 4$ years.
$10800 = \frac{22500 \times r \times 4}{100} \implies 10800 = 900 \times r \implies r = \frac{10800}{900} = 12 \%$.
Step $2$: Calculate the compound interest $(CI)$ for $t = 2$ years using the formula: $CI = P \left(1 + \frac{r}{100}\right)^t - P$.
$CI = 22500 \left(1 + \frac{12}{100}\right)^2 - 22500$.
$CI = 22500 \left(1.12\right)^2 - 22500$.
$CI = 22500 \times 1.2544 - 22500$.
$CI = 28224 - 22500 = 5724$.
Thus,the compound interest is ₹ $5724$.

(A) The formula for simple interest is $S.I. = \frac{P \times R \times T}{100}$.
Given,$S.I. = 1200$,$T = 4$ years,and $R = 8 \%$.
Substituting these values: $1200 = \frac{P \times 8 \times 4}{100}$.
$P = \frac{1200 \times 100}{32} = 3750$.
Now,we need to find the simple interest on thrice the principal $(3P)$ at $6 \%$ per annum for $3$ years.
New Principal $P' = 3 \times 3750 = 11250$.
New $S.I. = \frac{11250 \times 6 \times 3}{100} = 112.5 \times 18 = 2025$.
Thus,the simple interest is ₹ $2025$.

(C) Let the total capital invested be $₹ x$.
The capital is divided into three parts:
Part $1 = \frac{1}{3}x$ at $7 \%$ interest.
Part $2 = \frac{1}{4}x$ at $8 \%$ interest.
Part $3 = \left(1 - \frac{1}{3} - \frac{1}{4}\right)x = \left(\frac{12-4-3}{12}\right)x = \frac{5}{12}x$ at $10 \%$ interest.
The total annual interest is given by:
$\text{Interest} = \left(\frac{\frac{1}{3}x \times 7}{100}\right) + \left(\frac{\frac{1}{4}x \times 8}{100}\right) + \left(\frac{\frac{5}{12}x \times 10}{100}\right) = 561$
$\frac{7x}{300} + \frac{8x}{400} + \frac{50x}{1200} = 561$
Finding a common denominator $(1200)$:
$\frac{28x + 24x + 50x}{1200} = 561$
$\frac{102x}{1200} = 561$
$x = \frac{561 \times 1200}{102} = 5.5 \times 1200 = 6600$
Thus,the total capital invested is ₹ $6600$.

(D) Step $1$: Calculate the original principal $(P)$.
Simple Interest $(SI)$ = $\frac{P \times R \times T}{100}$
$1000 = \frac{P \times 5 \times 4}{100}$
$1000 = \frac{20P}{100} \implies 1000 = \frac{P}{5}$
$P = ₹ 5000$.
Step $2$: Calculate the new principal and compound interest.
New Principal $(P')$ = $2 \times P = 2 \times 5000 = ₹ 10000$.
Rate $(R)$ = $5 \%$,Time $(T)$ = $2$ years.
Compound Interest $(CI)$ = $P' \times [(1 + \frac{R}{100})^T - 1]$
$CI = 10000 \times [(1 + \frac{5}{100})^2 - 1]$
$CI = 10000 \times [(1.05)^2 - 1]$
$CI = 10000 \times [1.1025 - 1]$
$CI = 10000 \times 0.1025 = ₹ 1025$.