Simple Interest Questions in English

(C) Let the principal amount borrowed be ₹ $x$.
The total time period is $9$ years.
Interest for the first $2$ years at $6 \% \text{ p.a.} = x \times \frac{6}{100} \times 2 = \frac{12x}{100}$.
Interest for the next $3$ years at $9 \% \text{ p.a.} = x \times \frac{9}{100} \times 3 = \frac{27x}{100}$.
Interest for the remaining period ($9 - 2 - 3 = 4$ years) at $14 \% \text{ p.a.} = x \times \frac{14}{100} \times 4 = \frac{56x}{100}$.
Total interest = $\frac{12x + 27x + 56x}{100} = \frac{95x}{100}$.
Given that the total interest is ₹ $11,400$,we have:
$\frac{95x}{100} = 11400$.
$x = \frac{11400 \times 100}{95}$.
$x = 120 \times 100 = 12000$.
Therefore,the amount borrowed is ₹ $12,000$.

(B) Let the principal be $P$ and the rate of interest be $R \%$ per annum.
Since the sum doubles in $16$ years,the Simple Interest $(SI)$ earned will be $SI = 2P - P = P$.
The formula for Simple Interest is $SI = \frac{P \times R \times T}{100}$.
Substituting the values,we get $P = \frac{P \times R \times 16}{100}$.
Dividing both sides by $P$,we get $1 = \frac{16R}{100}$.
Solving for $R$,we get $R = \frac{100}{16} = 6.25 \%$.
Therefore,the rate of interest is $6.25 \%$ per annum.

(B) Let the amount lent at $8 \%$ be $x$ and the amount lent at $6 \%$ be $(1550 - x)$.
According to the problem,the total annual interest is ₹ $106$.
$\therefore \quad x \times \frac{8}{100} + (1550 - x) \times \frac{6}{100} = 106$
Multiply the entire equation by $100$:
$8x + 6(1550 - x) = 10600$
$8x + 9300 - 6x = 10600$
$2x = 10600 - 9300$
$2x = 1300$
$x = 650$
So,the amount lent at $8 \%$ is ₹ $650$.
The amount lent at $6 \%$ is $1550 - 650 = ₹ 900$.
Thus,the amounts are ₹ $650$ and ₹ $900$.

(B) Let the principal be $P$,the rate of interest be $R = x \%$,and the time be $T = x$ years.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$.
Given that $SI = \frac{4}{9} P$,we substitute the values:
$\frac{4}{9} P = \frac{P \times x \times x}{100}$
Dividing both sides by $P$:
$\frac{4}{9} = \frac{x^2}{100}$
Solving for $x^2$:
$x^2 = \frac{400}{9}$
$x = \sqrt{\frac{400}{9}} = \frac{20}{3} = 6 \frac{2}{3}$.
Therefore,the rate is $6 \frac{2}{3} \%$ and the time is $6 \frac{2}{3}$ years.
To convert $6 \frac{2}{3}$ years into years and months:
$6$ years $+ (\frac{2}{3} \times 12)$ months $= 6$ years and $8$ months.

(B) Let the principal amount be $₹ 100$.
Since the interest is calculated every six months,the rate for each six-month period is $10 \% / 2 = 5 \%$.
Step $1$: Interest for the first $6$ months $= 100 \times (5/100) = ₹ 5$.
Amount at the end of $6$ months $= 100 + 5 = ₹ 105$.
Step $2$: For the next $6$ months,the principal becomes $₹ 105$.
Interest for the second $6$ months $= 105 \times (5/100) = ₹ 5.25$.
Amount at the end of $12$ months $= 105 + 5.25 = ₹ 110.25$.
Step $3$: The effective interest earned over one year is $110.25 - 100 = 10.25$.
Therefore,the effective rate of interest is $10.25 \%$.

(B) To calculate the simple interest,we first determine the time period in years.
Time from $May$ $3^{rd}$ to $July$ $15^{th}$:
$May$ remaining days: $31 - 3 = 28$ days.
$June$ days: $30$ days.
$July$ days: $15$ days.
Total days = $28 + 30 + 15 = 73$ days.
Convert days into years: $T = \frac{73}{365} = \frac{1}{5}$ years.
Given: Principal $(P)$ = ₹ $500$,Rate $(R)$ = $6 \%$ per annum.
Simple Interest $(I)$ = $\frac{P \times R \times T}{100}$.
$I = \frac{500 \times 6 \times (1/5)}{100} = \frac{500 \times 6}{100 \times 5} = \frac{3000}{500} = ₹ 6$.
Thus,the correct option is $B$.

(A) Given: Principal $(P)$ = ₹ $10000$,Rate of interest $(R)$ = $8 \%$ per annum,Time $(T)$ = $6$ years.
First,calculate the Simple Interest $(I)$ using the formula: $I = \frac{P \times R \times T}{100}$.
$I = \frac{10000 \times 8 \times 6}{100} = 100 \times 48 = ₹ 4800$.
The total amount $(A)$ to be returned is the sum of the principal and the interest: $A = P + I$.
$A = 10000 + 4800 = ₹ 14800$.
Therefore,Mr. Irani returned ₹ $14800$ to the finance company.

(C) Given: Simple Interest $(I) = ₹ 60$,Rate $(R) = 6 \%$ per annum,Time $(T) = 5$ years.
The formula for Simple Interest is $I = \frac{P \times R \times T}{100}$,where $P$ is the principal.
Rearranging the formula to solve for $P$: $P = \frac{100 \times I}{R \times T}$.
Substituting the given values: $P = \frac{100 \times 60}{6 \times 5}$.
$P = \frac{6000}{30} = ₹ 200$.
Therefore,the principal is ₹ $200$.

(A) Given: Simple Interest $(I) = ₹ 1770$,Rate of interest $(R) = 8 \%$ per annum,Time $(T) = 7 \frac{1}{2} \text{ years} = \frac{15}{2} \text{ years}$.
The formula for Simple Interest is $I = \frac{P \times R \times T}{100}$,where $P$ is the Principal.
Rearranging the formula to find the Principal $(P)$:
$P = \frac{100 \times I}{R \times T}$
Substituting the given values:
$P = \frac{100 \times 1770}{8 \times \frac{15}{2}}$
$P = \frac{100 \times 1770}{4 \times 15}$
$P = \frac{100 \times 1770}{60}$
$P = \frac{10 \times 1770}{6} = \frac{17700}{6} = ₹ 2950$.
Therefore,the required sum of money is ₹ $2950$.

(B) Let the principal sum be $P = x$.
Given that the simple interest $(SI)$ is $\frac{3}{8}$ of the principal,so $SI = \frac{3}{8}x$.
The time period $(T)$ is given as $6 \frac{1}{4}$ years,which is equal to $\frac{25}{4}$ years.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$.
Rearranging the formula to find the rate of interest $(R)$: $R = \frac{100 \times SI}{P \times T}$.
Substituting the values: $R = \frac{100 \times (\frac{3}{8}x)}{x \times \frac{25}{4}}$.
$R = \frac{100 \times 3 \times 4}{8 \times 25} = \frac{1200}{200} = 6 \%$.
Thus,the rate of interest per annum is $6 \%$.

(A) Given:
Principal $(P)$ = ₹ $5000$
Simple Interest $(I)$ = ₹ $500$ (since he received ₹ $500$ more than the capital)
Time $(T)$ = $5$ years
The formula for Simple Interest is $I = \frac{P \times R \times T}{100}$.
Rearranging for the rate of interest $(R)$:
$R = \frac{100 \times I}{P \times T}$
Substituting the values:
$R = \frac{100 \times 500}{5000 \times 5}$
$R = \frac{50000}{25000}$
$R = 2 \%$
Therefore,the rate of interest per annum is $2 \%$.

(A) Given: Principal $(P) = ₹ 1200$,Amount $(A) = ₹ 1440$,Time $(T) = 4$ years.
First,calculate the Simple Interest $(I)$:
$I = A - P = 1440 - 1200 = ₹ 240$.
Now,use the formula for the rate of interest $(R)$:
$R = \frac{100 \times I}{P \times T}$
Substitute the values:
$R = \frac{100 \times 240}{1200 \times 4}$
$R = \frac{24000}{4800} = 5 \%$.
Therefore,the rate per cent per annum is $5 \%$.

(C) Let the principal sum be $P$ and the simple interest $I = ₹ 256$.
Let the rate of interest per annum be $R = x \%$ and the time period be $T = x$ years.
The formula for simple interest is $I = \frac{P \times R \times T}{100}$.
Substituting the given values: $256 = \frac{P \times x \times x}{100} \Rightarrow P = \frac{25600}{x^2}$.
Note: The problem implies the interest is calculated on a principal of $₹ 100$ (a standard assumption for such problems where the principal is not specified but the interest is absolute). If $P = 100$,then $256 = \frac{100 \times x \times x}{100}$.
$256 = x^2$.
$x = \sqrt{256} = 16$.
Therefore,the rate of interest is $16 \%$.

(B) Let the principal sum be $P = ₹ x$.
Given that the time $T = 2$ $years$.
The simple interest $I$ is one-fifth of the principal,so $I = \frac{x}{5}$.
The formula for the rate of interest $R$ is given by $R = \frac{100 \times I}{P \times T}$.
Substituting the values,we get $R = \frac{100 \times (x/5)}{x \times 2}$.
Simplifying the expression: $R = \frac{100}{5 \times 2} = \frac{100}{10} = 10 \%$.
Thus,the rate of interest is $10 \%$ per annum.

(C) Let the principal sum be $P = x$.
According to the problem,the simple interest $I = \frac{4}{25}x$.
Let the rate of interest be $r \%$ per annum and the time period be $T = r$ years.
The formula for simple interest is $I = \frac{P \times R \times T}{100}$.
Substituting the values: $\frac{4}{25}x = \frac{x \times r \times r}{100}$.
Dividing both sides by $x$: $\frac{4}{25} = \frac{r^2}{100}$.
Multiplying both sides by $100$: $r^2 = \frac{4 \times 100}{25} = 4 \times 4 = 16$.
Taking the square root: $r = \sqrt{16} = 4$.
Thus,the rate of interest is $4 \%$ per annum.

(A) Given: Amount $(A) = ₹ 1020$,Time $(T) = 4$ years,Rate $(R) = 5 \%$ per annum.
Let the principal sum borrowed be $P = ₹ x$.
We know that the Simple Interest $(I) = A - P = 1020 - x$.
Using the formula for Simple Interest: $I = \frac{P \times R \times T}{100}$.
Substituting the values: $1020 - x = \frac{x \times 5 \times 4}{100}$.
$1020 - x = \frac{20x}{100}$.
$1020 - x = \frac{x}{5}$.
Multiplying both sides by $5$: $5100 - 5x = x$.
$6x = 5100$.
$x = \frac{5100}{6} = 850$.
Therefore,the sum of money borrowed is ₹ $850$.

(C) Given:
Principal $(P) = ₹ 1200$
Amount $(A) = ₹ 1344$
Rate $(R) = 6 \% \text{ per annum}$
Step $1$: Calculate the Simple Interest $(I)$.
$I = A - P = 1344 - 1200 = ₹ 144$
Step $2$: Use the formula for Time $(T)$.
$T = \frac{100 \times I}{P \times R}$
$T = \frac{100 \times 144}{1200 \times 6}$
$T = \frac{14400}{7200} = 2 \text{ years}$.
Thus,the required time is $2 \text{ years}$.

(B) The interest earned on ₹ $225$ in $4$ years at $3 \%$ is calculated as:
Interest $= \frac{P \times R \times T}{100} = \frac{225 \times 3 \times 4}{100} = ₹ 27$.
Now,we need to find the time $T$ such that ₹ $8100$ earns the same interest (₹ $27$) at $3 \%$ rate.
Using the formula $T = \frac{100 \times I}{P \times R}$:
$T = \frac{100 \times 27}{8100 \times 3} = \frac{2700}{24300} = \frac{1}{9}$ years.

(A) The interest earned in $10$ years on ₹ $1000$ at $5 \%$ per annum is:
$I_1 = \frac{P \times R \times T}{100} = \frac{1000 \times 5 \times 10}{100} = ₹ 500$
The principal now becomes $P_2 = 1000 + 500 = ₹ 1500$.
We now find the time $T_2$ in which ₹ $1500$ becomes ₹ $2000$ at $5 \%$ per annum:
$I_2 = A - P_2 = 2000 - 1500 = ₹ 500$
$T_2 = \frac{100 \times I_2}{R \times P_2} = \frac{100 \times 500}{5 \times 1500} = \frac{100}{15} = 6 \frac{2}{3}$ years.
Total time = $10 + 6 \frac{2}{3} = 16 \frac{2}{3}$ years.

(B) Step $1$: Calculate the time period.
Simple Interest $(SI) = \text{Amount} - \text{Principal} = 725 - 500 = ₹ 225$.
Using the formula $SI = \frac{P \times R \times T}{100}$,we have $225 = \frac{500 \times 9 \times T}{100}$.
$225 = 45 \times T$,so $T = \frac{225}{45} = 5$ years.
Step $2$: Calculate the new amount for $P = ₹ 600$ at $R = 11 \%$ for $T = 5$ years.
$SI = \frac{600 \times 11 \times 5}{100} = 6 \times 55 = ₹ 330$.
Amount $(A) = P + SI = 600 + 330 = ₹ 930$.

(C) The principal amount $P = ₹ 10000$,rate $R = 20 \%$ per annum,and time $T = 1$ $year$.
The total amount after $1$ $year$ is calculated as:
$A_1 = P \left(1 + \frac{R \times T}{100}\right) = 10000 \left(1 + \frac{20 \times 1}{100}\right) = 10000 \times 1.2 = ₹ 12000$.
Sumit receives ₹ $6000$ after $1$ $year$. The remaining principal amount for the second year is:
$P_{new} = 12000 - 6000 = ₹ 6000$.
Now,the amount to be received at the end of the second year on the remaining principal is:
$A_2 = P_{new} \left(1 + \frac{R \times T}{100}\right) = 6000 \left(1 + \frac{20 \times 1}{100}\right) = 6000 \times 1.2 = ₹ 7200$.

(A) Given: Amount $(A)$ = ₹ $15000$,Rate $(R)$ = $10 \%$ per annum,Time $(T)$ = $5$ years.
The formula for the amount in simple interest is $A = P + SI$,where $SI = \frac{P \times R \times T}{100}$.
Substituting the formula for $SI$: $A = P + \frac{P \times R \times T}{100} = P \left(1 + \frac{R \times T}{100}\right)$.
Rearranging to solve for Principal $(P)$: $P = \frac{A \times 100}{100 + (R \times T)}$.
Substituting the given values: $P = \frac{15000 \times 100}{100 + (10 \times 5)} = \frac{1500000}{100 + 50} = \frac{1500000}{150}$.
$P = ₹ 10000$.

(C) Let the annual payment be $x$.
The debt is discharged in $3$ years,meaning the payments are made at the end of year $1$,year $2$,and year $3$.
The amount of the first payment $x$ will earn interest for $2$ years,the second payment $x$ will earn interest for $1$ year,and the third payment $x$ will earn no interest.
Total amount $= x(1 + \frac{R \times 2}{100}) + x(1 + \frac{R \times 1}{100}) + x(1 + \frac{R \times 0}{100}) = 47250$.
$x(1 + \frac{5 \times 2}{100}) + x(1 + \frac{5 \times 1}{100}) + x = 47250$.
$x(1 + 0.10) + x(1 + 0.05) + x = 47250$.
$1.10x + 1.05x + 1.00x = 47250$.
$3.15x = 47250$.
$x = \frac{47250}{3.15} = 15000$.
Thus,the annual payment is ₹ $15000$.

(A) Given: Total debt $A = ₹ 4200$,Time $T = 5$ years,Rate $R = 10 \%$ per annum.
The formula for the annual instalment to discharge a debt $A$ in $T$ years at $R \%$ simple interest is:
$\text{Instalment} = \frac{100 \times A}{100 \times T + \frac{R \times T(T-1)}{2}}$
Substituting the values:
$\text{Instalment} = \frac{100 \times 4200}{100 \times 5 + \frac{10 \times 5 \times (5-1)}{2}}$
$= \frac{420000}{500 + \frac{10 \times 5 \times 4}{2}}$
$= \frac{420000}{500 + 100}$
$= \frac{420000}{600}$
$= ₹ 700$
Thus,the annual instalment is ₹ $700$.

(A) Let the sum invested by Mahesh be $P_1 = ₹ 1500$,rate $R_1 = 8 \%$,and time $T_1 = 2.5$ years.
The amount $A_1$ obtained by Mahesh is $P_1 + \text{Simple Interest} = P_1(1 + \frac{R_1 T_1}{100})$.
$A_1 = 1500(1 + \frac{8 \times 2.5}{100}) = 1500(1 + \frac{20}{100}) = 1500(1.2) = ₹ 1800$.
Let the sum invested by Suresh be $P_2 = x$,rate $R_2 = 5 \%$,and time $T_2 = 2$ years.
The amount $A_2$ obtained by Suresh is $P_2(1 + \frac{R_2 T_2}{100})$.
$A_2 = x(1 + \frac{5 \times 2}{100}) = x(1 + \frac{10}{100}) = x(1.1) = 1.1x$.
Since $A_1 = A_2$,we have $1800 = 1.1x$.
$x = \frac{1800}{1.1} = \frac{18000}{11} = 1636 \frac{4}{11}$.
Thus,the sum invested by Suresh is ₹ $1636 \frac{4}{11}$.

(A) Let the amounts invested for Neeta,Sita,and Gita be $P_1, P_2,$ and $P_3$ respectively.
Given that the total amount $P_1 + P_2 + P_3 = 3965$.
Since the final amount received is the same for all,we use the formula $A = P(1 + \frac{RT}{100})$.
$P_1(1 + \frac{5 \times 2}{100}) = P_2(1 + \frac{5 \times 3}{100}) = P_3(1 + \frac{5 \times 4}{100})$
$P_1(1.10) = P_2(1.15) = P_3(1.20)$
$P_1 : P_2 : P_3 = \frac{1}{1.10} : \frac{1}{1.15} : \frac{1}{1.20} = \frac{1}{110} : \frac{1}{115} : \frac{1}{120}$
Multiplying by the $LCM$ of $110, 115, 120$ (which is $27600$):
$P_1 : P_2 : P_3 = 276 : 264 : 253$
Sum of ratios $= 276 + 264 + 253 = 793$.
Neeta's share $= \frac{276}{793} \times 3965 = 276 \times 5 = ₹ 1380$.
Sita's share $= \frac{264}{793} \times 3965 = 264 \times 5 = ₹ 1320$.
Gita's share $= \frac{253}{793} \times 3965 = 253 \times 5 = ₹ 1265$.

(B) Let the principal sum be $P$ and the rate of interest be $R \%$ per annum.
According to the problem,the amount becomes $4P$ in $24$ years.
Simple Interest $(SI) = \text{Amount} - \text{Principal} = 4P - P = 3P$.
Using the formula for simple interest: $SI = \frac{P \times R \times T}{100}$.
Substituting the values: $3P = \frac{P \times R \times 24}{100}$.
$3 = \frac{R \times 24}{100}$.
$R = \frac{3 \times 100}{24} = \frac{300}{24} = 12.5 \%$.
Thus,the rate of interest is $12.50 \%$ per annum.

(C) Let the principal amount be $P$.
Since the sum of money trebles itself,the final amount $A = 3P$.
The simple interest $SI = A - P = 3P - P = 2P$.
The rate of interest $R = 10 \%$ per annum.
Using the formula for simple interest: $SI = \frac{P \times R \times T}{100}$.
Substituting the values: $2P = \frac{P \times 10 \times T}{100}$.
$2 = \frac{10T}{100}$.
$2 = \frac{T}{10}$.
$T = 20$ years.
Thus,the sum of money will treble itself in $20$ years.

(A) Let the principal amount be $P$ and the rate of simple interest be $R$ percent per annum.
According to the problem,the amount doubles in $8$ years,so the simple interest $SI = P$.
Using the formula $SI = \frac{P \times R \times T}{100}$,we have $P = \frac{P \times R \times 8}{100}$,which gives $R = 12.5\%$.
Now,we want the sum to treble,which means the amount becomes $3P$,so the simple interest $SI = 3P - P = 2P$.
Using the formula $2P = \frac{P \times 12.5 \times T^{\prime}}{100}$,we get $2 = \frac{12.5 \times T^{\prime}}{100}$.
$T^{\prime} = \frac{200}{12.5} = 16$ years.
Thus,the sum will treble in $16$ years.

(B) Let the principal sum be $P$ and the original rate of interest be $R \%$.
Simple Interest $(SI)$ is given by the formula: $SI = \frac{P \times R \times T}{100}$.
According to the problem,the time $T = 4$ years.
If the rate is increased by $2 \%$,the new rate becomes $(R + 2) \%$.
The difference in simple interest is given as ₹ $56$.
So,$\frac{P \times (R + 2) \times 4}{100} - \frac{P \times R \times 4}{100} = 56$.
$\frac{4P}{100} \times (R + 2 - R) = 56$.
$\frac{4P}{100} \times 2 = 56$.
$\frac{8P}{100} = 56$.
$P = \frac{56 \times 100}{8} = 7 \times 100 = 700$.
Therefore,the sum is ₹ $700$.

(A) Given that the difference in Simple Interest $(SI)$ for the same time period is ₹ $40$.
Let the rate of interest be $R \%$ per annum.
Time $(T)$ = $2$ years.
Difference in principal amounts = $₹ 800 - ₹ 400 = ₹ 400$.
Using the formula for Simple Interest: $SI = \frac{P \times R \times T}{100}$.
The difference in interest is given by: $\Delta SI = \frac{(P_1 - P_2) \times R \times T}{100}$.
Substituting the values: $40 = \frac{400 \times R \times 2}{100}$.
$40 = 8 \times R$.
$R = \frac{40}{8} = 5 \%$.
Therefore,the rate of interest is $5 \%$ per annum.

(A) Let the principal sum be $P$.
Given time $T = 4$ years.
Rate $R_1 = 2 \frac{1}{2} \% = 2.5 \%$ per annum.
Rate $R_2 = 3 \%$ per annum.
The difference in simple interest is given by: $SI_2 - SI_1 = 60$.
Using the formula $SI = \frac{P \times R \times T}{100}$,we have:
$\frac{P \times 3 \times 4}{100} - \frac{P \times 2.5 \times 4}{100} = 60$
$\frac{12P}{100} - \frac{10P}{100} = 60$
$\frac{2P}{100} = 60$
$2P = 6000$
$P = 3000$.
Therefore,the sum is ₹ $3000$.

(B) Let the principal be $P$ and the rate of interest be $R \%$.
Simple interest for $3$ years ($5$ years - $2$ years) = $₹ 3250 - ₹ 2800 = ₹ 450$.
Simple interest for $1$ year = $₹ 450 / 3 = ₹ 150$.
Simple interest for $2$ years = $₹ 150 \times 2 = ₹ 300$.
Principal $(P)$ = Amount after $2$ years - Simple interest for $2$ years = $₹ 2800 - ₹ 300 = ₹ 2500$.
Rate of interest $(R)$ = $(SI \times 100) / (P \times T) = (150 \times 100) / (2500 \times 1) = 15000 / 2500 = 6 \%$.
Therefore,the rate of interest is $6 \%$ per annum.

(C) Let the principal be $P$ and the rate of interest be $R$ per annum.
Simple interest for $3$ years ($5$ years $- 2$ years) $= ₹ 2000 - ₹ 1760 = ₹ 240$.
Simple interest for $1$ year $= ₹ 240 / 3 = ₹ 80$.
Simple interest for $2$ years $= ₹ 80 \times 2 = ₹ 160$.
Principal $(P) = \text{Amount after } 2 \text{ years} - \text{Simple interest for } 2 \text{ years}$.
$P = ₹ 1760 - ₹ 160 = ₹ 1600$.
Thus,the sum is ₹ $1600$.

(B) Let the principal be $P$ and the time be $T$ years.
Using the formula for amount in simple interest: $A = P + \frac{P \times R \times T}{100} = P(1 + \frac{RT}{100})$.
For the first case: $450 = P(1 + \frac{7T}{100}) \implies 450 = P + \frac{7PT}{100} \implies \frac{7PT}{100} = 450 - P$ --- $(1)$
For the second case: $350 = P(1 + \frac{5T}{100}) \implies 350 = P + \frac{5PT}{100} \implies \frac{5PT}{100} = 350 - P$ --- $(2)$
Dividing $(1)$ by $(2)$: $\frac{7PT/100}{5PT/100} = \frac{450 - P}{350 - P} \implies \frac{7}{5} = \frac{450 - P}{350 - P}$.
Cross-multiplying: $7(350 - P) = 5(450 - P) \implies 2450 - 7P = 2250 - 5P$.
$2450 - 2250 = 7P - 5P \implies 200 = 2P \implies P = 100$.
The sum is ₹ $100$.

(C) Let the principal sum be $P$ and the time be $T$ years.
Using the simple interest formula $A = P + (P \times R \times T) / 100$,we have:
$400 = P + (P \times 10 \times T) / 100 \implies 400 = P(1 + 0.1T) \quad (i)$
$200 = P + (P \times 4 \times T) / 100 \implies 200 = P(1 + 0.04T) \quad (ii)$
Dividing equation $(i)$ by $(ii)$:
$400 / 200 = (1 + 0.1T) / (1 + 0.04T)$
$2 = (1 + 0.1T) / (1 + 0.04T)$
$2(1 + 0.04T) = 1 + 0.1T$
$2 + 0.08T = 1 + 0.1T$
$2 - 1 = 0.1T - 0.08T$
$1 = 0.02T$
$T = 1 / 0.02 = 50$ years.

(A) Given: Principal amount $P = ₹ 9$,Monthly instalment $a = ₹ 1$,Number of instalments $n = 10$,Time period for interest calculation $b = 12$ months (annual rate).
The formula for the total amount paid in instalments including simple interest is:
$z = na + \frac{R \times a}{100 \times 12} \times \frac{n(n-1)}{2}$
Here,$z$ is the total amount paid,which is $10 \times 1 = ₹ 10$. The principal lent is $₹ 9$.
Substituting the values:
$10 = 9 + \text{Interest}$
Interest $= 10 - 9 = ₹ 1$.
Using the formula for interest on instalments:
$1 = \frac{R \times 1}{100 \times 12} \times \frac{10 \times 9}{2}$
$1 = \frac{R \times 90}{2400}$
$R = \frac{2400}{9} = \frac{800}{3} = 266 \frac{2}{3} \%$.

(A) Let the shares of Vikas,Vijay,and Viraj be $P_1, P_2,$ and $P_3$ respectively.
Given that the simple interest on each part is equal at $R = 5 \%$ $p.a.$
Simple Interest $SI = \frac{P \times R \times T}{100}$.
For Vikas: $SI_1 = \frac{P_1 \times 5 \times 1}{100} = \frac{5P_1}{100}$.
For Vijay: $SI_2 = \frac{P_2 \times 5 \times 2}{100} = \frac{10P_2}{100}$.
For Viraj: $SI_3 = \frac{P_3 \times 5 \times 3}{100} = \frac{15P_3}{100}$.
Since $SI_1 = SI_2 = SI_3$,we have $5P_1 = 10P_2 = 15P_3$.
Dividing by $5$,we get $P_1 = 2P_2 = 3P_3$.
To find the ratio $P_1 : P_2 : P_3$,we take the $LCM$ of $1, 2,$ and $3$,which is $6$.
$P_1 = 6k, 2P_2 = 6k \implies P_2 = 3k, 3P_3 = 6k \implies P_3 = 2k$.
The ratio $P_1 : P_2 : P_3 = 6 : 3 : 2$.
Total sum $= 6k + 3k + 2k = 11k = 7700$.
$k = \frac{7700}{11} = 700$.
Share of Vikas $= 6 \times 700 = ₹ 4200$.
Share of Viraj $= 2 \times 700 = ₹ 1400$.
Difference $= 4200 - 1400 = ₹ 2800$.

(B) Let the required sum of money be $₹ x$.
Simple Interest $(SI)$ is calculated using the formula: $SI = \frac{P \times R \times T}{100}$.
According to the problem,the simple interest on $₹ x$ for $4$ years at $5 \%$ $p.a.$ is equal to the simple interest on $₹ 560$ for $10$ years at $4 \%$ $p.a$.
So,$\frac{x \times 5 \times 4}{100} = \frac{560 \times 4 \times 10}{100}$.
Canceling $100$ from both sides,we get: $x \times 20 = 560 \times 40$.
$x \times 20 = 22400$.
$x = \frac{22400}{20} = 1120$.
Therefore,the sum of money is $₹ 1120$.

(A) Let the second amount invested be $P_{2}$.
Given: $P_{1} = ₹ 12000$,$R_{1} = 10 \%$,$R_{2} = 20 \%$,and the effective rate of interest $R = 14 \%$.
The formula for the weighted average rate of interest is $R = \frac{P_{1}R_{1} + P_{2}R_{2}}{P_{1} + P_{2}}$.
Substituting the values: $14 = \frac{12000 \times 10 + P_{2} \times 20}{12000 + P_{2}}$.
$14(12000 + P_{2}) = 120000 + 20P_{2}$.
$168000 + 14P_{2} = 120000 + 20P_{2}$.
$6P_{2} = 48000$.
$P_{2} = ₹ 8000$.
Total amount invested $= P_{1} + P_{2} = 12000 + 8000 = ₹ 20000$.

(B) Given: $P_{1} = ₹ 3000, R_{1} = 10 \%, P_{2} = ₹ 5000, R_{2} = 8 \%$.
The total interest earned for one year is $I = (P_{1} \times R_{1} / 100) + (P_{2} \times R_{2} / 100)$.
$I = (3000 \times 0.10) + (5000 \times 0.08) = 300 + 400 = ₹ 700$.
The total principal is $P = P_{1} + P_{2} = 3000 + 5000 = ₹ 8000$.
The combined rate of interest $R$ is given by $R = (I / P) \times 100$.
$R = (700 / 8000) \times 100 = 70 / 8 = 8.75 \%$.
Thus,the correct option is $B$.

(C) Let the total capital be $x$.
The parts invested are $\frac{2}{3}x$,$\frac{1}{6}x$,and the remainder is $x - (\frac{2}{3}x + \frac{1}{6}x) = x - (\frac{4+1}{6})x = x - \frac{5}{6}x = \frac{1}{6}x$.
The annual income is the sum of interest from these three parts:
$\frac{2}{3}x \times \frac{3}{100} + \frac{1}{6}x \times \frac{6}{100} + \frac{1}{6}x \times \frac{12}{100} = 25$
$\frac{6x}{300} + \frac{6x}{600} + \frac{12x}{600} = 25$
$\frac{12x + 6x + 12x}{600} = 25$
$\frac{30x}{600} = 25$
$\frac{x}{20} = 25$
$x = 25 \times 20 = 500$.
Thus,the total capital is $₹ 500$.

(A) Let the principal be $P$ and the rate of interest be $R$ per annum.
Simple interest for $10$ years is $₹ 600$.
Simple interest for $5$ years $= (600 / 10) \times 5 = ₹ 300$.
After $5$ years,the principal becomes $3P$.
Since simple interest is directly proportional to the principal,the interest for the next $5$ years will be $3 \times 300 = ₹ 900$.
Total interest at the end of $10$ years $= 300 + 900 = ₹ 1200$.

(A) The simple interest on ₹ $1500$ invested at a rate of $10 \%$ per annum for $5$ years is:
Interest $= \frac{P \times R \times T}{100} = \frac{1500 \times 10 \times 5}{100} = ₹ 750$
After $5$ years,the interest is added to the principal. Therefore,the new principal $= ₹ 1500 + 750 = ₹ 2250$.
The target amount is ₹ $2500$. The remaining interest required is $₹ 2500 - 2250 = ₹ 250$.
Using the simple interest formula for the remaining time $T$:
$250 = \frac{2250 \times 10 \times T}{100}$
$250 = 225 \times T$
$T = \frac{250}{225} = \frac{10}{9}$ years.
Total time $= 5 + \frac{10}{9} = 5 + 1 \frac{1}{9} = 6 \frac{1}{9}$ years.

(B) Let the sum of money lent by Sumit to Mohit be $₹ x$.
Simple interest paid by Mohit to Sumit after $1$ year $= \frac{x \times 5 \times 1}{100} = ₹ \frac{5x}{100}$.
Simple interest received by Mohit from Birju after $1$ year $= \frac{x \times 8.5 \times 1}{100} = \frac{x \times 17/2}{100} = ₹ \frac{17x}{200}$.
According to the problem,the profit earned by Mohit is the difference between the interest received and the interest paid:
$\frac{17x}{200} - \frac{5x}{100} = 350$
$\frac{17x - 10x}{200} = 350$
$\frac{7x}{200} = 350$
$7x = 350 \times 200$
$7x = 70000$
$x = 10000$
Thus,the sum of money lent by Sumit to Mohit is ₹ $10000$.

(A) Simple interest paid by Brinda on ₹ $1000$ for $1$ year $= \frac{1000 \times 5 \times 1}{100} = ₹ 50$.
Rent received by Brinda from Ramu in $1$ year $= 12 \frac{1}{2} \times 12 = 12.5 \times 12 = ₹ 150$.
Net savings per year $= \text{Rent received} - \text{Interest paid} = 150 - 50 = ₹ 100$.
To clear the debt of ₹ $1000$,the time required $= \frac{\text{Total Debt}}{\text{Net savings per year}} = \frac{1000}{100} = 10 \text{ years}$.

(C) Let the principal sum be $₹ x$.
The simple interest is calculated as $SI = \frac{P \times R \times T}{100}$.
For the first $2$ years at $4 \%$ p.a.: $SI_1 = \frac{x \times 4 \times 2}{100} = \frac{8x}{100}$.
For the next $4$ years at $6 \%$ p.a.: $SI_2 = \frac{x \times 6 \times 4}{100} = \frac{24x}{100}$.
For the remaining period of $9 - (2 + 4) = 3$ years at $8 \%$ p.a.: $SI_3 = \frac{x \times 8 \times 3}{100} = \frac{24x}{100}$.
The total simple interest is $SI_1 + SI_2 + SI_3 = 1120$.
$\frac{8x + 24x + 24x}{100} = 1120$
$\frac{56x}{100} = 1120$
$56x = 112000$
$x = \frac{112000}{56} = 2000$.
Thus,the sum is ₹ $2000$.

(A) Let the principal sum be $P$.
The total time is $8$ years,split into two periods: $5$ years at $10 \%$ per annum and $3$ years at $15 \%$ per annum.
The formula for simple interest is $SI = \frac{P \times R \times T}{100}$.
Total Interest $= \frac{P \times 10 \times 5}{100} + \frac{P \times 15 \times 3}{100} = 47500$.
$\frac{50P}{100} + \frac{45P}{100} = 47500$.
$\frac{95P}{100} = 47500$.
$0.95P = 47500$.
$P = \frac{47500}{0.95} = 50000$.
Therefore,the sum is ₹ $50000$.

(A) Let the first part be $₹ x$ and the second part be $₹ (8000 - x)$.
Since the time period is the same (let it be $T = 1$ year),the simple interest formula is $SI = \frac{P \times R \times T}{100}$.
According to the problem,the simple interest on both parts is equal:
$\frac{x \times 21 \times 1}{100} = \frac{(8000 - x) \times 35 \times 1}{100}$
$21x = 35(8000 - x)$
$21x = 280000 - 35x$
$21x + 35x = 280000$
$56x = 280000$
$x = \frac{280000}{56} = 5000$.
So,the first part is $₹ 5000$ and the second part is $₹ (8000 - 5000) = ₹ 3000$.
The interest on the first part is $\frac{5000 \times 21 \times 1}{100} = ₹ 1050$.
The interest on the second part is $\frac{3000 \times 35 \times 1}{100} = ₹ 1050$.
Thus,the interest of each part is $₹ 1050$.

(C) Let the initial investment be $P$ ₹.
After the first year,the value becomes $P \times (1 + 12/100) = P \times 1.12$.
After the second year,the value becomes $(P \times 1.12) \times (1 + 9/100) = P \times 1.12 \times 1.09$.
Given that the final value is ₹ $97,664$,we have the equation: $P \times 1.12 \times 1.09 = 97664$.
$P \times 1.2208 = 97664$.
$P = 97664 / 1.2208$.
$P = 80000$.
Therefore,the initial investment was ₹ $80,000$.