Percentage Questions in English

(B) $x \%$ of $y + y \%$ of $x = \left(\frac{x}{100} \times y\right) + \left(\frac{y}{100} \times x\right)$
$= \frac{xy}{100} + \frac{xy}{100}$
$= \frac{2xy}{100}$
$= 2 \%$ of $xy$

(C) To find the decimal equivalent of a percentage,divide the percentage value by $100$.
$0.35 \% = \frac{0.35}{100} = 0.0035$.
Therefore,$0.35 \%$ of a number is equivalent to multiplying the number by $0.0035$.

(B) Given that $8 \%$ of $x = 4 \%$ of $y$.
This can be written as $\frac{8}{100} \times x = \frac{4}{100} \times y$.
Multiplying both sides by $100$,we get $8x = 4y$.
Dividing by $8$,we find $x = \frac{4y}{8} = \frac{y}{2}$.
Now,we need to find $20 \%$ of $x$.
$20 \%$ of $x = \frac{20}{100} \times x$.
Substituting $x = \frac{y}{2}$,we get $\frac{20}{100} \times \frac{y}{2} = \frac{10}{100} \times y$.
Thus,$20 \%$ of $x = 10 \%$ of $y$.

(B) Let the missing value be $a$.
The given equation is: $\frac{x}{100} \times y + \frac{a}{100} \times x = \frac{x}{100} \times (x + y)$.
Multiplying both sides by $100$,we get: $xy + ax = x(x + y)$.
Expanding the right side: $xy + ax = x^2 + xy$.
Subtracting $xy$ from both sides: $ax = x^2$.
Dividing by $x$ (assuming $x \neq 0$): $a = x$.
Therefore,the missing percentage is $x \%$.

(C) Given that $x = 125 \% \text{ of } y$.
This can be written as $x = \frac{125}{100} y$.
Simplifying the fraction,we get $x = \frac{5}{4} y$.
To find $y$ in terms of $x$,we multiply both sides by $\frac{4}{5}$.
$y = \frac{4}{5} x$.
Converting the fraction to a decimal,$\frac{4}{5} = 0.8$.
Therefore,$y = 0.8 x$.

(B) To find $25 \%$ of $25 \%$,we convert the percentages into fractions:
$25 \% = \frac{25}{100} = 0.25$
Now,multiply the two values:
$0.25 \times 0.25 = 0.0625$
Alternatively,using fractions:
$\frac{25}{100} \times \frac{25}{100} = \frac{625}{10000} = 0.0625$

(C) To find the number that is $60 \%$ less than $80$,we subtract $60 \%$ of $80$ from $80$.
Step $1$: Calculate $60 \%$ of $80$.
$60 \% \text{ of } 80 = \frac{60}{100} \times 80 = 0.6 \times 80 = 48$.
Step $2$: Subtract this value from $80$.
$80 - 48 = 32$.
Alternatively,if a number is $60 \%$ less than $80$,it is $(100 \% - 60 \%) = 40 \%$ of $80$.
$40 \% \text{ of } 80 = \frac{40}{100} \times 80 = 0.4 \times 80 = 32$.

(B) To find $20 \%$ of $30 \%$ of $20 \%$ of $₹ 850$,we convert the percentages into fractions:
$20 \% = \frac{20}{100} = \frac{1}{5}$
$30 \% = \frac{30}{100} = \frac{3}{10}$
$20 \% = \frac{20}{100} = \frac{1}{5}$
Now,multiply these fractions with the given amount:
$= \frac{20}{100} \times \frac{30}{100} \times \frac{20}{100} \times 850$
$= \frac{1}{5} \times \frac{3}{10} \times \frac{1}{5} \times 850$
$= \frac{3}{250} \times 850$
$= \frac{3 \times 85}{25} = \frac{3 \times 17}{5} = \frac{51}{5} = 10.20$
Thus,the result is $₹ 10.20$.

(C) To compare these values,we convert them all into decimal form:
$1$. $16 \frac{2}{3} \% = \frac{50}{3} \times \frac{1}{100} = \frac{50}{300} = \frac{1}{6} \approx 0.1667$
$2$. $6 \frac{2}{3} \% = \frac{20}{3} \times \frac{1}{100} = \frac{20}{300} = \frac{1}{15} \approx 0.0667$
$3$. $0.3 = 0.3000$
Comparing the decimal values: $0.3000 > 0.1667 > 0.0667$.
Therefore,$0.3$ is the greatest value.

(A) First,calculate each percentage term individually:
$40 \%$ of $20 \% = \frac{40}{100} \times \frac{20}{100} = \frac{800}{10000} = \frac{8}{100} = 8 \%$
$30 \%$ of $25 \% = \frac{30}{100} \times \frac{25}{100} = \frac{750}{10000} = \frac{7.5}{100} = 7.5 \%$
$50 \%$ of $28 \% = \frac{50}{100} \times \frac{28}{100} = \frac{1400}{10000} = \frac{14}{100} = 14 \%$
Now,add these values together:
$8 \% + 7.5 \% + 14 \% = 29.5 \%$
Therefore,the correct answer is $29.5 \%$.

(B) Given that $90 \%$ of $A = 30 \%$ of $B$.
This can be written as $\frac{90}{100} \times A = \frac{30}{100} \times B$.
Simplifying the equation,we get $90A = 30B$,which implies $B = \frac{90}{30} A = 3A$.
We are also given that $B = x \%$ of $A$,which means $B = \frac{x}{100} \times A$.
Equating the two expressions for $B$,we have $3A = \frac{x}{100} \times A$.
Dividing both sides by $A$ (assuming $A \neq 0$),we get $3 = \frac{x}{100}$.
Therefore,$x = 3 \times 100 = 300$.

(C) We know that $1$ metric tonne $= 1000$ $kg$.
Also,$1$ quintal $= 100$ $kg$.
Therefore,$1$ quintal $25$ $kg = 100$ $kg + 25$ $kg = 125$ $kg$.
To find the percentage,we calculate: $\frac{125 \text{ kg}}{1000 \text{ kg}} \times 100 \%$.
$= \frac{125}{10} \% = 12.5 \%$.
$12.5 \%$ can be written as $12 \frac{1}{2} \%$.

(B) Given that $12 \%$ of $x = 6 \%$ of $y$.
This can be written as $\frac{12}{100} x = \frac{6}{100} y$.
Simplifying the equation,we get $12x = 6y$,which implies $y = 2x$ or $x = 0.5y$.
We need to find what percentage of $y$ is equal to $18 \%$ of $x$.
$18 \%$ of $x = \frac{18}{100} x$.
Substituting $x = 0.5y$ into the expression:
$\frac{18}{100} \times (0.5y) = \frac{9}{100} y$.
Thus,$18 \%$ of $x$ is equal to $9 \%$ of $y$.

(B) Let the third number be $100$.
Since the first number is $30 \%$ less than the third number,the first number $= 100 - 30 = 70$.
Since the second number is $37 \%$ less than the third number,the second number $= 100 - 37 = 63$.
We need to find how much per cent the second number is less than the first number.
Difference between the first and second number $= 70 - 63 = 7$.
Percentage less $= \left( \frac{\text{Difference}}{\text{First number}} \times 100 \right) \% = \left( \frac{7}{70} \times 100 \right) \% = 10 \%$.

(A) Let the second number be $100$.
Then,the first number is $100 + 20 = 120$.
The difference between the two numbers is $120 - 100 = 20$.
To find how much per cent the second number is less than the first,we calculate the percentage decrease relative to the first number:
Required percentage $= \left( \frac{\text{Difference}}{\text{First Number}} \times 100 \right) \%$
$= \left( \frac{20}{120} \times 100 \right) \%$
$= \left( \frac{1}{6} \times 100 \right) \% = \frac{100}{6} \% = 16 \frac{2}{3} \%$.

(B) Let $B$'s income be $100$.
Since $A$'s income is $25 \%$ less than $B$'s income,$A$'s income $= 100 - 25 = 75$.
Now,we need to find how much percent $B$'s income is more than $A$'s income.
Difference $= 100 - 75 = 25$.
Percentage more $= (\text{Difference} / A\text{'s income}) \times 100$.
Percentage more $= (25 / 75) \times 100 = (1 / 3) \times 100 = 33 \frac{1}{3} \%$.

(A) Let the third number be $x$.
Then,the first number $= 0.07x$ and the second number $= 0.28x$.
We need to find what percentage the first number is of the second number.
Required percentage $= (\frac{\text{First number}}{\text{Second number}}) \times 100 \%$.
$= (\frac{0.07x}{0.28x}) \times 100 \%$.
$= (\frac{7}{28}) \times 100 \%$.
$= \frac{1}{4} \times 100 \% = 25 \%$.

(A) Let the third number be $100$.
Since the first number is $60 \%$ more than the third number,the first number $= 100 + 60 = 160$.
Since the second number is $20 \%$ more than the third number,the second number $= 100 + 20 = 120$.
We need to express the second number as a percentage of the first number.
Required percentage $= \left( \frac{\text{Second number}}{\text{First number}} \times 100 \right) \%$
$= \left( \frac{120}{160} \times 100 \right) \%$
$= \left( \frac{3}{4} \times 100 \right) \% = 75 \%$.

(A) Let the third number be $100$.
Since the first number is $20 \%$ more than the third,the first number $= 100 + 20 = 120$.
Since the second number is $10 \%$ more than the third,the second number $= 100 + 10 = 110$.
We need to find how much percent the first number is more than the second.
Difference $= 120 - 110 = 10$.
Percentage more $= \left( \frac{\text{Difference}}{\text{Second number}} \times 100 \right) \%$.
Percentage more $= \left( \frac{10}{110} \times 100 \right) \% = \frac{100}{11} \% = 9 \frac{1}{11} \%$.
Thus,the first number is $9 \frac{1}{11} \%$ more than the second.

(B) Let the initial price of cooking oil be $100$ units and the initial consumption be $100$ units.
Initial expenditure = $100 \times 100 = 10000$ units.
New price after $15 \%$ increase = $115$ units.
To keep the expenditure constant at $10000$ units,let the new consumption be $x$.
$115 \times x = 10000$
$x = \frac{10000}{115} = \frac{2000}{23} \approx 86.95$ units.
Reduction in consumption = $100 - 86.95 = 13.05$ units.
Using the formula for percentage reduction: $\text{Reduction} \% = \left( \frac{P}{100+P} \times 100 \right) \%$,where $P = 15$.
$\text{Reduction} \% = \left( \frac{15}{115} \times 100 \right) \% = \left( \frac{3}{23} \times 100 \right) \% = \frac{300}{23} \% = 13 \frac{1}{23} \%$.

(D) Let the initial number be $100$.
Step $1$: Increase the number by $20 \%$.
New value $= 100 + (20 \% \text{ of } 100) = 100 + 20 = 120$.
Step $2$: Decrease the new value by $20 \%$.
Final value $= 120 - (20 \% \text{ of } 120) = 120 - 24 = 96$.
Step $3$: Calculate the net percentage change.
Net change $= \text{Final value} - \text{Initial value} = 96 - 100 = -4$.
Percentage change $= \left( \frac{-4}{100} \right) \times 100 = -4 \%$.
The negative sign indicates a decrease of $4 \%$.

(B) Let the initial wage be $100$.
After a decrease of $50 \%$,the new wage becomes $100 - (50 \% \text{ of } 100) = 100 - 50 = 50$.
Now,the reduced wage is increased by $50 \%$,so the final wage becomes $50 + (50 \% \text{ of } 50) = 50 + 25 = 75$.
The total loss is $100 - 75 = 25$.
Therefore,the percentage loss is $\frac{25}{100} \times 100 = 25 \%$.
Alternatively,using the net percentage change formula: $\text{Net change} = \left(x + y + \frac{xy}{100}\right) \%$,where $x = -50$ and $y = 50$.
$\text{Net change} = -50 + 50 + \frac{(-50 \times 50)}{100} = 0 - 25 = -25 \%$.
The negative sign indicates a loss of $25 \%$.

(B) Let the initial population of the town be $100$.
After the first year,the population decreases by $20 \%$,so the new population is $100 - 20 = 80$.
After the second year,the population decreases by $25 \%$ of the remaining $80$. The decrease is $\frac{25}{100} \times 80 = 20$.
The final population is $80 - 20 = 60$.
The total decrease in population is $100 - 60 = 40$.
Therefore,the percentage decrease is $\frac{40}{100} \times 100 = 40 \%$.
Alternatively,using the formula for successive percentage change: $\text{Net } \% \text{ change} = \left(x + y + \frac{xy}{100}\right) \%$.
Here,$x = -20$ and $y = -25$.
Net change $= -20 - 25 + \frac{(-20) \times (-25)}{100} = -45 + 5 = -40 \%$.
The negative sign indicates a decrease of $40 \%$.

(B) Let the amount of the bill be $x$.
$A$ single discount of $35 \%$ means the customer pays $65 \%$ of $x$,i.e.,$0.65x$.
Two successive discounts of $20 \%$ and $20 \%$ are equivalent to a single discount of $D = \left(20 + 20 - \frac{20 \times 20}{100}\right) \% = (40 - 4) \% = 36 \%$.
This means the customer pays $(100 - 36) \% = 64 \%$ of $x$,i.e.,$0.64x$.
The difference between the two discount amounts is given as ₹ $22$.
Therefore,$36 \% \text{ of } x - 35 \% \text{ of } x = 22$.
$1 \% \text{ of } x = 22$.
$x = 22 \times 100 = 2200$.
Thus,the amount of the bill is ₹ $2200$.

(B) Let the original price of the goods be $100$.
Since the shopkeeper marks the price $25 \%$ higher,the marked price becomes $100 + 25 = 125$.
He allows a discount of $12 \%$ on the marked price.
Discount amount $= 12 \% \text{ of } 125 = \frac{12}{100} \times 125 = 15$.
Selling price $= \text{Marked price} - \text{Discount} = 125 - 15 = 110$.
Since the selling price $(110)$ is greater than the original price $(100)$,there is a profit.
Profit percentage $= \frac{\text{Selling price} - \text{Original price}}{\text{Original price}} \times 100 = \frac{110 - 100}{100} \times 100 = 10 \%$.
Alternatively,using the net percentage change formula: $\text{Net change} = x + y + \frac{xy}{100}$,where $x = 25$ and $y = -12$.
Net change $= 25 - 12 + \frac{25 \times (-12)}{100} = 13 - 3 = 10 \%$.
Since the result is positive,it is a $10 \%$ profit.

(A) The equivalent discount for two successive discounts of $x \%$ and $y \%$ is given by $\left(x + y - \frac{xy}{100}\right) \%$.
For the first shopkeeper,the successive discounts are $20 \%$ and $10 \%$.
Equivalent discount $= 20 + 10 - \frac{20 \times 10}{100} = 30 - 2 = 28 \%$.
Discount amount $= 28 \%$ of $₹ 1000 = \frac{28}{100} \times 1000 = ₹ 280$.
For the second shopkeeper,the successive discounts are $15 \%$ and $15 \%$.
Equivalent discount $= 15 + 15 - \frac{15 \times 15}{100} = 30 - 2.25 = 27.75 \%$.
Discount amount $= 27.75 \%$ of $₹ 1000 = \frac{27.75}{100} \times 1000 = ₹ 277.50$.
Difference in discounts $= ₹ 280 - ₹ 277.50 = ₹ 2.50$.

(B) Let the initial tax be $T$ and initial consumption be $C$. The initial revenue is $R_1 = T \times C$.
After the changes:
New tax $T' = T - 10\% \text{ of } T = 0.9T$.
New consumption $C' = C + 10\% \text{ of } C = 1.1C$.
New revenue $R_2 = T' \times C' = (0.9T) \times (1.1C) = 0.99 \times (T \times C) = 0.99R_1$.
Alternatively,using the net percentage change formula:
Net $\% \text{ change} = \left(x + y + \frac{xy}{100}\right) \%$,where $x = -10$ and $y = 10$.
Net $\% \text{ change} = \left(-10 + 10 + \frac{(-10)(10)}{100}\right) \% = (0 - 1) \% = -1 \%$.
Therefore,the revenue decreases by $1 \%$.

(A) The surface area $S$ of a sphere is given by the formula $S = 4 \pi r^2$,where $r$ is the radius.
Since $4$ and $\pi$ are constants,the surface area is directly proportional to the square of the radius $(S \propto r^2)$.
When a quantity changes by $x \%$ and then by $y \%$,the net percentage change is given by the formula $\left(x + y + \frac{xy}{100}\right) \%$.
Here,the radius increases by $10 \%$,so $x = 10$ and $y = 10$.
Substituting these values into the formula:
Net $\%$ change $= \left(10 + 10 + \frac{10 \times 10}{100}\right) \% = (20 + 1) \% = 21 \%$.

(B) We have,$\text{Receipts} = \text{Price} \times \text{Sales}$.
Let the initial price be $100$ and initial sales be $100$. Initial receipts $= 100 \times 100 = 10000$.
New price $= 100 - 15 = 85$.
New sales $= 100 + 35 = 135$.
New receipts $= 85 \times 135 = 11475$.
Change in receipts $= 11475 - 10000 = 1475$.
Percentage change $= \frac{1475}{10000} \times 100 = 14.75 \% = 14 \frac{3}{4} \%$.
Alternatively,using the net percentage change formula: $\left(x + y + \frac{xy}{100}\right) \%$.
Here,$x = -15$ and $y = 35$.
Net change $= -15 + 35 + \frac{(-15 \times 35)}{100} = 20 - 5.25 = 14.75 \% = 14 \frac{3}{4} \%$ increase.

(B) Let the original side of the square be $s$.
Original area $A_1 = s^2$.
If the side is increased by $30 \%$,the new side $s' = s + 0.30s = 1.3s$.
New area $A_2 = (1.3s)^2 = 1.69s^2$.
The increase in area is $A_2 - A_1 = 1.69s^2 - s^2 = 0.69s^2$.
Percentage increase in area $= \left( \frac{0.69s^2}{s^2} \right) \times 100 = 69 \%$.
Alternatively,using the net percentage change formula for two successive changes: $x + y + \frac{xy}{100}$,where $x = 30$ and $y = 30$.
Net change $= 30 + 30 + \frac{30 \times 30}{100} = 60 + 9 = 69 \%.$

(B) Let the side of the square be $s$. The area of the square is $s^2$.
When the length is increased by $30 \%$,the new length becomes $L = s(1 + 0.30) = 1.3s$.
When the breadth is increased by $20 \%$,the new breadth becomes $B = s(1 + 0.20) = 1.2s$.
The area of the rectangle formed is $L \times B = (1.3s) \times (1.2s) = 1.56s^2$.
The increase in area is $1.56s^2 - s^2 = 0.56s^2$.
The percentage increase in area is $\frac{0.56s^2}{s^2} \times 100 = 56 \%$.
Alternatively,using the net percentage change formula: $\text{Net } \% = x + y + \frac{xy}{100} = 30 + 20 + \frac{30 \times 20}{100} = 50 + 6 = 56 \%$.

(A) Let the original sides of the rectangle be $L$ and $W$. The original area is $A = L \times W$.
According to the problem,the new sides are $L' = L(1 + 0.10) = 1.1L$ and $W' = W(1 - 0.20) = 0.8W$.
The new area is $A' = L' \times W' = (1.1L) \times (0.8W) = 0.88 \times (L \times W) = 0.88A$.
The change in area is $A' - A = 0.88A - A = -0.12A$.
The error percentage is $\left( \frac{-0.12A}{A} \right) \times 100 = -12 \%$.
Alternatively,using the formula for successive percentage change: $\text{Net Change} = \left( x + y + \frac{xy}{100} \right) \%$.
Here,$x = 10$ and $y = -20$. Substituting these values: $\text{Net Change} = \left( 10 - 20 + \frac{10 \times (-20)}{100} \right) \% = (-10 - 2) \% = -12 \%$.
$A$ negative sign indicates a deficit. Thus,the error is $12 \%$ deficit.

(A) Let the original length be $L$ and the original breadth be $B$. The original area is $A = L \times B$.
After the increase,the new length $L' = L + 0.10L = 1.10L$ and the new breadth $B' = B + 0.20B = 1.20B$.
The new area $A' = L' \times B' = (1.10L) \times (1.20B) = 1.32 \times (L \times B) = 1.32A$.
The percentage increase in area is given by $\frac{A' - A}{A} \times 100 = \frac{1.32A - A}{A} \times 100 = 0.32 \times 100 = 32 \%$.
Alternatively,using the net percentage change formula for two successive changes $x$ and $y$:
Net $\%$ change $= (x + y + \frac{xy}{100}) \% = (10 + 20 + \frac{10 \times 20}{100}) \% = (30 + 2) \% = 32 \%$.

(A) We know that: $\text{Tax} \times \text{Consumption} = \text{Expenditure}$.
Let the initial tax be $T$ and initial consumption be $C$. Initial expenditure $= T \times C$.
New tax $= T + 20\% \text{ of } T = 1.2T$.
New consumption $= C - 20\% \text{ of } C = 0.8C$.
New expenditure $= 1.2T \times 0.8C = 0.96(T \times C)$.
Change in expenditure $= 0.96(T \times C) - (T \times C) = -0.04(T \times C)$.
Percentage change $= -0.04 \times 100 = -4\%$.
Alternatively,using the formula for net percentage change: $\left(x + y + \frac{xy}{100}\right) \%$.
Here,$x = 20$ and $y = -20$.
Net change $= \left(20 - 20 + \frac{20 \times (-20)}{100}\right) \% = (0 - 4) \% = -4 \%$.
Therefore,the expenditure decreases by $4 \%$.

(A) Let the original price be $P$ and the original number of units sold be $N$. The original revenue is $R_1 = P \times N$.
After the price decrease of $30 \%$,the new price is $P' = P - 0.30P = 0.70P$.
After the sale increase of $20 \%$,the new number of units sold is $N' = N + 0.20N = 1.20N$.
The new revenue is $R_2 = P' \times N' = (0.70P) \times (1.20N) = 0.84 \times (P \times N) = 0.84R_1$.
The percentage change in revenue is given by the formula: $\text{Net } \% \text{ change} = (x + y + \frac{xy}{100}) \%$,where $x = -30$ and $y = +20$.
Substituting the values: $(-30 + 20 + \frac{-30 \times 20}{100}) \% = (-10 - 6) \% = -16 \%$.
$A$ negative sign indicates a decrease. Therefore,the revenue decreases by $16 \%$.

(A) Let the population $3$ years ago be $P_0$.
Given,the present population $P = 90.51$ lacs,rate $r = 10 \%$,and time $n = 3$ years.
The formula for population growth is $P = P_0 \left(1 + \frac{r}{100}\right)^n$.
Substituting the values: $90.51 = P_0 \left(1 + \frac{10}{100}\right)^3$.
$90.51 = P_0 \left(1.1\right)^3$.
$90.51 = P_0 \times 1.331$.
$P_0 = \frac{90.51}{1.331} = 68$.
Therefore,the population $3$ years ago was $68$ lacs.

(B) Let the purchase price of the machine be $X$.
Given that the depreciation rate $r = 10 \%$ per annum and the time period $n = 3$ years.
The present value $P$ is given by the formula: $P = X \times (1 - \frac{r}{100})^n$.
Substituting the given values: $8748 = X \times (1 - \frac{10}{100})^3$.
$8748 = X \times (0.9)^3$.
$8748 = X \times 0.729$.
$X = \frac{8748}{0.729} = 12000$.
Therefore,the purchase price of the machine was $₹ 12000$.

(D) Let the income in the year $1997$ be $X$.
Given that the income increases by $20 \%$ per annum.
Income in $1999 = X \times (1 + \frac{20}{100})^2$.
$42664000 = X \times (1.2)^2$.
$42664000 = X \times 1.44$.
$X = \frac{42664000}{1.44}$.
$X = 29627777.78$.
Since this value is not among the options,the correct choice is $D$.

(C) Given: Initial population $P = 32000$,rate of increase $r = 15\%$,and time $n = 2$ years.
The formula for population growth is $A = P(1 + \frac{r}{100})^n$.
Substituting the values:
$A = 32000 \times (1 + \frac{15}{100})^2$
$A = 32000 \times (1.15)^2$
$A = 32000 \times 1.3225$
$A = 42320$.
Thus,the population after $2$ years will be $42320$.

(B) Initial value of the machine,$P = 6250$.
Rate of depreciation for the first year,$r_1 = 10 \%$.
Rate of depreciation for the second year,$r_2 = 20 \%$.
Rate of depreciation for the third year,$r_3 = 30 \%$.
The value of the machine after $3$ years is given by the formula:
$V = P \times (1 - \frac{r_1}{100}) \times (1 - \frac{r_2}{100}) \times (1 - \frac{r_3}{100})$
$V = 6250 \times (1 - \frac{10}{100}) \times (1 - \frac{20}{100}) \times (1 - \frac{30}{100})$
$V = 6250 \times (\frac{90}{100}) \times (\frac{80}{100}) \times (\frac{70}{100})$
$V = 6250 \times 0.9 \times 0.8 \times 0.7$
$V = 6250 \times 0.504$
$V = 3150$.
Therefore,the value of the machine after $3$ years will be $Rs$ $3150$.

(C) Let the population $2$ years ago be $P$.
Given that the population increases by $12 \%$ in the first year and decreases by $10 \%$ in the second year.
The present population $A$ is given by the formula:
$A = P \times (1 + \frac{12}{100}) \times (1 - \frac{10}{100})$
Substituting the given values:
$50400 = P \times (1.12) \times (0.90)$
$50400 = P \times (1.12 \times 0.90)$
$50400 = P \times 1.008$
$P = \frac{50400}{1.008}$
$P = 50000$
Therefore,the population $2$ years ago was $50000$.

(C) Let the total pocket money be $₹ A$.
After losing $20 \%$ of his money,the remaining amount is $A \times (1 - 0.20) = 0.80A$.
He then spends $25 \%$ of this remainder. The amount left after spending is $0.80A \times (1 - 0.25) = 0.80A \times 0.75$.
According to the problem,the final amount is $₹ 480$.
So,$0.80A \times 0.75 = 480$.
$0.60A = 480$.
$A = \frac{480}{0.60} = 800$.
Therefore,the total pocket money was $₹ 800$.

(B) Let the original strength of the army be $A$.
After losing $10 \%$ in war,the remaining strength is $A \times (1 - 0.10) = 0.9A$.
After losing $10 \%$ of the remaining due to diseases,the strength becomes $0.9A \times (1 - 0.10) = 0.9A \times 0.9 = 0.81A$.
After $10 \%$ of the rest were hurt,the final active strength is $0.81A \times (1 - 0.10) = 0.81A \times 0.9 = 0.729A$.
Given that the final strength is $729000$,we have:
$0.729A = 729000$
$A = \frac{729000}{0.729}$
$A = 1000000$
Thus,the original strength was $1000000$ men.

(A) Let the daily wage before the increase be $₹ A$.
According to the problem,the wage is increased by $25 \%$,so the new wage is $A + 0.25A = 1.25A$.
Given that the new wage is $₹ 25$,we have the equation:
$1.25A = 25$
Solving for $A$:
$A = \frac{25}{1.25}$
$A = \frac{2500}{125}$
$A = 20$
Therefore,the daily wage before the increase was $₹ 20$.

(B) Let the maximum marks for the examination be $M$.
According to the problem,the passing marks are $15\%$ of $M$.
The student obtained $80$ marks and failed by $70$ marks,which means the passing marks are $(80 + 70) = 150$.
Therefore,we can set up the equation: $0.15 \times M = 150$.
Solving for $M$: $M = \frac{150}{0.15} = \frac{15000}{15} = 1000$.
Thus,the maximum marks for the examination is $1000$.

(B) Let $H$ be the set of students who failed in History and $G$ be the set of students who failed in Geography.
Given: $n(H) = 30 \%$,$n(G) = 35 \%$,and $n(H \cap G) = 27 \%$.
Percentage of students who failed in at least one subject is given by the formula:
$n(H \cup G) = n(H) + n(G) - n(H \cap G)$
$n(H \cup G) = 30 \% + 35 \% - 27 \% = 38 \%$.
Percentage of students who passed the examination = $100 \% - n(H \cup G) = 100 \% - 38 \% = 62 \%$.
Let the total number of students who appeared be $x$.
Given that $62 \%$ of $x = 248$.
$\frac{62}{100} \times x = 248$
$x = \frac{248 \times 100}{62} = 4 \times 100 = 400$.
Therefore,the total number of students who appeared in the examination is $400$.

(C) Let the monthly rent be $₹x$.
The annual rent is $12x$.
Upkeep and repairs per year = $12.5 \%$ of $12x = 0.125 \times 12x = 1.5x$.
Annual taxes = $₹325$.
Net annual income = (Total annual rent) - (Upkeep and repairs) - (Taxes) = $12x - 1.5x - 325 = 10.5x - 325$.
The annual return on investment is $5.5 \%$ of $₹100000 = 0.055 \times 100000 = ₹5500$.
Equating the net annual income to the annual return:
$10.5x - 325 = 5500$
$10.5x = 5825$
$x = \frac{5825}{10.5} \approx ₹554.76$.
Thus,the monthly rent is approximately $₹554.76$.

(B) Total candidates $= 2000$.
Number of boys $= 900$.
Number of girls $= 2000 - 900 = 1100$.
Number of boys who passed $= 32 \% \text{ of } 900 = \frac{32}{100} \times 900 = 288$.
Number of girls who passed $= 38 \% \text{ of } 1100 = \frac{38}{100} \times 1100 = 418$.
Total number of passed candidates $= 288 + 418 = 706$.
Total number of failed candidates $= 2000 - 706 = 1294$.
Percentage of failed candidates $= \left( \frac{1294}{2000} \times 100 \right) \% = \frac{1294}{20} \% = 64.7 \%$.

(B) Suppose his salary is $₹ x$.
After deducting $10\%$ for house rent,the remaining balance is $x \times (1 - 0.10) = 0.90x$.
He spends $15\%$ of the remainder on education,so the balance left is $0.90x \times (1 - 0.15) = 0.90x \times 0.85 = 0.765x$.
He spends $10\%$ of this balance on clothes,so the final balance left is $0.765x \times (1 - 0.10) = 0.765x \times 0.90 = 0.6885x$.
Given that the final balance is $₹ 1377$,we have:
$0.6885x = 1377$
$x = \frac{1377}{0.6885}$
$x = 2000$
Therefore,his salary is $₹ 2000$.

(A) Let the initial price of gold be $100$ and the initial quantity be $100$ units.
Initial expenditure $= 100 \times 100 = 10000$.
New price of gold $= 100 + 30 = 130$.
Let the new quantity be $x$ units.
Since the expenditure remains the same,$130 \times x = 10000$.
$x = \frac{10000}{130} = \frac{1000}{13} = 76 \frac{12}{13}$ units.
Reduction in quantity $= 100 - 76 \frac{12}{13} = 23 \frac{1}{13} \%$.
Alternatively,using the formula: $\text{Reduction} = \frac{\text{Percentage Increase}}{100 + \text{Percentage Increase}} \times 100 \%$.
$\text{Reduction} = \frac{30}{100 + 30} \times 100 \% = \frac{30}{130} \times 100 \% = \frac{300}{13} \% = 23 \frac{1}{13} \%$.