Percentage Questions in English

(B) Let the original price of sugar be $P$ per quintal.
Total money available $= 18 \times P$.
The new price after a $10 \%$ fall is $P - 0.10P = 0.90P$.
Let $x$ be the quantity of sugar that can be bought with the same total money at the new price.
Then,$x \times 0.90P = 18 \times P$.
$x = \frac{18 \times P}{0.90P} = \frac{18}{0.9} = 20$.
Therefore,$20$ quintals can be bought at the lower price.

(A) Total number of candidates $= 1000 + 800 = 1800$.
Number of boys who failed $= (100 - 60) \% \text{ of } 1000 = 40 \% \text{ of } 1000 = 400$.
Number of girls who failed $= (100 - 40) \% \text{ of } 800 = 60 \% \text{ of } 800 = 480$.
Total number of failed candidates $= 400 + 480 = 880$.
Percentage of failed candidates $= \left( \frac{880}{1800} \times 100 \right) \% = \frac{880}{18} \% = 48.88 \%$.

(A) Let the original price of the article be $100$.
After a reduction of $20 \%$,the new price becomes $100 - 20 = 80$.
To restore the price back to $100$,the increase required is $100 - 80 = 20$.
The percentage increase required is calculated on the new price $(80)$:
$\text{Percentage Increase} = \left( \frac{\text{Increase}}{\text{New Price}} \times 100 \right) \% = \left( \frac{20}{80} \times 100 \right) \% = \left( \frac{1}{4} \times 100 \right) \% = 25 \%$.

(C) Let the original fraction be $\frac{x}{y}$.
According to the problem,the numerator is increased by $25 \%$ and the denominator is decreased by $10 \%$.
New numerator = $x + 0.25x = 1.25x$.
New denominator = $y - 0.10y = 0.90y$.
The new fraction is given as $\frac{5}{9}$,so:
$\frac{1.25x}{0.90y} = \frac{5}{9}$
To solve for $\frac{x}{y}$,multiply both sides by $\frac{0.90}{1.25}$:
$\frac{x}{y} = \frac{5}{9} \times \frac{0.90}{1.25}$
$\frac{x}{y} = \frac{5}{9} \times \frac{90}{125}$
$\frac{x}{y} = \frac{5}{9} \times \frac{18}{25} = \frac{1}{1} \times \frac{2}{5} = \frac{2}{5}$.
Thus,the original fraction is $\frac{2}{5}$.

(A) Let the side of the square be $x$.
Initially,the area of the square is $A = x^2$.
After increasing one side by $30 \%$,the new length is $x + 0.3x = 1.3x$.
Let the other side be $y$ such that the area remains $x^2$.
So,$1.3x \times y = x^2$.
$y = \frac{x^2}{1.3x} = \frac{x}{1.3} = \frac{10x}{13}$.
The decrease in the other side is $x - \frac{10x}{13} = \frac{3x}{13}$.
The percentage decrease is $\left( \frac{3x/13}{x} \right) \times 100 = \frac{300}{13} \% = 23 \frac{1}{13} \%$.

(A) To find $0.06 \%$ of $250 \%$ of $1600$,we perform the following calculation:
Step $1$: Calculate $250 \%$ of $1600$.
$250 \% \times 1600 = \frac{250}{100} \times 1600 = 2.5 \times 1600 = 4000$.
Step $2$: Calculate $0.06 \%$ of the result obtained in Step $1$.
$0.06 \% \times 4000 = \frac{0.06}{100} \times 4000 = 0.06 \times 40 = 2.4$.
Therefore,the final answer is $2.4$.

(D) Let the third number be $100$.
Since the first number is $90 \%$ less than the third number,the first number $= 100 - (90 \% \text{ of } 100) = 100 - 90 = 10$.
Since the second number is $75 \%$ less than the third number,the second number $= 100 - (75 \% \text{ of } 100) = 100 - 75 = 25$.
We need to increase the first number $(10)$ to make it equal to the second number $(25)$.
The required increase $= 25 - 10 = 15$.
The percentage increase required $= \left( \frac{\text{Increase}}{\text{Original Number}} \right) \times 100 = \left( \frac{15}{10} \right) \times 100 = 150 \%$.

(C) Let the number be $x$.
According to the problem,when $x$ is increased by $216$,it becomes $140\%$ of $x$.
So,the equation is: $x + 216 = 1.40x$.
Subtract $x$ from both sides: $216 = 1.40x - x$.
$216 = 0.40x$.
$x = \frac{216}{0.40}$.
$x = \frac{2160}{4} = 540$.
Therefore,the number is $540$.

(A) To find the percentage by which $X$ is less than $Y$,we use the formula:
Percentage less $= \frac{Y - X}{Y} \times 100$
Substituting the given values:
Percentage less $= \frac{800 - 600}{800} \times 100$
$= \frac{200}{800} \times 100$
$= \frac{1}{4} \times 100 = 25 \%$
Thus,$X$ is $25 \%$ less than $Y$.

(B) Let the total wealth of the man be $x$.
$1$. Donation to charity: $30\%$ of $x = 0.3x$.
Remaining wealth $= x - 0.3x = 0.7x$.
$2$. Share for wife: $30\%$ of the remaining wealth $= 0.3 \times 0.7x = 0.21x$.
$3$. Share for son: $25\%$ of the remaining wealth $= 0.25 \times 0.7x = 0.175x$.
$4$. Remaining wealth for daughters $= 0.7x - (0.21x + 0.175x) = 0.7x - 0.385x = 0.315x$.
$5$. Each daughter's share $= \frac{0.315x}{3} = 0.105x$.
Given that each daughter gets $Rs$ $42$ lakh:
$0.105x = 42$
$x = \frac{42}{0.105} = 400$.
Therefore,the man's total wealth was $Rs$ $400$ lakh.

(C) Let the initial annual income be $x$ lakhs.
The initial tax paid is $12 \%$ of $x = 0.12x$.
The new income is $(x + 5)$ lakhs.
The new tax paid is $10 \%$ of $(x + 5) = 0.10(x + 5) = 0.1x + 0.5$.
According to the problem,the new tax is ₹ $10,000$ (which is $0.1$ lakh) more than the initial tax.
So,$(0.1x + 0.5) - 0.12x = 0.1$.
$-0.02x + 0.5 = 0.1$.
$0.4 = 0.02x$.
$x = \frac{0.4}{0.02} = 20$ lakhs.
The question asks for the increased income,which is $x + 5 = 20 + 5 = 25$ lakhs.

(B) Let the maximum marks be $x$.
The passing marks required is $40 \%$ of $x$.
The student obtained $250$ marks and failed by $38$ marks,which means the passing marks is $250 + 38 = 288$.
Therefore,$0.40 \times x = 288$.
$x = \frac{288}{0.40} = \frac{28800}{40} = 720$.
Thus,the maximum marks is $720$.

(D) Let Surya's age be $x$ years.
Then,Ravi's age is $(x - 12)$ years.
According to the problem,Ravi's age is $40 \%$ of the sum of their ages:
$(x - 12) = 0.40 \times (x + x - 12)$
$(x - 12) = \frac{2}{5} \times (2x - 12)$
Multiply both sides by $5$:
$5(x - 12) = 2(2x - 12)$
$5x - 60 = 4x - 24$
$5x - 4x = 60 - 24$
$x = 36$
So,Surya's present age is $36$ years.
Surya's age $9$ years hence will be $36 + 9 = 45$ years.

(D) First,calculate the sum of $\frac{1}{4}$ and $\frac{1}{5}$:
$\frac{1}{4} + \frac{1}{5} = \frac{5+4}{20} = \frac{9}{20} = 0.45$.
Subtract this sum from unity $(1)$:
$1 - 0.45 = 0.55$.
This is the correct answer.
The boy's answer is $0.45$.
The error is $|0.55 - 0.45| = 0.10$.
The percentage error is calculated as:
$\text{Percentage Error} = \left( \frac{\text{Error}}{\text{Correct Value}} \right) \times 100$
$= \left( \frac{0.10}{0.55} \right) \times 100$
$= \left( \frac{10}{55} \right) \times 100 = \left( \frac{2}{11} \right) \times 100 = \frac{200}{11} \%$.

(D) The expression is given by: $\frac{a}{100} \times b + \frac{b}{100} \times a$.
This simplifies to $\frac{ab}{100} + \frac{ab}{100} = \frac{2ab}{100}$.
This can be rewritten as $2 \times (\frac{a}{100} \times b)$,which is $2 \times (a \% \text{ of } b)$.
Alternatively,it can be written as $\frac{2a}{100} \times b$,which is $2a \% \text{ of } b$.

(A) Let the number be $x$.
According to the problem,when the number is increased by $84$,it becomes $107 \%$ of itself.
So,$x + 84 = 1.07x$.
Rearranging the equation: $84 = 1.07x - x$.
$84 = 0.07x$.
$x = \frac{84}{0.07} = \frac{8400}{7} = 1200$.
Alternatively,the increase of $84$ represents $107 \% - 100 \% = 7 \%$ of the number.
$7 \%$ of the number $= 84$.
$\therefore 100 \%$ of the number $= \frac{84}{7} \times 100 = 1200$.

(A) Given that $20 \%$ of $a = b$.
This can be written as $\frac{20}{100} \times a = b$,which simplifies to $0.2a = b$.
We need to find $b \%$ of $20$.
Substituting $b = 0.2a$ into the expression:
$b \%$ of $20 = \frac{b}{100} \times 20 = \frac{0.2a}{100} \times 20$.
$= \frac{0.2 \times 20}{100} \times a = \frac{4}{100} \times a$.
$= 4 \%$ of $a$.

(C) Let the total wealth be $100$ units.
Wealth willed to the wife $= 40\% \text{ of } 100 = 40$ units.
Wealth willed to the children $= 100 - 40 = 60$ units.
We need to find what percentage of the wife's share the children's share represents.
Required percentage $= \left( \frac{\text{Wealth of children}}{\text{Wealth of wife}} \right) \times 100$
Required percentage $= \left( \frac{60}{40} \right) \times 100 = 1.5 \times 100 = 150\%$.

(A) Let the number be $x$.
Given that $18 \%$ of $x = 720$.
This can be written as $\frac{18}{100} \times x = 720$.
To find $x$,we calculate $x = \frac{720 \times 100}{18} = 40 \times 100 = 4000$.
Now,we need to find $81 \%$ of the same number $(x = 4000)$.
$81 \% \text{ of } 4000 = \frac{81}{100} \times 4000 = 81 \times 40 = 3240$.

(C) Let the third number be $100$.
Since the first number is $10\%$ less than the third number,the first number $= 100 - 10 = 90$.
Since the second number is $25\%$ less than the third number,the second number $= 100 - 25 = 75$.
To make the second number equal to the first number,the increase required is $90 - 75 = 15$.
The percentage increase required $= (\frac{15}{75}) \times 100$.
$= (\frac{1}{5}) \times 100 = 20\%$.

(C) Given that $Q = ₹ 150$.
$P$ is $20\%$ more than $Q$,so $P = Q + 0.20Q = 1.20Q$.
$P = 1.20 \times 150 = 180$.
It is also given that $P$ is $40\%$ less than $R$,which means $P = R - 0.40R = 0.60R$.
Substituting the value of $P = 180$,we get $180 = 0.60R$.
$R = \frac{180}{0.60} = \frac{18000}{60} = 300$.
Therefore,the value of $R$ is $₹ 300$.

(B) Let the initial number be $x$.
Given that $25 \%$ of $x = 6$.
$0.25 \times x = 6$
$x = \frac{6}{0.25} = 24$.
The initial number is $24$.
We need to find a number that is $50 \%$ more than the initial number.
New number $= x + 50 \% \text{ of } x = 1.5 \times x$.
New number $= 1.5 \times 24 = 36$.

(A) Given that $B$ scored $220$ marks.
$A$ got $18 \%$ more marks than $B$,so $A = B + 0.18B = 1.18 \times 220 = 259.6$ marks.
It is also given that $A$ got $12 \%$ less marks than $C$,which means $A = C - 0.12C = 0.88C$.
Equating the two expressions for $A$: $0.88C = 259.6$.
Solving for $C$: $C = \frac{259.6}{0.88} = 295$.
Therefore,$C$ got $295$ marks.

(A) Given that $A$ is $20 \%$ more than $360.$
First,calculate $20 \%$ of $360:$
$20 \% \text{ of } 360 = \frac{20}{100} \times 360 = 0.2 \times 360 = 72.$
Now,add this value to $360$ to find $A:$
$A = 360 + 72 = 432.$
Therefore,the value of $A$ is $432.$

(B) Given that $x$ is $30 \%$ more than $y$.
$x = y + 0.30y = 1.30y$.
Given $y = ₹ 300$,so $x = 1.30 \times 300 = ₹ 390$.
Also,$x$ is $25 \%$ less than $z$,which means $x = z - 0.25z = 0.75z$.
Substituting the value of $x$:
$390 = 0.75z$.
$z = \frac{390}{0.75} = \frac{39000}{75} = 520$.
Therefore,the value of $z$ is $₹ 520$.

(B) Let the third number be $100$.
Since the first number is $20 \%$ more than the third number,the first number $= 100 + 20 = 120$.
Since the second number is $80 \%$ more than the third number,the second number $= 100 + 80 = 180$.
We need to find what percent the first number is of the second number.
Required percentage $= (\frac{120}{180}) \times 100$.
$= (\frac{2}{3}) \times 100 = 66.66 \%$.

(D) Let the income of $P = 100$.
Since the income of $Q$ is $20 \%$ more than the income of $P$,the income of $Q = 100 + 20 \% \text{ of } 100 = 120$.
Since the income of $R$ is $30 \%$ more than the income of $Q$,the income of $R = 120 + 30 \% \text{ of } 120 = 120 + 36 = 156$.
The percentage increase of $R$'s income over $P$'s income is calculated as:
$\text{Required percentage} = \frac{156 - 100}{100} \times 100 = 56 \%$.

(C) Let the number of officers be $x$ and the number of remaining employees be $y$.
Using the method of alligation:
Average salary of officers = $15000$
Average salary of remaining employees = $8000$
Overall average salary = $8840$
Difference for officers = $|8840 - 8000| = 840$
Difference for remaining employees = $|15000 - 8840| = 6160$
The ratio of officers to remaining employees is $840 : 6160$.
Simplifying the ratio: $84 : 616 = 21 : 154 = 3 : 22$.
Total employees = $3 + 22 = 25$.
Percentage of officers = $\frac{3}{25} \times 100 = 12 \%$.

(C) Let the original annual salary be $x$.
After a $5 \%$ increase,the new annual salary is $105 \%$ of $x$.
Given,$1.05 \times x = 15120$.
$x = \frac{15120}{1.05} = 14400$.
So,the original annual salary was $₹ 14,400$.
To find the original monthly salary,divide the annual salary by $12$.
Original monthly salary $= \frac{14400}{12} = ₹ 1,200$.

(B) Total percentage of expenditure $= 25 \% + 15 \% + 45 \% = 85 \%$.
Percentage of savings $= 100 \% - 85 \% = 15 \%$.
Given that the annual savings is $₹ 14,400$,we have $15 \%$ of the total income $= 14,400$.
Let the total income be $x$.
$0.15 \times x = 14,400$.
$x = \frac{14,400}{0.15} = \frac{1,440,000}{15} = 96,000$.
Therefore,the person's total annual income is $₹ 96,000$.

(B) To pass the examination,a student needs to secure $25 \%$ of the maximum marks.
Given that the student obtained $47$ marks and failed by $43$ marks,the passing marks can be calculated as:
$\text{Passing Marks} = 47 + 43 = 90$.
Since $25 \%$ of the maximum marks is equal to $90$,we can set up the equation:
$0.25 \times \text{Maximum Marks} = 90$.
$\text{Maximum Marks} = \frac{90}{0.25} = 90 \times 4 = 360$.
Therefore,the maximum marks of the examination are $360$.

(A) Let the marks of the first student be $x$.
Then,the marks of the second student are $x + 19$.
The sum of their marks is $x + (x + 19) = 2x + 19$.
According to the problem,the marks of the second student are $60 \%$ of the sum of their marks:
$x + 19 = 60 \% \times (2x + 19)$
$x + 19 = \frac{60}{100} \times (2x + 19)$
$x + 19 = \frac{3}{5} \times (2x + 19)$
Multiply both sides by $5$:
$5(x + 19) = 3(2x + 19)$
$5x + 95 = 6x + 57$
$95 - 57 = 6x - 5x$
$x = 38$
Therefore,the marks of the first student are $38$,and the marks of the second student are $38 + 19 = 57$.
The marks obtained by them are $57$ and $38$.

(B) Let the maximum marks of the exam be $x$.
According to the problem,the student scores $45 \%$ of $x$ and fails by $40$ marks,meaning the passing marks are $0.45x + 40$.
The passing percentage is given as $55 \%$,so the passing marks are also $0.55x$.
Equating the two expressions for passing marks: $0.45x + 40 = 0.55x$.
Rearranging the terms: $0.55x - 0.45x = 40$.
$0.10x = 40$.
$x = \frac{40}{0.10} = 400$.
Therefore,the maximum marks of the exam is $400$.

(C) Let the initial price of electricity be $100$ and initial consumption be $100$ units.
Initial expenditure = $100 \times 100 = 10000$.
New price of electricity = $100 + 25 = 125$.
Let the new consumption be $x$ units.
Since the expenditure remains the same,$125 \times x = 10000$.
$x = \frac{10000}{125} = 80$ units.
Reduction in consumption = $100 - 80 = 20$ units.
Percentage reduction = $\frac{20}{100} \times 100 = 20 \%$.

(C) Let the original price of apples be $P$ and the original consumption be $C$.
Original expenditure = $P \times C$.
New price = $P - 10\% \text{ of } P = 0.9P$.
Let the new consumption be $C'$.
Since the expenditure remains the same,$P \times C = 0.9P \times C'$.
$C' = \frac{P \times C}{0.9P} = \frac{C}{0.9} = \frac{10}{9}C$.
Increase in consumption = $C' - C = \frac{10}{9}C - C = \frac{1}{9}C$.
Percentage increase in consumption = $\left( \frac{\frac{1}{9}C}{C} \times 100 \right) \% = \frac{100}{9} \% = 11 \frac{1}{9} \%$.

(C) Let the initial price be $P_1 = 80$ and the final price be $P_2 = 100$.
Let the initial consumption be $C_1$ and the final consumption be $C_2$.
Since the expenditure remains the same,$P_1 \times C_1 = P_2 \times C_2$.
Therefore,$\frac{C_2}{C_1} = \frac{P_1}{P_2} = \frac{80}{100} = \frac{4}{5}$.
The decrease in consumption is $C_1 - C_2 = 5 - 4 = 1$ unit.
The percentage decrease in consumption is $\left( \frac{1}{5} \right) \times 100 = 20\%$.

(D) Let the price of the house $2$ years ago be $P$.
Given that the price increases by $20 \%$ every year.
The price after $1$ year will be $P \times (1 + 20/100) = P \times 1.2$.
The price after $2$ years will be $P \times 1.2 \times 1.2 = P \times 1.44$.
Given that the present price is $Rs \ 720000$.
So,$P \times 1.44 = 720000$.
$P = 720000 / 1.44$.
$P = 72000000 / 144$.
$P = 500000$.
Therefore,the price of the house $2$ years ago was $Rs \ 500000$.

(C) Let the population $3$ years ago be $P$.
Given that the population increases at a rate of $5 \%$ per annum.
The formula for population after $n$ years is $A = P(1 + \frac{r}{100})^n$.
Here,$A = 3,70,440$,$r = 5$,and $n = 3$.
$3,70,440 = P(1 + \frac{5}{100})^3$
$3,70,440 = P(1 + \frac{1}{20})^3$
$3,70,440 = P(\frac{21}{20})^3$
$3,70,440 = P \times \frac{9261}{8000}$
$P = \frac{3,70,440 \times 8000}{9261}$
$P = 40 \times 8000 = 3,20,000$.
Thus,the population $3$ years ago was $3,20,000$.

(A) Let the parts given to $A, B,$ and $C$ be $P_A, P_B,$ and $P_C$ respectively.
Given that the total amount (Principal + Interest) is equal for all three.
Amount = $P(1 + \frac{RT}{100})$.
For $A$: $P_A(1 + \frac{5 \times 2}{100}) = P_A(1.10) = 110\% P_A$.
For $B$: $P_B(1 + \frac{5 \times 3}{100}) = P_B(1.15) = 115\% P_B$.
For $C$: $P_C(1 + \frac{5 \times 4}{100}) = P_C(1.20) = 120\% P_C$.
Equating them: $110 P_A = 115 P_B = 120 P_C = K$.
$P_A : P_B : P_C = \frac{1}{110} : \frac{1}{115} : \frac{1}{120} = \frac{1}{22} : \frac{1}{23} : \frac{1}{24}$.
Multiplying by the $LCM$ $(22 \times 23 \times 24 = 12144)$:
$P_A : P_B : P_C = (23 \times 24) : (22 \times 24) : (22 \times 23) = 552 : 528 : 506$.
Dividing by $2$: $276 : 264 : 253$.
Sum of ratios = $276 + 264 + 253 = 793$.
$P_A = \frac{276}{793} \times 7930 = 2760$.

(D) Let the present price of the table be $P$.
Given that the price depreciates by $20 \%$ every year,the value after $1$ year will be $P \times (1 - 0.20) = 0.8P$.
The value after $2$ years will be $0.8P \times 0.8 = 0.64P$.
According to the problem,$0.64P = 32,000$.
Therefore,$P = \frac{32,000}{0.64} = \frac{3,200,000}{64} = 50,000$.
The present price of the table is $Rs$ $50,000$.

(C) Let the population two years ago be $P$.
The population increases at a rate of $2\%$ per annum.
The formula for population after $n$ years is $A = P(1 + \frac{r}{100})^n$.
Given $A = 26010$,$r = 2$,and $n = 2$.
$26010 = P(1 + \frac{2}{100})^2$
$26010 = P(1 + 0.02)^2$
$26010 = P(1.02)^2$
$26010 = P(1.0404)$
$P = \frac{26010}{1.0404}$
$P = 25000$.
Therefore,the population of the town two years ago was $25000$.

(A) Let the population two years ago be $P$.
The population increases at a rate of $2\%$ per annum.
The formula for population after $n$ years is $A = P(1 + \frac{r}{100})^n$.
Given $A = 26010$,$r = 2$,and $n = 2$.
$26010 = P(1 + \frac{2}{100})^2$
$26010 = P(1 + 0.02)^2$
$26010 = P(1.02)^2$
$26010 = P(1.0404)$
$P = \frac{26010}{1.0404}$
$P = 25000$.
Therefore,the population of the town two years ago was $25000$.

(B) Let the initial number of $ODI$s played be $x$.
Initial number of wins $= 0.60x$.
After losing $30$ consecutive $ODI$s,the total number of $ODI$s becomes $x + 30$.
The number of wins remains $0.60x$ as they lost all the new matches.
According to the problem,the new success rate is $30 \%$.
So,$\frac{0.60x}{x + 30} = 0.30$.
$0.60x = 0.30(x + 30)$.
$0.60x = 0.30x + 9$.
$0.30x = 9$.
$x = \frac{9}{0.30} = 30$.
The total number of $ODI$s played initially was $30$.
The total number of $ODI$s played after losing $30$ more matches is $30 + 30 = 60$.

(C) Duty on laptops $= 10\% \text{ of } 210,000 = \frac{10}{100} \times 210,000 = 21,000$.
Duty on mobile phones $= 8\% \text{ of } 100,000 = \frac{8}{100} \times 100,000 = 8,000$.
Duty on television sets $= 5\% \text{ of } 150,000 = \frac{5}{100} \times 150,000 = 7,500$.
Total duty $= 21,000 + 8,000 + 7,500 = 36,500$.
Therefore,the total duty paid is $Rs$ $36,500$.

(A) To find the percentage weight of the first ball with respect to the second ball,we use the formula:
$\text{Percentage} = \left( \frac{\text{Weight of first ball}}{\text{Weight of second ball}} \right) \times 100$
Substituting the given values:
$\text{Percentage} = \left( \frac{3.5}{7.5} \right) \times 100$
$= \left( \frac{35}{75} \right) \times 100$
$= \left( \frac{7}{15} \right) \times 100$
$= \frac{700}{15} = \frac{140}{3} = 46 \frac{2}{3} \%$

(A) Let the larger number be $x$.
Since the smaller number is $25 \%$ of the larger number,the smaller number is $0.25x$.
According to the problem,the larger number is $12$ more than the smaller number:
$x = 0.25x + 12$
Subtract $0.25x$ from both sides:
$x - 0.25x = 12$
$0.75x = 12$
$x = \frac{12}{0.75}$
$x = \frac{1200}{75} = 16$
Therefore,the larger number is $16$.

(B) Let the initial number of students be $x$.
According to the problem,the number is increased by $20 \%$,so the new number is $x + 0.20x = 1.20x$.
We are given that the new number is $66$.
Therefore,$1.20x = 66$.
Solving for $x$: $x = \frac{66}{1.20} = \frac{660}{12} = 55$.
Thus,the initial number of students was $55$.

(B) Let the original number of goats be $x$.
After losing $12 \%$ of goats in the flood,the remaining goats are $(100 - 12) \% = 88 \%$ of $x$,which is $0.88x$.
Then,$5 \%$ of the remainder died from diseases. The remaining goats after this loss are $95 \%$ of the remainder.
So,the final number of goats is $0.95 \times 0.88x = 8360$.
$0.836x = 8360$.
$x = \frac{8360}{0.836} = 10000$.
Therefore,the original number of goats was $10000$.

(B) Given:
$A = 20\% \text{ of } B = 0.20 \times B$
$B = 25\% \text{ of } C = 0.25 \times C$
Substitute the value of $B$ in the equation for $A$:
$A = 0.20 \times (0.25 \times C)$
$A = 0.05 \times C$
To express $A$ as a percentage of $C$:
$A = (0.05 \times 100)\% \text{ of } C$
$A = 5\% \text{ of } C$
Therefore,$A$ is $5\%$ of $C$.

(C) Total number of students $= 1500$.
Number of girls $= 44 \% \text{ of } 1500 = \frac{44}{100} \times 1500 = 660$.
Number of boys $= 1500 - 660 = 840$.
Monthly fee of each boy $= ₹ 540$.
Monthly fee of each girl $= 540 - (25 \% \text{ of } 540) = 540 - 135 = ₹ 405$.
Total fees of boys $= 840 \times 540 = ₹ 453600$.
Total fees of girls $= 660 \times 405 = ₹ 267300$.
Sum of fees of boys and girls $= 453600 + 267300 = ₹ 720900$.