Percentage Questions in English

(B) Let the total number of students who appeared for the examination be $x$.
Given that $35 \%$ of the students passed,the percentage of students who failed is $(100 - 35) \% = 65 \%$.
We are given that the number of failed students is $455$.
Therefore,$65 \%$ of $x = 455$.
$\frac{65}{100} \times x = 455$.
$x = \frac{455 \times 100}{65}$.
$x = 7 \times 100 = 700$.
Thus,the total number of students who appeared for the examination is $700$.

(B) Let the total monthly income be $I$.
Given that the person spends $66 \frac{2}{3} \%$ of his income.
Therefore,the savings percentage is $100 \% - 66 \frac{2}{3} \% = 33 \frac{1}{3} \%$.
We know that $33 \frac{1}{3} \% = \frac{100}{3} \% = \frac{1}{3}$ of the income.
Given that the savings amount is $₹ 1200$.
So,$\frac{1}{3} \times I = 1200$,which implies $I = 1200 \times 3 = ₹ 3600$.
The monthly expenses are $66 \frac{2}{3} \%$ of the income.
Expenses $= \frac{2}{3} \times 3600 = 2 \times 1200 = ₹ 2400$.

(B) Let the total number of votes polled be $x$.
The winning candidate received $84 \%$ of the votes,so the losing candidate received $(100 - 84) \% = 16 \%$ of the votes.
The majority by which the candidate won is the difference between the votes of the winner and the loser.
Majority $= (84 \% - 16 \%) \text{ of } x = 68 \% \text{ of } x$.
Given that the majority is $476$ votes,we have:
$\frac{68}{100} \times x = 476$
Solving for $x$:
$x = \frac{476 \times 100}{68}$
$x = 7 \times 100 = 700$.
Therefore,the total number of votes polled is $700$.

(B) Given that $P \%$ of $P = 36$.
This can be written as $\frac{P}{100} \times P = 36$.
Multiplying both sides by $100$,we get $P^2 = 36 \times 100$.
$P^2 = 3600$.
Taking the square root of both sides,$P = \sqrt{3600} = 60$.
Therefore,the value of $P$ is $60$.

(D) Given that $\frac{x}{100} \times y = z$.
To find what percent of $z$ is $x$,we need to calculate $\frac{x}{z} \times 100$.
From the given equation,$\frac{x}{z} = \frac{100}{y}$.
Therefore,the required percentage is $\frac{x}{z} \times 100 = \frac{100}{y} \times 100 = \frac{100^{2}}{y} \%$.

(B) Let the total amount Aman had be $x$.
Amount given to Rohan $= 40 \% \text{ of } x = 0.4x$.
Amount given to Sahil by Rohan $= \frac{1}{4} \times (0.4x) = 0.1x$.
Sahil pays $₹ 200$ to the taxi driver,and he is left with $₹ 600$.
Therefore,the equation is: $0.1x - 200 = 600$.
Adding $200$ to both sides: $0.1x = 800$.
Solving for $x$: $x = \frac{800}{0.1} = 8000$.
Thus,Aman had $₹ 8000$.

(D) The total height of the pole is $192 \, m$.
In the first hour,the spider climbs $62 \frac{1}{2} \% = 62.5 \% = \frac{125}{200} = \frac{5}{8}$ of the total height.
Remaining height after the first hour $= 192 \times (1 - \frac{5}{8}) = 192 \times \frac{3}{8} = 72 \, m$.
In the second hour,the spider climbs $12 \frac{1}{2} \% = 12.5 \% = \frac{25}{200} = \frac{1}{8}$ of the remaining height.
Distance climbed in the second hour $= \frac{1}{8} \times 72 = 9 \, m$.

(C) Total inhabitants $= 1000$.
Number of males $= 1000 \times 60\% = 600$.
Number of literate males $= 600 \times 20\% = 120$.
Total literate inhabitants $= 1000 \times 25\% = 250$.
Number of females $= 1000 - 600 = 400$.
Number of literate females $= 250 - 120 = 130$.
Percentage of literate females $= (130 / 400) \times 100 = 32.5\%$.

(D) Let the total number of houses be $100$.
Given that $40 \%$ of houses contain two or more people,the number of houses with only one person is $100 - 40 = 60$.
Out of these $60$ houses,$25 \%$ have only a male. Therefore,the number of houses with only a male is $25 \% \text{ of } 60 = 0.25 \times 60 = 15$.
The remaining houses among those with only one person must contain exactly one female and no males.
Number of houses with exactly one female and no males = $60 - 15 = 45$.
Thus,the percentage of all houses that contain exactly one female and no males is $45 \%$.

(A) Let $x$ be the maximum marks.
The pass marks can be expressed as:
Pass marks $= 0.30x + 15$ (since $A$ failed by $15$ marks).
Pass marks $= 0.40x - 35$ (since $B$ got $35$ marks more than passing).
Equating the two expressions for pass marks:
$0.30x + 15 = 0.40x - 35$
$0.40x - 0.30x = 15 + 35$
$0.10x = 50$
$x = \frac{50}{0.10} = 500$
Now,calculate the pass marks:
Pass marks $= 0.30(500) + 15 = 150 + 15 = 165$.
Finally,calculate the pass percentage:
Pass percentage $= (\frac{165}{500}) \times 100 = 33\%$.

(C) Let the total number of students be $100$.
Total marks of the class $= 100 \times 80 = 8000$.
Marks of $10 \%$ students $= 10 \times 95 = 950$.
Marks of $20 \%$ students $= 20 \times 90 = 1800$.
Remaining students $= 100 - (10 + 20) = 70$.
Sum of marks of remaining students $= 8000 - (950 + 1800) = 8000 - 2750 = 5250$.
Average marks of remaining students $= \frac{5250}{70} = 75$.

(D) The total marks for the three papers are $150 + 150 + 180 = 480$.
To pass,the candidate needs $35 \%$ of the total marks.
Required passing marks $= 480 \times \frac{35}{100} = 4.8 \times 35 = 168$.
Let the marks obtained in the third paper be $x$.
The sum of marks obtained in the three papers is $62 + 35 + x = 97 + x$.
To just qualify for a pass,the sum of marks must equal the required passing marks:
$97 + x = 168$.
$x = 168 - 97 = 71$.
Therefore,the candidate must get $71$ marks in the third paper.

(C) Let the number of boys be $x$ and the number of girls be $y$.
Given that the total number of students is $x + y = 150$.
According to the problem,the number of girls is $x\%$ of the total number of students:
$y = \frac{x}{100} \times 150 = 1.5x$
Substitute $y = 1.5x$ into the equation $x + y = 150$:
$x + 1.5x = 150$
$2.5x = 150$
$x = \frac{150}{2.5} = 60$
Therefore,the number of boys is $60$.

(C) Let the total sales be $₹ x$.
The total earnings of the salesman consist of $5 \frac{1}{2} \%$ of total sales and a bonus of $\frac{1}{2} \%$ on sales exceeding $₹ 10000$.
Total earnings $= \frac{5.5}{100} \times x + \frac{0.5}{100} \times (x - 10000) = 1990$
$\Rightarrow \frac{5.5x}{100} + \frac{0.5(x - 10000)}{100} = 1990$
Multiply the entire equation by $100$:
$5.5x + 0.5x - 5000 = 199000$
$6x = 199000 + 5000$
$6x = 204000$
$x = \frac{204000}{6} = 34000$
Thus,the total sales are $₹ 34000$.

(B) Let the earning of Michael be $x$.
Albert earned $20\%$ less than Michael,so Albert's earning $= x - 0.20x = 0.80x$.
Peter earned $40\%$ more than Albert,so Peter's earning $= 0.80x + (0.40 \times 0.80x) = 0.80x + 0.32x = 1.12x$.
To find how much more Peter earned than Michael,we calculate the percentage difference: $\frac{1.12x - x}{x} \times 100 = 0.12 \times 100 = 12\%$.
Therefore,Peter earned $12\%$ more than Michael.

(B) Let $C$'s salary be $x$.
According to the problem,$B$'s salary is $25 \%$ of $C$'s salary:
$B = \frac{25}{100} \times x = 0.25x$
$A$'s salary is $40 \%$ of $B$'s salary:
$A = \frac{40}{100} \times B = 0.40 \times 0.25x$
Calculating the value:
$A = 0.10x$
To express $A$'s salary as a percentage of $C$'s salary:
$\frac{A}{C} \times 100 = \frac{0.10x}{x} \times 100 = 10 \%$
Thus,$A$'s salary is $10 \%$ of $C$'s salary.

(A) Let the original quantity of the mixture be $x \text{ kg}$.
Initially:
Pure ghee $= 0.6x \text{ kg}$
Vanaspati ghee $= 0.4x \text{ kg}$
After adding $10 \text{ kg}$ of pure ghee:
New total quantity $= (x + 10) \text{ kg}$
New quantity of vanaspati ghee remains $= 0.4x \text{ kg}$
According to the problem, the new concentration of vanaspati is $20 \%$:
$\frac{0.4x}{x + 10} = \frac{20}{100}$
$\frac{0.4x}{x + 10} = \frac{1}{5}$
Cross-multiplying:
$5 \times 0.4x = 1 \times (x + 10)$
$2x = x + 10$
$x = 10 \text{ kg}$
Thus, the original quantity was $10 \text{ kg}$.

(D) Let the third number be $100$.
Since the first number is $12 \frac{1}{2} \%$ more than the third number,the first number $= 100 + 12.5 = 112.5$.
Since the second number is $25 \%$ more than the third number,the second number $= 100 + 25 = 125$.
We need to find the first number as a percentage of the second number.
Required percentage $= (\frac{112.5}{125}) \times 100$.
$= 0.9 \times 100 = 90 \%$.

(D) Let the third number be $100$.
The first number is $30 \%$ less than $100$,so it is $100 - 30 = 70$.
The second number is $37 \%$ less than $100$,so it is $100 - 37 = 63$.
We need to find how much percent the second number $(63)$ is less than the first number $(70)$.
Difference $= 70 - 63 = 7$.
Required percentage $= \left( \frac{\text{Difference}}{\text{First number}} \right) \times 100 = \left( \frac{7}{70} \right) \times 100$.
$= 0.1 \times 100 = 10 \%$.

(D) Let $B$'s height be $100 \text{ units}$.
Since $A$'s height is $40 \%$ less than $B$'s height,$A$'s height $= 100 - 40 = 60 \text{ units}$.
We need to find how much percent $B$'s height is more than $A$'s height.
Difference in height $= 100 - 60 = 40 \text{ units}$.
Percentage more $= \left( \frac{\text{Difference}}{\text{A's height}} \right) \times 100 = \left( \frac{40}{60} \right) \times 100$.
Percentage more $= \left( \frac{2}{3} \right) \times 100 = \frac{200}{3} = 66 \frac{2}{3} \%$.

(C) Let $B$'s salary be $100$.
Since $A$'s salary is $50 \%$ more than $B$'s,$A$'s salary $= 100 + 50 = 150$.
The difference between $A$'s and $B$'s salary is $150 - 100 = 50$.
To find the percentage by which $B$'s salary is less than $A$'s,we use the formula: $\left( \frac{\text{Difference}}{\text{A's salary}} \times 100 \right) \%$.
Percentage $= \left( \frac{50}{150} \times 100 \right) \% = \left( \frac{1}{3} \times 100 \right) \% = \frac{100}{3} \% = 33 \frac{1}{3} \%$.

(D) Let the initial population of the town be $P$.
The net annual growth rate of the population is $4 \% - 0.5 \% = 3.5 \%$.
The population after $3$ years is given by the formula $A = P(1 + r/100)^n$,where $r = 3.5$ and $n = 3$.
Population after $3$ years $= P(1 + 3.5/100)^3 = P(1 + 0.035)^3 = P(1.035)^3$.
Calculating $(1.035)^3 = 1.035 \times 1.035 \times 1.035 \approx 1.108717875$.
The increase in population is $P(1.108717875) - P = 0.108717875P$.
The percentage increase is $(0.108717875P / P) \times 100 = 10.8717875 \%$.
Rounding to one decimal place,the increase is approximately $10.8 \%$.

(C) Let the number of males be $M$ and the number of females be $F$.
Given,$M + F = 5000$ --- $(1)$
After an increase of $10\%$ in males and $15\%$ in females,the new population is $5600$.
So,$1.10M + 1.15F = 5600$ --- $(2)$
From $(1)$,$F = 5000 - M$.
Substitute this into $(2)$:
$1.10M + 1.15(5000 - M) = 5600$
$1.10M + 5750 - 1.15M = 5600$
$5750 - 0.05M = 5600$
$0.05M = 5750 - 5600$
$0.05M = 150$
$M = \frac{150}{0.05} = 3000$
Therefore,the number of males in the village was $3000$.

(B) Let the initial value of the equipment be $x$.
The value after $3$ years of depreciation at a rate of $20 \%$ per annum is given by the formula: $V = x(1 - \frac{R}{100})^N$.
Substituting the values: $V = x(1 - \frac{20}{100})^3$.
$V = x(0.8)^3 = x \times 0.512 = 0.512x$.
The decrease in value is $x - 0.512x = 0.488x$.
To express this as a percentage: $0.488 \times 100 = 48.8 \%$.

(B) Population $2$ years ago $= 62500$.
The population decreases at a rate of $4\%$ per year.
Using the formula for depreciation: $\text{Present Population} = P \left(1 - \frac{R}{100}\right)^n$,where $P = 62500$,$R = 4$,and $n = 2$.
$\text{Present Population} = 62500 \left(1 - \frac{4}{100}\right)^2$
$= 62500 \left(\frac{96}{100}\right)^2 = 62500 \left(\frac{24}{25}\right)^2$
$= 62500 \times \frac{576}{625} = 100 \times 576 = 57600$.
Thus,the present population of the town is $57600$.

(B) The present population of the district is $P = 64000$.
The rate of increase is $R = 2 \frac{1}{2} \% = 2.5 \% = \frac{5}{2} \%$.
The time period is $n = 3$ years.
Using the formula for population growth: $A = P \left(1 + \frac{R}{100}\right)^n$.
Substituting the values: $A = 64000 \left(1 + \frac{2.5}{100}\right)^3$.
$A = 64000 \left(1 + \frac{25}{1000}\right)^3 = 64000 \left(1 + \frac{1}{40}\right)^3$.
$A = 64000 \left(\frac{41}{40}\right)^3$.
$A = 64000 \times \frac{41 \times 41 \times 41}{40 \times 40 \times 40}$.
$A = 64000 \times \frac{68921}{64000}$.
$A = 68921$.
Thus,the population after $3$ years will be $68921$.

(C) Let the purchase price of the machine be $P$.
The rate of depreciation $R = 10 \%$ per annum.
The time period $n = 3$ years.
The present value $A = ₹ 8748$.
The formula for depreciation is $A = P(1 - \frac{R}{100})^n$.
Substituting the values: $8748 = P(1 - \frac{10}{100})^3$.
$8748 = P(1 - \frac{1}{10})^3$.
$8748 = P(\frac{9}{10})^3$.
$8748 = P \times \frac{729}{1000}$.
$P = \frac{8748 \times 1000}{729}$.
$P = 12 \times 1000 = ₹ 12000$.
Thus,the purchase price of the machine was $₹ 12000$.

(D) The cost of an article increases due to inflation,which follows the compound interest formula: $A = P(1 + \frac{R}{100})^n$.
Here,the initial cost $P = ₹ 20$,the rate of inflation $R = 8 \%$,and the time period $n = 2$ years.
Substituting these values into the formula:
$A = 20 \times (1 + \frac{8}{100})^2$
$A = 20 \times (1 + 0.08)^2$
$A = 20 \times (1.08)^2$
$A = 20 \times 1.1664$
$A = 23.328$
Thus,the cost of the article after two years will be $₹ 23.328$,which lies between $₹ 23$ and $₹ 24$.

(B) Population $3$ years ago $= 160000$.
Present population $= 160000 \times (1 + \frac{3}{100}) \times (1 + \frac{2.5}{100}) \times (1 + \frac{5}{100})$.
Present population $= 160000 \times \frac{103}{100} \times \frac{102.5}{100} \times \frac{105}{100}$.
Present population $= 160000 \times \frac{103}{100} \times \frac{205}{200} \times \frac{105}{100}$.
Present population $= 16 \times 103 \times \frac{205}{2} \times \frac{21}{20} = 4 \times 103 \times 205 \times \frac{21}{10}$.
Present population $= 177366$.

(D) Let the population in $1998$ be $P$.
The population increases by $5 \%$ annually,so the population in $2001$ (after $3$ years) is given by the formula $A = P(1 + r/100)^n$.
Here,$A = 138915$,$r = 5$,and $n = 3$.
$138915 = P(1 + 5/100)^3$
$138915 = P(105/100)^3$
$138915 = P(21/20)^3$
$P = 138915 \times (20/21)^3$
$P = 138915 \times (8000 / 9261)$
$P = 15 \times 8000 = 120000$.
Thus,the population in $1998$ was $120000$.

(C) Let the initial production in the year $1998$ be $P$.
The production pattern is: increase by $15\%$ for two years,then decrease by $10\%$ in the third year.
For the years $1998$ to $2002$ (a span of $4$ years),the production changes as follows:
Year $1$: Increase by $15\% = P \times 1.15$
Year $2$: Increase by $15\% = (P \times 1.15) \times 1.15$
Year $3$: Decrease by $10\% = (P \times 1.15 \times 1.15) \times 0.90$
Year $4$: Increase by $15\% = (P \times 1.15 \times 1.15 \times 0.90) \times 1.15$
Resultant production $= P \times 1.15 \times 1.15 \times 0.90 \times 1.15$
$= P \times (1.15)^3 \times 0.90$
$= P \times 1.520875 \times 0.90$
$= P \times 1.3687875$
Percentage change $= (1.3687875 - 1) \times 100 = 36.87875\% \approx 37\%$ increase.

(D) In January,the number of ticketless travellers $= 4000$.
In February,the number of ticketless travellers increased by $5\%$:
$= 4000 \times (1 + 0.05) = 4000 \times 1.05 = 4200$.
In March,the number of ticketless travellers decreased by $5\%$ from the February count:
$= 4200 \times (1 - 0.05) = 4200 \times 0.95 = 3990$.
In April,the number of ticketless travellers decreased by $10\%$ from the March count:
$= 3990 \times (1 - 0.10) = 3990 \times 0.90 = 3591$.
Thus,the total number of ticketless travellers caught in April was $3591$.

(A) Let the initial number of bushes be $x$.
The population changes over three years as follows:
Year $1$: Increase of $10 \% \implies x \times (1 + 0.10) = 1.1x$
Year $2$: Increase of $8 \% \implies 1.1x \times (1 + 0.08) = 1.1x \times 1.08$
Year $3$: Decrease of $10 \% \implies (1.1x \times 1.08) \times (1 - 0.10) = 1.1x \times 1.08 \times 0.9$
Given that the final population is $26730$,we have:
$x \times 1.1 \times 1.08 \times 0.9 = 26730$
$x \times (\frac{11}{10}) \times (\frac{108}{100}) \times (\frac{9}{10}) = 26730$
$x \times \frac{10692}{1000} = 26730$
$x = \frac{26730 \times 1000}{10692}$
$x = 25000$
Thus,the number of bushes in the beginning was $25000$.

(C) Initial volume of shaving lotion = $9 \ ml$.
Alcohol concentration = $50 \%$.
Amount of alcohol = $9 \times 0.50 = 4.5 \ ml$.
Let the total volume of the new lotion be $V \ ml$ after adding water.
The amount of alcohol remains constant at $4.5 \ ml$.
In the new lotion,the concentration of alcohol is $30 \%$.
So,$30 \%$ of $V = 4.5 \ ml$.
$0.30 \times V = 4.5$.
$V = \frac{4.5}{0.30} = 15 \ ml$.
The quantity of water to be added = New total volume - Initial volume.
Water added = $15 \ ml - 9 \ ml = 6 \ ml$.

(C) Total money in the bag initially:
$600 \times 0.25 + 1200 \times 0.50 = 150 + 600 = ₹ 750$.
Amount of money removed:
Money removed from $25$ paise coins $= 12 \% \text{ of } 150 = \frac{12}{100} \times 150 = ₹ 18$.
Money removed from $50$ paise coins $= 24 \% \text{ of } 600 = \frac{24}{100} \times 600 = ₹ 144$.
Total money removed $= 18 + 144 = ₹ 162$.
Percentage of money removed:
$\text{Percentage} = \left( \frac{162}{750} \right) \times 100 = \frac{16200}{750} = 21.6 \%$.

(A) To convert a fraction or mixed number into a percentage,multiply the value by $100$.
Given the mixed number $5 \frac{1}{4}$.
First,convert the mixed number into an improper fraction: $5 \frac{1}{4} = \frac{(5 \times 4) + 1}{4} = \frac{21}{4}$.
Now,multiply by $100$ to find the percentage: $\frac{21}{4} \times 100 = 21 \times 25 = 525\%$.
Therefore,$5 \frac{1}{4}$ is equivalent to $525\%$.

(B) To convert a decimal to a percentage,we multiply the number by $100$ and add the percentage symbol $(\%)$.
$0.005 = 0.005 \times 100 \%$
$0.005 = 0.5 \%$
Since $0.5 = \frac{1}{2}$,we can write $0.5 \% = \frac{1}{2} \%$.

(B) To express $6 \frac{2}{3} \%$ as a fraction,first convert the mixed fraction to an improper fraction: $6 \frac{2}{3} = \frac{6 \times 3 + 2}{3} = \frac{20}{3}$.
Now,divide by $100$ to remove the percentage sign: $\frac{20}{3} \% = \frac{20}{3} \div 100$.
This is equivalent to $\frac{20}{3} \times \frac{1}{100}$.
Simplifying the expression: $\frac{20}{300} = \frac{1}{15}$.

(C) To convert a percentage into a fraction,divide the value by $100$.
$0.6 \% = \frac{0.6}{100}$
To remove the decimal from the numerator,multiply both the numerator and the denominator by $10$:
$\frac{0.6 \times 10}{100 \times 10} = \frac{6}{1000}$
Now,simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor,which is $2$:
$\frac{6 \div 2}{1000 \div 2} = \frac{3}{500}$
Therefore,the correct fraction is $3/500$.

(B) To convert a decimal into a percentage,we multiply the decimal by $100$.
$0.025 \times 100 = 2.5 \%$.
Therefore,$0.025$ in terms of rate per cent is $2.5 \%$.

(C) Let $x$ percent of $12$ be $84$.
According to the problem:
$\frac{x}{100} \times 12 = 84$
Solving for $x$:
$x = \frac{84 \times 100}{12}$
$x = 7 \times 100$
$x = 700$
Therefore,$700\%$ of $12$ is $84$.

(B) To express a fraction as a percentage,multiply the fraction by $100$ and add the percentage symbol $(\%)$.
$\frac{7}{8} = \left(\frac{7}{8} \times 100\right) \%$
$= \frac{700}{8} \%$
$= \frac{175}{2} \%$
$= 87 \frac{1}{2} \%$

(A) To express $8 \frac{1}{3} \%$ as a fraction,follow these steps:
$1$. Convert the mixed fraction $8 \frac{1}{3}$ into an improper fraction: $8 \frac{1}{3} = \frac{(8 \times 3) + 1}{3} = \frac{25}{3}$.
$2$. Since it is a percentage,divide by $100$ or multiply by $\frac{1}{100}$: $\frac{25}{3} \% = \frac{25}{3} \times \frac{1}{100}$.
$3$. Simplify the expression: $\frac{25}{300} = \frac{1}{12}$.

(C) To find $37 \frac{1}{2} \%$ of $₹ 48$,first convert the mixed fraction to an improper fraction:
$37 \frac{1}{2} = \frac{37 \times 2 + 1}{2} = \frac{75}{2} \%$.
Now,calculate the value:
$\frac{75}{2} \% \text{ of } 48 = \frac{75}{2 \times 100} \times 48$.
$= \frac{75}{200} \times 48$.
$= \frac{3}{8} \times 48$.
$= 3 \times 6 = 18$.
Therefore,$37 \frac{1}{2} \%$ of $₹ 48$ is $₹ 18$.

(C) Let $x \%$ of $\frac{2}{7} = \frac{1}{35}$.
This can be written as: $\frac{x}{100} \times \frac{2}{7} = \frac{1}{35}$.
Solving for $x$: $x = \frac{1}{35} \times \frac{100 \times 7}{2}$.
$x = \frac{700}{70} = 10$.
Therefore,$10 \%$ of $\frac{2}{7}$ is $\frac{1}{35}$.

(C) Given the equation: $75 \%$ of $480 = x \times 15$.
First,calculate $75 \%$ of $480$:
$\frac{75}{100} \times 480 = 0.75 \times 480 = 360$.
Now,substitute this value into the equation:
$360 = x \times 15$.
Solve for $x$:
$x = \frac{360}{15} = 24$.
Therefore,the value of $x$ is $24$.

(B) Let the number be $x$.
According to the problem,$200 \%$ of $x = 90$.
$\Rightarrow \frac{200}{100} \times x = 90$
$\Rightarrow 2x = 90$
$\Rightarrow x = \frac{90}{2} = 45$.
Now,we need to find $80 \%$ of $x$ (which is $45$).
$80 \%$ of $45 = \frac{80}{100} \times 45$
$= 0.8 \times 45 = 36$.
Therefore,the correct answer is $36$.

(C) Let the number be $x$.
Given that $37 \frac{1}{2} \%$ of $x = 45$.
Converting the percentage to a fraction: $37 \frac{1}{2} \% = \frac{75}{2} \% = \frac{75}{200} = \frac{3}{8}$.
So,$\frac{3}{8} x = 45$.
Solving for $x$: $x = \frac{45 \times 8}{3} = 15 \times 8 = 120$.
Now,we need to find $87 \frac{1}{2} \%$ of $120$.
Converting the percentage to a fraction: $87 \frac{1}{2} \% = \frac{175}{2} \% = \frac{175}{200} = \frac{7}{8}$.
Therefore,$\frac{7}{8} \times 120 = 7 \times 15 = 105$.

(A) Let the missing value be $x$.
The equation is $x \times 15 = 37.5 \%$ of $220$.
Convert the percentage to a fraction: $37.5 \% = \frac{37.5}{100}$.
Substitute this into the equation: $15x = \frac{37.5}{100} \times 220$.
Simplify the right side: $15x = 0.375 \times 220 = 82.5$.
Solve for $x$: $x = \frac{82.5}{15} = 5.5$.
Thus,the correct option is $A$.

(B) Let $x \%$ of $4 \text{ km} = 8 \text{ meters}$.
Since $1 \text{ km} = 1000 \text{ meters}$,$4 \text{ km} = 4000 \text{ meters}$.
According to the problem: $\frac{x}{100} \times 4000 = 8$.
Solving for $x$: $x = \frac{8 \times 100}{4000}$.
$x = \frac{800}{4000} = \frac{8}{40} = \frac{1}{5} = 0.2$.
Therefore,$0.2 \%$ of $4 \text{ km}$ is $8 \text{ meters}$.