Percentage Questions in English

(D) To convert a decimal number into a percentage,we multiply the number by $100$.
$3.5 \times 100 = 350$.
Therefore,$3.5$ expressed as a percentage is $350 \%$.

(D) To find $15$ percent of Rs. $34$,we use the formula:
$\text{Value} = \frac{\text{Percentage}}{100} \times \text{Total Value}$
$\text{Value} = \frac{15}{100} \times 34$
$\text{Value} = 0.15 \times 34$
$\text{Value} = 5.10$
Therefore,$15$ percent of Rs. $34$ is Rs. $5.10$.

(B) Let the missing value be $x$.
Given equation: $\frac{88}{100} \times 370 + \frac{24}{100} \times 210 - x = 118$
Calculate the first part: $0.88 \times 370 = 325.6$
Calculate the second part: $0.24 \times 210 = 50.4$
Substitute these values into the equation: $325.6 + 50.4 - x = 118$
$376 - x = 118$
$x = 376 - 118$
$x = 258$

(C) To solve the expression $860 \% \text{ of } 50 + 50 \% \text{ of } 860$,we use the property of percentages: $a \% \text{ of } b = b \% \text{ of } a$.
Step $1$: Calculate $860 \% \text{ of } 50 = \frac{860}{100} \times 50 = 8.6 \times 50 = 430$.
Step $2$: Calculate $50 \% \text{ of } 860 = \frac{50}{100} \times 860 = 0.5 \times 860 = 430$.
Step $3$: Add the results: $430 + 430 = 860$.
Alternatively,using the distributive property: $\frac{860}{100} \times 50 + \frac{50}{100} \times 860 = \frac{860 \times 50}{100} + \frac{50 \times 860}{100} = 430 + 430 = 860$.

(B) First,calculate the value of the given expression:
$60 \% \text{ of } 264 = \frac{60}{100} \times 264 = 0.6 \times 264 = 158.4$.
Now,evaluate the options:
$A) 10 \% \text{ of } 44 = 4.4$
$B) 15 \% \text{ of } 1056 = \frac{15}{100} \times 1056 = 0.15 \times 1056 = 158.4$
$C) 30 \% \text{ of } 132 = 0.3 \times 132 = 39.6$
$D) 17 \% \text{ of } 544 = 0.17 \times 544 = 92.48$
Since $158.4 = 158.4$,the correct option is $B$.

(A) To find the pass percentage,we use the formula: $\text{Pass Percentage} = \left( \frac{\text{Number of candidates passed}}{\text{Total number of candidates}} \right) \times 100$.
Given,$\text{Number of candidates passed} = 252$.
Total number of candidates = $270$.
$\text{Pass Percentage} = \left( \frac{252}{270} \right) \times 100$.
Simplifying the fraction: $\frac{252}{270} = \frac{252 \div 18}{270 \div 18} = \frac{14}{15}$.
Now,$\frac{14}{15} \times 100 = \frac{1400}{15} = \frac{280}{3} = 93.33\%$ or $93 \frac{1}{3} \%$.

(B) To find the percentage,we use the formula: $\text{Percentage} = \left( \frac{\text{Part}}{\text{Whole}} \right) \times 100$.
Here,the part is Rs. $1987.50$ and the whole is Rs. $2650$.
$\text{Percentage} = \left( \frac{1987.50}{2650} \right) \times 100$.
$\text{Percentage} = \frac{198750}{2650} = 75 \%$.

(B) day consists of $24$ hours.
To find the percentage of $3$ hours in a day,we use the formula:
$\text{Percentage} = \left( \frac{\text{Part}}{\text{Whole}} \right) \times 100$
$\text{Percentage} = \left( \frac{3}{24} \right) \times 100$
$\text{Percentage} = \frac{1}{8} \times 100 = 12.5 \%$

(C) To find the amount of pure acid in the solution,we multiply the total volume of the solution by the percentage of acid.
Total volume of solution $= 8$ litres.
Percentage of acid $= 20\%$.
Amount of pure acid $= 20\% \text{ of } 8 \text{ litres}$.
Amount of pure acid $= \frac{20}{100} \times 8 = 0.2 \times 8 = 1.6$ litres.

(B) To find the best percentage,we calculate the value of each fraction as a percentage:
$A) \frac{384}{540} \times 100 \approx 71.11\%$
$B) \frac{425}{500} \times 100 = 85.00\%$
$C) \frac{570}{700} \times 100 \approx 81.43\%$
$D) \frac{480}{660} \times 100 \approx 72.72\%$
Comparing these values,$85.00\%$ is the highest percentage.
Therefore,the correct option is $B$.

(B) First,convert the percentages into fractions:
$0.15 \% = \frac{0.15}{100} = \frac{15}{10000}$
$33 \frac{1}{3} \% = \frac{100}{3} \% = \frac{100}{3 \times 100} = \frac{1}{3}$
Now,calculate the value:
$\text{Value} = \frac{15}{10000} \times \frac{1}{3} \times 10000$
$\text{Value} = \frac{15}{3} = 5$
Therefore,the result is $Rs. 5$.

(D) Let the missing value be $x$.
Given equation: $\frac{45}{100} \times 1500 + \frac{35}{100} \times 1700 = \frac{x}{100} \times 3175$
Calculate the first part: $45 \times 15 = 675$
Calculate the second part: $35 \times 17 = 595$
Sum of the two parts: $675 + 595 = 1270$
Now,equate to the right side: $1270 = \frac{x}{100} \times 3175$
Solve for $x$: $x = \frac{1270 \times 100}{3175}$
$x = \frac{127000}{3175} = 40$
Therefore,the missing value is $40$.

(B) Let the total worth of the cloth sold be $x$.
Given that the agent receives a commission of $2.5 \%$ on the sales.
According to the problem,$2.5 \%$ of $x = 12.50$.
$x \times \frac{2.5}{100} = 12.50$
$x = \frac{12.50 \times 100}{2.5}$
$x = \frac{1250}{2.5}$
$x = 500$
Therefore,the worth of the cloth sold is Rs. $500$.

(B) Let the total worth of the house be $x$.
According to the problem,$\frac{2}{7}$ of $x = 2800$.
$x = 2800 \times \frac{7}{2}$.
$x = 1400 \times 7$.
$x = 9800$.
Therefore,the total worth of the house is Rs. $9800$.

(D) Let the number be $x$.
Given that $35 \%$ of $x = 175$.
$0.35 \times x = 175$
$x = \frac{175}{0.35} = 500$.
Now,we need to find what percent of $175$ is $500$.
Required percentage $= (\frac{500}{175}) \times 100$.
$= (\frac{20}{7}) \times 100 = \frac{2000}{7} \approx 285.71 \%$.

(D) Let the number be $x$.
According to the problem,$50 \%$ of $x$ minus $35 \%$ of $x$ equals $12$.
$\frac{50}{100}x - \frac{35}{100}x = 12$
$\frac{15}{100}x = 12$
$x = \frac{12 \times 100}{15}$
$x = \frac{1200}{15}$
$x = 80$
Therefore,the number is $80$.

(C) number's square ends in $1$ if the number itself ends in $1$ or $9$ (since $1^2 = 1$ and $9^2 = 81$).
In the range $1$ to $70$,the numbers ending in $1$ are: $1, 11, 21, 31, 41, 51, 61$ (Total $7$ numbers).
The numbers ending in $9$ are: $9, 19, 29, 39, 49, 59, 69$ (Total $7$ numbers).
Total count of such numbers $= 7 + 7 = 14$.
The total number of integers from $1$ to $70$ is $70$.
Required percentage $= (14 / 70) \times 100 = 0.2 \times 100 = 20 \%$.

(C) Let the number be $x$.
According to the problem,$75 \%$ of $x$ added to $75$ equals $x$.
Mathematically,this is expressed as: $\frac{75}{100}x + 75 = x$.
Simplify the fraction: $\frac{3}{4}x + 75 = x$.
Rearrange the equation to solve for $x$: $x - \frac{3}{4}x = 75$.
$\frac{1}{4}x = 75$.
Multiply both sides by $4$: $x = 75 \times 4 = 300$.
Therefore,the number is $300$.

(D) Let the two numbers be $x$ and $y$.
Given that the sum of the two numbers is $x + y = 2490$.
Also,$6.5\%$ of $x$ is equal to $8.5\%$ of $y$,which can be written as:
$\frac{6.5}{100} x = \frac{8.5}{100} y$
$6.5x = 8.5y$
$x = \frac{8.5}{6.5} y = \frac{17}{13} y$
Substitute the value of $x$ in the sum equation:
$\frac{17}{13} y + y = 2490$
$\frac{17y + 13y}{13} = 2490$
$\frac{30y}{13} = 2490$
$y = \frac{2490 \times 13}{30} = 83 \times 13 = 1079$
Now,find $x$:
$x = 2490 - 1079 = 1411$
Thus,the two numbers are $1411$ and $1079$.

(B) Let the first number be $x$ and the second number be $y$.
According to the problem,when $x$ is divided by $12$,the result is $\frac{1}{4}$ of $y$.
So,$\frac{x}{12} = \frac{1}{4}y$.
Multiplying both sides by $12$,we get $x = \frac{12}{4}y = 3y$.
This implies $\frac{x}{y} = \frac{3}{1}$,so $x = 3y$.
The percentage by which the first number $(x)$ is greater than the second number $(y)$ is given by the formula:
$\text{Percentage increase} = \frac{x - y}{y} \times 100 \%$.
Substituting $x = 3y$:
$\text{Percentage increase} = \frac{3y - y}{y} \times 100 \% = \frac{2y}{y} \times 100 \% = 2 \times 100 \% = 200 \%$.

(C) Let the two numbers be $x$ and $y$.
According to the problem,one number is $80 \%$ of the other:
$y = \frac{80}{100} x = \frac{4}{5} x$
This implies $\frac{x}{y} = \frac{5}{4}$. Let $x = 5k$ and $y = 4k$ for some constant $k$.
The sum of their squares is given as $656$:
$x^2 + y^2 = 656$
$(5k)^2 + (4k)^2 = 656$
$25k^2 + 16k^2 = 656$
$41k^2 = 656$
$k^2 = \frac{656}{41} = 16$
$k = 4$
Substituting the value of $k$ back into the expressions for $x$ and $y$:
$x = 5 \times 4 = 20$
$y = 4 \times 4 = 16$
Thus,the numbers are $16$ and $20$.

(A) Initial salary = $Rs. 7200$
Final salary = $Rs. 8100$
Increase in salary = $8100 - 7200 = Rs. 900$
Percentage increase = $\frac{\text{Increase}}{\text{Original Salary}} \times 100$
Percentage increase = $\frac{900}{7200} \times 100 = \frac{1}{8} \times 100 = 12.5\%$

(D) Let $B$'s salary be $= 100$.
Since $A$'s salary is $20 \%$ less than $B$'s salary,$A$'s salary $= 100 - 20 = 80$.
Now,we need to find how much more $B$'s salary is compared to $A$'s salary.
Difference $= 100 - 80 = 20$.
Percentage increase $= \left( \frac{\text{Difference}}{A's \text{ salary}} \right) \times 100$.
Percentage increase $= \left( \frac{20}{80} \right) \times 100 = \frac{1}{4} \times 100 = 25 \%$.

(A) Let the original price of petrol be $100$ per unit and the original consumption be $100$ units.
Original expenditure $= 100 \times 100 = 10000$.
After a $20 \%$ increase,the new price $= 120$ per unit.
Let the new consumption be $x$ units.
Since the expenditure remains unchanged,$120 \times x = 10000$.
$x = \frac{10000}{120} = \frac{250}{3} = 83.33 \text{ units}$.
Decrease in consumption $= 100 - 83.33 = 16.67 \text{ units}$.
Percentage decrease $= \frac{16.67}{100} \times 100 = 16.67 \%$,which is equal to $16 \frac{2}{3} \%$.

(B) Initial price $= 100$.
After the first increase of $10 \%$,the new price $= 100 + (10 \% \text{ of } 100) = 100 + 10 = 110$.
After the second increase of $10 \%$,the final price $= 110 + (10 \% \text{ of } 110) = 110 + 11 = 121$.
Total increase in price $= 121 - 100 = 21$.

(A) The total value of the goods sold is $Rs. 15000$.
The commission rate is $12 \frac{1}{2} \% = 12.5 \% = \frac{12.5}{100} = \frac{25}{200} = \frac{1}{8}$.
To find the commission amount,multiply the total value by the commission rate:
$\text{Commission} = 15000 \times \frac{1}{8} = 1875$.
Therefore,the commission amount is $Rs. 1875$.

(D) Original monthly income = $Rs. 5000$.
Increase percentage = $30\%$.
New monthly income = $\text{Original income} \times (1 + \frac{\text{Increase percentage}}{100})$.
New monthly income = $5000 \times (1 + \frac{30}{100}) = 5000 \times 1.3 = 6500$.
Therefore,the new monthly income is $Rs. 6500$.

(B) Original price of the article $= 15000 \ Rs.$
Decrease percentage $= 8 \%$.
Decrease in price $= 15000 \times \frac{8}{100} = 1200 \ Rs.$
New price $= \text{Original price} - \text{Decrease in price} = 15000 - 1200 = 13800 \ Rs.$
Alternatively, New price $= 15000 \times (1 - 0.08) = 15000 \times 0.92 = 13800 \ Rs.$

(D) Let the original amount of money be $x$.
After losing $20 \%$ of his money,the remaining amount is $x - 0.20x = 0.80x$.
He spends $25 \%$ of the remainder,so the amount left is $75 \%$ of the remainder.
Remaining amount $= 0.75 \times (0.80x) = 0.60x$.
Given that the remaining amount is $Rs. 480$,we have $0.60x = 480$.
$x = \frac{480}{0.60} = \frac{48000}{60} = 800$.
Therefore,the original amount of money he had was $Rs. 800$.

(D) To find the value,we convert the percentages into fractions and multiply them by the given number:
$15 \% = \frac{15}{100}$
$10 \% = \frac{10}{100}$
$20 \% = \frac{20}{100}$
Calculation:
$\frac{15}{100} \times \frac{10}{100} \times \frac{20}{100} \times 1000$
$= \frac{15 \times 10 \times 20 \times 1000}{100 \times 100 \times 100}$
$= \frac{3000000}{1000000}$
$= 3$

(B) Let the original fraction be $\frac{x}{y}$.
According to the problem,the numerator is increased by $120 \%$,so the new numerator is $x + 1.20x = 2.20x$.
The denominator is increased by $350 \%$,so the new denominator is $y + 3.50y = 4.50y$.
The new fraction is given as $\frac{2.20x}{4.50y} = \frac{11}{27}$.
Simplifying the equation: $\frac{22x}{45y} = \frac{11}{27}$.
Solving for $\frac{x}{y}$: $\frac{x}{y} = \frac{11}{27} \times \frac{45}{22} = \frac{1}{3} \times \frac{5}{2} = \frac{5}{6}$.
Converting to decimal: $\frac{5}{6} \approx 0.833$.

(C) Let the two numbers be $x$ and $y$.
According to the problem,$40 \%$ of $x$ is equal to $\frac{2}{3}$ of $y$.
This can be written as: $\frac{40}{100} x = \frac{2}{3} y$.
Simplifying the fraction $\frac{40}{100}$,we get $\frac{2}{5} x = \frac{2}{3} y$.
To find the ratio $x:y$,we rearrange the equation: $\frac{x}{y} = \frac{2}{3} \times \frac{5}{2}$.
Canceling the $2$ from the numerator and denominator,we get $\frac{x}{y} = \frac{5}{3}$.
Therefore,the ratio of the first number to the second number is $5:3$.

(B) Total runs scored by the batsman = $110$.
Runs scored through boundaries = $3 \times 4 = 12$.
Runs scored through sixes = $8 \times 6 = 48$.
Total runs scored through boundaries and sixes = $12 + 48 = 60$.
Runs scored by running between the wickets = $110 - 60 = 50$.
Percentage of runs scored by running between the wickets = $\frac{50}{110} \times 100 = \frac{500}{11} = 45 \frac{5}{11} \%$.

(B) Given that $\frac{50}{100}(x-y) = \frac{30}{100}(x+y)$.
Simplifying the equation: $50(x-y) = 30(x+y)$.
Dividing by $10$: $5(x-y) = 3(x+y)$.
Expanding the terms: $5x - 5y = 3x + 3y$.
Rearranging the terms to solve for $x$ and $y$: $5x - 3x = 3y + 5y$,which gives $2x = 8y$.
Therefore,$\frac{y}{x} = \frac{2}{8} = \frac{1}{4}$.
To find the percentage of $x$ that $y$ represents,we calculate $\frac{y}{x} \times 100 = \frac{1}{4} \times 100 = 25 \%$.

(B) The given length is $81.472 \, km$.
To express this with three significant digits,we round it to $81.5 \, km$.
The absolute error is $|81.5 - 81.472| = 0.028 \, km$.
The percentage error is calculated as $\frac{\text{Absolute Error}}{\text{Original Value}} \times 100$.
Percentage error $= \frac{0.028}{81.472} \times 100 \approx 0.03436 \%$.
Rounding to the nearest given option,the percentage error is $0.034 \%$.

(A) Let the two numbers be $x$ and $y$,where $x > y$.
According to the problem,the difference between the numbers is $x - y = 1600$.
Also,$7.5\%$ of $x = 12.5\%$ of $y$.
This can be written as $\frac{7.5}{100} x = \frac{12.5}{100} y$.
Multiplying both sides by $100$,we get $7.5x = 12.5y$.
Dividing both sides by $2.5$,we get $3x = 5y$,which implies $x = \frac{5}{3}y$.
Substitute $x = \frac{5}{3}y$ into the first equation: $\frac{5}{3}y - y = 1600$.
$\frac{2y}{3} = 1600$.
$2y = 4800$,so $y = 2400$.
Now,$x = 1600 + 2400 = 4000$.
Thus,the two numbers are $4000$ and $2400$.

(C) Let the number be $x$.
According to the problem,$65\%$ of $x$ is $21$ less than $\frac{4}{5}$ of $x$.
This can be written as the equation: $\frac{65}{100}x = \frac{4}{5}x - 21$.
Rearranging the terms,we get: $\frac{4}{5}x - \frac{65}{100}x = 21$.
Simplifying the fraction $\frac{65}{100}$ gives $\frac{13}{20}$.
So,$\frac{4}{5}x - \frac{13}{20}x = 21$.
To subtract,find a common denominator,which is $20$: $\frac{16}{20}x - \frac{13}{20}x = 21$.
$\frac{3}{20}x = 21$.
$x = 21 \times \frac{20}{3} = 7 \times 20 = 140$.
Thus,the number is $140$.

(C) Let the total number of meters examined be $x$.
Given that $0.08 \%$ of the total meters are rejected.
So,$0.08 \%$ of $x = 2$.
$\frac{0.08}{100} \times x = 2$.
$\frac{8}{10000} \times x = 2$.
$x = \frac{2 \times 10000}{8}$.
$x = \frac{20000}{8} = 2500$.
Therefore,the inspector must examine $2500$ meters to reject $2$ defective ones.

(A) To find the greatest number,we convert all values into decimal form:
$1$. $16 \frac{2}{3} = 16 + 0.666... = 16.666...$
$2$. $\frac{2}{15} \approx 0.1333...$
$3$. $\frac{1}{11} \approx 0.0909...$
$4$. $0.17 = 0.17$
Comparing these values: $16.666 > 0.17 > 0.1333 > 0.0909$.
Thus,the greatest number is $16 \frac{2}{3}$.

(C) First,convert the percentages into fractions:
$16 \frac{2}{3} \% = \frac{50}{3} \times \frac{1}{100} = \frac{1}{6}$
$33 \frac{1}{3} \% = \frac{100}{3} \times \frac{1}{100} = \frac{1}{3}$
Now,substitute these fractions into the expression:
$= (\frac{1}{6} \times 600 \, \text{gm}) - (\frac{1}{3} \times 180 \, \text{gm})$
$= 100 \, \text{gm} - 60 \, \text{gm}$
$= 40 \, \text{gm}$

(B) Let the total marks be $x$.
Passing marks required $= 35 \% \text{ of } x$.
Rishu obtained $216$ marks and failed by $5 \%$,which means his marks are $5 \%$ less than the passing marks.
So,Rishu's percentage $= 35 \% - 5 \% = 30 \%$.
According to the problem,$30 \% \text{ of } x = 216$.
$\frac{30}{100} \times x = 216$.
$x = \frac{216 \times 100}{30}$.
$x = 7.2 \times 100 = 720$.
Therefore,the total marks were $720$.

(D) Let the original price of petrol be $P$ and the original consumption be $C$. The original expenditure is $E = P \times C$.
If the price increases by $20 \%$,the new price $P' = P + 0.20P = 1.20P = \frac{6}{5}P$.
To keep the expenditure $E$ constant,let the new consumption be $C'$.
$P \times C = P' \times C'$
$P \times C = (\frac{6}{5}P) \times C'$
$C' = \frac{5}{6}C$.
The reduction in consumption (travel) is $C - C' = C - \frac{5}{6}C = \frac{1}{6}C$.
Percentage reduction $= (\frac{1/6}{1}) \times 100 = 16.67 \%$.

(A) Let the initial length be $L$ and breadth be $B$. The initial area is $A = L \times B$.
Let the percentage increase in length be $m = 20 \%$ and the percentage increase in breadth be $n = x \%$.
The total percentage increase in area is given by the formula: $\text{Total Increase} = m + n + \frac{m \times n}{100}$.
Given that the total increase in area is $50 \%$,we have:
$20 + x + \frac{20x}{100} = 50$
$x + \frac{x}{5} = 50 - 20$
$\frac{6x}{5} = 30$
$x = \frac{30 \times 5}{6} = 25 \%$.
Thus,the breadth was increased by $25 \%$.

(B) Let Amit's salary be $A$ and Rajiv's salary be $R = 30,000$.
Aditya's salary $(Ad)$ is $120\%$ of Rajiv's salary:
$Ad = 1.20 \times 30,000 = 36,000$.
We are given that Aditya's salary is $80\%$ of Amit's salary $(A)$:
$36,000 = 0.80 \times A$.
Solving for $A$:
$A = \frac{36,000}{0.80} = \frac{360,000}{8} = 45,000$.
Thus,Amit's salary is $45,000$.

(B) Let $H$ be the set of students who passed in Hindi and $S$ be the set of students who passed in Sanskrit.
Given: $n(H) = 60 \%$,$n(S) = 45 \%$,and $n(H \cap S) = 25 \%$.
The percentage of students who passed in at least one subject is given by the formula: $n(H \cup S) = n(H) + n(S) - n(H \cap S)$.
$n(H \cup S) = 60 \% + 45 \% - 25 \% = 80 \%$.
The percentage of students who failed in both subjects is the complement of those who passed in at least one subject.
Percentage failing in both $= 100 \% - n(H \cup S) = 100 \% - 80 \% = 20 \%$.

(C) Let the required quantity of the ore be $x \text{ kg}$.
Given that the ore contains $23 \%$ copper.
According to the problem, $23 \%$ of $x = 69 \text{ kg}$.
$\frac{23}{100} \times x = 69$
$x = 69 \times \frac{100}{23}$
$x = 3 \times 100 = 300 \text{ kg}$.
Therefore, the quantity of ore required is $300 \text{ kg}$.

(B) Let the initial consumption be $x$ $kg$ and the initial price be $Rs. 26$ per $kg$. The initial expenditure is $26x$.
To keep the expenditure fixed at $26x$ with the new price of $Rs. 30$ per $kg$,let the new consumption be $y$ $kg$.
$30y = 26x \implies y = \frac{26}{30}x = \frac{13}{15}x$.
The reduction in consumption is $x - y = x - \frac{13}{15}x = \frac{2}{15}x$.
The percentage reduction is $\frac{\text{Reduction}}{\text{Initial Consumption}} \times 100 = \frac{\frac{2}{15}x}{x} \times 100 = \frac{2}{15} \times 100 = \frac{40}{3} = 13 \frac{1}{3} \%$.

(C) Let the initial salary of Aditya be $100 \text{ Rs}$.
After an increase of $40 \%$,the new salary becomes $100 + (40 \% \text{ of } 100) = 140 \text{ Rs}$.
Now,this salary is decreased by $25 \%$. The decrease amount is $25 \% \text{ of } 140 = 0.25 \times 140 = 35 \text{ Rs}$.
The final salary after the decrease is $140 - 35 = 105 \text{ Rs}$.
The net effect on the salary is $105 - 100 = 5 \text{ Rs}$.
Since the final value is greater than the initial value,there is a $5 \% \text{ increase}$.

(D) Number of students aged $8$ years $= 48$.
Number of students above $8$ years of age $= 48 \times \frac{2}{3} = 32$.
Total number of students aged $8$ years or above $= 48 + 32 = 80$.
Since $20 \%$ of students are below $8$ years of age,the remaining $80 \%$ of students are $8$ years of age or above.
Let the total number of students be $x$.
$80 \% \text{ of } x = 80$.
$0.80 \times x = 80$.
$x = \frac{80}{0.80} = 100$.
Therefore,the total number of students in the school is $100$.

(D) Let the number be $x$.
The correct result should have been $x \times \frac{5}{3}$.
The actual result obtained is $x \times \frac{3}{5}$.
To simplify,let the number be the least common multiple of $3$ and $5$,which is $15$.
Correct result $= 15 \times \frac{5}{3} = 25$.
Wrong result $= 15 \times \frac{3}{5} = 9$.
Error $= 25 - 9 = 16$.
Percentage error $= \frac{\text{Error}}{\text{Correct result}} \times 100 = \frac{16}{25} \times 100 = 64 \%$.