Percentage Questions in English

(D) Let the total income be $x$.
Amount given to wife $= 0.40x$.
Remaining amount $= x - 0.40x = 0.60x$.
Amount given to son $= 25 \%$ of $0.60x = 0.25 \times 0.60x = 0.15x$.
Remaining amount after son $= 0.60x - 0.15x = 0.45x$.
Amount spent on education $= 60 \%$ of $0.45x = 0.60 \times 0.45x = 0.27x$.
Remaining amount with him $= 0.45x - 0.27x = 0.18x$.
Given that $0.18x = 2700$.
$x = \frac{2700}{0.18} = 15000$.
Amount given to wife $= 40 \%$ of $15000 = 0.40 \times 15000 = ₹ 6000$.

(C) Let the total number of candidates be $100$.
Percentage of candidates who passed in English $= 70 \%$.
Percentage of candidates who failed in English $= 100 \% - 70 \% = 30 \%$.
Percentage of candidates who passed in Mathematics $= 80 \%$.
Percentage of candidates who failed in Mathematics $= 100 \% - 80 \% = 20 \%$.
Percentage of candidates who failed in both subjects $= 10 \%$.
Using the principle of inclusion-exclusion for failures:
Percentage of candidates who failed in at least one subject $= (\text{Failed in English}) + (\text{Failed in Mathematics}) - (\text{Failed in both})$
$= 30 \% + 20 \% - 10 \% = 40 \%$.
Therefore,the percentage of candidates who passed in both subjects $= 100 \% - 40 \% = 60 \%$.
Given that $60 \%$ of the total candidates $= 144$.
Let the total number of candidates be $x$.
$0.60 \times x = 144$
$x = \frac{144}{0.60} = 240$.
Thus,the total number of candidates is $240$.

(D) Let the total number of voters be $100x$.
Number of voters who did not cast their votes $= 8\% \text{ of } 100x = 8x$.
Number of votes polled $= 100x - 8x = 92x$.
The winner obtained $48\% \text{ of the total votes} = 48x$.
Number of votes obtained by the loser $= 92x - 48x = 44x$.
The difference between the winner and the loser $= 48x - 44x = 4x$.
Given that the difference is $1100$ votes,we have $4x = 1100$.
$x = \frac{1100}{4} = 275$.
Total number of voters $= 100x = 100 \times 275 = 27500$.

(A) The area of a regular hexagon with side length $s$ is given by $A = \frac{3\sqrt{3}}{2}s^2$.
Since the area is proportional to the square of the side length $(A \propto s^2)$,if the side length increases by $x \%$,the percentage increase in area is given by the formula $(2x + \frac{x^2}{100}) \%$.
Here,$x = 7$.
Percentage increase in area $= (2 \times 7 + \frac{7^2}{100}) \% = (14 + \frac{49}{100}) \% = (14 + 0.49) \% = 14.49 \%$.

(D) Let the monthly income of Manoj be $₹ 100$.
Since Ashok's income is $10 \%$ less than Manoj's,Ashok's income $= 100 - (10 \% \text{ of } 100) = ₹ 90$.
Anil's income is $30 \%$ more than Ashok's,so Anil's income $= 90 + (30 \% \text{ of } 90) = 90 + 27 = ₹ 117$.
The difference between the monthly income of Anil and Manoj is $117 - 100 = 17$ units.
Given that the difference is $₹ 1360$,we have $17 \text{ units} = ₹ 1360$.
Therefore,$1 \text{ unit} = \frac{1360}{17} = ₹ 80$.
Ashok's monthly income is $90 \text{ units}$,so Ashok's income $= 90 \times 80 = ₹ 7200$.

(A) Let the number be $x$.
According to the problem,$75 \%$ of $x$ is $380$ more than $35 \%$ of $x$.
So,the equation is: $0.75x = 0.35x + 380$.
Subtracting $0.35x$ from both sides,we get: $(0.75 - 0.35)x = 380$.
$0.40x = 380$.
We need to find $20 \%$ of the number,which is $0.20x$.
Since $0.40x = 380$,we can divide both sides by $2$ to get $0.20x = 380 / 2$.
$0.20x = 190$.
Therefore,$20 \%$ of the number is $190$.

(D) Let the original length be $L$ and the original breadth be $B$. The original perimeter is $P_1 = 2(L + B)$.
After the increase,the new length $L' = 1.3L$ and the new breadth $B' = 1.2B$.
The new perimeter is $P_2 = 2(1.3L + 1.2B) = 2.6L + 2.4B$.
The percentage increase in perimeter is given by $\frac{P_2 - P_1}{P_1} \times 100 = \frac{(2.6L + 2.4B) - 2(L + B)}{2(L + B)} \times 100$.
This simplifies to $\frac{0.6L + 0.4B}{2(L + B)} \times 100 = \frac{0.3L + 0.2B}{L + B} \times 100$.
Since the ratio of $L$ to $B$ is not provided,the exact percentage increase cannot be determined. Therefore,the data is inadequate.

(B)
1. Initial volume of solution $= 10$ litres.  
2. Concentration of salt $= 10\%$.  
3. Amount of salt $= 10\% \text{ of } 10 = 0.10 \times 10 = 1$ litre (or unit of mass).  
4. After evaporation of $2$ litres of water, the new volume of solution  
$= 10 - 2 = 8$ litres.  
5. The amount of salt remains constant at $1$ litre.  
6. New percentage of salt  
$= \left( \frac{\text{Amount of salt}}{\text{New volume of solution}} \right) \times 100$  
$= \left( \frac{1}{8} \right) \times 100 = 12.5\%$.

(B) Let $P(A)$ be the probability that $A$ speaks the truth and $P(B)$ be the probability that $B$ speaks the truth.
Given: $P(A) = 0.8$,$P(B) = 0.9$.
Therefore,the probability that $A$ lies is $P(A') = 1 - 0.8 = 0.2$,and the probability that $B$ lies is $P(B') = 1 - 0.9 = 0.1$.
They contradict each other if $A$ speaks the truth and $B$ lies,$OR$ $A$ lies and $B$ speaks the truth.
Required probability $= P(A) \times P(B') + P(A') \times P(B)$
$= (0.8 \times 0.1) + (0.2 \times 0.9)$
$= 0.08 + 0.18 = 0.26$.
Converting to percentage: $0.26 \times 100 = 26 \%$.

(D) Given equation: $\frac{125}{100} \times 320 + \frac{x}{100} \times 125 = 440$
Step $1$: Calculate the first part: $\frac{125}{100} \times 320 = 1.25 \times 320 = 400$.
Step $2$: Substitute this into the equation: $400 + \frac{x}{100} \times 125 = 440$.
Step $3$: Subtract $400$ from both sides: $\frac{x}{100} \times 125 = 440 - 400 = 40$.
Step $4$: Solve for $x$: $x = \frac{40 \times 100}{125} = \frac{4000}{125} = 32$.

(C) Let the total number of students be $100$.
Students who read Hindi $= 70$.
Students who do not read Hindi $= 100 - 70 = 30$.
Number of boys $= 40$.
Number of girls $= 100 - 40 = 60$.
Number of girls who do not read Hindi $= 20 \% \text{ of } 60 = \frac{20}{100} \times 60 = 12$.
Since the total number of students who do not read Hindi is $30$,the number of boys who do not read Hindi $= 30 - 12 = 18$.
Number of boys who read Hindi $= \text{Total boys} - \text{Boys who do not read Hindi} = 40 - 18 = 22$.
Percentage of boys who read Hindi $= \frac{22}{40} \times 100 = 55 \%$.

(A) The area of a triangle is given by $A = \frac{1}{2} \times \text{base} \times \text{height}$.
Let the original base be $b$ and the original height be $h$. The original area is $A_1 = \frac{1}{2}bh$.
After the changes,the new base $b' = b + 0.30b = 1.3b$ and the new height $h' = h - 0.40h = 0.6h$.
The new area is $A_2 = \frac{1}{2} \times (1.3b) \times (0.6h) = 0.78 \times (\frac{1}{2}bh) = 0.78A_1$.
The percentage change in area is $\frac{A_2 - A_1}{A_1} \times 100 = (0.78 - 1) \times 100 = -22 \%$.
Alternatively,using the net percentage change formula: $\text{Net effect} = (x + y + \frac{xy}{100}) \% = (-40 + 30 + \frac{-40 \times 30}{100}) \% = (-10 - 12) \% = -22 \%$.
Thus,the area decreases by $22 \%$.

(B) To evaluate the expression $28 \%$ of $450 + 45 \%$ of $280$,we calculate each part separately:
First,$28 \%$ of $450 = \frac{28}{100} \times 450 = 0.28 \times 450 = 126$.
Next,$45 \%$ of $280 = \frac{45}{100} \times 280 = 0.45 \times 280 = 126$.
Finally,adding the two results: $126 + 126 = 252$.

(C) Let the original price of the product be $P$ and the original quantity sold be $Q$. The original revenue is $R_1 = P \times Q$.
After a $10 \%$ decrease in price,the new price is $P' = P - 0.10P = 0.90P$.
After a $30 \%$ increase in quantity sold,the new quantity is $Q' = Q + 0.30Q = 1.30Q$.
The new revenue is $R_2 = P' \times Q' = (0.90P) \times (1.30Q) = 1.17PQ = 1.17R_1$.
The percentage change in revenue is $\frac{R_2 - R_1}{R_1} \times 100 = \frac{1.17R_1 - R_1}{R_1} \times 100 = 0.17 \times 100 = 17 \%$.
Alternatively,using the net percentage change formula: $\text{Net Change} = a + b + \frac{ab}{100} = -10 + 30 + \frac{(-10)(30)}{100} = 20 - 3 = 17 \%$.

(C) Let the original salary be $100$.
After a reduction of $10 \%$,the new salary becomes $100 - 10 = 90$.
To bring the salary back to $100$,we need an increase of $10$.
The percentage increase required on the reduced salary $(90)$ is calculated as:
$\text{Percentage Increase} = \left( \frac{\text{Increase}}{\text{Reduced Salary}} \right) \times 100$
$= \left( \frac{10}{90} \right) \times 100 = \frac{100}{9} = 11.11 \%$.
Thus,the reduced salary must be raised by $11.11 \%$ to restore the original salary.

(D) Let the initial population at the beginning of the first year be $x$.
After the first year,the population increases by $5 \%$,so the population becomes $x(1 + 0.05) = 1.05x$.
After the second year,the population decreases by $5 \%$ from the population at the end of the first year,so the population becomes $1.05x(1 - 0.05) = 1.05x(0.95)$.
Given that the final population is $9975$,we have the equation: $1.05x \times 0.95 = 9975$.
$x \times (1.05 \times 0.95) = 9975$.
$x \times 0.9975 = 9975$.
$x = \frac{9975}{0.9975} = 10000$.
Therefore,the population at the beginning of the first year was $10000$.

(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 $350 \%$.

(B) Given equation: $88 \%$ of $370 + 24 \%$ of $210 - ? = 118$
Step $1$: Calculate $88 \%$ of $370$.
$\frac{88}{100} \times 370 = 0.88 \times 370 = 325.6$
Step $2$: Calculate $24 \%$ of $210$.
$\frac{24}{100} \times 210 = 0.24 \times 210 = 50.4$
Step $3$: Substitute the values into the equation.
$325.6 + 50.4 - ? = 118$
Step $4$: Simplify the sum.
$376 - ? = 118$
Step $5$: Solve for $?$.
$? = 376 - 118 = 258$

(D) To solve the expression $860\%$ of $50 + 50\%$ of $860$,we calculate each part separately:
First part: $860\% \text{ of } 50 = \frac{860}{100} \times 50 = 8.6 \times 50 = 430$.
Second part: $50\% \text{ of } 860 = \frac{50}{100} \times 860 = 0.5 \times 860 = 430$.
Adding both parts: $430 + 430 = 860$.

(B) To solve the expression $45 \%$ of $750 - 25 \%$ of $480$,we calculate each part separately:
First,calculate $45 \%$ of $750$: $\frac{45}{100} \times 750 = 0.45 \times 750 = 337.50$.
Next,calculate $25 \%$ of $480$: $\frac{25}{100} \times 480 = 0.25 \times 480 = 120$.
Finally,subtract the second result from the first: $337.50 - 120 = 217.50$.

(C) Given equation: $40 \%$ of $1640 + ? = 35 \%$ of $980 + 150 \%$ of $850$
Let the missing value be $x$.
$x = (35 \% \text{ of } 980) + (150 \% \text{ of } 850) - (40 \% \text{ of } 1640)$
Calculate each term:
$35 \% \text{ of } 980 = \frac{35}{100} \times 980 = 343$
$150 \% \text{ of } 850 = \frac{150}{100} \times 850 = 1275$
$40 \% \text{ of } 1640 = \frac{40}{100} \times 1640 = 656$
Substitute the values:
$x = 343 + 1275 - 656$
$x = 1618 - 656$
$x = 962$

(B) To find the value of $60 \%$ of $264$,we calculate:
$0.60 \times 264 = 158.4$
Now,let us check the options:
$A) 10 \% \text{ of } 44 = 0.10 \times 44 = 4.4$
$B) 15 \% \text{ of } 1056 = 0.15 \times 1056 = 158.4$
$C) 32 \% \text{ of } 132 = 0.32 \times 132 = 42.24$
Since $158.4 = 158.4$,the correct option is $B$.

(B) To find what percent $0.01$ is of $0.1$,we use the formula: $\text{Percentage} = (\frac{\text{Part}}{\text{Whole}}) \times 100$.
Substituting the given values: $\text{Percentage} = (\frac{0.01}{0.1}) \times 100$.
Simplifying the fraction: $\frac{0.01}{0.1} = \frac{1}{10} = 0.1$.
Now,multiplying by $100$: $0.1 \times 100 = 10$.
Therefore,$0.01$ is $10\%$ of $0.1$.

(B) To find the percentage,we use the formula: $\text{Percentage} = (\frac{\text{Part}}{\text{Whole}}) \times 100$.
Given,$\text{Part} = 1987.50$ and $\text{Whole} = 2650$.
$\text{Percentage} = (\frac{1987.50}{2650}) \times 100$.
$\text{Percentage} = 0.75 \times 100 = 75\%$.
Therefore,$1987.50$ is $75\%$ of $2650$.

(A) The original price of the goods is $₹ 6650$.
First,calculate the rebate amount: $6 \% \text{ of } 6650 = \frac{6}{100} \times 6650 = ₹ 399$.
Now,calculate the net price after the rebate: $6650 - 399 = ₹ 6251$.
Next,calculate the sales tax at $10 \%$ on the net price: $10 \% \text{ of } 6251 = \frac{10}{100} \times 6251 = ₹ 625.10$.
Finally,the total amount to be paid is the net price plus the sales tax: $6251 + 625.10 = ₹ 6876.10$.

(B) Let the required number be $x$.
According to the problem,$95 \%$ of $x = 4598$.
This can be written as: $\frac{95}{100} \times x = 4598$.
To solve for $x$,multiply both sides by $100$ and divide by $95$:
$x = \frac{4598 \times 100}{95}$.
First,divide $4598$ by $19$ (since $95 = 19 \times 5$):
$4598 \div 19 = 242$.
Now,divide $100$ by $5$:
$100 \div 5 = 20$.
Finally,multiply the results:
$x = 242 \times 20 = 4840$.
Therefore,$4598$ is $95 \%$ of $4840$.

(A) Let the required percentage be $x\%$.
According to the problem,$x\%$ of $360 = 129.6$.
This can be written as: $\frac{x}{100} \times 360 = 129.6$.
Now,solve for $x$:
$x = \frac{129.6 \times 100}{360}$.
$x = \frac{12960}{360}$.
$x = \frac{1296}{36} = 36$.
Therefore,$36\%$ of $360 = 129.6$.

(B) Let the unknown value be $x$.
Given equation: $65 \% \text{ of } x = 20 \% \text{ of } 422.50$
$\frac{65}{100} \times x = \frac{20}{100} \times 422.50$
$x = \frac{20}{100} \times 422.50 \times \frac{100}{65}$
$x = \frac{20 \times 422.50}{65}$
$x = \frac{8450}{65}$
$x = 130$

(B) Let the given expression be $\frac{x}{100} \times 932 + 30 = 309.6$.
Subtract $30$ from both sides: $\frac{x}{100} \times 932 = 309.6 - 30$.
$\frac{x}{100} \times 932 = 279.6$.
Now,solve for $x$: $x = \frac{279.6 \times 100}{932}$.
$x = \frac{27960}{932}$.
$x = 30$.

(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 correct answer is $40$,which is not listed in the options.

(B) Let the total value of the house be $₹ x$.
According to the problem,$\frac{2}{7}$ percent of $x$ is equal to $₹ 2800$.
Mathematically,this is expressed as: $\frac{2}{7} \times \frac{1}{100} \times x = 2800$.
Simplifying the equation: $\frac{2}{700} \times x = 2800$.
Solving for $x$: $x = \frac{2800 \times 700}{2}$.
$x = 1400 \times 700 = 980000$.
Therefore,the worth of the house is $₹ 980000$.

(D) Let the number be $x$.
Given that $20 \%$ of $x = 120$.
This can be written as: $\frac{20}{100} \times x = 120$.
Solving for $x$: $x = \frac{120 \times 100}{20} = 6 \times 100 = 600$.
Now,we need to find $120 \%$ of this number $x$.
$120 \%$ of $600 = \frac{120}{100} \times 600$.
$= 120 \times 6 = 720$.

(D) Let the number be $x$.
According to the problem,$\frac{2}{5} \times \frac{1}{3} \times \frac{3}{7} \times x = 15$.
Simplifying the equation: $\frac{2}{5} \times \frac{1}{7} \times x = 15$.
$\frac{2}{35} \times x = 15$.
$x = 15 \times \frac{35}{2} = \frac{525}{2} = 262.5$.
Now,we need to find $40$ per cent of $x$.
$40\% \text{ of } x = \frac{40}{100} \times 262.5$.
$= 0.4 \times 262.5 = 105$.

(D) Let the number be $x$.
According to the problem,$35 \%$ of $x$ is $12$ less than $50 \%$ of $x$.
$\therefore \frac{50}{100}x - \frac{35}{100}x = 12$
$\frac{15}{100}x = 12$
$x = \frac{12 \times 100}{15}$
$x = 4 \times 20 = 80$
Thus,the number is $80$.

(A) Let the number be $x$.
According to the problem,the number exceeds $16 \%$ of itself by $42$.
Therefore,the equation is: $x - 0.16x = 42$.
$0.84x = 42$.
$x = \frac{42}{0.84}$.
$x = \frac{4200}{84} = 50$.
Thus,the number is $50$.

(A) Given that $y$ is $10 \%$ more than $125.$
So,$y = 125 + (10 \% \text{ of } 125) = 125 + 12.5 = 137.5.$
Given that $x$ is $10 \%$ less than $y.$
So,$x = y - (10 \% \text{ of } y) = 0.9y.$
Substituting the value of $y$:
$x = 0.9 \times 137.5 = 123.75.$
Therefore,$x = 123.75.$

(D) Let the number be $x$.
According to the problem,when $35$ is subtracted from $x$,it becomes $80\%$ of $x$.
$x - 35 = 0.8x$
Rearranging the terms to solve for $x$:
$x - 0.8x = 35$
$0.2x = 35$
$x = \frac{35}{0.2} = 175$
We need to find four-fifths $(4/5)$ of the number $x$:
$\frac{4}{5} \times 175 = 4 \times 35 = 140$.

(D) Let the two numbers be $x$ and $y$.
Given that the sum of the numbers is $2490$,so $x + y = 2490$ (Equation $1$).
Given that $6.5\%$ of $x$ is equal to $8.5\%$ of $y$,so $\frac{6.5}{100}x = \frac{8.5}{100}y$.
This simplifies to $6.5x = 8.5y$,or $x = \frac{8.5}{6.5}y = \frac{17}{13}y$ (Equation $2$).
Substituting the value of $x$ from Equation $2$ into Equation $1$:
$\frac{17}{13}y + y = 2490$
$\frac{17y + 13y}{13} = 2490$
$\frac{30}{13}y = 2490$
$y = \frac{2490 \times 13}{30} = 83 \times 13 = 1079$.
Now,find $x$ using Equation $1$:
$x = 2490 - 1079 = 1411$.
Thus,the two numbers are $1411$ and $1079$.

(D) Let the number be $x$.
Correct value $= x \times \frac{5}{3} = \frac{5x}{3}$.
Incorrect value $= x \times \frac{3}{5} = \frac{3x}{5}$.
Error $= \text{Incorrect value} - \text{Correct value} = \frac{3x}{5} - \frac{5x}{3} = \frac{9x - 25x}{15} = -\frac{16x}{15}$.
Percentage error $= \frac{\text{Error}}{\text{Correct value}} \times 100 = \frac{-16x / 15}{5x / 3} \times 100$.
$= -\frac{16}{15} \times \frac{3}{5} \times 100 = -\frac{16}{25} \times 100 = -64\%$.
The magnitude of the percentage error is $64\%$.

(B) Total increase in population = $2,62,500 - 1,75,000 = 87,500$.
Percentage increase over a decade = $\frac{87,500}{1,75,000} \times 100 = 50 \%$.
Since the time period is $10$ years,the average percentage increase per year = $\frac{50 \%}{10} = 5 \%$ per year.

(A) Given that the sum of $6\%$ of $A$ and $4\%$ of $B$ is $\frac{2}{3}$ of the sum of $6\%$ of $A$ and $8\%$ of $B$.
Mathematically,this can be written as:
$\frac{6}{100}A + \frac{4}{100}B = \frac{2}{3} \left( \frac{6}{100}A + \frac{8}{100}B \right)$
Multiply both sides by $100$ to simplify:
$6A + 4B = \frac{2}{3}(6A + 8B)$
Multiply by $3$ to remove the fraction:
$3(6A + 4B) = 2(6A + 8B)$
$18A + 12B = 12A + 16B$
Rearrange the terms to group $A$ and $B$:
$18A - 12A = 16B - 12B$
$6A = 4B$
Therefore,the ratio $A:B$ is:
$\frac{A}{B} = \frac{4}{6} = \frac{2}{3}$

(A) Let the larger number be $x$ and the smaller number be $y$.
Given that the smaller number $y = 20$.
According to the problem,the difference between the two numbers is $20 \%$ of the larger number $x$.
So,$x - y = 0.20x$.
Substituting $y = 20$ into the equation:
$x - 20 = 0.20x$
$x - 0.20x = 20$
$0.80x = 20$
$x = \frac{20}{0.80}$
$x = \frac{2000}{80} = 25$.
Therefore,the larger number is $25$.

(C) Let the total number of people eligible to vote be $x$.
Number of people between $18$ and $21$ years of age $= 8 \% \text{ of } x = \frac{8}{100}x$.
Number of people between $18$ and $21$ years of age who actually voted $= 85 \% \text{ of } (\frac{8}{100}x)$.
$= \frac{85}{100} \times \frac{8}{100}x = \frac{680}{10000}x = \frac{6.8}{100}x$.
Therefore,the number of persons who actually voted is $6.8 \%$ of the total people eligible to vote.

(D) Let the total number of students be $x$.
Number of students below $8$ years of age $= 20\% \text{ of } x = 0.2x$.
Number of students of $8$ years of age $= 48$.
Number of students above $8$ years of age $= \frac{2}{3} \times 48 = 32$.
The sum of students in all three categories equals the total number of students $x$:
$0.2x + 48 + 32 = x$
$0.2x + 80 = x$
$80 = x - 0.2x$
$80 = 0.8x$
$x = \frac{80}{0.8} = 100$.
Therefore,the total number of students in the school is $100$.

(B) Let the amount paid to tailor $X$ be $x$ and the amount paid to tailor $Y$ be $y$.
According to the problem,the total payment is $x + y = 550$.
It is given that $X$ is paid $120$ percent of the sum paid to $Y$,so $x = 1.2y$.
Substituting the value of $x$ in the first equation:
$1.2y + y = 550$
$2.2y = 550$
$y = \frac{550}{2.2} = 250$.
Therefore,$Y$ is paid ₹ $250$ per week.

(B) Given that $\frac{x}{100} \times y = 100$ (Equation $1$)
Given that $\frac{y}{100} \times z = 200$ (Equation $2$)
From Equation $1$,we have $xy = 10000$,so $y = \frac{10000}{x}$.
Substitute $y$ into Equation $2$:
$\frac{10000}{x} \times \frac{z}{100} = 200$
$\frac{100z}{x} = 200$
$100z = 200x$
$z = 2x$

(D) Given that $A = x \%$ of $y = \frac{x}{100} \times y = \frac{xy}{100}$.
Given that $B = y \%$ of $x = \frac{y}{100} \times x = \frac{yx}{100}$.
Since $xy = yx$,it follows that $\frac{xy}{100} = \frac{yx}{100}$.
Therefore,$A = B$.

(B) Given that $x = \frac{80}{100} y = 0.8y$.
We need to find $P$ such that $P \%$ of $2x = y$.
This can be written as $\frac{P}{100} \times 2x = y$.
Substituting $x = 0.8y$ into the equation:
$\frac{P}{100} \times 2(0.8y) = y$.
$\frac{P}{100} \times 1.6y = y$.
Dividing both sides by $y$ (assuming $y \neq 0$):
$\frac{P \times 1.6}{100} = 1$.
$P = \frac{100}{1.6} = \frac{1000}{16} = 62.5$.
Thus,$y$ is $62.5 \%$ or $62 \frac{1}{2} \%$ of $2x$.

(C) Given that $x = \frac{90}{100} \times y$.
Let $P \%$ of $x = y$.
This implies $\frac{P}{100} \times x = y$.
From the first equation,we have $\frac{y}{x} = \frac{100}{90} = \frac{10}{9}$.
Substituting this into the second equation: $P = 100 \times \frac{y}{x}$.
$P = 100 \times \frac{10}{9} = \frac{1000}{9} = 111 \frac{1}{9} \%$.
Therefore,$y$ is $111 \frac{1}{9} \%$ of $x$.

(A) First,calculate the marks obtained in each test:
Test $1$: $90\%$ of $100 = 90$ marks.
Test $2$: $60\%$ of $150 = 90$ marks.
Test $3$: $54\%$ of $200 = 108$ marks.
Total marks obtained $= 90 + 90 + 108 = 288$.
Total maximum marks $= 100 + 150 + 200 = 450$.
Aggregate percentage $= (\text{Total marks obtained} / \text{Total maximum marks}) \times 100$.
Aggregate percentage $= (288 / 450) \times 100 = 64\%$.