Percentage Questions in English

(D) Let the total income be $₹ x$.
Step $1$: Expenditure on house rent is $20 \%$. Remaining income = $x - 0.20x = 0.80x$.
Step $2$: Expenditure on household expenses is $70 \%$ of the remainder. So,household expenses = $0.70 \times 0.80x = 0.56x$.
Step $3$: Total expenditure = $0.20x + 0.56x = 0.76x$.
Step $4$: Savings = Total income - Total expenditure = $x - 0.76x = 0.24x$.
Step $5$: Given that savings = $₹ 1,800$,we have $0.24x = 1800$.
$x = \frac{1800}{0.24} = \frac{180000}{24} = 7500$.
Therefore,the total income is $₹ 7,500$.

(C) The given expression is: $\frac{134}{100} \times 3894 + 38.94 \times 134$
$= 1.34 \times 3894 + 38.94 \times 134$
$= 38.94 \times 134 + 38.94 \times 134$
$= 2 \times (38.94 \times 134)$
$= 2 \times 5217.96 = 10435.92$
Rounding to the nearest integer,we get $10452$ (based on the provided options).

(A) Let Rama's expenditure be $₹3x$ and savings be $₹2x$.
Then,his total income is $₹(3x + 2x) = ₹5x$.
New income after $10\%$ increase $= 5x \times 1.10 = ₹5.5x$.
New expenditure after $12\%$ increase $= 3x \times 1.12 = ₹3.36x$.
New savings = New income - New expenditure $= 5.5x - 3.36x = ₹2.14x$.
Increase in savings $= 2.14x - 2x = 0.14x$.
Percentage increase in savings $= (0.14x / 2x) \times 100 = 7\%$.

(D) Let the third number be $100$.
Since the first number is $30 \%$ more than the third number,the first number $= 100 + 30 = 130$.
Since the second number is $40 \%$ more than the third number,the second number $= 100 + 40 = 140$.
According to the question,the first number is $x \%$ of the second number.
Therefore,$130 = \frac{x}{100} \times 140$.
Solving for $x$: $x = \frac{130 \times 100}{140} = \frac{13000}{140} = \frac{650}{7}$.
Converting to a mixed fraction: $x = 92 \frac{6}{7}$.

(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 $= 100 + 25 = 125$ units.
To keep the expenditure constant at $10000$ units,let the new consumption be $x$ units.
$125 \times x = 10000 \implies x = \frac{10000}{125} = 80$ units.
Reduction in consumption $= 100 - 80 = 20$ units.
Percentage reduction $= \frac{\text{Reduction}}{\text{Initial consumption}} \times 100 = \frac{20}{100} \times 100 = 20 \%$.
Alternatively,using the formula: $\text{Percentage decrease} = \frac{r}{100+r} \times 100 \% = \frac{25}{125} \times 100 = 20 \%$.

(A) Let $E$ be the set of candidates who failed in English and $M$ be the set of candidates who failed in Mathematics.
Given: $n(E) = 52 \%$,$n(M) = 42 \%$,and $n(E \cap M) = 17 \%$.
The percentage of candidates who failed in at least one subject is given by the formula: $n(E \cup M) = n(E) + n(M) - n(E \cap M)$.
$n(E \cup M) = 52 + 42 - 17 = 77 \%$.
The percentage of candidates who passed in both subjects is the complement of those who failed in at least one subject.
Percentage of passed candidates $= 100 \% - n(E \cup M) = 100 - 77 = 23 \%$.

(C) Let the total number of votes polled be $x$.
The candidate who lost secured $40\%$ of the votes.
Therefore,the winning candidate secured $(100\% - 40\%) = 60\%$ of the votes.
The difference in the percentage of votes between the two candidates is $(60\% - 40\%) = 20\%$.
According to the problem,this difference is equal to $298$ votes.
So,$20\%$ of $x = 298$.
$\Rightarrow \frac{20}{100} \times x = 298$
$\Rightarrow \frac{1}{5} \times x = 298$
$\Rightarrow x = 298 \times 5 = 1490$.
Thus,the total number of votes polled is $1490$.

(D) To find the value of $23 \% \text{ of } 6783 + 57 \% \text{ of } 8431$,we calculate each part separately:
$23 \% \text{ of } 6783 = \frac{23}{100} \times 6783 = 0.23 \times 6783 = 1560.09$
$57 \% \text{ of } 8431 = \frac{57}{100} \times 8431 = 0.57 \times 8431 = 4805.67$
Adding these two results together:
$1560.09 + 4805.67 = 6365.76$
Rounding to the nearest whole number as per the given options,we get $6366$,which is closest to $6360$.

(D) Let the three consecutive numbers be $x$,$x+1$,and $x+2$.
Then,$x + (x+1) + (x+2) = 2262$.
Simplifying the equation,$3x + 3 = 2262$.
$3x = 2262 - 3 = 2259$.
$x = \frac{2259}{3} = 753$.
The three numbers are $753$,$754$,and $755$.
The highest number is $755$.
$41\%$ of $755 = \frac{41}{100} \times 755 = 0.41 \times 755 = 309.55$.

(B) Akash scored $73$ marks in subject $A$.
Marks in subject $B = \frac{56}{100} \times 150 = 84$ marks.
Total maximum marks for all three subjects $= 150 \times 3 = 450$.
Total marks obtained by Akash in all three subjects $= \frac{54}{100} \times 450 = 243$ marks.
Let the marks in subject $C$ be $x$.
According to the problem,$A + B + C = 243$.
$73 + 84 + x = 243$.
$157 + x = 243$.
$x = 243 - 157 = 86$ marks.

(B) Let the original price of sugar be $₹ x$ per $\text{kg}$.
The new price of sugar after a $10 \%$ decrease is $x - 0.10x = 0.9x$ per $\text{kg}$.
According to the problem,the difference in quantity purchased for $₹ 270$ at the original price and the new price is $1 \text{ kg}$.
$\frac{270}{0.9x} - \frac{270}{x} = 1$
$\frac{300}{x} - \frac{270}{x} = 1$
$\frac{30}{x} = 1$
$x = 30$
Therefore,the original price of sugar is $₹ 30$ per $\text{kg}$.

(D) 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 less the second number is than the first number in terms of percentage.
Difference $= 70 - 63 = 7$.
Required percentage $= \left( \frac{\text{Difference}}{\text{First number}} \right) \times 100 = \left( \frac{7}{70} \right) \times 100$.
$= 0.1 \times 100 = 10 \%$.

(B) Let the initial consumption be $C_1$ and the initial price be $P_1 = 6$. The initial expenditure is $E = P_1 \times C_1 = 6C_1$.
The new price is $P_2 = 7.50$. Let the new consumption be $C_2$. Since the expenditure remains constant,$E = P_2 \times C_2 = 7.50 \times C_2$.
Equating the expenditures: $6C_1 = 7.50C_2$.
Solving for $C_2$: $C_2 = \frac{6}{7.50} C_1 = \frac{60}{75} C_1 = 0.8C_1$.
The reduction in consumption is $C_1 - C_2 = C_1 - 0.8C_1 = 0.2C_1$.
The percentage reduction is $\frac{0.2C_1}{C_1} \times 100 = 20\%$.

(A) Let the total marks of the exam be $x$.
According to the problem,Seeta's marks are $54 \%$ of $x$,which is $24$ marks less than Raman's marks $(456)$.
$x \times \frac{54}{100} = 456 - 24$
$x \times \frac{54}{100} = 432$
$x = \frac{432 \times 100}{54}$
$x = 800$
So,the total marks of the examination is $800$.
Now,calculate the minimum passing marks:
Minimum passing marks $= 34 \%$ of $800 = \frac{34}{100} \times 800 = 272$.
Finally,calculate how many more marks Raman scored than the minimum passing marks:
Marks difference $= 456 - 272 = 184$.
Therefore,Raman scored $184$ marks more than the minimum passing marks.

(D) Maximum marks in the examination $= 875$.
Ritu's marks $= 875 \times \frac{56}{100} = 490$.
Smita's marks $= 875 \times \frac{92}{100} = 805$.
Rina's marks $= 634$.
Total marks scored by all three girls $= 490 + 805 + 634 = 1929$.
Average marks $= \frac{\text{Total marks}}{3} = \frac{1929}{3} = 643$.

(C) Let the maximum marks be $A$.
The percentage of marks secured by the candidate is constant in both cases.
In the first case,the percentage is $\frac{468}{A} \times 100$.
In the second case,the percentage is $\frac{336}{700} \times 100$.
Equating both percentages:
$\frac{468}{A} = \frac{336}{700}$
$A = \frac{468 \times 700}{336}$
$A = \frac{468 \times 25}{12}$
$A = 39 \times 25 = 975$.
Thus,the maximum marks of the test were $975$.

(D) Total number of employees $= 4800$.
Number of males $= 4800 \times \frac{45}{100} = 2160$.
Given that $60 \%$ of the males are $25$ years or older,the percentage of males younger than $25$ years is $(100 - 60) \% = 40 \%$.
Therefore,the number of males younger than $25$ years $= 2160 \times \frac{40}{100} = 216 \times 4 = 864$.

(D) Given,the third number $= 2400$.
The second number $= \frac{1}{4} \times 2400 = 600$.
Let the first number be $x$. According to the problem,$\frac{6}{11}x = 22 \%$ of $600$.
$\frac{6}{11}x = \frac{22}{100} \times 600$.
$\frac{6}{11}x = 22 \times 6 = 132$.
$x = 132 \times \frac{11}{6} = 22 \times 11 = 242$.
Now,we need to find $45 \%$ of the first number $(242)$:
$45 \% \text{ of } 242 = \frac{45}{100} \times 242 = 0.45 \times 242 = 108.9$.

(C) Let the number of red marbles to be added be $x$.
Initially,the number of red marbles is $10$ and green marbles is $30$,so the total number of marbles is $10 + 30 = 40$.
After adding $x$ red marbles,the new number of red marbles becomes $10 + x$ and the new total number of marbles becomes $40 + x$.
According to the problem,the percentage of red marbles should be $60 \%$.
Therefore,$\frac{10+x}{40+x} = \frac{60}{100}$.
Simplifying the fraction,$\frac{10+x}{40+x} = \frac{3}{5}$.
Cross-multiplying gives $5(10 + x) = 3(40 + x)$.
$50 + 5x = 120 + 3x$.
$5x - 3x = 120 - 50$.
$2x = 70$.
$x = 35$.
Thus,$35$ red marbles must be added.

(A) Let the number be $x$.
According to the problem,the number multiplied by $25 \%$ of itself is equal to the number plus $200 \%$ of the number.
$25 \%$ of $x = \frac{25}{100}x = \frac{x}{4}$.
$200 \%$ more than $x = x + 200 \% \text{ of } x = x + 2x = 3x$.
Setting up the equation:
$x \times \frac{x}{4} = 3x$.
Since $x$ cannot be $0$ (as the result is $200 \%$ more than the number),we can divide both sides by $x$:
$\frac{x}{4} = 3$.
$x = 3 \times 4 = 12$.
Therefore,the number is $12$.

(B) Let the initial price of onions be $P$ and initial consumption be $C$.
Initial expenditure $= P \times C$.
New price $= P + 0.50P = 1.50P$.
Let the new consumption be $C'$.
Since the expenditure remains the same: $1.50P \times C' = P \times C$.
$C' = \frac{P \times C}{1.50P} = \frac{C}{1.5} = \frac{2}{3}C$.
Reduction in consumption $= C - \frac{2}{3}C = \frac{1}{3}C$.
Percentage reduction $= (\frac{1/3C}{C}) \times 100 = 33 \frac{1}{3} \%$.
Alternatively,using the formula: $\text{Required reduction } \% = \frac{x}{100+x} \times 100 = \frac{50}{100+50} \times 100 = \frac{50}{150} \times 100 = 33 \frac{1}{3} \%$.

(C) To find the approximate value,we round the percentages and the numbers to the nearest convenient values.
$39.897 \%$ is approximately $40 \%$.
$58.779 \%$ is approximately $59 \%$.
$4331$ is approximately $4330$.
$5003$ is approximately $5000$.
Now,calculate the expression:
$40 \% \text{ of } 4330 + 59 \% \text{ of } 5000 = \left(\frac{40}{100} \times 4330\right) + \left(\frac{59}{100} \times 5000\right)$
$= (40 \times 43.3) + (59 \times 50)$
$= 1732 + 2950$
$= 4682$
Rounding $4682$ to the nearest option,we get $4700$.

(B) Ruchira's monthly income $= ₹ 32,000$.
Ravina's monthly income $= 32,000 + (15\% \text{ of } 32,000) = 32,000 \times \frac{115}{100} = ₹ 36,800$.
Ramola's monthly income $= 3 \times 36,800 = ₹ 1,10,400$.
Ramola's annual income $= 12 \times 1,10,400 = ₹ 13,24,800$.

(C) The percentage of marks obtained by the candidate is calculated based on the converted marks and the converted maximum marks.
Percentage of marks $= \frac{336}{700} \times 100 = 48 \%$.
Since the candidate's performance percentage remains constant,$468$ marks represent $48 \%$ of the original maximum marks $A$.
Therefore,$0.48 \times A = 468$.
$A = \frac{468}{0.48} = 975$.
Thus,the maximum marks $A$ of the test was $975$.

(D) Let the first number be $x$,the second number be $y$,and the third number be $z$.
Given,$z = 2400$.
According to the problem,$y = \frac{1}{4} \times z = \frac{1}{4} \times 2400 = 600$.
Also,$\frac{6}{11} \times x = 22 \% \text{ of } y$.
$\frac{6}{11} \times x = \frac{22}{100} \times 600$.
$\frac{6}{11} \times x = 22 \times 6 = 132$.
$x = \frac{132 \times 11}{6} = 22 \times 11 = 242$.
We need to find $45 \%$ of the first number $(x)$:
$45 \% \text{ of } 242 = \frac{45}{100} \times 242 = 0.45 \times 242 = 108.9$.

(D) To solve $32.05 \% \text{ of } 259.99$,we can approximate the values for a quick calculation.
$32.05 \% \approx 32 \% = \frac{32}{100}$.
$259.99 \approx 260$.
Now,calculate: $\frac{32}{100} \times 260 = 32 \times 2.6 = 83.2$.
Rounding $83.2$ to the nearest whole number gives $83$.

(A) Let the initial investment in Debt and Equity funds be $4x$ and $5x$ respectively. Total initial investment $= 4x + 5x = 9x$.
After one year,he earned a $30\%$ dividend. The total amount becomes $9x \times 1.3 = 11.7x$.
This total amount is reinvested in the ratio $6:7$. The amount reinvested in Equity funds is $\frac{7}{13} \times 11.7x = 94,500$.
Calculating $x$: $\frac{7}{13} \times 11.7x = 94,500 \implies 7 \times 0.9x = 94,500 \implies 6.3x = 94,500$.
$x = \frac{94,500}{6.3} = 15,000$.
The original amount invested in Equity funds was $5x = 5 \times 15,000 = 75,000$.

(B) Let the original numbers be $x$ and $y$. Their product is $xy$.
One-third of the first number is $\frac{x}{3}$.
$150 \%$ of the second number is $\frac{150}{100} \times y = \frac{3}{2}y$.
The product of these two values is $\frac{x}{3} \times \frac{3}{2}y = \frac{xy}{2}$.
To find what percent this product is of the original product $xy$,we calculate:
$\frac{\frac{xy}{2}}{xy} \times 100 = \frac{1}{2} \times 100 = 50 \%$.

(D) Let the salary in June $2009$ be $x$.
Since the salary increases by $10 \%$ every year,the salary in June $2010$ will be $x \times (1 + 0.10) = 1.1x$.
The salary in June $2011$ will be $1.1x \times 1.1 = 1.21x$.
Given that the salary in June $2011$ is $Rs$ $22,385$,we have:
$1.21x = 22385$
$x = \frac{22385}{1.21}$
$x = 18500$
Therefore,his salary in June $2009$ was $Rs$ $18,500$.

(D) From statement $II$,the number of students who passed in second class is $750$.
From statement $III$,the number of students who passed in pass class is $28 \%$ of $750$,which is $0.28 \times 750 = 210$.
From statement $I$,we know that $85 \%$ of the total students who appeared passed in either first,second,or pass class. However,we do not know the total number of students who appeared for the examination.
Even if we knew the total number of students,we have no information regarding the distribution or the specific count of students who passed in the first class relative to the total or other classes.
Therefore,the information provided in all three statements is insufficient to determine the number of students who passed in the first class.

(B) Calculate the percentage values:
$34.5 \% \text{ of } 1800 = \frac{34.5}{100} \times 1800 = 34.5 \times 18 = 621$
$12.4 \% \text{ of } 1500 = \frac{12.4}{100} \times 1500 = 12.4 \times 15 = 186$
Substitute these values into the equation:
$621 + 186 = (?)^3 + 78$
$807 = (?)^3 + 78$
$(?)^3 = 807 - 78$
$(?)^3 = 729$
Taking the cube root on both sides:
$? = \sqrt[3]{729} = 9$

(A) Given equation: $0.67 \times 801 - 231.17 = ? - 0.23 \times 789$
Calculate $67 \%$ of $801$: $0.67 \times 801 = 536.67$
Substitute the value: $536.67 - 231.17 = ? - (0.23 \times 789)$
$305.5 = ? - 181.47$
$? = 305.5 + 181.47$
$? = 486.97$
Rounding to the nearest integer,we get $? \approx 490$.

(B) Let the first,second,and third numbers be $x$,$y$,and $z$ respectively.
Given that $z = 2960$.
The second number is one-fourth of the third number: $y = \frac{1}{4} \times 2960 = 740$.
According to the problem,five-ninths of the first number is equal to $25$ per cent of the second number:
$\frac{5}{9} x = \frac{25}{100} \times y$
$\frac{5}{9} x = \frac{1}{4} \times 740$
$\frac{5}{9} x = 185$
$x = 185 \times \frac{9}{5} = 37 \times 9 = 333$.
Now,$30$ per cent of the first number is:
$30\% \text{ of } 333 = \frac{30}{100} \times 333 = 0.3 \times 333 = 99.9$.

(B) Given,Jyoti's monthly income $= ₹ 22,000$.
Suresh's monthly income is $20\%$ more than Jyoti's monthly income.
Suresh's monthly income $= 22000 + (20\% \text{ of } 22000) = 22000 + 4400 = ₹ 26,400$.
Dinesh's monthly income is four times Suresh's monthly income.
Dinesh's monthly income $= 4 \times 26400 = ₹ 105,600$.

(B) Total number of girls $= \frac{12}{100} \times 250 = 30$.
Total number of boys $= 250 - 30 = 220$.
Monthly fee of each girl $= ₹ 450$.
Monthly fee of each boy $= 450 + (24\% \text{ of } 450) = 450 + (0.24 \times 450) = 450 + 108 = ₹ 558$.
Total monthly fee of girls $= 30 \times 450 = ₹ 13500$.
Total monthly fee of boys $= 220 \times 558 = ₹ 122760$.
Total monthly fee of girls and boys together $= 13500 + 122760 = ₹ 136260$.

(D) Let $C$'s share be $x$.
Since $B$ receives $25\%$ less than $C$,$B = x - 0.25x = 0.75x$.
Since $A$ receives $25\%$ more than $B$,$A = 0.75x + 0.25(0.75x) = 1.25 \times 0.75x = 0.9375x$.
The total sum is $A + B + C = 731$.
Substituting the values: $0.9375x + 0.75x + x = 731$.
$2.6875x = 731$.
$x = \frac{731}{2.6875} = 272$.
Therefore,$C$'s share is $₹ 272$.

(A) Let the investment of Raghu be $x$.
Since Mohit invested $10\%$ less than Raghu,Mohit's investment $= x - 0.10x = 0.9x$.
Pradeep invested $20\%$ more than Mohit,so Pradeep's investment $= 0.9x + 0.20(0.9x) = 0.9x + 0.18x = 1.08x$.
The total sum of their investments is $x + 0.9x + 1.08x = 2.98x$.
Given that the total sum is $₹ 17,880$,we have $2.98x = 17880$.
Solving for $x$: $x = \frac{17880}{2.98} = 6000$.
Therefore,Raghu invested $₹ 6000$.

(A) Let the original fraction be $\frac{x}{y}$.
According to the problem,the numerator is increased by $150 \%$,so the new numerator becomes $x + 1.5x = 2.5x$.
The denominator is increased by $300 \%$,so the new denominator becomes $y + 3y = 4y$.
The resulting fraction is given as $\frac{2.5x}{4y} = \frac{5}{18}$.
To simplify,multiply the numerator and denominator by $10$ to remove the decimal: $\frac{25x}{40y} = \frac{5}{18}$.
Simplifying the fraction $\frac{25}{40}$ gives $\frac{5}{8}$,so $\frac{5x}{8y} = \frac{5}{18}$.
Multiplying both sides by $\frac{8}{5}$,we get $\frac{x}{y} = \frac{5}{18} \times \frac{8}{5} = \frac{8}{18} = \frac{4}{9}$.
Thus,the original fraction is $\frac{4}{9}$.

(D) Let the original price of the article be $x$.
First,the price is increased by $10 \%$,so the new price becomes $x \times (1 + 0.10) = 1.1x$.
Then,this price is increased by $20 \%$,so the final price becomes $1.1x \times (1 + 0.20) = 1.1x \times 1.2 = 1.32x$.
Given that the final price is $₹ 33$,we have the equation:
$1.32x = 33$
$x = \frac{33}{1.32}$
$x = \frac{3300}{132} = 25$.
Therefore,the original price was $₹ 25$.

(C) Let the total amount of the electricity bill be $₹ x$.
According to the problem,a reduction of $4 \%$ on the bill amount is equal to $₹ 13$.
Therefore,we can write the equation as:
$\frac{4}{100} \times x = 13$
Multiplying both sides by $100$,we get:
$4x = 1300$
Dividing both sides by $4$,we get:
$x = \frac{1300}{4} = 325$
Thus,the total amount of the electricity bill was $₹ 325$.

(D) Let the total maximum marks of the test be $x$.
According to the problem,a boy needs $40 \%$ of $x$ to pass.
The boy scored $483$ marks and failed by $117$ marks,which means the passing marks are $483 + 117 = 600$.
So,$40 \%$ of $x = 600$.
$\frac{40}{100} \times x = 600$
$x = \frac{600 \times 100}{40} = 1500$.
The minimum passing percentage for girls is $35 \%$ of the total marks.
Passing marks for girls $= 35 \%$ of $1500 = \frac{35}{100} \times 1500 = 35 \times 15 = 525$.

(C) Let the original fraction be $\frac{x}{y}$.
According to the problem,the numerator is increased by $150 \%$,so the new numerator is $x + 1.5x = 2.5x$.
The denominator is increased by $350 \%$,so the new denominator is $y + 3.5y = 4.5y$.
The resultant fraction is $\frac{2.5x}{4.5y} = \frac{25}{51}$.
Simplifying the ratio: $\frac{25x}{45y} = \frac{25}{51}$.
$\frac{5x}{9y} = \frac{25}{51}$.
$\frac{x}{y} = \frac{25}{51} \times \frac{9}{5} = \frac{5 \times 3}{17} = \frac{15}{17}$.

(A) Let the original price of one toy be $₹ 100$ and the original number of toys sold be $100$.
Original total sales $= 100 \times 100 = ₹ 10000$.
New price after $20 \%$ increase $= 100 + 20 = ₹ 120$.
New number of toys sold after $15 \%$ decrease $= 100 - 15 = 85$.
New total sales $= 120 \times 85 = ₹ 10200$.
Increase in sales $= 10200 - 10000 = ₹ 200$.
Percentage increase $= \frac{200}{10000} \times 100 \% = 2 \%$ increase.

(C) Initial monthly income $= ₹ 15000$.
Initial expenditure $= 80 \%$ of $₹ 15000 = \frac{80}{100} \times 15000 = ₹ 12000$.
Initial savings $= ₹ 15000 - ₹ 12000 = ₹ 3000$.
New monthly income after $20 \%$ increase $= 15000 + (20 \% \text{ of } 15000) = 15000 + 3000 = ₹ 18000$.
New expenditure after $20 \%$ increase $= 12000 + (20 \% \text{ of } 12000) = 12000 + 2400 = ₹ 14400$.
New savings $= \text{New income} - \text{New expenditure} = 18000 - 14400 = ₹ 3600$.

(A) Let the third number be $100$.
The first number is $12 \frac{1}{2} \%$ more than the third number,so it is $100 + 12.5 = 112.5$.
The second number is $25 \%$ more than the third number,so it is $100 + 25 = 125$.
We need to find what percentage the first number is of the second number.
Required percentage $= \left( \frac{112.5}{125} \right) \times 100 \%$.
$= 0.9 \times 100 \% = 90 \%$.

(B) The net annual growth rate is $2.5 \% - 0.5 \% = 2.0 \%$.
Let the initial population be $P = 100$.
After $1$ year,the population becomes $100 \times (1 + 0.02) = 102$.
After $2$ years,the population becomes $102 \times (1 + 0.02) = 102 \times 1.02 = 104.04$.
The total increase in population is $104.04 - 100 = 4.04$.
Therefore,the net percentage increase after $2$ years is $4.04 \%$.

(D) The cost of one shirt is $₹ 320$.
The rebate percentage is $25 \%$.
Discount on one shirt $= 25 \% \text{ of } ₹ 320 = \frac{25}{100} \times 320 = ₹ 80$.
Let the number of shirts to be purchased be $x$.
The total rebate is given as $₹ 400$.
Therefore,$80 \times x = 400$.
Solving for $x$,we get $x = \frac{400}{80} = 5$.
Thus,the purchaser should buy $5$ shirts.

(C) Let the original price of tea be $x$ per $kg$.
Total amount available $= Rs \ 22500$.
Original quantity of tea purchased $= \frac{22500}{x} \ kg$.
Reduced price of tea $= 0.9x$ per $kg$.
New quantity of tea purchased $= \frac{22500}{0.9x} \ kg$.
According to the problem,the difference in quantity is $25 \ kg$:
$\frac{22500}{0.9x} - \frac{22500}{x} = 25$
$\frac{25000}{x} - \frac{22500}{x} = 25$
$\frac{2500}{x} = 25$
$x = 100$.
The original price is $Rs \ 100$ per $kg$.
The reduced price $= 0.9 \times 100 = Rs \ 90$ per $kg$.

(D) Let Ram's total income be $x$.
Ram donates $4 \%$ of his income to charity,so the remaining amount is $x - 0.04x = 0.96x$.
He then deposits $10 \%$ of the remaining amount in a bank,which is $0.10 \times 0.96x = 0.096x$.
The amount left with him is $0.96x - 0.096x = 0.864x$.
Given that the amount left is $Rs$ $8640$,we have the equation: $0.864x = 8640$.
Solving for $x$: $x = \frac{8640}{0.864} = 10000$.
Therefore,Ram's total income is $Rs$ $10000$.

(D) Let $K$,$N$,and $S$ be the monthly salaries of Kaushal,Nandini,and Suresh respectively.
Given that $12\% \text{ of } K = 16\% \text{ of } N$,which implies $0.12K = 0.16N$,or $K = \frac{16}{12}N = \frac{4}{3}N$.
Given that Suresh's monthly salary is half of Nandini's,so $S = \frac{N}{2}$,which means $N = 2S$.
Suresh's annual salary is $Rs$ $1.08$ Lakhs,so his monthly salary $S = \frac{1.08}{12} = 0.09$ Lakhs.
Now,find $N$: $N = 2 \times 0.09 = 0.18$ Lakhs.
Finally,find $K$: $K = \frac{4}{3} \times 0.18 = 4 \times 0.06 = 0.24$ Lakhs.
$0.24$ Lakhs is equal to $24,000$.