Percentage Questions in English

(A) To find the value,we calculate the expression step by step:
$40 \% = \frac{40}{100} = 0.4$
$50 \% = \frac{50}{100} = 0.5$
Now,calculate the final value:
$= 0.4 \times 0.5 \times \frac{3}{4} \times 3200$
$= 0.2 \times \frac{3}{4} \times 3200$
$= 0.2 \times 3 \times 800$
$= 0.2 \times 2400$
$= 480$

(C) Let the number be $x$.
Given that one-fifth of the number is $62$,we have $\frac{1}{5}x = 62$.
Multiplying both sides by $5$,we get $x = 62 \times 5 = 310$.
Now,we need to find $73\%$ of this number $x$.
$73\% \text{ of } 310 = \frac{73}{100} \times 310 = 0.73 \times 310 = 226.3$.

(A) Let the number be $x$.
According to the problem,$\frac{2}{3} \times \frac{3}{4} \times \frac{1}{5} \times x = 15$.
Simplifying the left side: $\frac{2 \times 3 \times 1}{3 \times 4 \times 5} \times x = 15$.
$\frac{6}{60} \times x = 15$.
$\frac{1}{10} \times x = 15$.
$x = 15 \times 10 = 150$.
Now,we need to find $30$ per cent of $150$.
$30 \% \text{ of } 150 = \frac{30}{100} \times 150 = 30 \times 1.5 = 45$.

(D) Let the number be $x$.
According to the problem,$\frac{3}{4} \times \frac{2}{3} \times \frac{1}{5} \times x = 249.6$.
Simplifying the left side: $\frac{3 \times 2 \times 1}{4 \times 3 \times 5} \times x = \frac{6}{60} \times x = \frac{1}{10} \times x$.
So,$\frac{x}{10} = 249.6$.
Multiplying both sides by $10$,we get $x = 2496$.
We need to find $50\%$ of the number $x$.
$50\% \text{ of } x = \frac{50}{100} \times 2496 = \frac{1}{2} \times 2496 = 1248$.
Thus,the required value is $1248$.

(C) Total amount spent by Ishan on the bike and television $= 35645 + 24355 = Rs. 60000$.
Since he kept $20\%$ of the total amount as cash,the amount spent represents $100\% - 20\% = 80\%$ of the total amount.
Let the total amount be $x$.
Then,$80\% \text{ of } x = 60000$.
$\frac{80}{100} \times x = 60000$.
$x = \frac{60000 \times 100}{80}$.
$x = 750 \times 100 = 75000$.
Therefore,the total amount Ishan had was $Rs. 75000$.

(B) Let the total amount Sonal had be $x$.
Sonal spent $Rs. 45760$ on interior decoration and $Rs. 27896$ on an air conditioner.
Total amount spent $= 45760 + 27896 = Rs. 73656$.
The remaining amount is $28 \%$ of the total amount,which means the amount spent is $(100 - 28) \% = 72 \%$ of the total amount.
Therefore,$72 \%$ of $x = 73656$.
$x = \frac{73656}{72} \times 100$.
$x = 1023 \times 100 = Rs. 102300$.
Thus,the total amount was $Rs. 102300$.

(B) Let the total amount be $x$.
Rajesh spent $Rs. 44620$ and $Rs. 32764$.
Total spent $= 44620 + 32764 = 77384$.
The remaining amount is $32\%$ of the total amount.
Therefore,the amount spent is $(100 - 32)\% = 68\%$ of the total amount.
According to the problem,$68\%$ of $x = 77384$.
$0.68x = 77384$.
$x = \frac{77384}{0.68} = 113800$.
Thus,the total amount was $Rs. 113800$.

(D) Harjeet's total monthly expenditure percentage $= 50 \% + 20 \% + 5 \% = 75 \%$.
Harjeet's savings percentage $= 100 \% - 75 \% = 25 \%$.
Given that the savings amount is $Rs. 11250$.
Therefore,$25 \%$ of the total monthly income $= 11250$.
Let the total monthly income be $x$.
$0.25 \times x = 11250$.
$x = \frac{11250}{0.25} = 11250 \times 4 = 45000$.
Thus,Harjeet's total monthly income is $Rs. 45000$.

(B) Let the total monthly income be $x$.
Expenditure on household items $= 50\% \text{ of } x = 0.5x$.
Remaining income $= x - 0.5x = 0.5x$.
Now,he spends $50\% + 25\% + 10\% = 85\%$ of the remaining income on transport,entertainment,and sports.
Therefore,the savings percentage of the remaining income $= 100\% - 85\% = 15\%$.
Given that the savings amount is $RS. 900$:
$15\% \text{ of } (0.5x) = 900$
$0.15 \times 0.5x = 900$
$0.075x = 900$
$x = \frac{900}{0.075}$
$x = 12000$.
Thus,Mr. Giridhar's monthly income is $RS. 12000$.

(A) Let Shruti's salary be $S$.
Initially,she decided to donate $12 \%$ of $S$,which is $0.12S$.
On the day of donation,she donated $75 \%$ of the initially decided amount.
So,the actual donation $= 75 \% \text{ of } (0.12S) = 0.75 \times 0.12S = 0.09S$.
Given that the actual donation is $Rs. 3150$,we have $0.09S = 3150$.
$S = \frac{3150}{0.09} = \frac{315000}{9} = 35000$.
Therefore,Shruti's salary is $Rs. 35000$.

(A) Given,Deepak's monthly income $= \text{Rs. } 78000$.
Asha's monthly income $= 60 \% \text{ of } 78000 = \frac{60}{100} \times 78000 = \text{Rs. } 46800$.
It is given that Asha's income is $120 \% \text{ of Maya's income}$.
Let Maya's income be $M$.
$120 \% \text{ of } M = 46800$
$\frac{120}{100} \times M = 46800$
$M = \frac{46800 \times 100}{120} = \text{Rs. } 39000$.
Therefore,Maya's monthly income is Rs. $39000$.

(C) Let the share of $B$ be $x$ Rs.
According to the problem:
Share of $C = x \times (1 - 0.25) = 0.75x = \frac{3}{4}x$.
Share of $A = C + 25\% \text{ of } C = \frac{3}{4}x \times (1 + 0.25) = \frac{3}{4}x \times \frac{5}{4} = \frac{15}{16}x$.
Total sum = $A + B + C = 2236$.
$\frac{15}{16}x + x + \frac{3}{4}x = 2236$.
Multiply by $16$ to clear the denominator: $15x + 16x + 12x = 2236 \times 16$.
$43x = 35776$.
$x = \frac{35776}{43} = 832$.
Therefore,$A$'s share = $\frac{15}{16} \times 832 = 15 \times 52 = 780$ Rs.

(C) Pooja's total monthly investment percentage $= 13 \% + 23 \% + 8 \% = 44 \%$.
Given that $13 \%$ of her monthly salary $= Rs. 8554$.
Therefore,$1 \%$ of her monthly salary $= \frac{8554}{13} = Rs. 658$.
Total monthly investment $= 44 \% = 44 \times 658 = Rs. 28952$.
To find the total annual investment,multiply the monthly investment by $12$.
Annual investment $= 28952 \times 12 = Rs. 347424$.

(B) Given that $6 \%$ of the monthly salary is equal to $Rs.$ $2,100$.
Let the monthly salary be $S$.
$0.06 \times S = 2100 \Rightarrow S = \frac{2100}{0.06} = 35000$.
Total percentage of monthly salary invested $= (6 + 8 + 9) \% = 23 \%$.
Total monthly investment $= 23 \% \text{ of } 35000 = 0.23 \times 35000 = 8050$.
Total annual investment $= 8050 \times 12 = 96600$.

(D) Step $1$: Calculate the total percentage of monthly salary invested.
Total percentage $= 14 \% + 21 \% + 6.5 \% = 41.5 \%$.
Step $2$: Find the total monthly investment.
Given that $14 \%$ of the monthly salary $= Rs. 7014$.
Therefore,$1 \%$ of the monthly salary $= \frac{7014}{14} = Rs. 501$.
Total monthly investment $= 41.5 \% \times 501 = Rs. 20791.5$.
Step $3$: Calculate the total annual investment.
Annual investment $= 20791.5 \times 12 = Rs. 249498$.

(D) Abhinav's investment $= Rs. 6000$.
Sunil's investment $= 6000 \times (1 - 0.30) = 6000 \times 0.70 = Rs. 4200$.
Rita's investment $= 4200 \times (1 + 0.25) = 4200 \times 1.25 = Rs. 5250$.
Total investment $= 6000 + 4200 + 5250 = Rs. 15450$.
Required ratio $= \frac{\text{Rita's investment}}{\text{Total investment}} = \frac{5250}{15450}$.
Dividing both by $150$,we get $\frac{35}{103}$.
Thus,the ratio is $35:103$.

(A) Given that $11 \%$ of the monthly salary is $Rs. 5236$.
Let the monthly salary be $S$.
$0.11 \times S = 5236 \Rightarrow S = \frac{5236}{0.11} = 47600$.
Deepti's total monthly investment percentage $= (11 + 19 + 7) \% = 37 \%$.
Total monthly investment $= 37 \% \text{ of } 47600 = 0.37 \times 47600 = Rs. 17612$.
Total annual investment $= 17612 \times 12 = Rs. 211344$.

(D) Number of sweets each student received $= 80 \times \frac{15}{100} = 12$.
Total number of sweets $= \text{Total students} \times \text{Sweets per student} = 80 \times 12 = 960$.

(D) Number of students $= 50$.
Number of teachers $= 5$.
Sweets received by each student $= 12 \% \text{ of } 50 = \frac{12}{100} \times 50 = 6$.
Sweets received by each teacher $= 20 \% \text{ of } 50 = \frac{20}{100} \times 50 = 10$.
Total number of sweets $= (\text{Number of students} \times \text{Sweets per student}) + (\text{Number of teachers} \times \text{Sweets per teacher})$.
Total number of sweets $= (50 \times 6) + (5 \times 10) = 300 + 50 = 350$.

(D) Number of students $= 80$
Number of teachers $= 5$
Sweets received by each student $= 15 \% \text{ of } 80 = \frac{15}{100} \times 80 = 12$
Sweets received by each teacher $= 25 \% \text{ of } 80 = \frac{25}{100} \times 80 = 20$
Total number of sweets $= (\text{Number of students} \times \text{Sweets per student}) + (\text{Number of teachers} \times \text{Sweets per teacher})$
Total number of sweets $= (80 \times 12) + (5 \times 20) = 960 + 100 = 1060$

(C) Let the maximum marks of the test be $x$.
To pass the examination,a candidate needs $35 \%$ of the maximum marks.
Passing marks $= 35 \% \text{ of } x = 0.35x$.
The candidate secured $40$ marks and failed by $30$ marks,which means the passing marks are $40 + 30 = 70$.
Equating the two expressions for passing marks:
$0.35x = 70$
$x = \frac{70}{0.35}$
$x = \frac{7000}{35} = 200$.
Therefore,the maximum marks of the test is $200$.

(B) Let the maximum marks of the test be $M$.
Since Raja obtained $76 \%$ of the maximum marks,Seema obtained $(100 - 76) \% = 24 \%$ of the maximum marks.
Given that Seema obtained $480$ marks,we have $24 \% \text{ of } M = 480$.
$0.24 \times M = 480 \Rightarrow M = \frac{480}{0.24} = 2000$.
The maximum marks of the test is $2000$.
Since the maximum marks is equal to the sum of marks obtained by Raja and Seema,and Seema got $480$,the marks obtained by Raja $= 2000 - 480 = 1520$.

(A) Total number of questions $= 150$.
Target grade is $60 \%$,so total correct answers required $= 150 \times \frac{60}{100} = 90$.
Correct answers from the first $75$ questions $= 75 \times \frac{40}{100} = 30$.
Remaining correct answers needed $= 90 - 30 = 60$.
Percentage of the remaining $75$ questions needed $= \frac{60}{75} \times 100 = 80 \%$.

(C) Total number of votes $= 8400$.
Invalid votes $= 25 \%$ of $8400 = 8400 \times \frac{25}{100} = 2100$.
Total valid votes $= 8400 - 2100 = 6300$.
One candidate received $52 \%$ of the valid votes,so the other candidate received $(100 - 52) \% = 48 \%$ of the valid votes.
Number of valid votes the other person got $= 6300 \times \frac{48}{100} = 63 \times 48 = 3024$.

(D) Total number of votes $= 15,200$.
Percentage of invalid votes $= 15 \%$.
Percentage of valid votes $= 100 \% - 15 \% = 85 \%$.
Total number of valid votes $= 15,200 \times \frac{85}{100} = 152 \times 85 = 12,920$.
One candidate received $55 \%$ of the valid votes.
Therefore,the other candidate received $(100 \% - 55 \%) = 45 \%$ of the valid votes.
Number of valid votes the other candidate got $= 12,920 \times \frac{45}{100} = 129.2 \times 45 = 5,814$.

(D) Initial population = $48600$.
Population after the first year (increase of $25\%$): $48600 \times (1 + \frac{25}{100}) = 48600 \times \frac{125}{100} = 48600 \times 1.25 = 60750$.
Population after the second year (decrease of $8\%$): $60750 \times (1 - \frac{8}{100}) = 60750 \times \frac{92}{100} = 60750 \times 0.92 = 55890$.
Therefore,the population at the end of $2$ years is $55890$.

(D) Initial weight of the mixture $= 60 \text{ gm}$.
Weight of water in the initial mixture $= 60 \times \frac{75}{100} = 45 \text{ gm}$.
When $15 \text{ gm}$ of water is added,the new weight of water $= 45 + 15 = 60 \text{ gm}$.
The new total weight of the mixture $= 60 + 15 = 75 \text{ gm}$.
Required percentage of water $= \left( \frac{\text{New weight of water}}{\text{New total weight of mixture}} \right) \times 100 = \left( \frac{60}{75} \right) \times 100 = 0.8 \times 100 = 80 \%$.

(B) Let the initial number be $100$.
First increase of $10 \%$: $100 \times (1 + 0.10) = 110$.
Second increase of $20 \%$ on the new value: $110 \times (1 + 0.20) = 110 \times 1.2 = 132$.
Third increase of $25 \%$ on the resulting value: $132 \times (1 + 0.25) = 132 \times 1.25 = 165$.
The final value is $165$.
The single equivalent increase is $(165 - 100) = 65 \%$.

(B) Let the original price of rice be $100$ units and the original consumption be $100$ units.
Original expenditure $= 100 \times 100 = 10000$ units.
After a $20 \%$ increase,the new price of rice is $120$ units.
To keep the expenditure the same ($10000$ units),let the new consumption be $C$.
$120 \times C = 10000 \implies C = \frac{10000}{120} = \frac{250}{3} = 83.33$ units.
Reduction in consumption $= 100 - 83.33 = 16.67$ units.
Percentage reduction $= \left( \frac{16.67}{100} \times 100 \right) = 16.67 \% = 16 \frac{2}{3} \%$.
Alternatively,using the formula: $\text{Required percentage} = \left( \frac{x}{100+x} \times 100 \right) \%$,where $x = 20$.
$= \left( \frac{20}{100+20} \times 100 \right) = \frac{20}{120} \times 100 = \frac{1}{6} \times 100 = 16 \frac{2}{3} \%$.

(D) Let the marked price of the article be $x$ Rs.
Selling price at a discount of $20 \% = x - \frac{20x}{100} = \frac{4x}{5}$ Rs.
In order to gain $60 \%$ on the marked price,the new selling price should be:
$= x \left( \frac{100 + 60}{100} \right) = \frac{160x}{100} = \frac{8x}{5}$ Rs.
The increase required over the discounted price is:
$= \frac{8x}{5} - \frac{4x}{5} = \frac{4x}{5}$ Rs.
Required percentage increase $= \left( \frac{\text{Increase}}{\text{Discounted Price}} \right) \times 100$
$= \left( \frac{\frac{4x}{5}}{\frac{4x}{5}} \right) \times 100 = 100 \%$.

(C) Let the maximum aggregate marks be $x$.
The passing marks required are $256$.
The marks obtained by the student are $192$.
The difference between the passing marks and the marks obtained is $256 - 192 = 64$.
According to the problem,this difference is equal to $10 \%$ of the maximum aggregate marks $x$.
Therefore,$10 \% \text{ of } x = 64$.
$\frac{10}{100} \times x = 64$.
$0.1x = 64$.
$x = \frac{64}{0.1} = 640$.
Thus,the maximum aggregate marks are $640$.

(C) Let the total number of enrolled voters be $100x$.
Given that $60 \%$ of the voters cast their vote,so the number of votes cast $= 60x$.
Out of these,$4 \%$ were invalid,so the number of valid votes $= 60x - (4 \% \text{ of } 60x) = 60x - 2.4x = 57.6x$.
$A$ candidate received $7344$ votes,which is $75 \%$ of the total valid votes.
Therefore,$75 \% \text{ of } 57.6x = 7344$.
$0.75 \times 57.6x = 7344$.
$43.2x = 7344$.
$x = \frac{7344}{43.2} = 170$.
Total number of enrolled votes $= 100x = 100 \times 170 = 17000$.

(D) Let the total monthly salary be $100 \%$.
Total expenditure percentage $= 52 \% + 23 \% = 75 \%$.
Percentage of money left $= 100 \% - 75 \% = 25 \%$.
Given that the remaining amount is $Rs. 4500$,we have $25 \% \text{ of salary} = 4500$.
Therefore,$\text{Total salary} = \frac{4500 \times 100}{25} = 18000$.
Thus,his monthly salary is $Rs. 18000$.

(C) Number of students who speak only Hindi $= 60 \times \frac{40}{100} = 24$.
Number of students who speak only English $= 60 \times \frac{25}{100} = 15$.
Number of students who speak both languages $= 60 - (24 + 15) = 60 - 39 = 21$.
Total number of students who can speak English $= (\text{Students who speak only English}) + (\text{Students who speak both languages}) = 15 + 21 = 36$.

(D) The profit earned by partners in a business depends on both the capital invested and the time period of investment.
In this question,the ratio of investment is given as $3:2:5$,but the total profit amount is not provided.
Additionally,the relationship between profit and investment is not explicitly defined (e.g.,whether the time period is the same for all).
Since the total profit is missing,it is impossible to calculate the specific numerical share of $B$.
Therefore,the correct answer is $\text{Cannot be determined}$.

(A) Let the total salary be $S$.
Given that $10 \%$ of the salary is spent on education,which is equal to $Rs. 1950$.
So,$0.10 \times S = 1950$.
$S = \frac{1950}{0.10} = 19500$.
The total salary is $Rs. 19500$.
Now,the amount spent on conveyance is $15 \%$ of the total salary.
Amount spent on conveyance $= 0.15 \times 19500 = 2925$.
Alternatively,using the ratio method:
Since $10 \% = 1950$,then $1 \% = \frac{1950}{10} = 195$.
Therefore,$15 \% = 15 \times 195 = 2925$.
Thus,Rajiv spends $Rs. 2925$ on conveyance.

(B) The initial food production in $2009$ was $5.5$ million tonnes.
The final food production in $2010$ was $4.4$ million tonnes.
The decrease in production is $5.5 - 4.4 = 1.1$ million tonnes.
The percentage decrease is calculated as: $\frac{\text{Decrease}}{\text{Initial Value}} \times 100$.
Percentage decrease $= \frac{1.1}{5.5} \times 100 = \frac{1}{5} \times 100 = 20 \%$.

(A) Let the original price of the ticket be $P$ and the original number of visitors be $N$. The original revenue is $R_1 = P \times N$.
After a $25 \%$ discount,the new price is $P' = P - 0.25P = 0.75P$.
The number of visitors increases by $30 \%$,so the new number of visitors is $N' = N + 0.30N = 1.30N$.
The new revenue is $R_2 = P' \times N' = (0.75P) \times (1.30N) = 0.975 \times (P \times N) = 0.975R_1$.
The change in revenue is $R_2 - R_1 = 0.975R_1 - R_1 = -0.025R_1$.
This represents a decrease of $0.025 \times 100 \% = 2.5 \%$,which is $2 \frac{1}{2} \% \text{ decrease}$.

(D) Let the initial income of the man be $100$ Rs.
Initial expenditure $= 75$ Rs and initial saving $= 100 - 75 = 25$ Rs.
New income $= 100 + (20 \% \text{ of } 100) = 120$ Rs.
New expenditure $= 75 + (10 \% \text{ of } 75) = 75 + 7.5 = 82.5$ Rs.
New saving $= 120 - 82.5 = 37.5$ Rs.
Increase in saving $= 37.5 - 25 = 12.5$ Rs.
Percentage increase in saving $= (12.5 / 25) \times 100 = 50 \%$.

(C) The percentage of votes not received by the elected candidate is $100 \% - 63 \% = 37 \%$.
Given that $54982$ voters did not vote in favour of the elected candidate,we have $37 \% \text{ of the total votes} = 54982$.
Let the total number of votes be $x$.
$\frac{37}{100} \times x = 54982$
$x = \frac{54982 \times 100}{37}$
$x = 1486 \times 100 = 148600$.
Therefore,the total number of votes polled was $148600$.

(B) Let the capacity of the tank be $27$ litres.

Step	Ordinary Petrol (litres)	High-speed Petrol (litres)
$I$. Initial fill	$27$	$0$
$II$. $1/3$ empty ($18$ left),add $9$ high-speed	$18$	$9$
$III$. $1/3$ empty ($18$ left),add $9$ ordinary	$12 + 9 = 21$	$6$
$IV$. $1/3$ empty ($18$ left),add $9$ high-speed	$14$	$4 + 9 = 13$

The total volume is $27$ litres,and the volume of high-speed petrol is $13$ litres.
Percentage of high-speed petrol = $\frac{13}{27} \times 100 = 48 \frac{4}{27} \%$.

(D) Let the initial length be $l$ and breadth be $b$. The initial perimeter is $P_1 = 2(l + b)$.
After the increase,the new length $l' = l + 0.20l = 1.2l$ and the new breadth $b' = b + 0.30b = 1.3b$.
The new perimeter is $P_2 = 2(1.2l + 1.3b) = 2.4l + 2.6b$.
The percentage increase in perimeter is given by $\frac{P_2 - P_1}{P_1} \times 100$.
Substituting the values: $\frac{(2.4l + 2.6b) - 2(l + b)}{2(l + b)} \times 100 = \frac{0.4l + 0.6b}{2(l + b)} \times 100 = \frac{0.2l + 0.3b}{l + b} \times 100$.
Since the ratio of $l$ to $b$ is not provided,the percentage increase depends on the values of $l$ and $b$. Therefore,the data is inadequate.

(C) Let the total number of workers be $x$.
Skilled workers $= 0.75x$.
Unskilled workers $= 0.25x$.
Permanent skilled workers $= 80\% \text{ of } 0.75x = 0.80 \times 0.75x = 0.60x$.
Permanent unskilled workers $= 20\% \text{ of } 0.25x = 0.20 \times 0.25x = 0.05x$.
Total permanent workers $= 0.60x + 0.05x = 0.65x$.
Temporary workers $= \text{Total workers} - \text{Permanent workers} = x - 0.65x = 0.35x$.
Given that the number of temporary workers is $126$,we have:
$0.35x = 126$
$x = \frac{126}{0.35} = \frac{12600}{35} = 360$.
Therefore,the total number of workers is $360$.

(C) 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 original height be $h$. The original area is $A_1 = \frac{1}{2}bh$.
New base $b' = b + 0.40b = 1.4b$.
New height $h' = h - 0.40h = 0.6h$.
New area $A_2 = \frac{1}{2} \times (1.4b) \times (0.6h) = 0.84 \times (\frac{1}{2}bh) = 0.84A_1$.
The change in area is $A_2 - A_1 = 0.84A_1 - A_1 = -0.16A_1$.
This represents a $16 \%$ decrease in the area.
Alternatively,using the net percentage change formula: $\text{Change} = x + y + \frac{xy}{100} = -40 + 40 + \frac{(-40)(40)}{100} = -16 \%$.

(C) Let the initial population be $P = 123,456,789$.
After the first year,the population becomes $P \times (1 + 0.10) = 1.1P$.
After the second year,the population becomes $1.1P \times (1 + 0.20) = 1.1P \times 1.2 = 1.32P$.
After the third year,the population becomes $1.32P \times (1 + 0.30) = 1.32P \times 1.3 = 1.716P$.
The total percentage increase over three years is $(1.716P - P) / P \times 100 = 71.6\%$.
The average percentage increase per year is the total percentage increase divided by $3$.
Average percentage increase $= 71.6 / 3 = 716 / 30 = 23 \frac{26}{30} = 23 \frac{13}{15}\%$.

(C) Initial monthly income $= ₹ 13500$
Initial monthly expenditure $= ₹ 9000$
Initial savings $= 13500 - 9000 = ₹ 4500$
New monthly income $= 13500 \times (1 + \frac{14}{100}) = 13500 \times 1.14 = ₹ 15390$
New monthly expenditure $= 9000 \times (1 + \frac{7}{100}) = 9000 \times 1.07 = ₹ 9630$
New savings $= 15390 - 9630 = ₹ 5760$
Increase in savings $= 5760 - 4500 = ₹ 1260$
Percentage increase in savings $= (\frac{1260}{4500}) \times 100 = \frac{126}{450} \times 100 = 28 \%$

(D) Let the missing value be $x$.
Given equation: $60 = x \% \text{ of } 400$.
This can be written as: $60 = \frac{x}{100} \times 400$.
Simplifying the expression: $60 = x \times 4$.
Solving for $x$: $x = \frac{60}{4} = 15$.
Therefore,$60$ is $15 \%$ of $400$.

(D) Let the unknown value be $x$.
Given: $40 \%$ of $x = 240$.
$\Rightarrow \frac{40}{100} \times x = 240$
$\Rightarrow x = \frac{240 \times 100}{40}$
$\Rightarrow x = 6 \times 100 = 600$.
Thus,the required value is $600$.

(D) Let the unknown value be $x$.
Given equation: $23 \%$ of $8040 + 42 \%$ of $545 = x \%$ of $3000$
$\Rightarrow \left(\frac{23}{100} \times 8040\right) + \left(\frac{42}{100} \times 545\right) = \frac{x}{100} \times 3000$
$\Rightarrow \frac{184920}{100} + \frac{22890}{100} = 30x$
$\Rightarrow 1849.2 + 228.9 = 30x$
$\Rightarrow 2078.1 = 30x$
$\Rightarrow x = \frac{2078.1}{30}$
$\Rightarrow x = 69.27$

(B) Let Sunil's salary be $₹ x$.
Sunil decided to donate $5 \%$ of his salary,which is $\frac{5x}{100} = \frac{x}{20}$.
He donated $₹ 1687.50$,which is $75 \%$ of the amount he initially decided to donate.
So,$1687.50 = 75 \% \text{ of } (\frac{x}{20})$.
$1687.50 = \frac{75}{100} \times \frac{x}{20}$.
$1687.50 = \frac{3}{4} \times \frac{x}{20} = \frac{3x}{80}$.
$3x = 1687.50 \times 80$.
$3x = 135000$.
$x = \frac{135000}{3} = 45000$.
Therefore,Sunil's salary is $₹ 45000$.