\displaystyle \textbf{Question 1: }36\%\text{ of the students in a school are girls. If the}
\displaystyle \text{number of boys is }1440,\text{ find the total strength of the school.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the total strength of the school be }x.
\displaystyle \text{Percentage of boys}=100\%-36\%=64\%.
\displaystyle \frac{64}{100}\times x=1440.
\displaystyle x=1440\times\frac{100}{64}=2250.
\displaystyle \therefore \text{the total strength of the school is }2250\text{ students}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Geeta saves }18\%\text{ of her monthly salary. If she spends}
\displaystyle \text{Rs. }10250\text{ per month, what is her monthly salary?}
\displaystyle \text{Answer:}
\displaystyle \text{Let Geeta's monthly salary be Rs. }x.
\displaystyle \text{Percentage of salary spent}=100\%-18\%=82\%.
\displaystyle \frac{82}{100}\times x=10250.
\displaystyle x=10250\times\frac{100}{82}=12500.
\displaystyle \therefore \text{Geeta's monthly salary is Rs. }12500.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In an examination, a student must secure }40\%\text{ marks}
\displaystyle \text{to pass. Raul gets }178\text{ marks and fails by }32\text{ marks. Find the}
\displaystyle \text{maximum marks of the examination.}
\displaystyle \text{Answer:}
\displaystyle \text{Passing marks}=178+32=210.
\displaystyle \text{Let the maximum marks be }x.
\displaystyle \frac{40}{100}\times x=210.
\displaystyle x=210\times\frac{100}{40}=525.
\displaystyle \therefore \text{the maximum marks of the examination are }525.
\displaystyle \\

\displaystyle \textbf{Question 4: }8\%\text{ of the students in a school remained absent on a day.}
\displaystyle \text{If }1633\text{ students attended school that day, how many remained absent?}
\displaystyle \text{Answer:}
\displaystyle \text{Percentage of students present}=100\%-8\%=92\%.
\displaystyle \text{Let the total number of students in the school be }x.
\displaystyle \frac{92}{100}\times x=1633.
\displaystyle x=1633\times\frac{100}{92}=1775.
\displaystyle \text{Number of students absent} =\frac{8}{100}\times1775=142.
\displaystyle \therefore 142\text{ students remained absent}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{On increasing the price of an article by }14\%\text{, it}
\displaystyle \text{becomes Rs. }1995.\text{ What was its original price?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original price of the article be Rs. }x.
\displaystyle x\left(1+\frac{14}{100}\right)=1995.
\displaystyle \frac{114}{100}x=1995.
\displaystyle x=1995\times\frac{100}{114}=1750.
\displaystyle \therefore \text{the original price of the article was Rs. }1750.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{On decreasing the price of an article by }6\%\text{, it}
\displaystyle \text{becomes Rs. }1551.\text{ What was its original price?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original price of the article be Rs. }x.
\displaystyle x\left(1-\frac{6}{100}\right)=1551.
\displaystyle \frac{94}{100}x=1551.
\displaystyle x=1551\times\frac{100}{94}=1650.
\displaystyle \therefore \text{the original price of the article was Rs. }1650.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Reenu reduced her weight by }15\%.\text{ If she now weighs}
\displaystyle 52.7\text{ kg, what was her original weight?}
\displaystyle \text{Answer:}
\displaystyle \text{Let Reenu's original weight be }x\text{ kg}.
\displaystyle x\left(1-\frac{15}{100}\right)=52.7.
\displaystyle \frac{85}{100}x=52.7.
\displaystyle x=52.7\times\frac{100}{85}=62.
\displaystyle \therefore \text{Reenu's original weight was }62\text{ kg}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Two candidates contest an election. One of them secures }58\%
\displaystyle \text{of the votes and wins the election by a margin of }2560\text{ votes. How many}
\displaystyle \text{votes were polled in all?}
\displaystyle \text{Answer:}
\displaystyle \text{Percentage of votes secured by the other candidate} =100\%-58\%=42\%.
\displaystyle \text{Difference in their percentage of votes} =58\%-42\%=16\%.
\displaystyle \text{Let the total number of votes polled be }x.
\displaystyle \frac{16}{100}x=2560.
\displaystyle x=2560\times\frac{100}{16}=16000.
\displaystyle \therefore \text{the total number of votes polled was }16000.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In an examination, Preety scored }60\text{ out of }75\text{ in}
\displaystyle \text{Science, }84\text{ out of }100\text{ in Mathematics, }36\text{ out of }50\text{ in Hindi}
\displaystyle \text{and }30\text{ out of }45\text{ in English. Find:}
\displaystyle \text{(i) In which subject is her performance the worst?}
\displaystyle \text{(ii) In which subject is her performance the best?}
\displaystyle \text{(iii) What is her aggregate percentage of marks?}
\displaystyle \text{Answer:}
\displaystyle \text{First, we calculate the percentage of marks in each subject}.
\displaystyle \text{Science percentage} =\frac{60}{75}\times100=80\%.
\displaystyle \text{Mathematics percentage} =\frac{84}{100}\times100=84\%.
\displaystyle \text{Hindi percentage} =\frac{36}{50}\times100=72\%.
\displaystyle \text{English percentage} =\frac{30}{45}\times100=66\frac{2}{3}\%.

\displaystyle \text{(i) The lowest percentage is }66\frac{2}{3}\%.
\displaystyle \therefore \text{Preety's performance is the worst in English}.

\displaystyle \text{(ii) The highest percentage is }84\%.
\displaystyle \therefore \text{Preety's performance is the best in Mathematics}.

\displaystyle \text{(iii) Total marks obtained}=60+84+36+30=210.
\displaystyle \text{Total maximum marks}=75+100+50+45=270.
\displaystyle \text{Aggregate percentage} =\frac{210}{270}\times100=77\frac{7}{9}\%\approx77.78\%.
\displaystyle \therefore \text{Preety's aggregate percentage is approximately }77.78\%.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The price of an article is increased by }25\%.\text{ By what}
\displaystyle \text{percentage must the new price be decreased to restore its original value?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original price of the article be }x.
\displaystyle \text{New price} =x\left(1+\frac{25}{100}\right)=1.25x.
\displaystyle \text{Let the required percentage decrease be }y\%.
\displaystyle 1.25x\left(1-\frac{y}{100}\right)=x.
\displaystyle 1.25\left(1-\frac{y}{100}\right)=1.
\displaystyle 1-\frac{y}{100}=\frac{1}{1.25}=\frac{4}{5}.
\displaystyle \frac{y}{100}=1-\frac{4}{5}=\frac{1}{5}.
\displaystyle y=20.
\displaystyle \therefore \text{the new price must be decreased by }20\%.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The price of an article is reduced by }10\%.\text{ By what}
\displaystyle \text{percentage must the new price be increased to restore its original value?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original price of the article be }x.
\displaystyle \text{New price} =x\left(1-\frac{10}{100}\right)=0.9x.
\displaystyle \text{Let the required percentage increase be }y\%.
\displaystyle 0.9x\left(1+\frac{y}{100}\right)=x.
\displaystyle 0.9\left(1+\frac{y}{100}\right)=1.
\displaystyle 1+\frac{y}{100}=\frac{1}{0.9}=\frac{10}{9}.
\displaystyle \frac{y}{100}=\frac{10}{9}-1=\frac{1}{9}.
\displaystyle y=\frac{100}{9}=11\frac{1}{9}.
\displaystyle \therefore \text{the new price must be increased by }11\frac{1}{9}\%.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The price of tea is increased by }20\%.\text{ By what}
\displaystyle \text{percentage should a housewife reduce the consumption of tea so that her}
\displaystyle \text{expenditure on tea does not increase?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original price of tea per unit be Rs. }x.
\displaystyle \text{New price per unit} =x\left(1+\frac{20}{100}\right)=1.2x.
\displaystyle \text{Let the original consumption be }q\text{ units}.
\displaystyle \text{Original expenditure}=xq.
\displaystyle \text{Let the consumption be reduced by }y\%.
\displaystyle \text{New consumption} =q\left(1-\frac{y}{100}\right).
\displaystyle \text{Since the expenditure remains unchanged,}
\displaystyle 1.2xq\left(1-\frac{y}{100}\right)=xq.
\displaystyle 1.2\left(1-\frac{y}{100}\right)=1.
\displaystyle 1-\frac{y}{100}=\frac{1}{1.2}=\frac{5}{6}.
\displaystyle \frac{y}{100}=\frac{1}{6}.
\displaystyle y=16\frac{2}{3}.
\displaystyle \therefore \text{the consumption of tea should be reduced by }16\frac{2}{3}\%.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{A man gave }35\%\text{ of his money to his elder son and}
\displaystyle 40\%\text{ of the remainder to his younger son. He is now left with Rs. }11700.
\displaystyle \text{How much money did he have originally?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the amount of money he had originally be Rs. }x.
\displaystyle \text{Money given to the elder son} =\frac{35}{100}x=0.35x.
\displaystyle \text{Money remaining}=x-0.35x=0.65x.
\displaystyle \text{Money given to the younger son} =\frac{40}{100}\times0.65x=0.26x.
\displaystyle \text{Money left after giving money to both sons} =0.65x-0.26x=0.39x.
\displaystyle 0.39x=11700.
\displaystyle x=\frac{11700}{0.39}=30000.
\displaystyle \therefore \text{the man originally had Rs. }30000.
\displaystyle \\

\displaystyle \textbf{Question 14: }5\%\text{ of the population of a town was killed in an}
\displaystyle \text{earthquake and }8\%\text{ of the remainder left the town. If the population}
\displaystyle \text{of the town is now }43700,\text{ what was the population at the beginning?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the initial population of the town be }x.
\displaystyle \text{Population remaining after the earthquake} =\frac{95}{100}x=0.95x.
\displaystyle \text{Population that left the town} =\frac{8}{100}\times0.95x=0.076x.
\displaystyle \text{Population remaining} =0.95x-0.076x=0.874x.
\displaystyle 0.874x=43700.
\displaystyle x=\frac{43700}{0.874}=50000.
\displaystyle \therefore \text{the population of the town at the beginning was }50000.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A and B are two towers. The height of tower A is }20\%
\displaystyle \text{less than that of tower B. By what percentage is the height of tower B}
\displaystyle \text{more than that of tower A?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the height of tower B be }x.
\displaystyle \text{Height of tower A} =x\left(1-\frac{20}{100}\right)=0.8x.
\displaystyle \text{Difference in their heights} =x-0.8x=0.2x.
\displaystyle \text{Required percentage} =\frac{0.2x}{0.8x}\times100=25\%.
\displaystyle \therefore \text{the height of tower B is }25\%\text{ more than that of tower A}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In an examination, }30\%\text{ of the candidates failed}
\displaystyle \text{in English, }35\%\text{ failed in G.K. and }27\%\text{ failed in both subjects.}
\displaystyle \text{If }310\text{ candidates passed in both subjects, how many candidates appeared}
\displaystyle \text{in the examination?}
\displaystyle \text{Answer:}
\displaystyle \text{Percentage who failed in English}=30\%.
\displaystyle \text{Percentage who failed in G.K.}=35\%.
\displaystyle \text{Percentage who failed in both subjects}=27\%.
\displaystyle \text{Percentage who failed in English only}=30\%-27\%=3\%.
\displaystyle \text{Percentage who failed in G.K. only}=35\%-27\%=8\%.
\displaystyle \text{Percentage who failed in one or both subjects}=3\%+27\%+8\%=38\%.
\displaystyle \text{Percentage who passed in both subjects}=100\%-38\%=62\%.
\displaystyle \text{Let the total number of candidates be }x.
\displaystyle \frac{62}{100}x=310.
\displaystyle x=310\times\frac{100}{62}=500.
\displaystyle \therefore \text{the number of candidates who appeared was }500.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The value of a car depreciates annually by }10\%.\text{ If its}
\displaystyle \text{present value is Rs. }650000,\text{ find its value after two years.}
\displaystyle \text{Answer:}
\displaystyle \text{Present value of the car}=\text{Rs. }650000.
\displaystyle \text{Value after one year} =650000\left(1-\frac{10}{100}\right).
\displaystyle =650000\times0.9=\text{Rs. }585000.
\displaystyle \text{Value after two years} =585000\left(1-\frac{10}{100}\right).
\displaystyle =585000\times0.9=\text{Rs. }526500.
\displaystyle \therefore \text{the value of the car after two years is Rs. }526500.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The population of a village increases by }7\%\text{ every year.}
\displaystyle \text{If the present population is }80000,\text{ what will be the population after}
\displaystyle 2\text{ years?}
\displaystyle \text{Answer:}
\displaystyle \text{Population after one year} =80000\left(1+\frac{7}{100}\right).
\displaystyle =80000\times1.07=85600.
\displaystyle \text{Population after two years} =85600\left(1+\frac{7}{100}\right).
\displaystyle =85600\times1.07=91592.
\displaystyle \therefore \text{the population of the village after two years will be }91592.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{A student was asked to multiply a number by }\frac{5}{3},
\displaystyle \text{but he multiplied it by }\frac{3}{5}\text{ instead. Find the percentage error}
\displaystyle \text{in the calculation.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number be }x.
\displaystyle \text{Correct answer}=\frac{5}{3}x.
\displaystyle \text{Wrong answer}=\frac{3}{5}x.
\displaystyle \text{Error} =\frac{5}{3}x-\frac{3}{5}x =\frac{25x-9x}{15}=\frac{16x}{15}.
\displaystyle \text{Percentage error} =\frac{\text{Error}}{\text{Correct answer}}\times100.
\displaystyle =\frac{\frac{16x}{15}}{\frac{5x}{3}}\times100 =\frac{16}{25}\times100=64\%.
\displaystyle \therefore \text{the percentage error in the calculation is }64\%.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In an election between two candidates, }10\%\text{ of the}
\displaystyle \text{voters did not cast their votes. Of the votes polled, }10\%\text{ were found}
\displaystyle \text{invalid. The successful candidate got }54\%\text{ of the valid votes and won}
\displaystyle \text{by a majority of }1620\text{ votes. Find the number of voters enrolled.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of voters enrolled be }x.
\displaystyle \text{Number of voters who cast their votes} =\frac{90}{100}x=0.9x.
\displaystyle \text{Number of invalid votes} =\frac{10}{100}\times0.9x=0.09x.
\displaystyle \text{Number of valid votes} =0.9x-0.09x=0.81x.
\displaystyle \text{Successful candidate's votes} =\frac{54}{100}\times0.81x.
\displaystyle \text{Other candidate's votes} =\frac{46}{100}\times0.81x.
\displaystyle \text{Difference in their percentage of valid votes} =54\%-46\%=8\%.
\displaystyle \frac{8}{100}\times0.81x=1620.
\displaystyle 0.0648x=1620.
\displaystyle x=\frac{1620}{0.0648}=25000.
\displaystyle \therefore \text{the number of voters enrolled was }25000.
\displaystyle \\

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