\displaystyle \textbf{Question 1: }\text{Find the sum of }\frac{4}{7}\text{ of Rs. }588\text{ and }\frac{5}{9}\text{ of Rs. }432.
\displaystyle \text{Answer:}
\displaystyle \text{Sum}=\frac{4}{7}\times588+\frac{5}{9}\times432
\displaystyle =4\times84+5\times48=336+240=576.
\displaystyle \therefore \text{The required sum is Rs. }576.
\displaystyle \\

\displaystyle \textbf{Question 2: }\frac{16}{25}\text{ of a number is }816.\text{ Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number be }x.
\displaystyle \therefore \frac{16}{25}x=816
\displaystyle \Rightarrow x=\frac{25}{16}\times816
\displaystyle =25\times51=1275.
\displaystyle \therefore \text{The required number is }1275.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{One-fifth of a number exceeds its one-ninth by }16.\text{ Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number be }x.
\displaystyle \therefore \frac{1}{5}x-\frac{1}{9}x=16
\displaystyle \Rightarrow \frac{9x-5x}{45}=16
\displaystyle \Rightarrow \frac{4}{45}x=16
\displaystyle \Rightarrow x=\frac{45}{4}\times16=180.
\displaystyle \therefore \text{The required number is }180.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }\frac{3}{8}\text{ of an estate is worth Rs. }3546000,\text{ what is the value}
\displaystyle \text{of the whole estate?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the value of the whole estate be Rs. }x.
\displaystyle \therefore \frac{3}{8}x=3546000
\displaystyle \Rightarrow x=\frac{8}{3}\times3546000
\displaystyle =8\times1182000=9456000.
\displaystyle \therefore \text{The value of the whole estate is Rs. }9456000.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The product of two fractions is }\frac{3}{20}.\text{ If one of the fractions is }\frac{13}{25},
\displaystyle \text{find the other fraction.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the other fraction be }x.
\displaystyle \therefore x\times\frac{13}{25}=\frac{3}{20}
\displaystyle \Rightarrow x=\frac{3}{20}\times\frac{25}{13}
\displaystyle =\frac{15}{52}.
\displaystyle \therefore \text{The other fraction is }\frac{15}{52}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\frac{4}{15}\text{ of }\frac{5}{7}\text{ of a number is greater than }\frac{4}{9}\text{ of }\frac{2}{5}\text{ of the same}
\displaystyle \text{number by }8.\text{ Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number be }x.
\displaystyle \therefore \frac{4}{15}\times\frac{5}{7}x-\frac{4}{9}\times\frac{2}{5}x=8
\displaystyle \Rightarrow \frac{4}{21}x-\frac{8}{45}x=8
\displaystyle \Rightarrow \frac{60-56}{315}x=8
\displaystyle \Rightarrow \frac{4}{315}x=8
\displaystyle \Rightarrow x=8\times\frac{315}{4}=630.
\displaystyle \therefore \text{The required number is }630.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the fraction which is as much greater than }\frac{5}{8}\text{ as it is less than }\frac{6}{7}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the fraction be }x.
\displaystyle \therefore x-\frac{5}{8}=\frac{6}{7}-x
\displaystyle \Rightarrow 2x=\frac{5}{8}+\frac{6}{7}
\displaystyle \Rightarrow 2x=\frac{35+48}{56}=\frac{83}{56}
\displaystyle \Rightarrow x=\frac{83}{112}.
\displaystyle \therefore \text{The required fraction is }\frac{83}{112}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{On a certain day, }810\text{ people visited the school exhibition. Out of these, }\frac{7}{15}
\displaystyle \text{were men, }\frac{11}{30}\text{ were ladies and the rest were children. How many children visited the exhibition?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of children be }x.
\displaystyle \therefore \frac{7}{15}\times810+\frac{11}{30}\times810+x=810
\displaystyle \Rightarrow \left(\frac{14}{30}+\frac{11}{30}\right)\times810+x=810
\displaystyle \Rightarrow \frac{25}{30}\times810+x=810
\displaystyle \Rightarrow x=810-\frac{25}{30}\times810
\displaystyle =\left(1-\frac{25}{30}\right)\times810
\displaystyle =\frac{5}{30}\times810=135.
\displaystyle \therefore 135\text{ children visited the exhibition.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{A cake weighs }2\text{ kg. If }\frac{2}{7}\text{ of its weight is flour, }\frac{1}{8}\text{ of its weight is sugar,}
\displaystyle \frac{3}{14}\text{ of its weight is milk and the rest is nuts and plums, find the weight of nuts and plums in the cake.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the weight of nuts and plums be }x\text{ kg.}
\displaystyle \therefore \frac{2}{7}\times2+\frac{1}{8}\times2+\frac{3}{14}\times2+x=2
\displaystyle \Rightarrow \frac{4}{7}+\frac{1}{4}+\frac{3}{7}+x=2
\displaystyle \Rightarrow 1+\frac{1}{4}+x=2
\displaystyle \Rightarrow x=2-\frac{5}{4}=\frac{3}{4}\text{ kg}.
\displaystyle \frac{3}{4}\text{ kg}=\frac{3}{4}\times1000\text{ g}=750\text{ g}.
\displaystyle \therefore \text{The weight of nuts and plums is }\frac{3}{4}\text{ kg or }750\text{ g}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\frac{1}{10}\text{ of a pole is colored white, }\frac{1}{20}\text{ is colored red, }\frac{1}{30}\text{ is colored green}
\displaystyle \text{and the rest is colored black. If the length of the black portion is }14\text{ m }70\text{ cm, find the length of the pole.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length of the pole be }x\text{ m.}
\displaystyle 14\text{ m }70\text{ cm}=14.70\text{ m}.
\displaystyle \therefore \frac{1}{10}x+\frac{1}{20}x+\frac{1}{30}x+14.70=x
\displaystyle \Rightarrow \left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}\right)x+14.70=x
\displaystyle \Rightarrow \frac{11}{60}x+14.70=x
\displaystyle \Rightarrow x-\frac{11}{60}x=14.70
\displaystyle \Rightarrow \frac{49}{60}x=14.70
\displaystyle \Rightarrow x=\frac{60}{49}\times14.70=18.
\displaystyle \therefore \text{The length of the pole is }18\text{ m}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A man had Rs. }121200.\text{ He gave }\frac{1}{3}\text{ of it to his wife, }\frac{3}{10}\text{ of the remainder}
\displaystyle \text{to his daughter and }\frac{4}{7}\text{ of the remainder to his elder son. The rest of the money he gave to his younger son.}
\displaystyle \text{How much money did the younger son get?}
\displaystyle \text{Answer:}
\displaystyle \text{Wife's share}=\frac{1}{3}\times121200=\text{Rs. }40400.
\displaystyle \text{Money remaining}=121200-40400=\text{Rs. }80800.
\displaystyle \text{Daughter's share}=\frac{3}{10}\times80800=\text{Rs. }24240.
\displaystyle \text{Money remaining}=80800-24240=\text{Rs. }56560.
\displaystyle \text{Elder son's share}=\frac{4}{7}\times56560=\text{Rs. }32320.
\displaystyle \text{Younger son's share}=56560-32320=\text{Rs. }24240.
\displaystyle \therefore \text{The younger son received Rs. }24240.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }\frac{1}{8}\text{ of a pencil is black, }\frac{1}{2}\text{ of the remaining is white and the remaining}
\displaystyle 3\frac{1}{2}\text{ cm is blue, find the total length of the pencil.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length of the pencil be }x\text{ cm}.
\displaystyle \text{Black portion}=\frac{1}{8}x.
\displaystyle \text{Remaining portion}=x-\frac{1}{8}x=\frac{7}{8}x.
\displaystyle \text{White portion}=\frac{1}{2}\times\frac{7}{8}x=\frac{7}{16}x.
\displaystyle \therefore \frac{1}{8}x+\frac{7}{16}x+3\frac{1}{2}=x
\displaystyle \Rightarrow \frac{2}{16}x+\frac{7}{16}x+\frac{7}{2}=x
\displaystyle \Rightarrow \frac{9}{16}x+\frac{7}{2}=x
\displaystyle \Rightarrow \frac{7}{2}=\frac{7}{16}x
\displaystyle \Rightarrow x=\frac{7}{2}\times\frac{16}{7}=8.
\displaystyle \therefore \text{The total length of the pencil is }8\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{A man spends }\frac{3}{7}\text{ of his monthly income on food and clothes and }\frac{7}{10}\text{ of the remainder}
\displaystyle \text{on house rent. What fraction of his income is left with him? If the money left is Rs. }2400,
\displaystyle \text{what is his monthly income?}
\displaystyle \text{Answer:}
\displaystyle \text{Fraction of income remaining after food and clothes}=1-\frac{3}{7}=\frac{4}{7}.
\displaystyle \text{Fraction spent on house rent}=\frac{7}{10}\times\frac{4}{7}=\frac{2}{5}.
\displaystyle \text{Fraction of income left}=1-\frac{3}{7}-\frac{2}{5}
\displaystyle =1-\frac{15}{35}-\frac{14}{35}=\frac{6}{35}.
\displaystyle \text{Let the monthly income be Rs. }x.
\displaystyle \therefore \frac{6}{35}x=2400
\displaystyle \Rightarrow x=2400\times\frac{35}{6}=14000.
\displaystyle \therefore \text{The fraction of income left is }\frac{6}{35}\text{ and the monthly income is Rs. }14000.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{A student was asked to multiply a given number by }\frac{8}{17}.\text{ Instead, he divided the given}
\displaystyle \text{number by }\frac{8}{17}.\text{ His answer was }225\text{ more than the correct answer. What was the given number?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the given number be }x.
\displaystyle \text{Incorrect answer}=x\div\frac{8}{17}=\frac{17}{8}x.
\displaystyle \text{Correct answer}=\frac{8}{17}x.
\displaystyle \therefore \frac{17}{8}x-\frac{8}{17}x=225
\displaystyle \Rightarrow \frac{17\times17-8\times8}{8\times17}x=225
\displaystyle \Rightarrow \frac{289-64}{136}x=225
\displaystyle \Rightarrow \frac{225}{136}x=225
\displaystyle \Rightarrow x=136.
\displaystyle \therefore \text{The given number was }136.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The highest score in an inning was }\frac{3}{11}\text{ of the total and the next highest was}
\displaystyle \frac{3}{11}\text{ of the remainder. If the scores differed by }63,\text{ find the total score.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the total score be }x.
\displaystyle \text{Highest score}=\frac{3}{11}x.
\displaystyle \text{Remainder}=x-\frac{3}{11}x=\frac{8}{11}x.
\displaystyle \text{Next highest score}=\frac{3}{11}\times\frac{8}{11}x=\frac{24}{121}x.
\displaystyle \therefore \frac{3}{11}x-\frac{24}{121}x=63
\displaystyle \Rightarrow \frac{33}{121}x-\frac{24}{121}x=63
\displaystyle \Rightarrow \frac{9}{121}x=63
\displaystyle \Rightarrow x=\frac{121}{9}\times63=847.
\displaystyle \therefore \text{The total score was }847.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In a school, }\frac{1}{5}\text{ of the girls and }\frac{1}{8}\text{ of the boys took part in a social camp.}
\displaystyle \text{What fraction of the total strength took part in the camp?}
\displaystyle \text{Answer:}
\displaystyle \text{The required fraction cannot be determined from the given information.}
\displaystyle \text{Let the number of girls be }G\text{ and the number of boys be }B.
\displaystyle \text{Number of girls who took part}=\frac{G}{5}.
\displaystyle \text{Number of boys who took part}=\frac{B}{8}.
\displaystyle \therefore \text{Fraction of the total strength that took part}
\displaystyle =\frac{\frac{G}{5}+\frac{B}{8}}{G+B}
\displaystyle =\frac{8G+5B}{40(G+B)}.
\displaystyle \therefore \text{The ratio of girls to boys is required to find a numerical answer.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{A train starts full of passengers. At the first station, it drops one-third of the passengers}
\displaystyle \text{and takes }280\text{ more. At the second station, it drops one-half of the new total and takes }12\text{ more.}
\displaystyle \text{On arriving at the third station, it is found to have }248\text{ passengers. Find the number of passengers in the beginning.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the initial number of passengers be }x.
\displaystyle \text{After the first station, the number of passengers}=\frac{2}{3}x+280.
\displaystyle \text{After the second station, the number of passengers}
\displaystyle =\frac{1}{2}\left(\frac{2}{3}x+280\right)+12.
\displaystyle \therefore \frac{1}{2}\left(\frac{2}{3}x+280\right)+12=248
\displaystyle \Rightarrow \frac{1}{3}x+140+12=248
\displaystyle \Rightarrow \frac{1}{3}x=96
\displaystyle \Rightarrow x=288.
\displaystyle \therefore \text{The train had }288\text{ passengers in the beginning.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In a fraction, the denominator exceeds the numerator by }4.\text{ If the numerator and}
\displaystyle \text{the denominator are both increased by }9,\text{ the fraction becomes }\frac{7}{8}.\text{ Find the original fraction.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the numerator be }x.
\displaystyle \therefore \text{The denominator}=x+4.
\displaystyle \therefore \text{The original fraction}=\frac{x}{x+4}.
\displaystyle \text{On increasing both the numerator and denominator by }9,
\displaystyle \frac{x+9}{x+13}=\frac{7}{8}.
\displaystyle \Rightarrow 8(x+9)=7(x+13)
\displaystyle \Rightarrow 8x+72=7x+91
\displaystyle \Rightarrow x=19.
\displaystyle \therefore \text{The original fraction}=\frac{19}{19+4}=\frac{19}{23}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{A water tank was }\frac{5}{8}\text{ filled with water. }\frac{7}{10}\text{ of this water was consumed and then}
\displaystyle 125\text{ litres of water was added through a filling pipe. If the tank was now half-full, find the capacity of the tank.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the capacity of the tank be }x\text{ litres}.
\displaystyle \therefore \frac{5}{8}x-\frac{7}{10}\times\frac{5}{8}x+125=\frac{1}{2}x
\displaystyle \Rightarrow \frac{5}{8}x-\frac{7}{16}x+125=\frac{1}{2}x
\displaystyle \Rightarrow \frac{10}{16}x-\frac{7}{16}x+125=\frac{8}{16}x
\displaystyle \Rightarrow \frac{3}{16}x+125=\frac{8}{16}x
\displaystyle \Rightarrow 125=\frac{5}{16}x
\displaystyle \Rightarrow x=125\times\frac{16}{5}=400.
\displaystyle \therefore \text{The capacity of the tank is }400\text{ litres}.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{A car covers }313\frac{3}{5}\text{ km in }3\frac{11}{15}\text{ hours. Find the speed of the car in km/hr.}
\displaystyle \text{Answer:}
\displaystyle \text{Speed}=\frac{\text{Distance covered}}{\text{Time taken}}
\displaystyle =313\frac{3}{5}\div3\frac{11}{15}
\displaystyle =\frac{1568}{5}\div\frac{56}{15}
\displaystyle =\frac{1568}{5}\times\frac{15}{56}
\displaystyle =84.
\displaystyle \therefore \text{The speed of the car is }84\text{ km/hr}.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Find the least fraction that must be subtracted from the sum of }2\frac{1}{2},\ 3\frac{2}{3},
\displaystyle 4\frac{3}{4}\text{ and }5\frac{4}{5}\text{ so that the result is a whole number.}
\displaystyle \text{Answer:}
\displaystyle \text{Sum}=2\frac{1}{2}+3\frac{2}{3}+4\frac{3}{4}+5\frac{4}{5}
\displaystyle =\frac{5}{2}+\frac{11}{3}+\frac{19}{4}+\frac{29}{5}
\displaystyle =\frac{150+220+285+348}{60}
\displaystyle =\frac{1003}{60}=16\frac{43}{60}.
\displaystyle \therefore \text{The least fraction that must be subtracted is }\frac{43}{60}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Rajan earns twice as much in the month of June as in each of the other months of the year.}
\displaystyle \text{What part of his entire annual earnings was earned in June?}
\displaystyle \text{Answer:}
\displaystyle \text{Let Rajan's monthly income in each month other than June be Rs. }x.
\displaystyle \therefore \text{His income in June}=\text{Rs. }2x.
\displaystyle \text{Income during the other }11\text{ months}=11x.
\displaystyle \therefore \text{Total annual income}=11x+2x=13x.
\displaystyle \text{Fraction of annual income earned in June}=\frac{2x}{13x}=\frac{2}{13}.
\displaystyle \therefore \frac{2}{13}\text{ of his entire annual earnings was earned in June.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{A certain amount is distributed among }A,\ B\text{ and }C.\ A\text{ gets }\frac{3}{16}\text{ and }B\text{ gets }\frac{1}{4}
\displaystyle \text{of the whole amount. If }C\text{ gets Rs. }8100,\text{ find the amount which }B\text{ gets.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the total amount be Rs. }x.
\displaystyle \therefore \frac{3}{16}x+\frac{1}{4}x+8100=x
\displaystyle \Rightarrow \frac{3}{16}x+\frac{4}{16}x+8100=x
\displaystyle \Rightarrow \frac{7}{16}x+8100=x
\displaystyle \Rightarrow 8100=\frac{9}{16}x
\displaystyle \Rightarrow x=8100\times\frac{16}{9}=14400.
\displaystyle \therefore B\text{'s share}=\frac{1}{4}\times14400=\text{Rs. }3600.
\displaystyle \therefore B\text{ gets Rs. }3600.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{In a family, the father took }\frac{1}{4}\text{ of the cake and he had as much as each of the other}
\displaystyle \text{members had. Find the total number of family members.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the total amount of cake be }x\text{ and the number of family members be }n.
\displaystyle \text{Since each family member gets an equal share, each member gets }\frac{x}{n}.
\displaystyle \text{The father gets }\frac{1}{4}\text{ of the cake, i.e., }\frac{x}{4}.
\displaystyle \therefore \frac{x}{n}=\frac{x}{4}
\displaystyle \Rightarrow n=4.
\displaystyle \therefore \text{The total number of family members is }4.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{A sum of Rs. }20060\text{ has been divided among }A,\ B\text{ and }C\text{ such that }A\text{ gets }\frac{2}{3}
\displaystyle \text{of what }B\text{ gets and }B\text{ gets }\frac{1}{4}\text{ of what }C\text{ gets. Find }B\text{'s share.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }C\text{'s share be Rs. }x.
\displaystyle \therefore B\text{'s share}=\frac{1}{4}x.
\displaystyle \text{and }A\text{'s share}=\frac{2}{3}\times\frac{1}{4}x=\frac{1}{6}x.
\displaystyle \therefore x+\frac{1}{4}x+\frac{1}{6}x=20060
\displaystyle \Rightarrow \frac{12x+3x+2x}{12}=20060
\displaystyle \Rightarrow \frac{17}{12}x=20060
\displaystyle \Rightarrow x=20060\times\frac{12}{17}=14160.
\displaystyle \therefore B\text{'s share}=\frac{1}{4}\times14160=\text{Rs. }3540.
\displaystyle \therefore B\text{ gets Rs. }3540.
\displaystyle \\

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