\displaystyle \textbf{Question 1: }\text{17 less than four times a number is 11.}
\displaystyle \text{Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number be }x.
\displaystyle \therefore 4x-17=11
\displaystyle \Rightarrow 4x=28
\displaystyle \Rightarrow x=\frac{28}{4}=7
\displaystyle \therefore \text{The number is }7.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If 10 is added to four times a certain}
\displaystyle \text{number, the result is 5 less than five times the number.}
\displaystyle \text{Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number be }x.
\displaystyle \therefore 4x+10=5x-5
\displaystyle \Rightarrow 10+5=5x-4x
\displaystyle \Rightarrow x=15
\displaystyle \therefore \text{The number is }15.
\displaystyle \\

\displaystyle \textbf{Question 3: }\frac{2}{3}\text{ of a number is 20 less than}
\displaystyle \text{the original number. Find the original number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original number be }x.
\displaystyle \therefore \frac{2}{3}x=x-20
\displaystyle \Rightarrow x-\frac{2}{3}x=20
\displaystyle \Rightarrow \frac{1}{3}x=20
\displaystyle \Rightarrow x=60
\displaystyle \therefore \text{The original number is }60.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{A number is 25 more than }\frac{5}{6} \text{ of itself.}
\displaystyle \text{Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number be }x.
\displaystyle \therefore x=25+\frac{5}{6}x
\displaystyle \Rightarrow x-\frac{5}{6}x=25
\displaystyle \Rightarrow \frac{1}{6}x=25
\displaystyle \Rightarrow x=150
\displaystyle \therefore \text{The number is }150.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{A number is as much greater than 21 as it}
\displaystyle \text{is less than 71. Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number be }x.
\displaystyle \therefore x-21=71-x
\displaystyle \Rightarrow x+x=71+21
\displaystyle \Rightarrow 2x=92
\displaystyle \Rightarrow x=46
\displaystyle \therefore \text{The number is }46.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{6 more than one-fourth of a number is}
\displaystyle \text{two-fifths of the number. Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number be }x.
\displaystyle \therefore \frac{1}{4}x+6=\frac{2}{5}x
\displaystyle \Rightarrow 6=\left(\frac{2}{5}-\frac{1}{4}\right)x
\displaystyle \Rightarrow 6=\left(\frac{8-5}{20}\right)x
\displaystyle \Rightarrow 6=\frac{3}{20}x
\displaystyle \Rightarrow x=40
\displaystyle \therefore \text{The number is }40.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{One-third of a number exceeds one-fourth}
\displaystyle \text{of the number by 15. Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number be }x.
\displaystyle \therefore \frac{1}{3}x-\frac{1}{4}x=15
\displaystyle \Rightarrow \left(\frac{4-3}{12}\right)x=15
\displaystyle \Rightarrow \frac{1}{12}x=15
\displaystyle \Rightarrow x=180
\displaystyle \therefore \text{The number is }180.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If one-fifth of a number decreased by 5}
\displaystyle \text{is 16, find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number be }x.
\displaystyle \therefore \frac{1}{5}x-5=16
\displaystyle \Rightarrow \frac{1}{5}x=21
\displaystyle \Rightarrow x=105
\displaystyle \therefore \text{The number is }105.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{A number when divided by 6 is diminished}
\displaystyle \text{by 40. Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number be }x.
\displaystyle \therefore \frac{x}{6}=x-40
\displaystyle \Rightarrow x-\frac{x}{6}=40
\displaystyle \Rightarrow \frac{5}{6}x=40
\displaystyle \Rightarrow x=48
\displaystyle \therefore \text{The number is }48.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Four-fifths of a number exceeds two-thirds}
\displaystyle \text{of the number by 10. Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number be }x.
\displaystyle \therefore \frac{4}{5}x-\frac{2}{3}x=10
\displaystyle \Rightarrow \left(\frac{12-10}{15}\right)x=10
\displaystyle \Rightarrow \frac{2}{15}x=10
\displaystyle \Rightarrow x=75
\displaystyle \therefore \text{The number is }75.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Two numbers are in the ratio }3:4 \text{ and their sum is }84.
\displaystyle \text{Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two numbers be }3x\text{ and }4x.
\displaystyle \therefore 3x+4x=84
\displaystyle \Rightarrow 7x=84
\displaystyle \Rightarrow x=12
\displaystyle \therefore 3x=3\times12=36
\displaystyle \text{and }4x=4\times12=48.
\displaystyle \therefore \text{The two numbers are }36\text{ and }48.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Three numbers are in the ratio }4:5:6 \text{ and their sum is }135.
\displaystyle \text{Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three numbers be }4x,\ 5x\text{ and }6x.
\displaystyle \therefore 4x+5x+6x=135
\displaystyle \Rightarrow 15x=135
\displaystyle \Rightarrow x=9
\displaystyle \therefore 4x=36,\qquad 5x=45,\qquad 6x=54.
\displaystyle \therefore \text{The three numbers are }36,\ 45\text{ and }54.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Two numbers are in the ratio }3:5.
\displaystyle \text{If each number is increased by }10,\text{ the ratio of the new}
\displaystyle \text{numbers is }5:7.\text{ Find the original numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original numbers be }3x\text{ and }5x.
\displaystyle \text{After increasing each number by }10,\text{ the numbers are}
\displaystyle 3x+10\text{ and }5x+10.
\displaystyle \therefore \frac{3x+10}{5x+10}=\frac{5}{7}
\displaystyle \Rightarrow 7(3x+10)=5(5x+10)
\displaystyle \Rightarrow 21x+70=25x+50
\displaystyle \Rightarrow 20=4x
\displaystyle \Rightarrow x=5
\displaystyle \therefore 3x=15\text{ and }5x=25.
\displaystyle \therefore \text{The original numbers are }15\text{ and }25.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The sum of three consecutive odd numbers}
\displaystyle \text{is }75.\text{ Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three consecutive odd numbers be }x,\ x+2,\ x+4.
\displaystyle \therefore x+(x+2)+(x+4)=75
\displaystyle \Rightarrow 3x+6=75
\displaystyle \Rightarrow 3x=69
\displaystyle \Rightarrow x=23
\displaystyle \therefore x+2=25\text{ and }x+4=27.
\displaystyle \therefore \text{The three numbers are }23,\ 25\text{ and }27.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Divide 25 into two parts such that 7 times}
\displaystyle \text{the first part added to 5 times the second part is 139.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two parts be }x\text{ and }y.
\displaystyle \therefore x+y=25
\displaystyle \text{Also, }7x+5y=139.
\displaystyle \text{From }x+y=25,\qquad x=25-y.
\displaystyle \text{Substituting in }7x+5y=139,
\displaystyle 7(25-y)+5y=139
\displaystyle \Rightarrow 175-7y+5y=139
\displaystyle \Rightarrow 175-2y=139
\displaystyle \Rightarrow 2y=36
\displaystyle \Rightarrow y=18
\displaystyle \therefore x=25-18=7.
\displaystyle \therefore \text{The two parts are }7\text{ and }18.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Divide 180 into two parts such that the first}
\displaystyle \text{part is 12 less than twice the second part.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first part be }x\text{ and the second part be }y.
\displaystyle \therefore x+y=180
\displaystyle \text{Also, }x=2y-12.
\displaystyle \text{From }x+y=180,\qquad y=180-x.
\displaystyle \text{Substituting in }x=2y-12,
\displaystyle x=2(180-x)-12
\displaystyle \Rightarrow x=360-2x-12
\displaystyle \Rightarrow 3x=348
\displaystyle \Rightarrow x=116
\displaystyle \therefore y=180-116=64.
\displaystyle \therefore \text{The two parts are }116\text{ and }64.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The denominator of a fraction is 4 more than}
\displaystyle \text{its numerator. On subtracting 1 from both the numerator and}
\displaystyle \text{denominator, the fraction becomes }\frac{1}{2}. \text{ Find the original fraction.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the numerator be }x.
\displaystyle \therefore \text{The denominator is }x+4.
\displaystyle \therefore \text{The original fraction is }\frac{x}{x+4}.
\displaystyle \text{After subtracting 1 from the numerator and denominator,}
\displaystyle \frac{x-1}{x+3}=\frac{1}{2}
\displaystyle \Rightarrow 2(x-1)=x+3
\displaystyle \Rightarrow 2x-2=x+3
\displaystyle \Rightarrow x=5
\displaystyle \therefore x+4=9.
\displaystyle \therefore \text{The original fraction is }\frac{5}{9}.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The denominator of a fraction is 1 more than}
\displaystyle \text{twice its numerator. On adding 2 to the numerator and subtracting}
\displaystyle \text{3 from the denominator, the fraction becomes 1.}
\displaystyle \text{Find the original fraction.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the numerator be }x.
\displaystyle \therefore \text{The denominator is }2x+1.
\displaystyle \therefore \text{The original fraction is }\frac{x}{2x+1}.
\displaystyle \text{According to the given condition,}
\displaystyle \frac{x+2}{2x+1-3}=1
\displaystyle \Rightarrow \frac{x+2}{2x-2}=1
\displaystyle \Rightarrow x+2=2x-2
\displaystyle \Rightarrow x=4
\displaystyle \therefore 2x+1=9.
\displaystyle \therefore \text{The original fraction is }\frac{4}{9}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The sum of the digits of a two-digit number}
\displaystyle \text{is }5.\text{ On adding }27\text{ to the number, its digits are reversed.}
\displaystyle \text{Find the original number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the tens digit be }x\text{ and the units digit be }y.
\displaystyle \therefore \text{The original number is }10x+y.
\displaystyle \text{The number obtained by reversing the digits is }10y+x.
\displaystyle \text{Given, }x+y=5.
\displaystyle \text{Also, }10x+y+27=10y+x.
\displaystyle \Rightarrow 9x+27=9y
\displaystyle \Rightarrow x+3=y
\displaystyle \text{Since }x+y=5,\qquad y=5-x.
\displaystyle \therefore x+3=5-x
\displaystyle \Rightarrow 2x=2
\displaystyle \Rightarrow x=1
\displaystyle \therefore y=5-1=4.
\displaystyle \therefore \text{The original number is }14.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{What number should be added to each of the}
\displaystyle \text{numbers }15,\ 23,\ 29\text{ and }44\text{ so that the resulting}
\displaystyle \text{numbers are in proportion?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number added to each of the four numbers be }x.
\displaystyle \therefore \frac{15+x}{23+x}=\frac{29+x}{44+x}
\displaystyle \Rightarrow (15+x)(44+x)=(23+x)(29+x)
\displaystyle \Rightarrow 660+59x+x^2=667+52x+x^2
\displaystyle \Rightarrow 7x=7
\displaystyle \Rightarrow x=1
\displaystyle \therefore \text{The required number is }1.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{The sum of two numbers is }110.
\displaystyle \text{One-fifth of the larger number is }8\text{ more than one-ninth}
\displaystyle \text{of the smaller number. Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the larger number be }x\text{ and the smaller number be }y.
\displaystyle \therefore x+y=110
\displaystyle \text{Also, }\frac{1}{5}x=\frac{1}{9}y+8.
\displaystyle \text{From }x+y=110,\qquad y=110-x.
\displaystyle \text{Substituting,}
\displaystyle \frac{1}{5}x=\frac{1}{9}(110-x)+8
\displaystyle \Rightarrow \left(\frac{1}{5}+\frac{1}{9}\right)x=\frac{110}{9}+8
\displaystyle \Rightarrow \frac{14}{45}x=\frac{182}{9}
\displaystyle \Rightarrow x=\frac{182}{9}\times\frac{45}{14}=65
\displaystyle \therefore y=110-65=45.
\displaystyle \therefore \text{The two numbers are }65\text{ and }45.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{A number is subtracted from the numerator}
\displaystyle \text{of the fraction }\frac{12}{13}\text{ and six times that number is}
\displaystyle \text{added to the denominator. If the new fraction is }\frac{1}{11},
\displaystyle \text{find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number be }x.
\displaystyle \therefore \frac{12-x}{13+6x}=\frac{1}{11}
\displaystyle \Rightarrow 11(12-x)=13+6x
\displaystyle \Rightarrow 132-11x=13+6x
\displaystyle \Rightarrow 132-13=6x+11x
\displaystyle \Rightarrow 119=17x
\displaystyle \Rightarrow x=7
\displaystyle \therefore \text{The required number is }7.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{A right-angled triangle has a perimeter}
\displaystyle \text{of }120\text{ cm. Its two perpendicular sides are in the ratio }5:12.
\displaystyle \text{Find the lengths of its sides.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two perpendicular sides be }5x\text{ cm and }12x\text{ cm.}
\displaystyle \text{By Pythagoras' theorem,}
\displaystyle \text{Hypotenuse}=\sqrt{(5x)^2+(12x)^2}
\displaystyle =\sqrt{25x^2+144x^2}=\sqrt{169x^2}=13x.
\displaystyle \text{Since the perimeter is }120\text{ cm,}
\displaystyle 5x+12x+13x=120
\displaystyle \Rightarrow 30x=120
\displaystyle \Rightarrow x=4
\displaystyle \therefore 5x=20,\qquad 12x=48,\qquad 13x=52.
\displaystyle \therefore \text{The lengths of the sides are }20\text{ cm, }48\text{ cm} \text{ and }52\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{The sum of the digits of a two-digit number}
\displaystyle \text{is }9.\text{ If }9\text{ is added to the number formed by reversing}
\displaystyle \text{the digits, the result is thrice the original number.}
\displaystyle \text{Find the original number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the tens digit be }x\text{ and the units digit be }y.
\displaystyle \therefore \text{The original number is }10x+y.
\displaystyle \text{The number obtained by reversing the digits is }10y+x.
\displaystyle \text{Given, }x+y=9.
\displaystyle \text{Also, }10y+x+9=3(10x+y).
\displaystyle \Rightarrow 10y+x+9=30x+3y
\displaystyle \Rightarrow 7y+9=29x
\displaystyle \text{Since }x+y=9,\qquad y=9-x.
\displaystyle \therefore 7(9-x)+9=29x
\displaystyle \Rightarrow 63-7x+9=29x
\displaystyle \Rightarrow 72=36x
\displaystyle \Rightarrow x=2
\displaystyle \therefore y=9-2=7.
\displaystyle \therefore \text{The original number is }27.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{The length of a rectangular plot of land}
\displaystyle \text{exceeds its breadth by }23\text{ m. If the length is decreased by}
\displaystyle 15\text{ m and the breadth is increased by }7\text{ m, the area is reduced}
\displaystyle \text{by }360\text{ m}^2.\text{ Find the length and breadth of the plot.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length be }l\text{ m and the breadth be }b\text{ m.}
\displaystyle \text{Since the length exceeds the breadth by }23\text{ m,}
\displaystyle l=b+23.
\displaystyle \text{Original area}=lb.
\displaystyle \text{New area}=(l-15)(b+7).
\displaystyle \text{Since the area is reduced by }360\text{ m}^2,
\displaystyle (l-15)(b+7)=lb-360.
\displaystyle \text{Substituting }l=b+23,
\displaystyle (b+23-15)(b+7)=(b+23)b-360
\displaystyle \Rightarrow (b+8)(b+7)=b^2+23b-360
\displaystyle \Rightarrow b^2+15b+56=b^2+23b-360
\displaystyle \Rightarrow 416=8b
\displaystyle \Rightarrow b=52
\displaystyle \therefore l=b+23=52+23=75.
\displaystyle \therefore \text{The length is }75\text{ m and the breadth is }52\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{The length of a rectangular park is twice}
\displaystyle \text{its breadth. If the perimeter of the park is }186\text{ m, find its}
\displaystyle \text{length and breadth.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length be }l\text{ m and the breadth be }b\text{ m.}
\displaystyle \text{Given, }l=2b.
\displaystyle \text{Perimeter}=2(l+b)=186.
\displaystyle \Rightarrow 2l+2b=186
\displaystyle \Rightarrow 2(2b)+2b=186
\displaystyle \Rightarrow 6b=186
\displaystyle \Rightarrow b=31
\displaystyle \therefore l=2b=62.
\displaystyle \therefore \text{The length is }62\text{ m and the breadth is }31\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{The length of a rectangle is }7\text{ cm more}
\displaystyle \text{than its breadth. If the perimeter of the rectangle is }90\text{ cm,}
\displaystyle \text{find its length and breadth.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length be }l\text{ cm and the breadth be }b\text{ cm.}
\displaystyle \therefore l=b+7
\displaystyle \text{Perimeter}=2(l+b)=90.
\displaystyle \Rightarrow 2l+2b=90
\displaystyle \Rightarrow 2(b+7)+2b=90
\displaystyle \Rightarrow 2b+14+2b=90
\displaystyle \Rightarrow 4b=76
\displaystyle \Rightarrow b=19
\displaystyle \therefore l=19+7=26.
\displaystyle \therefore \text{The length is }26\text{ cm and the breadth is }19\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{The length of a rectangle is }7\text{ cm less}
\displaystyle \text{than twice its breadth. If the length is decreased by }2\text{ cm}
\displaystyle \text{and the breadth is increased by }3\text{ cm, the perimeter of the}
\displaystyle \text{resulting rectangle is }66\text{ cm. Find the length and breadth}
\displaystyle \text{of the original rectangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length be }l\text{ cm and the breadth be }b\text{ cm.}
\displaystyle \therefore l=2b-7
\displaystyle \text{The new length is }l-2\text{ and the new breadth is }b+3.
\displaystyle \therefore 2(l-2)+2(b+3)=66
\displaystyle \Rightarrow 2l-4+2b+6=66
\displaystyle \Rightarrow 2l+2b=64
\displaystyle \text{Substituting }l=2b-7,
\displaystyle 2(2b-7)+2b=64
\displaystyle \Rightarrow 4b-14+2b=64
\displaystyle \Rightarrow 6b=78
\displaystyle \Rightarrow b=13
\displaystyle \therefore l=2(13)-7=19.
\displaystyle \therefore \text{The length is }19\text{ cm and the breadth is }13\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{A man is five times as old as his son.}
\displaystyle \text{In two years' time, he will be four times as old as his son.}
\displaystyle \text{Find their present ages.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the son's present age be }x\text{ years.}
\displaystyle \therefore \text{The man's present age is }5x\text{ years.}
\displaystyle \text{After two years,}
\displaystyle \text{Son's age}=x+2
\displaystyle \text{Man's age}=5x+2
\displaystyle \therefore 5x+2=4(x+2)
\displaystyle \Rightarrow 5x+2=4x+8
\displaystyle \Rightarrow x=6
\displaystyle \therefore 5x=5\times6=30.
\displaystyle \therefore \text{The son's present age is }6\text{ years and the man's} latex \displaystyle \text{present age is }30\text{ years.} $
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{A man is twice as old as his son. Twelve years}
\displaystyle \text{ago, the man was thrice as old as his son. Find their present ages.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the son's present age be }x\text{ years.}
\displaystyle \therefore \text{The man's present age is }2x\text{ years.}
\displaystyle \text{Twelve years ago,}
\displaystyle \text{Son's age}=x-12
\displaystyle \text{Man's age}=2x-12
\displaystyle \therefore 2x-12=3(x-12)
\displaystyle \Rightarrow 2x-12=3x-36
\displaystyle \Rightarrow 36-12=3x-2x
\displaystyle \Rightarrow x=24
\displaystyle \therefore 2x=2\times24=48.
\displaystyle \therefore \text{The son's present age is }24\text{ years and the man's}
\displaystyle \text{present age is }48\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{Seema is }10\text{ years older than Rekha.}
\displaystyle \text{The ratio of their ages is }5:3.\text{ Find their ages.}
\displaystyle \text{Answer:}
\displaystyle \text{Let Rekha's age be }x\text{ years.}
\displaystyle \therefore \text{Seema's age is }x+10\text{ years.}
\displaystyle \text{Given, }\frac{x+10}{x}=\frac{5}{3}.
\displaystyle \Rightarrow 3(x+10)=5x
\displaystyle \Rightarrow 3x+30=5x
\displaystyle \Rightarrow 2x=30
\displaystyle \Rightarrow x=15
\displaystyle \therefore \text{Rekha's age is }15\text{ years.}
\displaystyle \text{Seema's age}=15+10=25\text{ years.}
\displaystyle \therefore \text{Their ages are }15\text{ years and }25\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{Five years ago, Parvati's age was four times}
\displaystyle \text{the age of her son. The sum of their present ages is }55\text{ years.}
\displaystyle \text{Find Parvati's present age.}
\displaystyle \text{Answer:}
\displaystyle \text{Let Parvati's present age be }x\text{ years and her son's present}
\displaystyle \text{age be }y\text{ years.}
\displaystyle \therefore x+y=55
\displaystyle \text{Five years ago, their ages were }x-5\text{ and }y-5.
\displaystyle \therefore x-5=4(y-5)
\displaystyle \Rightarrow x-5=4y-20
\displaystyle \Rightarrow x-4y=-15
\displaystyle \text{From }x+y=55,\qquad y=55-x.
\displaystyle \text{Substituting,}
\displaystyle x-4(55-x)=-15
\displaystyle \Rightarrow x-220+4x=-15
\displaystyle \Rightarrow 5x=205
\displaystyle \Rightarrow x=41
\displaystyle \therefore y=55-41=14.
\displaystyle \therefore \text{Parvati's present age is }41\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{A man is }56\text{ years old and his son is}
\displaystyle 24\text{ years old. In how many years will the father be twice as old}
\displaystyle \text{as his son at that time?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required number of years be }x.
\displaystyle \text{After }x\text{ years, the father's age will be }56+x.
\displaystyle \text{After }x\text{ years, the son's age will be }24+x.
\displaystyle \therefore 56+x=2(24+x)
\displaystyle \Rightarrow 56+x=48+2x
\displaystyle \Rightarrow x=8
\displaystyle \therefore \text{After }8\text{ years, the father will be twice as old}
\displaystyle \text{as his son.}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{Nine years hence, a girl will be three times}
\displaystyle \text{as old as she was nine years ago. How old is she now?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the present age of the girl be }x\text{ years.}
\displaystyle \text{Nine years hence, her age will be }x+9\text{ years.}
\displaystyle \text{Nine years ago, her age was }x-9\text{ years.}
\displaystyle \therefore x+9=3(x-9)
\displaystyle \Rightarrow x+9=3x-27
\displaystyle \Rightarrow 36=2x
\displaystyle \Rightarrow x=18
\displaystyle \therefore \text{The girl's present age is }18\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{A man made a trip of }480\text{ km in }9 \text{ hours.}
\displaystyle \text{Part of the trip was covered at }45\text{ km/hr and the remaining}
\displaystyle \text{part at }60\text{ km/hr. Find the distance covered at }60\text{ km/hr.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the distance covered at }60\text{ km/hr be }x\text{ km.}
\displaystyle \therefore \text{The distance covered at }45\text{ km/hr is }480-x\text{ km.}
\displaystyle \text{Using }\text{Time}=\frac{\text{Distance}}{\text{Speed}},
\displaystyle \frac{480-x}{45}+\frac{x}{60}=9
\displaystyle \text{Multiplying throughout by }180,
\displaystyle 4(480-x)+3x=1620
\displaystyle \Rightarrow 1920-4x+3x=1620
\displaystyle \Rightarrow x=300
\displaystyle \therefore \text{The distance covered at }60\text{ km/hr is }300\text{ km.}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{A motorist travelled from town }A\text{ to town }B
\displaystyle \text{at an average speed of }54\text{ km/hr. On his return journey, his}
\displaystyle \text{average speed was }60\text{ km/hr. If the total time taken was}
\displaystyle 9\text{ hours, find the distance between the two towns.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the distance between towns }A\text{ and }B\text{ be }x\text{ km.}
\displaystyle \text{Time taken from }A\text{ to }B=\frac{x}{54}\text{ hours.}
\displaystyle \text{Time taken from }B\text{ to }A=\frac{x}{60}\text{ hours.}
\displaystyle \therefore \frac{x}{54}+\frac{x}{60}=9
\displaystyle \text{Multiplying throughout by }540,
\displaystyle 10x+9x=4860
\displaystyle \Rightarrow 19x=4860
\displaystyle \Rightarrow x=\frac{4860}{19}
\displaystyle \Rightarrow x=255\frac{15}{19}\text{ km}
\displaystyle \therefore \text{The distance between the two towns is } 255\frac{15}{19}\text{ km.}
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{The distance between two stations is }300 \text{ km.}
\displaystyle \text{Two motorcyclists start simultaneously from these stations and}
\displaystyle \text{move towards each other. The speed of one is }7\text{ km/hr faster}
\displaystyle \text{than that of the other. If the distance between them after }2 \text{ hours}
\displaystyle \text{is }34\text{ km, find the speed of each motorcyclist.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the speed of the slower motorcyclist be }x\text{ km/hr.}
\displaystyle \therefore \text{The speed of the faster motorcyclist is }x+7\text{ km/hr.}
\displaystyle \text{Distance covered by the slower motorcyclist in }2\text{ hours}=2x.
\displaystyle \text{Distance covered by the faster motorcyclist in }2\text{ hours}
\displaystyle =2(x+7).
\displaystyle \text{The remaining distance between them is }34\text{ km.}
\displaystyle \therefore 2x+2(x+7)+34=300
\displaystyle \Rightarrow 2x+2x+14+34=300
\displaystyle \Rightarrow 4x+48=300
\displaystyle \Rightarrow 4x=252
\displaystyle \Rightarrow x=63
\displaystyle \therefore x+7=70.
\displaystyle \therefore \text{Their speeds are }63\text{ km/hr and }70\text{ km/hr.}
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{A boat travels }30\text{ km upstream in a river}
\displaystyle \text{in the same time as it travels }50\text{ km downstream. If the speed}
\displaystyle \text{of the stream is }5\text{ km/hr, find the speed of the boat in still}
\displaystyle \text{water.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the speed of the boat in still water be }x\text{ km/hr.}
\displaystyle \text{Speed of the stream}=5\text{ km/hr.}
\displaystyle \therefore \text{Upstream speed}=x-5\text{ km/hr.}
\displaystyle \text{Downstream speed}=x+5\text{ km/hr.}
\displaystyle \text{Since the time taken in both cases is the same,}
\displaystyle \frac{30}{x-5}=\frac{50}{x+5}
\displaystyle \Rightarrow 30(x+5)=50(x-5)
\displaystyle \Rightarrow 30x+150=50x-250
\displaystyle \Rightarrow 400=20x
\displaystyle \Rightarrow x=20
\displaystyle \therefore \text{The speed of the boat in still water is }20\text{ km/hr.}
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{The length of each of the equal sides of an}
\displaystyle \text{isosceles triangle is }4\text{ cm longer than its base. If the perimeter}
\displaystyle \text{of the triangle is }62\text{ cm, find the lengths of its sides.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length of the base be }x\text{ cm.}
\displaystyle \therefore \text{Each of the two equal sides is }x+4\text{ cm.}
\displaystyle \text{Since the perimeter is }62\text{ cm,}
\displaystyle x+(x+4)+(x+4)=62
\displaystyle \Rightarrow 3x+8=62
\displaystyle \Rightarrow 3x=54
\displaystyle \Rightarrow x=18
\displaystyle \therefore x+4=22.
\displaystyle \therefore \text{The lengths of the sides are }22\text{ cm, }22\text{ cm} \text{ and }18\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{A certain number of candidates appeared for}
\displaystyle \text{an examination. One-fifth of the total plus }16\text{ secured first}
\displaystyle \text{division, one-fourth of the total plus }15\text{ secured second}
\displaystyle \text{division, and one-fourth of the total minus }25\text{ secured third}
\displaystyle \text{division. If the remaining }60\text{ candidates failed, find the total}
\displaystyle \text{number of candidates who appeared for the examination.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the total number of candidates be }x.
\displaystyle \therefore \left(\frac{1}{5}x+16\right)+\left(\frac{1}{4}x+15\right)
\displaystyle +\left(\frac{1}{4}x-25\right)+60=x
\displaystyle \Rightarrow \left(\frac{1}{5}+\frac{1}{4}+\frac{1}{4}\right)x+66=x
\displaystyle \Rightarrow \frac{7}{10}x+66=x
\displaystyle \Rightarrow x-\frac{7}{10}x=66
\displaystyle \Rightarrow \frac{3}{10}x=66
\displaystyle \Rightarrow x=\frac{66\times10}{3}=220
\displaystyle \therefore \text{The total number of candidates is }220.
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{Raman has three times as much money as Kamal.}
\displaystyle \text{If Raman gives Rs. 750 to Kamal, Kamal will have twice as much}
\displaystyle \text{money as Raman has left. How much money did each have originally?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the amount of money with Kamal be Rs. }x.
\displaystyle \therefore \text{The amount of money with Raman is Rs. }3x.
\displaystyle \text{After Raman gives Rs. 750 to Kamal,}
\displaystyle \text{Kamal has Rs. }(x+750)\text{ and Raman has Rs. }(3x-750).
\displaystyle \therefore x+750=2(3x-750)
\displaystyle \Rightarrow x+750=6x-1500
\displaystyle \Rightarrow 5x=2250
\displaystyle \Rightarrow x=450
\displaystyle \therefore 3x=3\times450=1350.
\displaystyle \therefore \text{Kamal had Rs. 450 and Raman had Rs. 1350 originally.}
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{The angles of a triangle are in the ratio }2:3:4.
\displaystyle \text{Find the angles.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the angles be }2x,\ 3x\text{ and }4x.
\displaystyle \text{Since the sum of the angles of a triangle is }180^\circ,
\displaystyle 2x+3x+4x=180^\circ
\displaystyle \Rightarrow 9x=180^\circ
\displaystyle \Rightarrow x=20^\circ
\displaystyle \therefore 2x=40^\circ,\qquad3x=60^\circ,\qquad4x=80^\circ.
\displaystyle \therefore \text{The angles are }40^\circ,\ 60^\circ\text{ and }80^\circ.
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{A certain number of men can finish a piece of}
\displaystyle \text{work in }50\text{ days. If there are }7\text{ more men, the work can}
\displaystyle \text{be completed }10\text{ days earlier. How many men were originally there?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original number of men be }x.
\displaystyle \text{Total work}=50x\text{ man-days.}
\displaystyle \text{With }7\text{ more men, the number of men is }x+7.
\displaystyle \text{The work is then completed in }50-10=40\text{ days.}
\displaystyle \therefore 50x=40(x+7)
\displaystyle \Rightarrow 50x=40x+280
\displaystyle \Rightarrow 10x=280
\displaystyle \Rightarrow x=28
\displaystyle \therefore \text{Originally, there were }28\text{ men.}
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{Divide }600\text{ into two parts such that }40\%
\displaystyle \text{of one part exceeds }60\%\text{ of the other part by }120.
\displaystyle \text{Find the two parts.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two parts be }x\text{ and }y.
\displaystyle \therefore x+y=600
\displaystyle \text{Also, }40\%\text{ of }x-60\%\text{ of }y=120.
\displaystyle \therefore 0.4x-0.6y=120
\displaystyle \text{From }x+y=600,\qquad y=600-x.
\displaystyle \text{Substituting,}
\displaystyle 0.4x-0.6(600-x)=120
\displaystyle \Rightarrow 0.4x-360+0.6x=120
\displaystyle \Rightarrow x=480
\displaystyle \therefore y=600-480=120.
\displaystyle \therefore \text{The two parts are }480\text{ and }120.
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{A workman is paid Rs. 150 for each day he}
\displaystyle \text{works and is fined Rs. 50 for each day he is absent. In a month of}
\displaystyle 30\text{ days, he earned Rs. 2100. For how many days did he remain absent?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of days he worked be }x.
\displaystyle \therefore \text{The number of days he was absent is }30-x.
\displaystyle \text{Amount earned for working}=150x.
\displaystyle \text{Fine for absence}=50(30-x).
\displaystyle \therefore 150x-50(30-x)=2100
\displaystyle \Rightarrow 150x-1500+50x=2100
\displaystyle \Rightarrow 200x=3600
\displaystyle \Rightarrow x=18
\displaystyle \therefore \text{Number of days absent}=30-18=12.
\displaystyle \therefore \text{The workman remained absent for }12\text{ days.}
\displaystyle \\

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