Question 1: $17$ less than four times a number is $11$. Find the number.

Let the number $=x$

$4x-17=11$

$x=$ $\frac{28}{4}$ $=7$

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Question 2: If $10$ be added to four times a certain number, the result is $5$ less than five times the number. Find the number.

Let the number $=x$

$4x+10=5x-5$

$x=15$

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Question 3: $\frac{2}{3}$ Of a number is $20$ less than the original number. Find the original number.

Let the original number $= x$

$\frac{2}{3}$ $x=x-20$

$\frac{1}{3}$ $x=20$

$x=60$

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Question 4: A number is $25$ more than its part. Find the number.

Let the number $= x$

$x=25+$ $\frac{5}{6}$ $x$

$\frac{1}{6}$ $x=25$

$x=150$

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Question 5: A number is as much greater than $21$ as is less than $71$. Find the number.

Let the number $= x$

$x-21=71-x$

$2x=92$

$x=46$

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Question 6: 6 more than one-fourth of the number is two-fifth of the number. Find the number.

Let the number $= x$

$\frac{1}{4} x+6=$ $\frac{2}{5} x$

$6=($ $\frac{2}{5}$ $-$ $\frac{1}{4}$ $)x=$ $\frac{30}{20}$ $x$

$x=40$

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Question 7: One-third of a number exceeds one-fourth of the number by $15$. Find the number.

Let the number $= x$

$\frac{1}{3}$ $x-$ $\frac{1}{4}$ $x=15$

$\frac{1}{12}$ $x=15 \ or \ x = 180$

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Question 8: If one-fifth of a number decreased by $5$ is $16$, find the number.

Let the number $= x$

$\frac{1}{5}$ $x-5=16$

$x=105$

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Question 9: A number when divided by $6$ is diminished by $40$. Find the number.

Let the number $=x$

$\frac{x}{6}$ $=x-40$

$\frac{5}{6}$ $x=40$

$x=48$

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Question 10: Four-fifths of a number exceeds two-third of the number by $10$. Find the number.

Let the number $=x$

$\frac{4}{3} x=$ $\frac{2}{3}$ $x+10$

or $\frac{2}{15}$ $x=10$

or $x=75$

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Question 11: Two numbers are in the ratio $3:4$ and their sum is $84$. Find the number.

Let the two numbers be $x$and $y$

Therefore

$3x=4y$

$x+y=84$

Solving

$\frac{4}{3}$ $y+y=84$

$\frac{7y}{3}$ $=84$

$y=3\times 12=36$

Hence, $x=$ $\frac{4}{3}$ $\times 36=48$

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Question 12: Three numbers are in ratio $4:5:6$ and their sum is $135$. Find the numbers.

Let the three numbers be $x, \ y, \ z$

Therefore,

$4x:5x:6x$

$4x+5x+6x=135$

Solving,

$15x=135$

$x=9$

Therefore

The three numbers are $36, \ 45, \ 54$

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Question 13: Two numbers are in the ratio $3:5$. If each is increased by $10$ , then ratio between the new numbers so formed is $5:7$ , Find the original numbers.

Let the two numbers be $x$ and $y$

Given,

$\frac{x}{y}$ $=$ $\frac{3}{5}$ $...i)$

$\frac{x+10}{y+10}$ $=$ $\frac{5}{7}$ $...ii)$

solving,

From $i) x=$ $\frac{3}{5}$ $y$

Substituting in ii)

$\frac{\frac{3}{5} y+10}{y+10}$ $=$ $\frac{5}{7}$

$\frac{21}{5}$ $y+70=5y+50$

$20=$ $\frac{4}{5}$ $y$

or $y=25$

$x=$ $\frac{3}{5}$ $\times 25=15$

Two numbers are $15$ and $25$

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Question 14: The sum of three consecutive odd numbers is $75$. Find the numbers.

Let the three consecutive numbers be

$x, x+2, x+4$

therefore,

$x+x+2+x+4=75$

$3x+6=75$

$3x=69$

$x=23$

Therefore, the three numbers are $23, \ 25 \ and \ 27$

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Question 15: Divide $25$ into two parts such that $7$ times the first part added to $5$ times the second part makes $139$.

Let the two parts be $x$ and $y$

Therefore

$x+y=25$

$7x+5y+139$

Solving we get

$x=25-y$

$7(25-y)+5y=139$

$175-2y=139$

$2y=175-139=36$

or $y=18$

The other part $=7$

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Question 16: Divide $180$ into two parts such that the first part is $12$ less than twice the second part.

Let the two parts be $x$ and $2y$

Therefore

$x+y=180$

$x+12=2y$

Solving

$y=180-x$

$x+12=2(180-x)$

$3x=360-12=348$

$x=116$

Therefore $y=180-116=64$

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Question 17: The denominator of the fraction is $4$ more than its numerator. On subtracting $1$ from each numerator and denominator the fraction becomes. Find the original fraction.

Let the fraction be $\frac{x}{y}$

Given

$y=x+4$

Therefore, the fraction $=$ $\frac{x}{x+4}$

Given,

$\frac{x-1}{x+4-1}=$ $\frac{1}{2}$

$2x-2=x+3$

$x=5 \ and\ y=9$

Therefore fraction $=$ $\frac{5}{9}$

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Question 18: The denominator of the fraction is $1$ more than the double the numerator. On adding $2$ to the numerator and subtracting $3$ from denominator, we obtain $1$. Find the original fraction.

Let the fraction be $\frac{x}{2x+1}$

Given

$\frac{x+2}{2x+1-3}$ $=1$

$x+2=2x-2$

$x=4$

Fraction $=$ $\frac{4}{9}$

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Question 19: The sum of the digits of a two-digit number is $5$. On adding $27$ to the number, its digits are reversed. Find the original number.

Let the two digit number be $xy$

Given

$x+y=5 ...i)$

$xy+27=yx$

$10x+y+27=10y+x$

$9x+27=9y$

or $x+3=y ...ii)$

Solving i) and ii) together.

$x+3=(5-x)$

$2x=2$

$x=1$

$y=4$

Hence the number $= 14$

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Question 20: What same numbers should be added to each one of the number $15,23,29,44$ to obtain numbers which are in proportion?

Let the number added to each one of $15, 23, 29, 44$ be $x$

$\frac{15+x}{23+x}$ $=$ $\frac{29+x}{44+x}$

$660+59x+ x^2=667+52x+ x^2$

$7x=7$

$x=1$

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Question 21: The sum of two numbers is $110$. One-fifth of the larger number is $8$ more than one-ninth of the smaller number. Find the numbers.

Let the two numbers be $x$ and $y$

Given

$x+y 110$

$\frac{1}{5}$ $x=$ $\frac{1}{9}$ $y+8$

Solving

$\frac{1}{5}$ $x=$ $\frac{1}{9}$ $(110-x)+ 8$

$(\frac{1}{5}$ $+$ $\frac{1}{9}$ $)x=$ $\frac{110}{9}$ $+8=$ $\frac{182}{9}$

$x=$ $\frac{182\times 45}{9\times 14}$ $=65$

$y=10-65=45$

Two numbers are $45$ and $65$

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Question 22: A number is subtracted from the numerator of the fraction $\frac{12}{13}$ and six times that number is added to the denominator. If the new fraction is $\frac{1}{11}$ then find the number.

Let the number subtracted from the numerator $= x$

$\frac{12-x}{13+6x}= \frac{1}{11}$

$132-11x=13+6x$

$17x=119$

or $x=7$

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Question 23: A right angled triangle having perimeter $120\ cm$ has its two-side perpendicular side in the ratio $5:12$. Find the lengths of its sides.

Perimeter of right angled triangle $= 120$

Perpendicular sides $= 5x \ and \ 12x$

Hypotenuse $=\sqrt{(5x)^2+(12x)^2}=13x$

Therefore

$5x+12x+13x=120$

$30x=120$

$x=4$

Therefore, length of side $= 20, 48, 52$

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Question 24: The sum of the digits of a two-digit number is $9$. If $9$ is added to the number formed by reversing the digits, then the result is thrice the original number. Find the original number.

Let the two-digit number $= xy$

$x+y=9 ...i)$

$yx+9=3(xy)$

$10y+x+9=3(10x+y)$

$10y+x+9=30x+3y$

$y+9=29x ...ii)$

Solving i) and ii)

$7(9-x)+9=29x$

$63-7x+9=29x$

$72x=36$

Or $x=2$

$y=9-2=7$

Therefore, the number $= 27$

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Question 25: The lengths of a rectangle plot of land exceeds its breadth by $23 m$ if the length is decreased by $15 m$. and the breadth is increased by $7 m$. the area is reduced by $360 m^2$.

Find the length and the breadth of the plot.

Let the length $= l$ and breadth $= b$

$l= 23 +b$

Given

$(l-15)(b+7)= lb-360$

$(23+b-15)(b+7)=(23+b)b-360$

$(b+8)(b+7)= 23b+b^2-360$

$b^2+15b+56=23b+b^2-360$

$416=8b$

or $b=52m$

Therefore

$l=b+23=52+23=75m$

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Question 26: The length of the rectangular park is twice its breadth. If the perimeter of the park is 186 m, find its length and breadth.

Let the length $= l$ and breadth $= b$

$l=2b$

$2l+2b=186$

$4b+2b=186$

$6b=186$

or $b=31$

$l=62$

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Question 27: The length of the rectangle is $7 cm$ more than its breadth. If the perimeter of the rectangle is $90\ cm$, find its length and breadth.

Let the length $= l$ breadth $= b$

$l= b+7$

Given

$21l +2b = 90$

$2(b+7) +2b = 90$

$4b = 76$

Or $b= 19 cm$

$l= 19+7 = 26 cm$

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Question 28: The length of a rectangle is $7 cm$ less than twice its breadth. If the length is decreased by $2\ cm$ and breadth increased by $3\ cm$, the perimeter of the resulting rectangle is $66 cm$. find the length and the breadth of the original rectangle.

Let the length $= l$ and breadth $= b$

$l+ 7 =2b$

Given,

$2(l-2)+ 2(b+3)= 66$

$2l-4+2b+6=66$

$2l+2b=64$

Solving,

$2(2b-7) + 2b = 64$

$6b=78$

$b=13,$

$l=2\times 13-7=19$

breadth $=13 cm$

length $=19cm$

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Question 29: A man is five times as old as his son. In two years’ time, he will be four times as old as his son. Find their present ages.

Let the man’s age $= 5x$

If son’s age $= x$

Two years letter

Man’s age $= 5x +2$

Son’s age $= x+ 2$

$5x +2 = 4(x+2)$

$x + 6 years = son's \ age$

Man’s age $= 30 yrs.$

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Question 30: A man is twice as old as his son. Twelve years ago, the man was thrice as old as his son. Find their present ages.

Let the son’s age $= x$

Man’s age $= 2x$

$12 \ years \ ago$

Son’s age $= x-12$

Man’s age $= 2x-12$

$2x-12 = 3 (x-12)$

$2x-12 = 3x-36$

$x= 24 = son's \ age$

Man’s age $= 48 \ years$

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Question 31: Seema is $10 \ years$ elder than Rekha. The ratio of their ages is $5:3$. Find their ages.

Let Rekha’s age $= x$

Seema’s age $= x + 10$

given

$\frac{x+10}{x}$ $=$ $\frac{5}{3}$

$3x+30=5x$

$2x=30$

or $x = 15$

Rekha’s sage $=15 yrs.$

Seema’s sage $=25 yrs.$

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Question 32: $5 \ years$ ago, the age of Parvati was $4$ times the age of her son. The sum of their present ages is $55 years$. Find Parvati’s age.

Let the present age of Parvati $= x yrs$

age of son $= y yrs.$

$x+y=55 ...i)$

Five years before

Parvati $=x-5 yrs.$

son $=y-5 yrs.$

Given,

$(x-5)= 4(y-5)$

$x-4y= -15 ...ii)$

solving i) and ii)

$x-4 (55-x)= -15$

$5x=205$

or $x = 44 =$ Parvati’s age

Son’s age $= 55-44 = 11 years$

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Question 33: A man is $56$ years old and his son is $24$ years old. In how many years, the father will be twice as old as his son at that time?

Man’s age $= 56 years$

Son’s age $= 24 years$

Let in $x years$, man would be twice the age of son

$56 +x=2(24=x)$

$56+x=48+2x$

or $x=8 years$

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Question 34: 9 years hence, a girl will be $3$ times as old as she was $9$ years ago. How old is she now?

Let the current age of the girl $= x$

Given,

$x + 9 = 3(x- 9)$

$x+9=3x-27$

$2x=36$

$x=18 \ years =age \ of \ the \ girl$

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Question 35: A man made a trip of $480\ km$ in $9$ hours. Some part of trip was covered at $45 km/hr$ and the remaining at $60\ km/hr$. find the part of the trip covered by him at $60\ km/hr$.

Let the distance covered at $45 km/ hr = x$

Let the distance covered at $60 km/ hr = y$

Total distance $= 480 km.$

$x+y=480$

$\frac{x}{45}$ $+$ $\frac{y}{60}$ $=9$

Solving

$\frac{480-y}{45}$ $+$ $\frac{y}{60}$ $=9$

or $y=300 \ km \ and \ x=180 \ km$

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Question 36: A motorist traveled from town $A$ to town $B$ at an average speed of $54\ km/hr$. on his return journey, his average speed was $60\ km/hr$. if the total time taken is $9 hours$, find the distance between the two towns.

Let the distance between town A and B $= x$

Therefore $\frac{x}{54}$ $+$ $\frac{x}{60}$ $=9.5$

or $x=270 km.$

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Question 37: The distance between two stations is $300\ km$. two motor-cyclist start simultaneously from these stations and move towards each other. The speed of one of them is $7\ km/hr$ faster than that of other. If the distance between them after $2 hours$ is $34\ km$, find the speed of each motor-cycle

Distance $= 300 km$

Let the speed of 1st cyclist $= x$

Then speed of 2nd cyclist $= x + 7$

Distance covered by 1s cyclist in 2hr $=2x$

Distance covered by 2nd cyclist in 2 hr $= 2( x+7)$

Therefore

$2x+34+2(x+7)= 300$

$4x + 48 = 300$

$x= \frac{252}{4}=63 km/hr$

Speed of 1st cyclist $=63km/hr$

Speed of 2nd cyclist $=63+7=70 km/hr$

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Question 38: A boat travels $30\ km$ upstream in river in the same period of time as it travels $50\ km$ downstream. If the ratio of stream be $5 km/hr$, find the speed of the boat in still water.

Let the speed of boat $= x km/hr$

Speed of stream $= 5 km/hr$

Speed of boat upstream $= x-5 km/hr$

Speed of the boat downstream $= x + 5 km/ hr$

Therefore $\frac{30}{x-5}$ $=$ $\frac{50}{x+5}$

$30x + 150 = 50x-250$

$400 = 20 x$

Or $x = 20 km/hr$

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Question 39: The length of each of the equal sides of an isosceles triangle is $4\ cm$ longer than the base. If the perimeter of the triangle is $62\ cm$ , find the lengths of the sides of the triangle.

Let the base $= x cm$

Sides $= (x+ 4)$

Perimeter $= 62 cm$

Therefore

$(x+4) + (x) + (x+ 4) = 62$

$3x+8=62$

$3x=54$

or $x=18$

Base $=18 cm$, Sides $=22cm$

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Question 40: A certain number of candidates appeared for an examination in which one-fifth of the whole plus $16$ secured first division, one-fourth plus $15$ secured second division and one-fourth minus $25$ secured third division, if the remaining $60$ candidates failed, find the total number of candidates appeared.

Let the number of candidates $= x$

$($ $\frac{1}{5}$ $x+6)+ ($ $\frac{1}{4}$ $x+15 )+($ $\frac{1}{4}$ $x-25)+ 60=x$

$6+60=x-($ $\frac{1}{5}$ $latex+$ $\frac{1}{4}$ $latex+$ $\frac{1}{4}$ $)x$

$x(1- 14/20)=66$

$x=$ $\frac{20\times 66}{6}$ $=220$

No. of candidates $=220$

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Question 41: Raman has $3$ times as much money as Kamal. If Raman gives $Rs. 750$ to Kamal, then Kamal will have twice as much as left with Raman. How much had each originally?

Let money with Kamal $= x$

Then money with Raman $= 3x$

$2(3x-750) = (x+750)$

Or $x= 450 Rs.$

Kamal has $450 Rs.$ And Raman $1350 Rs.$

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Question 42: The angles of triangle are in ratio $2:3:4$. Find the angles.

Ratio of angle $= 2: 3: 4$

Therefore
$2x+3x+4x=180$

$9x=180$

$x=20$

Therefore, angles are $40, 60, 80$ degrees.

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Question 43: A certain number man can finish a piece of work in $50$ days. If there are $7$ more men, the work can be completed $10$ days earlier. How many men were originally there?

Let $x$ men finish work in $50 days$

Total work $= 50x \ man \ days$

$x + 7 \ men \ finish \ work \ in \ 49 \ days$

Total work $= (x + 7)\times 40$

Therefore

$50x=40 (x+7)$

$5x=4x+28$

$x=28$

Original no of men $=28$

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Question 44: Divide $600$ in two parts such that $495$ of one exceeds $60$ of the other by $120$.

Let the two parts $= x \ and\ y$

$x + y = 600$

$0.4x-0.6y = 120$

Solving we get

$x= 480$

$y = 120$

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Question 45: A workman is paid $Rs. 150$ for each day he works and is fined $Rs. 50$ for each day he is absent. In a month of $30$ days he earned $Rs. 2100$. For how many days did he remain absent?

Salary $= 150 \ Rs./day$

Fine $= 50 \ Rs./day$

Let $x$ be the number of days worked

Therefore

$150x-(30-x)50 = 2100$

Or $x = 18$ days.

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