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

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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 .

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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?

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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 .

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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?

Answer:

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?

Answer:

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 .

Answer:

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.

Answer:

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

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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}+ \frac{1}{4}+ \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?

Answer:

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.

Answer:

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?

Answer:

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 .

Answer:

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?

Answer:

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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