Class 8 MAHARASHTRA STATE TEXTBOOK BUREAU Solutions Maths Chapter 12: Equations in one variable

x - 4 = 3

Put x = ̵ 1

LHS = x - 4

= ̵1 - 4

= ̵5

̵5 ≠ 3

x = -1 is not the solution of the given equation.

Put x = 7

LHS = 7 - 4 = 3

x = 7 is the solution of the given equation.

Put x = ̵7

LHS = ̵7 - 4 = ̵11

x = ̵7 is not the solution of given equation.

Put m = 3

LHS = 9 × 3 = 27

m = 3 is not the solution of the given equation.

Put m = 9

LHS = 9 × 9 = 81

m = 9 is the solution of the given equation.

Put m = ̵3

LHS = 9 × (-3) = ̵27

m = ̵3 is not the solution of the given equation.

Put a = 2

LHS = 2 × 2 + 4 = 8

a = 2 is not the solution of the given equation.

Put a = ̵2

LHS = 2 × (-2) + 4 =  0

a = -2 is the solution of the given equation.

Put a = 1

LHS = 2 × 1 + 4 = 6

a = 1 is not the solution of the given equation.

Put y = -1

LHS = 3 - (-1) = 3 + 1 = 4

y = -1 is the solution of the given equation.

Put y = 1

LHS = 3 - 1 = 2

y = 1 is not the solution of the given equation. 

Put y = 2

LHS = 3 - 2 = 1

y = 2 is not the solution of the given equation.

17p - 2 = 49

∴ 17p - 2 + 2 = 49 + 2

∴ 17p = 51

∴   

∴ p = 3

2m + 7 = 9

∴ 2m + 7 - 7 = 9 - 7

∴ 2m = 2

∴   

∴ m = 1

3x + 12 = 2x - 4

∴ 3x + 12 - 12 = 2x - 4 - 12

∴ 3x = 2x - 16

∴ 3x - 2x = 2x - 2x - 16

∴ x = - 16

5(x - 3) = 3(x + 2)

∴ 5x - 15 = 3x + 6

∴ 5x - 15 + 15 = 3x + 6 + 15

∴ 5x = 3x + 21

∴ 5x - 3x = 3x - 3x + 21

∴ 2x = 21

∴   

∴  

∴   

∴  

∴   

∴  

∴   

∴ 3y + 7(y - 4) = 2 × 21

∴ 3y + 7y - 28 = 42

∴ 10y - 28 = 42

∴ 10y - 28 + 28 = 42 + 28

∴ 10y = 70

∴ y = 7

13x - 5 =  

∴ 13x - 5 + 5 =  + 5

∴ 13x =  

∴  

3(y + 8) = 10(y - 4) + 8

∴ 3y + 24 = 10y - 40 + 8

∴ 3y + 24 - 24 = 10y - 32 - 24

∴ 3y = 10y - 56

∴ 3y + 56 = 10y - 56 + 56

∴ 3y + 56 = 10y

∴ 3y - 3y + 56 = 10y - 3y

∴ 56 = 7y

∴ y = 8

∴ 7(x - 9) = 5(x - 5)

∴7x - 63 = 5x - 25

∴ 7x - 63 + 63 = 5x - 25 + 63

∴ 7x = 5x + 38

∴ 7x - 5x = 5x - 5x + 38

∴ 2x = 38

∴   

∴ x = 19

∴  

∴ y - 4 + 9y = 4× 3

∴ y + 9y - 4 = 12

∴ 10y - 4 = 12

∴ 10y - 4 + 4 = 12 + 4

∴ 10y = 16

∴  

∴ b + b + 1 + b + 2 = 21 × 4

∴ 3b + 3 = 84

∴ 3b + 3 - 3 = 84 - 3

∴ 3b = 81

∴  

∴ b = 27

Let the age of son be x years.

Mother is 25 year older than her son. Hence, age of mother is x + 25 .

After 8 years the ages of son and mother will be x + 8 and x + 25 + 8 = x + 33 years respectively.

The ratio of their ages is .

∴ 9(x + 8) = 4(x + 33)

∴ 9x + 72 = 4x + 132

∴ 9x + 72 - 72 = 4x + 132 - 72

∴ 9x = 4x + 0

∴ 9x - 4x = 4x - 4x + 60

∴ 5x = 60

∴ x = 12

Son's age is 12 years.

The numerator of a fraction be x.

Hence, denominator will be 12 greater than its numerator.

Hence, the fraction will be  

Now,

The numerator is decreased by 2 and the denominator is increased by 7,

The new fraction is .

.

∴ 2(x - 2) = x + 19

∴ 2x - 4 = x + 19

∴2x - 4 + 4 = x + 19 + 4

∴ 2x = x + 23

∴ 2x - x = x - x + 23

∴ x = 23

The numerator of a fraction is 23 and denominator is 23 + 12 = 35.

The fraction is  

The weight of brass utensil is 700 gm.

Now,

Let the weight of copper be x gm.

Hence, the weight of zinc be (700 - x) gm

The ratio of weights of copper and zinc in brass is 13:7.

∴ 7x = 13(700 - x)

∴ 7x = 9100 - 13x

∴ 7x + 13x = 9100 - 13x + 13x

∴ 20x = 9100

∴  

∴ x = 455

The weight of copper is 455 gm and

weight of zinc is 700 - x = 700 - 455 = 245 gm

Let the three consecutive numbers be x, x + 1, x + 2.

The sum of them is x + x + 1 + x + 2 = 3x + 3 which is greater than 45.

3x + 3 > 45

∴ 3x + 3 - 3 > 45 - 3

∴ 3x > 42

∴ x > 14 ….(i)

The sum of numbers is less than 54.

3x + 3 < 54

∴ 3x + 3 - 3 < 54 - 3

∴ 3x < 51

∴ x < 17 ….(ii)

x > 14 and x < 17

Hence, x = 15 or x = 16.

The numbers are 15, 16, 17 or 16, 17, 18.

Let the x be unit digit in two digit number.

Ten's place digit is twice of unit digit i.e. 2x.

Original number will be 10 × 2x + x = 21x

Interchanging digits the new number will be = 10 × x + 2x = 12x

Given that the sum of original number and new number is 66.

21x + 12x = 66

∴ 33x = 66

∴ x = 2

Hence, the original number is 21x = 21 ∴ 2 = 42.

Let the number of tickets of Rs. 100 sold be x.

The number of tickets sold of Rs. 200 is x + 20.

The total amount received in transaction is Rs. 37000.

The total amount is 100x + 200(x + 20)

∴ 100x + 200(x + 20) = 37000

∴ 100x + 200x + 4000 = 37000

∴ 300x + 4000 - 4000 = 37000 - 4000

∴ 300x = 33000

∴ x = 110

The number of tickets sold of Rs. 100 was 110.

Let the three consecutive natural numbers be x, x + 1, x + 2.

The smallest number is x and greatest number is x + 2.

Five times smallest number is 5x and four times the greatest number is 4(x + 2)

According to the given information,

5× smallest number = 4 × greatest number + 9

∴ 5x = 4(x + 2) + 9

∴ 5x = 4x + 8 + 9

∴ 5x - 4x = 4x - 4x + 17

∴ x = 17

The consecutive numbers are 17, 18, 19.

Let the cost price of a bicycle be Rs. x.

Selling price for Raju = C. P. + Profit

= x +  

=  

Cost price for Amit is selling price for Raju =  

Amit spent Rs. 54 on repairing bicycle.

Total cost price for Amit = + 54

Amit sold bicycle to Nikhil at no profit and no loss at Rs. 1134.

+ 54 = 1134

∴ + 54 - 54 = 1134 - 54

∴  

∴ x =  = Rs. 1000

The cost price of a bicycle which is Raju purchased is Rs. 1000.

Let the number of runs a cricket player scores in the third match be x.

A Cricket player scored 180 runs in the first match and 257 runs in second match.

The average score of the three matches is 230.

Hence,

∴ 437 + x = 690

∴ 437 + x = 690 

∴ 437 + x - 437 = 690 - 437

∴ x = 253

A cricket player scored 253 runs in the third match.

Let the present age of Viru is x years.

Sudhir's present age is 5 more than three times the age of Viru.

Hence, the age of Sudhir is 3 × Viru's age + 5 = 3x + 5.

Anil's age is half the age of Sudhir i.e. .

The ratio of the sum of Sudhir's and Viru's age to three times Anil's age is 5:6,

∴  

∴ 12(4x + 5) = 15(3x + 5)

∴ 48x + 60 = 45x + 75

∴ 48x - 45x + 60 = 45x + 75 - 45x

∴ 3x + 60 = 75

∴ 3x + 60 - 60 = 75 - 60

∴ 3x = 15

∴ x = 5

Hence, Viru's present age is 5 years.