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    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable

    Edited By Ramraj Saini | Updated on Oct 09, 2023 02:17 PM IST

    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable

    Linear Equations in One Variable Class 8 Questions And Answers provided here. These NCERT Solutions are prepared by subject matter expert of Careers360 considering latest syllabus and pattern of CBSE 2023-24. this chapter created a strong foundation for algebra unit. NCERT solutions for Class 8 Maths chapter 2 Linear Equations in One Variable is comprehensively covering each problem related to this particular topic.

    In this chapter of NCERT Syllabus for Class 8 Maths, there will be an expression consisting of variables as well as numbers and you have to find the value of that variable. NCERT solutions for Class 8 Maths chapter 2 Linear Equations in One Variable are covering solutions to word problems using the linear equations of one variable. This chapter has a total of 6 exercises consisting of a total of 65 questions in all. NCERT solutions for Class 8 Maths chapter 2 Linear Equations in One Variable has the solution to all 65 questions to boost your overall preparation level.

    Linear Equations in One Variable Class 8 Questions And Answers PDF Free Download

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    Linear Equations in One Variable Class 8 Solutions - Important Formulae

    Linear equations with a linear expression on one side:

    • Transpose the number to the side with all the numbers, maintaining the sign.

    • Solve (add/subtract) the equation on both sides to simplify and find the variable's value.

    Linear equations with variables on both sides:

    • Transpose both the number and the variable to one side, maintaining the sign of the number.

    • Solve (add/subtract) the equation on both sides to simplify and find the variable's value.

    Linear equations with a number in the denominator and variables on both sides:

    • Find the LCM (Least Common Multiple) of the denominators on both sides.

    • Multiply both sides by the LCM to simplify the equation.

    • Solve it like a linear equation with variables on both sides to find the variable's value.

    Linear equations that are reducible to the linear form:

    • If the equation is of the form (x + a / x + b) = c / d, cross-multiply the numerator and denominator to simplify it to a linear form, such as (x + a) * d = c * (x + b).

    • Solve this linear equation with variables on both sides to find the variable's value.

    Free download NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable for CBSE Exam.

    Linear Equations in One Variable Class 8 NCERT Solutions (Intext Questions and Exercise)

    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable - Excercise: 2.1

    Q1 Solve the following equations x – 2 = 7

    Answer: 1. Transposing -2 to the RHS, we get

    x = 7 + 2 = 9

    Q2 Solve the following equation y + 3 = 10

    Answer: Transposing 3 to the RHS, we get

    y = 10 - 3 = 7

    y = 7

    Q3 Solve the following equation 6 = z + 2

    Answer: Transposing 2 to the LHS, we get

    6 - 2 = z => z= 4

    Thus z = 4

    Q4 Solve the following equations

    \frac{3}{7} + x = \frac{17}{7}

    Answer: 4. Transposing \frac{3}{7} to the RHS, we get

    x = \frac{17}{7}-\frac{3}{7} = \frac{14}{7} = 2

    Thus x =2

    Q5 Solve the following equation 6x = 12

    Answer: Dividing both sides by 6, we get

    x = \frac{12}{6} => 2

    Thus x = 2

    Q6 Solve the following equation.

    \frac{t}{5} = 10

    Answer: Multiplying both sides by 5, we get

    t = 10\times 5 = 50

    Thus t = 50

    Q 7 Solve the following equations.

    \frac{2x}{3} = 18

    Answer: 7.

    \frac{2x}{3} = 18

    Multiplying both sides by 3, we get

    2x = 18\times 3 = 54

    Now, dividing both sides by 2, we get

    x = \frac{54}{2} = 27

    Q8 Solve the following equation.

    1.6 = \frac{y}{1.5}

    Answer: Multiplying both sides by 1.5, we get

    1.6\times 1.5 = y = 2.4

    Thus y = 2.4

    Q9 Solve the following equation 7x – 9 = 16

    Answer: Transposing 9 to the RHS, we get

    7x = 16 + 9 = 25

    Now, dividing both sides by :

    x = 25\times \frac{1}{7} = \frac{25}{7}

    Q10 Solve the following equation 14y – 8 = 13

    Answer: Transposing -8 to the RHS, we get

    14y = 13 + 8 = 21

    Now dividing both sides by 14:

    y = \frac{21}{14} = \frac{3}{2}

    Q11 Solve the following equation 17 + 6p = 9

    Answer: At first, transposing 17 to the RHS:

    6p = 9 - 17 = -8

    Now, dividing both sides by 6, we get

    p = \frac{-8}{6} = \frac{-4}{3}

    Q12 Solve the following equations.

    \frac{x}{3} +1 = \frac{7}{15}

    Answer: 12.

    \frac{x}{3} + 1 = \frac{7}{15}

    Transposing 1 to the RHS, we get

    \frac{x}{3} = \frac{7}{15} -1 = \frac{7-15}{15} = \frac{-8}{15}

    Now multiplying both sides by 3, it becomes

    x = \frac{-8}{15}\times 3 = \frac{-8}{5}

    Class 8 maths chapter 2 question answer - Exercise: 2.2

    Q1 If you subtract \frac{1}{2} from a number and multiply the result by \frac{1}{2} , you get \frac{1}{8} . What is the number?

    Answer: Assume the number to be x.

    Thus according to the question,

    \left ( x-\frac{1}{2} \right )\times \frac{1}{2} = \frac{1}{8}

    Multiplying both sides by 2, we get

    \left ( x-\frac{1}{2} \right ) = \frac{1}{8}\times 2 = \frac{1}{4}

    Now transposing -\frac{1}{2} to the RHS, we get

    x = \frac{1}{4} + \frac{1}{2} = \frac{1+2}{4} = \frac{3}{4}

    Thus the number was x = 3/4

    Q 2 The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and breadth of the pool?

    Answer: Let the breadth of the pool be x m.

    According to the question, the length of the pool = 2x + 2 m.

    Perimeter of recatangle = 2(l + b) = 154 m .

    i.e., 2(2x + 2 + x) = 154

    2(3x + 2) = 154

    Dividing both sides by 2, we get

    3x + 2 = 77

    Now transposing 2 to the RHS, we get

    3x = 77 - 2 = 75

    Dividing both sides by 3, we get

    x = \frac{75}{3} = 25

    Thus breadth of pool = 25 m

    and lenght of the pool = 2 \times 25 + 2 = 52 m

    Q3 The base of an isosceles triangle is \frac{4}{3} cm. The perimeter of the triangle is 4\frac{2}{15} cm . What is the length of either of the remaining equal sides?

    Answer: In isosceles triangles, we have 2 sides of equal length.

    Given that its perimeter is \frac{62}{15} cm.

    Let's assume the length of the equal side is x cm.

    Also,

    Perimeter = x + x + \frac{4}{3} = \frac{62}{15}

    Transposing \frac{4}{3} to the RHS side it becomes,

    2x = \frac{62}{15}-\frac{4}{3} = \frac{62-20}{15} = \frac{42}{15} = \frac{14}{5}

    Now dividing both sides by 2, we get

    x = \frac{14}{5}\times \frac{1}{2} = \frac{7}{5}

    Hence, the length of the equal sides of the isosceles triangle is \frac{7}{5} cm.

    Q4 Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

    Answer: Assume one number to be x. Then the other number will be x + 15.

    Now it is given that sum of the two numbers is 95.

    Thus quation becomes x + x + 15 = 95

    or 2x +15 = 95

    Transposing 15 to the RHS, it becomes

    2x = 95 - 15 = 80

    Dividng both sides by 2, we get

    x = \frac{80}{2} = 40

    Thus two numbers are 40 and 55

    Q5 Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

    Answer: Given that numbers are in the ratio of 5:3, so we can assume numbers to be 5x and 3x.

    Also, the difference between these numbers is 18, so the equation becomes

    5x - 3x = 18

    2x = 18

    Dividing both sides by 2, we get

    x = 9

    Hence the two numbers are 5 \times 9 = 45 and 3 \times 9 = 27

    Q6 Three consecutive integers add up to 51. What are these integers?

    Answer: Let the consecutive integers be x, x+1, x+2.

    Sum of these integers is given to be 51.

    Thus equation becomes : x + x + 1 + x + 2 = 51

    3x + 3 = 51

    Transposing 3 to the RHS,

    3x = 48

    Now dividing both sides by 3, we get

    x = \frac{48}{3} = 16

    Hence the consecutive integers are 16, 17, 18.

    Q7 The sum of three consecutive multiples of 8 is 888. Find multiples.

    Answer: Let x be the multiple of 8.

    Then the three consecutive integers (multiple of 8) are x, x+8, x+16.

    Given their sum is 888,

    thus the equation becomes: x + x + 8 + x + 16 = 888

    or 3x + 24 = 888

    Transposing 24 to the RHS

    3x = 888 - 24 = 864

    Now dividing both sides by 3, we get

    x = \frac{864}{3} = 288

    Thus the required consecutive integers are 288, 296, 304.

    Q8 Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

    Answer: Let the consecutive integers be x, x+1, x+2.

    Now according to the question equation becomes,

    2x + 3(x+1) + 4(x+2) = 74

    or 2x + 3x + 3 + 4x + 8 = 74

    or 9x + 11 = 74

    Transposing 11 to the RHS we get,

    9x = 74 - 11 = 63

    Divinding both sides by 9.

    x = \frac{63}{9} = 7

    Therefore required consecutive integers are 7, 8, 9.

    Q9 The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?

    Answer: Let the current age of Rahul and Haroon be 5x and 7x respectively(Since there age ratio is given as 5:7).

    Four years later their ages become 5x + 4 and 7x + 4 respectively.

    According to the question,

    5x + 4 +7x + 4 = 56

    or 12x + 8 = 56

    Transposing 8 to the RHS we get,

    12x = 48

    Dividing both sides by 12:

    x = \frac{48}{12} = 4

    Thus current age of Rahul = 5 \times 4 = 20

    Haroon = 7 \times 4 = 28

    Q10 The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

    Answer: Let us assume the number of boys and the number of girls is 7x and 5x.

    According to the given data in the question,

    7x = 5x + 8

    Transposing 5x to the LHS we get,

    7x - 5x = 8

    2x = 8

    Dividing both sides by 2

    x = 4

    Hence number of boys = 7 \times 4 = 28

    and number of girls = 5 \times 4 = 20

    So total number of students = 28 + 20 = 48.

    Q11 Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

    Answer: Let the age of Baichung be x years.

    Then according to question age of Baichung's father = x + 29 years

    and age of Baichung's grandfather = x + 29 + 26 years

    The sum of their ages is given 135 years. According to that, x + x + 29 + x + 29 + 26 = 135

    3x + 84 = 135

    Transposing 81 on the RHS we get,

    3x = 135 - 84 = 51

    Dividing both sides by 3

    x =\frac{51}{3} = 17

    Thus, the age of Baichung = 17 years

    The age of Baichung's father = 17 + 29 = 46 years

    The age of Baichung's grandfather = 46 + 26 = 72 years

    Q12 Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

    Answer: Let us assume Ravi's present age to be x years.

    According to the question,

    x + 15 = 4x

    Transposing x to the RHS

    15 = 3x

    Now dividing both sides by 3, we get

    x = \frac{15}{3} = 5

    Hence Ravi's current age is 5 years.

    Q13 A rational number is such that when you multiply it by \frac{5}{2} and add \frac{2}{3} to the product, you get \frac{-7}{12} . What is the number?

    Answer: Let the rational number be x.

    According to the question,

    x\times \frac{5}{2} + \frac{2}{3} = -\frac{7}{12}

    Transposing \frac{2}{3} to the RHS:

    x\times \frac{5}{2} = -\frac{7}{12} - \frac{2}{3} = \frac{-15}{12} = \frac{-5}{4}

    Multiplication by 2 in both sides, we get

    5x = \frac{-5}{4}\times 2 = \frac{-5}{2}

    Dividing both sides by 5, we get

    x = \frac{-5}{2}\times \frac{1}{5} = \frac{-1}{2}


    Q14 Lakshmi is a cashier in a bank. She has currency notes of denominations ?100, ?50 and ?10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ?4,00,000. How many notes of each denomination does she have?

    Answer: Since the ratio of currency notes is 2:3:5. Therefore,

    Let the number of currency notes of Rs.100, Rs.50 and Rs.10 be 2x, 3x, 5x respectively.

    Hence according to the question equation becomes,

    100 \times 2x + 50 \times 3x + 10 \times 5x = 400000

    or 200x + 150x + 50x = 400000

    or 400x = 400000

    Dividing both sides by 400 we get,

    x = 1000 .

    No. of Rs.100 notes = 2 \times 1000 = 2000 notes

    No. of Rs.50 notes = 3 \times 1000 = 3000 notes

    No. of Rs.10 notes = 5 \times 1000 = 5000 notes

    Q15 I have a total of Rs 300 in coins of denomination Rs1, Rs2 and Rs5. The number of Rs2 coins is 3 times the number of Rs5 coins. The total number of coins is 160. How many coins of each denomination are with me?

    Answer: Let the no. of Rs.1 coin be y.

    And the no. of Rs.5 coin be x.

    Thus according to question no. of Rs.2 coin will be 3x.

    Also, the total no. of coins = 160

    This implies : y + x + 3x = 160

    or y + 4x = 160

    Transposing 4x to the RHS

    y = 160 - 4x = No. of Rs.1 coin.

    Now, it is given that total amount is Rs.300.

    i.e., 1(160-4x) + 2(3x) + 5(x) = 300

    or 160 - 4x + 6x + 5x = 300

    or 160 + 7x = 300

    Transposing 160 to the RHS :

    7x = 300 - 160 = 140

    Dividing both sides by 7:

    x = \frac{140}{7} = 20

    x = 20

    Thus number of Rs.1 coin = 160 - 4x = 160 - 80 = 80

    number of Rs.2 coin = 3x = 60

    number of Rs.5 coin = x = 20

    Q16 The organizers of an essay competition decide that a winner in the competition gets a prize of Rs100 and a participant who does not win gets a prize of Rs25. The total prize money distributed is Rs3,000. Find the number of winners, if the total number of participants is 63.

    Answer: Let the no. of winners be x.

    Since total no. of participants is 63, thus no. of participants who does not win = 63 - x

    According to the question, equation becomes:

    x(100) + (63-x)(25) = 3000

    or 100x + 1575 - 25x = 3000

    or 75x + 1575 = 3000

    Transposing 1575 to the RHS :

    75x = 3000 - 1575

    or 75x = 1425

    Dividing both sides by 75, we get

    x = \frac{1425}{75} = 19

    x = 19

    Therefore no. of winners = 19

    Class 8 maths chapter 2 ncert solutions - Exercise: 2.3

    Q1 Solve the following equations and check your results 3x = 2x + 18

    Answer: We have

    3x = 2x + 18

    Subtracting 2x from both sides

    3x - 2x = 2x - 2x + 18

    or x = 18

    Check :- Put x = 18 in both LHS and the RHS.

    LHS = 3x = 3(18) = 54

    RHS = 2x + 18 = 2(18) + 18 = 36 + 18 = 54

    Thus, LHS = RHS

    Q2 Solve the following equation and check your results 5t – 3 = 3t – 5

    Answer: We have

    5t - 3 = 3t - 5

    Transposing 3t to the LHS and -3 to the RHS, we get:

    5t - 3t = -5 + 3

    or 2t = -2

    Dividing both sides by 2 :

    t = -1

    Check :- Put t = -1 in the LHS we have,

    5t - 3 = 5(-1) - 3 = -5 -3 = -8

    Similarly put t = -1 in the RHS:

    3t - 5 = 3(-1) - 5 = -3 - 5 = -8

    Hence, LHS = RHS

    Q3 Solve the following equation and check your result 5x + 9 = 5 + 3x

    Answer: We have

    5x + 9 = 5 + 3x

    Transposing 3x to the LHS and 9 to the RHS, we get:

    5x - 3x = 5 - 9

    or 2x = -4

    Dividing both sides by 2 :

    x = -2

    Check :- Put x = - 2 in both LHS and RHS

    LHS :- 5x + 9 = 5(-2) + 9 = -10 + 9 = -1

    RHS :- 5 + 3x = 5 + 3(-2) = 5 - 6 = -1

    Hence, LHS = RHS

    Q4 Solve the following equation and check your result. 4z + 3 = 6 + 2z

    Answer: We have

    4z + 3 = 6 +2z

    Transposing 2z to the LHS and 3 to the RHS, we get:

    4z - 2z = 6 - 3

    or 2z = 3

    DIviding both sides by 2,

    z = \frac{3}{2}

    Check :- Put z = \frac{3}{2} in LHS as well as RHS, we have

    LHS :- 4z + 3

    = 4\times \frac{3}{2} + 3 = 9

    RHS :- 6 + 2z

    = 6 + 2\times \frac{3}{2} = 6 + 3 = 9

    Thus, LHS = RHS

    Q5 Solve the following equation and check your result 2x – 1 = 14 – x

    Answer: We have

    2x - 1 = 14 - x

    Transposing -x to the LHS and -1 to the RHS

    2x + x = 14 + 1

    or 3x = 15

    Dividing both sides by 3, we get

    x = 5

    Check :- Put x = 5 in both the LHS and the RHS

    LHS :- 2x - 1 = 2(5) - 1 = 10 - 1 = 9

    RHS :- 14 - x = 14 - 5 = 9

    Hence, LHS = RHS

    Q6 Solve the following equation and check your result 8x + 4 = 3 (x – 1) + 7

    Answer: We have

    8x + 4 = 3 (x - 1) + 7

    or 8x + 4 = 3x - 3 + 7 = 3x + 4

    Transposing 3x to the LHS and 4 to the RHS, we get

    8x - 3x = 4 - 4 = 0

    5x = 0

    Dividing both sides by 5:

    x = 0

    Check :- Putting x = 0 in both LHS and RHS:

    LHS :- 8x + 4 = 8(0) + 4 = 4

    RHS :- 3(x-1) + 7 = 3(-1) + 7 = -3 + 7 = 4

    Hence, LHS = RHS

    Q7 Solve the following equations and check your results.

    x =\frac{4}{5}\times \left ( x+10 \right )

    Answer: We have

    x = \frac{4}{5}\left ( x + 10 \right )

    = \frac{4}{5}x + 8

    Transposing \frac{4}{5}x to the left-hand side:

    x - \frac{4}{5}x = 8

    or \frac{5x-4x}{5} = 8

    or \frac{x}{5} = 8

    Multiplying both sides by 5

    x = 40

    Check :- Put x = 40 in both the LHS and the RHS.

    LHS :- x = 40

    RHS :- \frac{4}{5}\left ( x + 10 \right ) = \frac{4}{5}\left ( 40 + 10 \right )

    = \frac{4}{5}\left ( 50 \right ) = 40

    Thus, LHS = RHS

    Q8 Solve the following equation and check your result.

    \frac{2x}{3}+1 = \frac{7x}{15}+3

    Answer: We have

    \frac{2x}{3} + 1 = \frac{7x}{15} + 3

    Transposing \frac{7x}{15} to the LHS and 1 to the RHS:

    \frac{2x}{3} - \frac{7x}{15} = 3 - 1 = 2

    or \frac{10x - 7x}{15} = 2

    or \frac{3x}{15} = 2

    or x = 10

    Check :- Put x = 10 in both the LHS and the RHS

    LHS :

    \frac{2x}{3} + 1 = \frac{2}{3}\times 10 + 1 = \frac{23}{3}

    RHS :

    \frac{7x}{15} + 3 = \frac{7}{15}\times 10 + 3 = \frac{14}{3} + 3 = \frac{23}{3}

    Thus, LHS = RHS

    Q9 Solve the following equations and check your results.

    2y+\frac{5}{3} = \frac{26}{3} -y

    Answer: We have

    2y+ \frac{5}{3} = \frac{26}{3} - y

    Transposing -y to the LHS and \frac{5}{3} to the RHS, we get

    2y + y = \frac{26}{3}- \frac{5}{3} = \frac{21}{3} = 7

    or 3y = 7

    Dividing both sides by 3:

    y = \frac{7}{3} i

    Check :- Put y = \frac{7}{3} in both the LHS and the RHS:

    LHS :

    2y + \frac{5}{3} = 2\times \frac{7}{3} + \frac{5}{3} = \frac{19}{3}

    RHS :

    \frac{26}{3} - \frac{7}{3} = \frac{19}{3}

    Hence, LHS = RHS

    Q10 Solve the following equation and check your result.

    3m = 5m - \frac{8}{5}

    Answer: We have

    3m = 5m - \frac{8}{5}

    Transposing 5m to the LHS

    3m -5m = - \frac{8}{5}

    or -2m = - \frac{8}{5}

    Dividing both sides by -2, we get

    m = \frac{4}{5}

    Check :- Put m = \frac{4}{5} in the LHS and the RHS:

    LHS :-

    3m = 3\times \left ( \frac{4}{5} \right ) = \frac{12}{5}

    RHS :-

    5m - \frac{8}{5} = 5 \times \frac{4}{5} - \frac{8}{5} = \frac{12}{5}

    Hence, LHS = RHS

    Class 8 linear equations in one variable NCERT solutions - Exercise: 2.4

    Q1 Amina thinks of a number and subtracts \frac{5}{2 } from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?

    Answer: Let us assume that the number Amina thought was x.

    As per the question,

    \left ( x-\frac{5}{2} \right )\times8 = 3x

    or 8x - 20 = 3x

    Transposing 3x to the LHS and -20 to the RHS, we get

    8x - 3x = 20

    or 5x = 20

    Dividing both sides by 5.

    x = 4

    Therefore the number Amina thought was 4.

    Q2 A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

    Answer: Let the No.1 = x

    Then the second no. No.2 = 5x.

    Now the question states that if we add 21 to both numbers one number becomes twice of the other.

    i.e., 2(x + 21) = 5x + 21

    or 2x + 42 = 5x + 21

    Transposing 2x to the RHS and 21 to the LHS:

    21 = 5x - 2x = 3x

    or x = 7

    Thus the two numbers are 7 and 35 . (Since No.2 = 5x)

    Q3 Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?

    Answer: Given that sum of the two digits is 9.

    Let us assume the digit of units place be x.

    Then the digit of tens place will be 9-x.

    Thus the two digit number is 10(9-x) + x

    Now if we reverse the digits, the number becomes 10x + (9-x).

    As per the question,

    10x + (9-x) = 10(9-x) + x + 27

    or 9x + 9 = 90 - 10x + x + 27

    or 9x + 9 = 117 - 9x

    Transposing -9x to the LHS and 9 to the RHS:

    9x + 9x = 117 - 9

    or 18x = 108

    x = 6

    Thus two digit number is 36.

    Q4 One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

    Answer: Let us assume that the units place of a two digit number is x.

    Then as per the question, tens place of the number is 3x.

    Thus the two digit number is 10(3x) + x.

    If we interchange the digits the number becomes : 10x + 3x

    According to the question,

    10(3x) + x + 10x + 3x = 88

    or 30x + 14x = 88

    or 44x = 88

    Dividing both sides by 44, we get

    x = 2

    Hence the two digit number is 10(6) + 2 = 62

    Note that the two-digit number can also be found if we reverse the digits. So one more possible number is 26.

    Q5 Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?

    Answer: Let us assume that the Shobo's present age be x.

    So, Shobo's mother's present age = 6x.

    Shobo's age five years from now will be = x + 5.

    It is given that Shobo’s age five years from now will be one-third of his mother’s present age.

    So,

    x + 5 = \left ( \frac{1}{3} \right )\times6x = 2x

    or x + 5 = 2x

    Transposing x to the RHS,

    5 = 2x - x = x

    Hence Shobo's present age is x = 5 years

    and Shobo's mother's present age is 6x = 30 years

    Q6 There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate Rs100 per metre it will cost the village panchayat Rs75000 to fence the plot. What are the dimensions of the plot?

    Answer: Let the length and breadth of the plot be 11x and 4x respectively.

    (Since they are in the ratio11:4)

    We know that the fencing will be done on the boundary of the plot, so we need to calculate its perimeter.

    Perimeter of rectangle = 2(lenght + breadth)

    Total cost to fence the plot = Rs.75000 = Rs.100(Perimeter of plot)

    Thus the equation becomes,

    100 [2(11x + 4x)] = 75000

    or 200(15x) = 75000

    or 3000x = 75000

    Dividing both sides by 3000:

    x = 25

    Hence length of plot is 11x = 275 m

    and breadth of the plot is 4x = 100m

    Q7 Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs50 per meter and trouser material that costs him Rs90 per meter. For every 3 meters of the shirt material, he buys 2 meters of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs36,600. How much trouser material did he buy?

    Answer: Let the total shirt material that Hasan bought be x metre.

    and the total trouser material is y metre.

    As per the question,

    \frac{x}{3} = \frac{y}{2}

    Thus trouser material y = \frac{2}{3}x

    Now we will make another equation by using cost of the materials.

    Profit for shirt is 12% i.e., selling price of shirt is

    =50 +\frac{12}{100}\times 50 = 50 + 6 = 56

    and selling price for trouser is

    =90 +\frac{10}{100}\times 90 = 90 + 9 = 99

    Hence equation becomes:

    x\left ( 56 \right ) + \frac{2}{3}x\left ( 99 \right ) = 36600

    or 56x + 66x = 36600

    or 122x = 36600

    Dividing both sides by 122:

    x = 300

    Thus trouser material that Hasan bought was = 200 meter

    \left ( \frac{2}{3}x = \frac{2}{3}\times300 = 200 \right )


    Q8 Half of a herd of deer are grazing in the field and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

    Answer: Let us assume the number of deer in the herd to be x.

    Half of a herd of deer are grazing in the field i.e., \frac{x}{2} .

    Now half are remaining.

    Three-fourths of the remaining are playing nearby, implies :

    \frac{3}{4}\times \frac{x}{2} = \frac{3x}{8}

    The remaining number of deer will be :

    \frac{x}{2} - \frac{3x}{8} = \frac{4x - 3x}{8} = \frac{x}{8}

    Given that rest is 9, so

    \frac{x}{8} = 9

    Hence the number of deer in the herd = 72

    Q9 A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

    Answer: Let the age of granddaughter is x years.

    So the age of grandfather is x + 54.

    Also, grandfather is ten times older than his granddaughter.

    i.e., x + 54 = 10x

    Transposing x to the RHS :

    54 = 10x - x

    or 54 = 9x

    or x = 6

    Therefore age of granddaughter = 6 years

    and the age of grandfather = 6 + 54 = 60 years

    Q10 Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

    Answer: Let the age of Aman's son is x years.

    So the age of Aman will be 3x years.

    Ten years ago, the age of Aman's son = x - 10 and Aman's age = 3x - 10.

    So according to question,

    5(x - 10) = 3x - 10

    or 5x - 50 = 3x - 10

    Transposing 3x to the LHS and -50 to the RHS:

    5x - 3x = -10 + 50 = 40

    or 2x = 40

    or x = 20

    So Aman's age = 3x = 60 years and

    Aman's son's age = x = 20 years.

    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Excercise: 2.5

    Q1 Solve the following linear equations.

    \frac{x}{2}-\frac{1}{5} = \frac{x}{3}+\frac{1}{4}

    Answer: The detailed solution of the above-written question is here,

    We have

    \frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}

    or \frac{x}{2} - \frac{x}{3} = \frac{1}{4} + \frac{1}{5}


    or \frac{3x - 2x}{6} = \frac{4 +5}{20}


    or \frac{x}{6} = \frac{9}{20}

    Multiplying both sides by 6:

    x = \frac{9}{20}\times 6 = \frac{27}{10}

    Thus x = \frac{27}{10}


    Q2 Solve the following linear equations.

    \frac{n}{2}- \frac{3n}{4}+\frac{5n}{6} = 21

    Answer: We have

    \frac{n}{2}- \frac{3n}{4}+\frac{5n}{6} = 21

    Multiplying both sides 12.

    (We multiplied both sides by 12 because it is the lowest common factor.)

    6n - 9n + 10n = 252

    or 7n = 252

    or n = 36

    Q3 Solve the following linear equations.

    x+7 -\frac{8x}{3} = \frac{17}{6}- \frac{5x}{2}

    Answer: We have

    x+7 -\frac{8x}{3} = \frac{17}{6}- \frac{5x}{2}

    Multiplying both sides by 6, we get

    6x + 42 - 16x = 17 - 15x

    or -10x +42 = 17 - 15x

    Transposing -15x to the LHS and 42 to the RHS :

    -10x + 15x = 17 - 42

    or 5x = -25

    or x = -5

    Q4 Solve the following linear equations.

    \frac{x-5}{3} = \frac{x-3}{5}

    Answer: We have

    \frac{x-5}{3} = \frac{x-3}{5}

    Multiplying both sides by 15 . (because the LCM of denomenator is 15 )

    5x - 25 = 3x - 9

    Transposing 3x to the LHS and -25 to the RHS.

    5x - 3x = -9 + 25

    or 2x = 16

    x = 8

    Q5 Solve the following linear equations.

    \frac{3t-2}{4} -\frac{2t+3}{3} = \frac{2}{3} - t

    Answer: We have

    \frac{3t-2}{4} -\frac{2t+3}{3} = \frac{2}{3} - t

    Multiplying both sides by 12, we get

    9t - 6 - ( 8t + 12 ) = 8 - 12t

    or 9t - 6 - 8t - 12 = 8 - 12t

    or t - 18 = 8 - 12t

    Transposing -18 to the RHS and -12t to the LHS:

    12t + t = 8 + 18 = 26

    or 13t = 26

    Thus, t = 2

    Q6 Solve the following linear equation.

    m - \frac{m-1}{2} = 1 - \frac{m-2}{3}

    Answer: We have

    m - \frac{m-1}{2} = 1 - \frac{m-2}{3}

    Multiplying both sides by 6 (as it is LCM of the denominator.)

    6m - 3(m - 1) = 6 - 2(m - 2)

    or 6m - 3m + 3 = 6 - 2m + 4

    or 3m + 3 = 10 - 2m

    Transposing -2m to the LHS and 3 to the RHS:

    3m + 2m = 10 - 3

    or 5m = 7

    m = \frac{7}{5}

    Q7 Simplify and solve the following linear equation. 3(t – 3) = 5(2t + 1)

    Answer: Let us open the brackets :

    LHS : 3(t - 3) = 3t - 9

    RHS : 5(2t + 1) = 10t + 5

    The equation is 3t - 9 = 10t + 5

    or -9 = 10t - 3t + 5

    or -9 - 5 = 7t

    or -14 = 7t

    Therefore t = -2

    Q8 Simplify and solve the following linear equation . 15(y – 4) –2(y – 9) + 5(y + 6) = 0

    Answer: Let us open the brackets

    Equation becomes: 15y - 60 - 2y + 18 + 5y + 30 = 0

    or 18y - 12 = 0

    or 18y = 12

    y = \frac{12}{18} = \frac{2}{3}

    Q9 Simplify and solve the following linear equation. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

    Answer: Let us open the brackets.

    LHS: 3(5z - 7) - 2(9z - 11)

    = 15z - 21 - 18z + 22

    = -3z + 1

    RHS: 4(8z - 13) - 17

    = 32z - 52 - 17

    = 32z - 69

    So the equation becomes : -3z + 1 = 32z - 69

    or 69 + 1 = 32z + 3z

    or 70 = 35z

    z = 2

    Q10 Simplify and solve the following linear equations. 0.25(4f – 3) = 0.05(10f – 9)

    Answer: We have

    0.25(4f – 3) = 0.05(10f – 9)

    Let us open the brackets:

    f - 0.75 = 0.5f - 0.45

    or f - 0.5f = - 0.45 + 0.75

    or 0.5f = 0.30

    Dividing both sides by 0.5, we get

    f = 0.6

    NCERT linear equations in one variable class 8 questions and answers - Exercise: 2.6

    Q1 Solve the following equation.

    \frac{8x-3}{3x} = 2

    Answer: We observe that the equation is not a linear equation since LHS is not linear.

    So we multiply 3x to both the sides to make it linear.

    We get, 8x - 3 = 2 \times 3x

    or 8x - 3 = 6x

    or 8x - 6x = 3

    or 2x = 3

    x = \frac{3}{2}

    Q2 Solve the following equation.

    \frac{9x}{7-6x} = 15

    Answer: We will convert the given equation into a linear equation by multiplying

    (7 - 6x) to both sides.

    Equation becomes: 9x = 15(7 - 6x)

    or 9x = 105 - 90x

    or 99x = 105

    x = \frac{105}{99} = \frac{35}{33}

    Q3 Solve the following equation.

    \frac{z}{z+15} = \frac{4}{9}

    Answer: We have

    \frac{z}{z+15} = \frac{4}{9}

    Cross-multiplication gives:

    9z = 4z + 60

    or 9z - 4z = 60

    or 5z = 60

    or z = 12

    Q4 Solve the following equation.

    \frac{3y+4}{2-6y} = \frac{-2}{5}

    Answer:

    We have

    \frac{3y+4}{2-6y} = \frac{-2}{5}

    By cross-multiplication we get:

    5(3y + 4) = -2(2 - 6y)

    or 15y + 20 = -4 + 12y

    or 15y - 12y = -4 - 20

    or 3y = -24

    y = -8

    Q5 Solve the following equation.

    \frac{7y+4}{y+2} = \frac{-4}{3}

    Answer: We have

    \frac{7y+4}{y+2} = \frac{-4}{3}

    Cross-multiplication gives,

    3(7y + 4) = -4(y + 2)

    or 21y + 12 = -4y - 8

    or 21y + 4y = -8 - 12

    or 25y = -20

    y = \frac{-20}{25} = \frac{-4}{5}

    Q6 The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.

    Answer: Let the present ages of Hari and Harry be 5x and 7x respectively.

    (Since there are in the ratio of 5:7)

    Four years from age of Hari will be = 5x + 4

    and Harry's age will be = 7x + 4.

    According to question four years from now, the ratio of their ages will be 3:4.

    So the equation becomes :

    \frac{5x + 4}{7x + 4} = \frac{3}{4}

    Cross-multiplication gives:

    4(5x + 4) = 3(7x + 4)

    or 20x + 16 = 21x + 12

    or x = 4

    Hence present age of Hari = 5x = 20 and

    present age of Harry = 7x = 28

    Q7 The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. Find the rational number.

    Answer: Let the numerator of the rational number be x.

    Then denominator will be x + 8.

    Further, according to question,

    \frac{x + 17}{x+ 8 - 1} = \frac{3}{2}

    or \frac{x + 17}{x+ 7} = \frac{3}{2}

    Cross-multiplication gives:

    2(x + 17) = 3(x + 7)

    or 2x + 34 = 3x + 21

    or x = 13

    Hence the rational number is \frac{13}{21} .

    Linear Equations In One Variable Class 8 NCERT Solutions - Topics

    • Solving Equations which have Linear Expressions on one Side and Numbers on the other Side
    • Some Applications
    • Solving Equations having the Variable on both Sides
    • Some More Applications
    • Reducing Equations to Simpler Form
    • Equations Reducible to the Linear Form

    NCERT Solutions for Class 8 Maths - Chapter Wise

    Key Features Of Linear Equations In One Variable Class 8 Solutions

    Step-by-Step Solutions: Detailed step-by-step solutions of maths chapter 2 class 8 to each problem, ensuring clarity and understanding for students.

    Variety of Problems: A diverse set of problems covering different aspects of linear equations in one variable. Practice class 8 maths ch 2 question answer to command the concepts.

    Concept Clarity: A focus on explaining the fundamental concepts and principles related to linear equations in a clear and concise manner provided in this ch 2 maths class 8.

    NCERT Solutions for Class 8 -Subject Wise

    How to use NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable

    • Go through some examples to get an idea about how to solve a question.
    • Once you have identified the process, do the practice problems given in the exercises.
    • For better preparation and time saving, you must use NCERT solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable.
    • You can also practice previous year papers' questions to get perfection over the chapter.

    Keep working hard and happy learning!

    Also Check NCERT Books and NCERT Syllabus here

    Frequently Asked Question (FAQs)

    1. What are the important topics of chapter Linear Equations in One Variable ?

    Multiplication and division of algebraic equations, identities, factorization, and the method of solving linear equations in one variable are the important topics of this chapter.

    2. Does CBSE class 8 maths is tough ?

    CBSE class 8 maths is base and foundation for the upcoming classes. Most of the topics are related to previous classes. It is basic maths and very easy.

    3. Does CBSE provides the solutions of NCERT for class 8 ?

    No, CBSE doesn’t provide NCERT solutions for any class or subject.

    4. Where can I find the complete solutions of NCERT for class 8 ?

    Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

    5. Where can I find the complete solutions of NCERT for class 8 maths ?

    Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.

    6. Which is best book for CBSE class 8 maths ?

    NCERT textbook is the best book for CBSE class 8 maths. You don't need to buy any supplementary book. All you need to do is a rigorous practice of NCERT problems.

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    Get answers from students and experts

    A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

    Option 1)

    0.34\; J

    Option 2)

    0.16\; J

    Option 3)

    1.00\; J

    Option 4)

    0.67\; J

    A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

    Option 1)

    2.45×10−3 kg

    Option 2)

     6.45×10−3 kg

    Option 3)

     9.89×10−3 kg

    Option 4)

    12.89×10−3 kg

     

    An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

    Option 1)

    2,000 \; J - 5,000\; J

    Option 2)

    200 \, \, J - 500 \, \, J

    Option 3)

    2\times 10^{5}J-3\times 10^{5}J

    Option 4)

    20,000 \, \, J - 50,000 \, \, J

    A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

    Option 1)

    K/2\,

    Option 2)

    \; K\;

    Option 3)

    zero\;

    Option 4)

    K/4

    In the reaction,

    2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

    Option 1)

    11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

    Option 2)

    6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

    Option 3)

    33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

    Option 4)

    67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

    How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

    Option 1)

    0.02

    Option 2)

    3.125 × 10-2

    Option 3)

    1.25 × 10-2

    Option 4)

    2.5 × 10-2

    If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

    Option 1)

    decrease twice

    Option 2)

    increase two fold

    Option 3)

    remain unchanged

    Option 4)

    be a function of the molecular mass of the substance.

    With increase of temperature, which of these changes?

    Option 1)

    Molality

    Option 2)

    Weight fraction of solute

    Option 3)

    Fraction of solute present in water

    Option 4)

    Mole fraction.

    Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

    Option 1)

    twice that in 60 g carbon

    Option 2)

    6.023 × 1022

    Option 3)

    half that in 8 g He

    Option 4)

    558.5 × 6.023 × 1023

    A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

    Option 1)

    less than 3

    Option 2)

    more than 3 but less than 6

    Option 3)

    more than 6 but less than 9

    Option 4)

    more than 9

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    An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

    2 Jobs Available
    Orthotist and Prosthetist

    Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

    6 Jobs Available
    Veterinary Doctor

    A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

    5 Jobs Available
    Pathologist

    A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

    5 Jobs Available
    Gynaecologist

    Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

    4 Jobs Available
    Surgical Technologist

    When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications. 

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    3 Jobs Available
    Oncologist

    An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

    3 Jobs Available
    Chemical Pathologist

    Are you searching for a chemical pathologist job description? A chemical pathologist is a skilled professional in healthcare who utilises biochemical laboratory tests to diagnose disease by analysing the levels of various components or constituents in the patient’s body fluid. 

    2 Jobs Available
    Biochemical Engineer

    A Biochemical Engineer is a professional involved in the study of proteins, viruses, cells and other biological substances. He or she utilises his or her scientific knowledge to develop products, medicines or ways to improve quality and refine processes. A Biochemical Engineer studies chemical functions occurring in a living organism’s body. He or she utilises the observed knowledge to alter the composition of products and develop new processes. A Biochemical Engineer may develop biofuels or environmentally friendly methods to dispose of waste generated by industries. 

    2 Jobs Available
    Actor

    For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

    4 Jobs Available
    Acrobat

    Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

    3 Jobs Available
    Video Game Designer

    Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

    3 Jobs Available
    Talent Agent

    The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

    If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

    3 Jobs Available
    Radio Jockey

    Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

    A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

    3 Jobs Available
    Producer

    An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

    They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

    2 Jobs Available
    Fashion Blogger

    Fashion bloggers use multiple social media platforms to recommend or share ideas related to fashion. A fashion blogger is a person who writes about fashion, publishes pictures of outfits, jewellery, accessories. Fashion blogger works as a model, journalist, and a stylist in the fashion industry. In current fashion times, these bloggers have crossed into becoming a star in fashion magazines, commercials, or campaigns. 

    2 Jobs Available
    Photographer

    Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

    2 Jobs Available
    Copy Writer

    In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

    5 Jobs Available
    Editor

    Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

    3 Jobs Available
    Journalist

    Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

    3 Jobs Available
    Publisher

    For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

    3 Jobs Available
    Vlogger

    In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

    3 Jobs Available
    Travel Journalist

    The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

    2 Jobs Available
    Videographer

    Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production. 

    2 Jobs Available
    SEO Analyst

    An SEO Analyst is a web professional who is proficient in the implementation of SEO strategies to target more keywords to improve the reach of the content on search engines. He or she provides support to acquire the goals and success of the client’s campaigns. 

    2 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    Quality Controller

    A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

    A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

    3 Jobs Available
    Production Manager

    Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers. 

    Resource Links for Online MBA 

    3 Jobs Available
    QA Manager

    Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes. 

    2 Jobs Available
    QA Lead

    A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

    2 Jobs Available
    Reliability Engineer

    Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment. 

    2 Jobs Available
    Safety Manager

    A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

    2 Jobs Available
    Corporate Executive

    Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

    2 Jobs Available
    Information Security Manager

    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

    3 Jobs Available
    Computer Programmer

    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    ITSM Manager

    ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks. 

    3 Jobs Available
    .NET Developer

    .NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of  creating, designing and developing applications using .NET languages such as VB and C#. 

    2 Jobs Available
    Corporate Executive

    Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

    2 Jobs Available
    DevOps Architect

    A DevOps Architect is responsible for defining a systematic solution that fits the best across technical, operational and and management standards. He or she generates an organised solution by examining a large system environment and selects appropriate application frameworks in order to deal with the system’s difficulties. 

    2 Jobs Available
    Cloud Solution Architect

    Individuals who are interested in working as a Cloud Administration should have the necessary technical skills to handle various tasks related to computing. These include the design and implementation of cloud computing services, as well as the maintenance of their own. Aside from being able to program multiple programming languages, such as Ruby, Python, and Java, individuals also need a degree in computer science.

    2 Jobs Available
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