**CBSE Important Questions about Maths Chapter 2 – Linear equations in one variable** is provided here for all the Class 8 students. To score well in the final exams, students can practice these important questions. All the questions given here are as per the CBSE syllabus and NCERT curriculum. All these questions have been prepared based on the exam perspective.

Linear equations in one variable are the algebraic expressions which have only one variable. These equations are easy to solve. Let us see some of the questions here to practice well.

## Class 8 Maths Chapter 2 Important Questions With Answers

**Question. 1**:** Solve the following linear equations:**

**(i) x – 11 =7**

**(ii) z + 8 = 9**

**(iii) 11x = 121**

Solution:

(i) x – 11 = 7

x = 7 + 11

x = 18

(ii) z + 8 = 9

z = 9 -8

z =1

(iii) 11x = 121

x = 121/11

x = 11

**Question. 2**: Solve: y/11 = 11

Solution: y/11 = 11

y = 11 × 11

y = 121

**Question. 3**: Solve: 23x/2 = 46

Solution: 23x/2 = 46

x = (46 × 2)/23

x = 2 × 2

x = 4

**Question.4**: Solve 1.2 = z/1.4

Solution: 1.2 = z/1.4

z = 1.2 × 1.4

z = 1.68

**Question.5**: Solve 7x – 12 = 16

Solution: 7x -12 = 16

7x = 16 + 12

x = 28/7

x = 4

**Question.6**: Solve 7z – 5 = 16

Solution: 7z – 5 = 16

7z = 21

z = 21/7

z =3

**Question.7**: 10 + 6p = 22

Solution: 10 + 6p = 22

6p = 22 – 10

6p = 12

p = 12/6

p = 2

**Question.8**: 11 – 5x + 3x + 4x =18

Solution: 11 – 5x + 3x + 4x = 18

11 – 5x + 7x = 18

2x = 18 -11

x = 7/2

**Question.9**: (x – 2) + (x – 3) + (x – 9) = 0

Solution: (x – 2) + (x – 3) + (x – 9) = 0

x – 2 + x – 3 + x – 9 = 0

3x – 2 – 3 – 9 = 0

3x – 14 = 0

x = 14/3

**Question.10:** (2x-2)+(3x-3)+(9x-9) = 1

Solution: (2x-2)+(3x-3)+(9x-9) = 1

2x – 2 + 3x – 3 + 9x – 9 = 1

14x – 14 = 1

14x = 1 + 14

x = 15/14

**Question.11**:** Solve each of the following equations:**

**(i)x+2 = -11**

Solution: x + 2 = -11

x = -11 -2

x = -13

**(ii) 2x – 1/6 = 3**

Solution: 2x – 1/6 = 3

2x = 3 + 1/6

2x = 19/6

x = 19/12

**(iii)7x – 7 = 21**

Solution: 7x – 7 = 21

7x = 21 + 7

7x = 28

x = 4

**(iv) -7x = 84**

Solution: -7x = 84

-x = 84/7

x = -12

**(v) 18+ 7x = -3 **

Solution: 18+ 7x = -3

7x = -3 -18

7x = -21

x = -3

**(vi) 3(x-4) = 21**

Solution: 3(x-4) = 21

3x – 12 = 21

3x = 21 + 12

3x = 33

x = 11

**(vii) 3x/2 – 2x/3 = 8**

Solution: 3x/2 – 2x/3 = 8

(9x – 4x)/6 = 8

5x = 48

x = 48/5

**(Viii) 3x-9 = 4x – 3**

Solution: 3x-9 = 4x – 3

3x – 4x = -3 + 9

-x = 6

x = – 6

**(ix) 3(2x – 3) = 4(2x + 4)**

Solution: 3(2x – 3) = 4(2x + 4)

6x – 9 = 8x + 16

6x – 8x = 16 + 9

-2x = 25

x = -25/2

**Question.12: Solve each of the following equations and check your solution by substituting in the equation.**

**(i) x/2-10=1/2**

Solution: x/2 – 10 = 1/2

(x – 20)/2 = 1/2

x – 20 = 1

x = 21

**(ii) x/3-x/2=6**

x/3 – x/2 =6

(2x – 3x)/6 = 6

-x/6 = 6

-x = 36

x = – 36

**(iii) 6x-9-2(1+x)=x-9**

Solution: 6x – 9 -2(1+x) = x -9

6x -9 -2 – 2x = x – 9

6x – 2x – x = -9 + 9 + 2

3x = 2

x = 2/3

**(iv) 2(x+2)+5(x+5)=4(x-8)+2(x-2)**

Solution: 2(x+2)+5(x+5) = 4(x-8)+2(x-2)

2x + 4 + 5x + 25 = 4x – 32 + 2x – 4

2x + 5x -4x – 2x = -32 – 4 -4 – 25

7x – 6x = -65

x = – 65

**(v) (3+x)/(2x-3) = -1/2**

Solution: (3+x)/(2x-3) = -1/2

2 (3 + x) = -1 (2x – 3)

6 + 2x = -2x + 3

4x = 3 -6

x = -3/4

**(vi) (x-7)/3 = (x-1)/5**

Solution: (x-7)/3 = (x-1)/5

5(x-7) = 3(x-1)

5x – 35 = 3x – 3

5x – 3x = -3 + 35

2x = 32

x = 16

### Word Problems

**Question.13: A positive number is 5 time another number. If 21 is added to both the numbers, then one of the new numbers become twice of other new numbers. Find the original numbers.**

Solution: Let the smaller number be x

And another number = 5x

If 21 is added to both the numbers then as per the given condition;

5x+21 = 2×(x+21)

5x + 21 = 2x + 42

5x – 2x = 42 – 21

3x =. 21

x = 21/3= 7

Therefore, the positive number is = 5×7 = 35

**Question.14: The sum of three consecutive multiples of 8 is 888. Find the multiple.**

Solution: Let the multiples be x, x+8, x+16

Given, the sum of three consecutive multiples of 8 is 888

Thus,

x+x+8+x+16=888

3x+24=888

3x=888-24

3x =864

x =864/3

x =288

Therefore the multiples are:

x =288

x +8=296

x+16=304

**Question.15: Five years ago, Anu was thrice as old as Sonu. After ten years, Anu will be twice as old as Sonu. How old are Anu and Sonu?**

Solution:

Let us assume the present ages of Anu is x and Sonu is y.

According to the question:

x – 5 = 3(y – 5)

x – 5 = 3y – 15

x= 3y-15+5

x= 3y -10 ……………….. (1)

Again, as per question;

x + 10 = 2(y + 10)

x + 10 = 2y + 20

x – 2y = 10

(3y – 10) – 2y = 10 (from equation 1)

3y-2y =10+10

y= 20

Substitute, y=20 in equation 1,

x = 3(20) -10

x =60 – 10

x = 50

Hence, the present ages of Anu is 50 years and of Sonu is 20 years.

**Question 16: Three consecutive integers are as such they are taken in increasing order and multiplied by 2, 3, and 4, respectively, they add up to 56. Find these numbers.**

Solution: Let us say the three consecutive numbers be x, x+1 and x+2.

Now as per the given question;

2×(x)+3×(x+1)+4×(x+2)=56

9x + 11 = 56

9x = 56-11

9x = 45

x = 45/9

x= 5

Therefore, the three consecutive numbers are 5, 6 and 7.

**Question 17: The perimeter of a rectangular swimming pool is 154 meters. Its length is 2m more than twice its breadth. What are the length and the breadth of the pool?**

Solution: Let the breadth of the pool =x

Length of the pool will be 2+2x.

Given, Perimeter of the pool = 154 m

We know, for a given rectangle,

Perimeter=2(length+breath)

By using this formula, we get;

154=2(2+2x+x)

77=2+3x

75=3x

x=25m

Hence, the breadth of the pool is 25m

and Length=2+2×25=52m

**Question 18: Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.**

Solution: Let first number be x

So, second number will be x+15

As per the given question;

x+x+15=95

x+x=95-15

2x=80

x=80/2

x=40

x=40

First number = 40

Second number = x+15=40+15=55

Hence, 40 + 55 = 95

**Question 19: The numbers are in the ratio 4:3. If they differ by 18, find these numbers.**

**Solution: **Let the numbers be 4x and 3x

According to the question, the two numbers are differed by 18. Thus,

4x – 3x = 18

=> x = 18

4x = 4 × 18 = 72

3x = 3 × 18 = 54

Therefore, the two numbers are 72 and 54

**Question 20: Three consecutive integers add up to 57. What are these integers?**

Solution:

Let us say the three consecutive numbers be x-1, x and x+1

Now, as per the given question,

x-1+x+x+1=57

3x=57

x=57/3

x=19

Hence,

x-1=19-1=18

x=19

x+1=19+1=20

The three consecutive numbers are 18,19 and 20.

**Question 21: There is a narrow rectangular plot. The length and breadth of the plot are in the ratio of 11:4. At the rate of Rs. 100 per meter it will cost the village panchayat Rs. 75000 to fence the plot. What are the dimensions of the plot?**

Solution: Let the common ratio be x.

Thus, the length and breadth of a rectangular plot be 11x and 4x.

We know that perimeter of rectangal = 2(length + breadth)

Therefore, the perimeter of the plot here is:

= 2(11x + 4x)

= 2(15x)

= 30x.

Given that Cost of fencing the plot at the rate of rs.100 per metre is 75000.

⇒ 100 × Perimeter = 75000

⇒ 100 × 30x = 75000

⇒ 3000x = 75000

⇒ x = 25

So,

Length of rectangular plot = 11x = 275m

Breadth of rectangular plot = 4x = 100m

Hence, the dimensions of the rectangular plot are 275m and 100m respectively.

**Question 22: Convert the following statements into equations.**

**(a) 3 added to a number is 11**

**(b) 2 subtracted from a number is equal to 15.**

**(c) 3 times a number decreased by 2 is 4.**

**(d) 2 times the sum of the number x and 7 is 13.**

Solution:

(a) x +3 = 11

(b) x – 2 = 15

(c) 3x – 2 = 4

(d) = 2(x+7) = 13

**Question 23: Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. the final result is 3 times her original number. Find the number.**

Solution: Let the number thought by Amina be x

She subtracts 5/2 from the number.

x – 5/2

Now, she multiplies the result by 8

8(x – 5/2)

The result obtained is three times her original number;

8(x – 5/2) = 3x

8(x – 5/2) = 3x

8x – 20 = 3x

5x = 20

x = 4

Hence, the number is 4.

**Question 24: A number is 12 more than the other. Find the numbers if their sum is 48.**

Solution: Let the first number be x and the second number be x + 12

As per the given question;

Sum of the two numbers = 48

x + x + 12 = 48

2x + 12 = 48

2x = 48 – 12

2x = 36

x = 36/2

x = 18

Therefore, x = 18 and x+12= 18+12=30

So the answer is 18 and 30.

**Question 25: The sum of three consecutive odd numbers is 51. Find the numbers.**

Solution: Let the three consecutive odd numbers are x, x+2 and x+4.

As per the given question, the sum all the three numbers is 51

Therefore,

x+x+2+x+4=51

3x+6=51

x=(51-6)/3=45/3 =15

Hence the numbers are:

x=15

x+2=17

x+4=19

Therefore, the required three consecutive odd numbers are 15,17 and 19.

**Question 26: Jane is 6 years older than her younger sister. After 10 years, the sum of their ages will be 50 years. Find their present ages.**

Solution: Let the age of Jane’s younger sister is x.

Age of Jane will be x + 6

As per the question, after 10 years the sum of their ages will be 50.

After 10 years,

age of Jane = x + 16

age of her younger sister = x + 10

Therefore,

x + 16 + x +10 = 50

2x + 26 = 50

2x = 24

x = 12

Hence, the present age of Jane’s younger sister is 12 years.

And of Jane’s is 12+6 = 18 years.

**Question 27: The denominator of a fraction is greater than the numerator by 8. If the numerator is increased by 17 and denominator is decreased by 1, the number obtained is 3/2. Find the fraction.**

Solution: Let numerator of a fraction be x.

Denominator = ( x + 8 )

Fraction = Numerator/Denominator = x/x+8

According to the question,

(x + 17) / (x + 8 – 1) = 3/2

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

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

2x + 34 = 3x + 21

3x – 2x = 34 – 21

x = 13

Numerator = x = 13

And

Denominator = X + 8 = 13+8 = 21

Fraction = 13/21

**Question 28: A sum of Rs.2700 is to be given in the form of 63 prizes. If the prize is of either Rs.100 or Rs.25, find the number of prizes of each type.**

Solution: Let the number of prizes of Rs. 100 be =x, and worth rs.25 be y.

Given that,

100x+25y= 2700

Simplify the above equation, we get

4x+y= 108 ….(1)

As per the given question;

x+y= 63 ….(2)

Now, solve equation (1) and (2), we get

3x= 45

x= 15

Now, substitute x= 15 in equation (2), we get

15+y= 63

y=63-15

y=48

Hence, the number of prizes worth Rs. 1oo is 15 and the number of prizes worth Rs. 25 is 48.

**Question 29: In an isosceles triangle, the base angles are equal, and the vertex angle is 80 degrees. Find the measure of the base angles.**

Solution: Let the base angle of the isosceles triangle is x.

Since the two sides of the isosceles triangle are equal, therefore its base angles are also equal.

By angle sum property of triangle, we know;

2x+80=180

2x=180-80

2x=100

x=50

Hence, the measure of both the base angles is equal to 50 degrees.

### Extra Practice Questions For Class 8 Maths Chapter 2 Linear Equations in One Variable

1. The tens and unit digits of a number are the same. By adding the number to its reverse, you get the sum 110. Calculate the number.

2. A man sold his bicycle for an amount that is over Rs 988 by half the price that he paid for it. He made a profit of Rs, 300. Now, what was the original cost of the bicycle?

3. Sum of two numbers were given as 2490. If 6.5% of one number is equal to 8.5% of another number, then find the value of both the numbers.