NCERT Solutions | Class 8 Maths Chapter 2

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

CBSE Solutions | Maths Class 8

Check the below NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Pdf free download. NCERT Solutions Class 8 Maths  were prepared based on the latest exam pattern. We have Provided Linear Equations in One Variable Class 8 Maths NCERT Solutions to help students understand the concept very well.

NCERT | Class 8 Maths

NCERT Solutions Class 8 Maths

Book:	National Council of Educational Research and Training (NCERT)
Board:	Central Board of Secondary Education (CBSE)
Class:	8th
Subject:	Maths
Chapter:	2
Chapters Name:	Linear Equations in One Variable
Medium:	English

Linear Equations in One Variable | Class 8 Maths | NCERT Books Solutions

You can refer to MCQ Questions for Class 8 Maths Chapter 2 Linear Equations in One Variable to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable Exercise 2.1

Solve the following equations:

Ex 2.1 Class 8 Maths Question 1.

x – 2 = 7

Solution:

Ex 2.1 Class 8 Maths Question 2.

y + 3 = 10

Solution:

Ex 2.1 Class 8 Maths Question 3.

6 = z + 2

Solution:

Ex 2.1 Class 8 Maths Question 4.

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

Solution:

Ex 2.1 Class 8 Maths Question 5.

6x = 12

Solution:

Ex 2.1 Class 8 Maths Question 6.

\(\frac { t }{ 5 } =10\)

Solution:

Ex 2.1 Class 8 Maths Question 7.

\(\frac { 2x }{ 3 } =18\)

Solution:

Ex 2.1 Class 8 Maths Question 8.

\(1.6=\frac { y }{ 1.5 } \)

Solution:

Ex 2.1 Class 8 Maths Question 9.

7x – 9 = 16

Solution:

Ex 2.1 Class 8 Maths Question 10.

14y – 8 = 13

Solution:

Ex 2.1 Class 8 Maths Question 11.

17 + 6p = 9

Solution:

Ex 2.1 Class 8 Maths Question 12.

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

Solution:

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable Exercise 2.2

Ex 2.2 Class 8 Maths Question 1.

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

Solution:

Let the required number be x. Then,

Thus, the required number is \(\frac { 3 }{ 4 } \)

Ex 2.2 Class 8 Maths Question 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 the breadth of the pool ?

Solution:

Ex 2.2 Class 8 Maths Question 3.

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 ?

Solution:

Let each equal side of an isosceles triangle ABC be x cm.

Ex 2.2 Class 8 Maths Question 4.

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

Solution:

Let the number be x and (x + 15).

Hence, the numbers are 40 and 55.

Ex 2.2 Class 8 Maths Question 5.

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

Solution:

Let the number br 5x and 3x. Then,

Hence, the numbers are 45 and 27.

Ex 2.2 Class 8 Maths Question 6.

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

Solution:

Let the consective numbers be x, (x + 1) and (x + 2). Then,

Hence, the consective numbers are 16, 17 and 18.

Ex 2.2 Class 8 Maths Question 7.

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

Solution:

Let the three consective multiples of 8 be 8x, 8(x + 1) and 8(x + 2).

Hence, the three consective multiples of 8 are 8 x 36, 8 x (36 + 1) and 8 x (36 + 2) i.e., 288, 296 and 3.4.

Ex 2.2 Class 8 Maths Question 8.

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.

Solution:

let the three consective integers be x, (x + 1) and (x + 2). It is given that

Hence, the numbers are 7, 8 and 9

Ex 2.2 Class 8 Maths Question 9.

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 ?

Solution:

Let the ages of Rahul and haroom be 5x and 7x years.

Ex 2.2 Class 8 Maths Question 10.

The numbers 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 ?

Solution:

Let the number of boys and girls in a class be 7x and 5x.
Since the number of boys is 8 more than the number of girls. Therefore,
7x = 5x + 8
⇒ 7x – 5x = 8
⇒ 2x = 8
⇒ x = 4
∴ The number of boys = 7 x 4 = 28
and the number of girls = 5 x 4 = 20
Thus, the total strength of the class is 28 + 20, i. e., 48.

Ex 2.2 Class 8 Maths Question 11.

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 ?

Solution:

Let Baichung’s age be x years. Then,

Ex 2.2 Class 8 Maths Question 12.

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

Solution:

Let Ravi’s present age be x years.
After 15 years his age will be (x +15) years.
It is given that after 15 years, Ravi’s age will be four times his present age.

Ex 2.2 Class 8 Maths Question 13.

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 ?

Solution:

Let the required number be x. Then,

Ex 2.2 Class 8 Maths Question 14.

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 ?

Solution:

Let notes of denominations ₹ 100, ₹ 50 and ₹ 10 be 2x, 3x and 5x respectively.

Ex 2.2 Class 8 Maths Question 15.

I have a total oft 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me ?

Solution:

Let the number of ₹ 5 coins be x. Then,

Ex 2.2 Class 8 Maths Question 16.

The organisers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3,000. Find the number of winners, if the total number of participants is 63.

Solution:

Let x be the number of the winners: So, the number of participants who does not win is (63 – x).

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable Exercise 2.3

Solve the following equations and check your results:

Ex 2.3 Class 8 Maths Question 1.

3x = 2x + 18

Solution:

Ex 2.3 Class 8 Maths Question 2.

5t – 3 = 3t – 5

Solution:

Ex 2.3 Class 8 Maths Question 3.

5x + 9 = 5 + 3x

Solution:

Ex 2.3 Class 8 Maths Question 4.

4z + 3 = 6 + 2z

Solution:

Ex 2.3 Class 8 Maths Question 5.

2x – 1 = 14 – x

Solution:

Ex 2.3 Class 8 Maths Question 6.

8x + 4 = 3(x – 1) + 7

Solution:

Ex 2.3 Class 8 Maths Question 7.

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

Solution:

Ex 2.3 Class 8 Maths Question 8.

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

Solution:

Ex 2.3 Class 8 Maths Question 9.

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

Solution:

Ex 2.3 Class 8 Maths Question 10.

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

Solution:

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable Exercise 2.4

Ex 2.4 Class 8 Maths Question 1.

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 ?

Solution:

Let Amina thinks of a number x. It is given that

Hence, the number thought is 4.

Ex 2.4 Class 8 Maths Question 2.

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 ?

Solution:

Let the numbers be x and 5x. it is given that

Hence, the required numbers are 7 and 35.

Ex 2.4 Class 8 Maths Question 3.

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 ?

Solution:

Let the unit’s digit be x.

∴ Unit’s digit = 6 and ten’s digit = 3
Hence, the number = 36.

Ex 2.4 Class 8 Maths Question 4.

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 ?

Solution:

Let one’s digit be x. Then, ten’s digit 3x.

Ex 2.4 Class 8 Maths Question 5.

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 ?

Solution:

Let the present age of Shobo be x years. Then,

∴ Present age of Shobo = 5 years ,
∴ Present age of his mother = 30 years.

Ex 2.4 Class 8 Maths Question 6.

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 of ₹ 100 per metre it will cost the village panchayat ₹ 75000 to fence the plot. What are the dimensions of the plot ?

Solution:

Let the length and breadth of the rectangular plot be 11x and 4x metres. Then,

Ex 2.4 Class 8 Maths Question 7.

Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per metre and trouser material that costs him ₹ 90 per metre. For every 2 meters of the trouser material he buys 3 metres of the shirt material. He sells the materials at 12% and 10% profit respectively. His total sale is ₹ 36,600. How much trouser material did he buy ?

Solution:

Let material bought for trousers and shirts be 2x metres and 3x metres respectively.
Cost price of 2x metres of trouser material @ ₹ 90 per metre = ₹ (90 x 2x) = ₹ 180x
Cost price of 3x metres of shirt material @ ₹ 50 per metre = ₹ (50 x 3x) = ₹ 150x

Ex 2.4 Class 8 Maths Question 8.

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.

Solution:

Let the number of deer in the herd be x. Then,

Ex 2.4 Class 8 Maths Question 9.

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

Solution:

Let the present age of grand daughter be x years. Then, the present age of grandfather is 10x years
According to the question.
10x – x = 54 ⇒ 9x = 54
⇒ \(x=\frac { 54 }{ 9 } =6\)
∴ Granddaughter’s age = 6 years
and grandfather’s age = 10 x 6 = 60 years.

Ex 2.4 Class 8 Maths Question 10.

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

Solution:

Let the present age of son be x years. Then, the age of his father Aman is 3x.
10 years ago : Their ages will be (x – 10) years and (3x – 10) years respectively.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable Exercise 2.5

Solve the following linear equations:

Ex 2.5 Class 8 Maths Question 1.

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

Solution:

Ex 2.5 Class 8 Maths Question 2.

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

Solution:

Ex 2.5 Class 8 Maths Question 3.

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

Solution:

Ex 2.5 Class 8 Maths Question 4.

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

Solution:

Ex 2.5 Class 8 Maths Question 5.

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

Solution:

Ex 2.5 Class 8 Maths Question 6.

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

Solution:

Simplify and solve the following linear equations:

Ex 2.5 Class 8 Maths Question 7.

3(t – 3) = 5(2t + 1)

Solution:

Ex 2.5 Class 8 Maths Question 8.

15(y – 4) – 2 (y – 9) + 5(y + 6) = 0

Solution:

Ex 2.5 Class 8 Maths Question 9.

3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Solution:

Ex 2.5 Class 8 Maths Question 10.

0.25(4f – 3) = 0.05(10f – 9).

Solution:

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable Exercise 2.6

Solve the following equations:

Ex 2.6 Class 8 Maths Question 1.

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

Solution:

Ex 2.6 Class 8 Maths Question 2.

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

Solution:

Ex 2.6 Class 8 Maths Question 3.

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

Solution:

Ex 2.6 Class 8 Maths Question 4.

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

Solution:

Ex 2.6 Class 8 Maths Question 5.

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

Solution:

Ex 2.6 Class 8 Maths Question 6.

The ages ofHari 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.

Solution:

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

∴ Present age of Hari =5 x 4 years = 20 years
∴ Present age of Harry =7 x 4 years = 28 years.

Ex 2.6 Class 8 Maths Question 7.

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 \(\frac { 3 }{ 2 } \). Find the rational number.

Solution:

Let the numerator of the rational number be x. Then, the denominator of the rational number = x + 8.
∴ The rational number = \(\frac { x }{ x+8 } \)
If the numerator is increased by 17 and the denominator is decreased by 1, the number becomes \(\frac { 3 }{ 2 } \).

Hence, the required rational number = \(\frac { 13 }{ 21 } \).

NCERT Class 8 Maths

Class 8 Maths Chapters | Maths Class 8 Chapter 2

NCERT Solutions for Class 8 Maths

NCERT Solutions of Maths Class 8 Chapter-wise

Chapter-wise NCERT Solutions for Class 8 Maths

NCERT Solutions For Class 8 Maths Chapter 1 Rational Numbers
NCERT Solutions For Class 8 Maths Chapter 2 Linear Equations in One Variable
NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals
NCERT Solutions For Class 8 Maths Chapter 4 Practical Geometry
NCERT Solutions For Class 8 Maths Chapter 5 Data Handling
NCERT Solutions For Class 8 Maths Chapter 6 Squares and Square Roots
NCERT Solutions For Class 8 Maths Chapter 7 Cubes and Cube Roots
NCERT Solutions For Class 8 Maths Chapter 8 Comparing Quantities
NCERT Solutions For Class 8 Maths Chapter 9 Algebraic Expressions and Identities
NCERT Solutions For Class 8 Maths Chapter 10 Visualising Solid Shapes
NCERT Solutions For Class 8 Maths Chapter 11 Mensuration
NCERT Solutions For Class 8 Maths Chapter 12 Exponents and Powers
NCERT Solutions For Class 8 Maths Chapter 13 Direct and Indirect proportions
NCERT Solutions For Class 8 Maths Chapter 14 Factorisation
NCERT Solutions For Class 8 Maths Chapter 15 Introduction to Graphs
NCERT Solutions For Class 8 Maths Chapter 16 Playing with Numbers

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