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

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable.

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 2
Chapter Name	Linear Equations in One Variable
Exercise	Ex 2.1, Ex 2.2, Ex 2.3, Ex 2.4, Ex 2.5, Ex 2.6
Number of Questions Solved	65
Category	NCERT Solutions

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

Chapter 2 Linear Equations In One Variable Exercise 2.1

Solve the following equations:

Question 1.
x – 2 = 7
Solution:

Question 2.
y + 3 = 10
Solution:

Question 3.
6 = z + 2
Solution:

Question 4.

Solution:

Question 5.
6x = 12
Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.
7x – 9 = 16
Solution:

Question 10.
14y – 8 = 13
Solution:

Question 11.
17 + 6p = 9
Solution:

Question 12.

Solution:

Chapter 2 Linear Equations In One Variable Exercise 2.2

Question 1.
If you subtract from a number and multiply the result by you get What is the number ?
Solution.
Let the required number be x. Then,

Thus, the required number is

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.

Question 3.
The base of an isosceles triangle is cm. The perimeter of the triangle is 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.

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.

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.

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.

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.

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

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.

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.

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,

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.

Question 13.
A rational number is such that when you multiply it by and add to the product, you get. What is the number ?
Solution.
Let the required number be x. Then,

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.

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,

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).

Chapter 2 Linear Equations In One Variable Exercise 2.3

Solve the following equations and check your results:

Question 1.
3x = 2x + 18
Solution:

Question 2.
5t – 3 = 3t – 5
Solution:

Question 3.
5x + 9 = 5 + 3x
Solution:

Question 4.
4z + 3 = 6 + 2z
Solution:

Question 5.
2x – 1 = 14 – x
Solution:

Question 6.
8x + 4 = 3(x – 1) + 7
Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Chapter 2 Linear Equations In One Variable Exercise 2.4

Question 1.
Amina thinks of a number and subtracts 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.

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.

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.

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.

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.

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,

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

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,

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
⇒ 
∴ Granddaughter’s age = 6 years
and grandfather’s age = 10 x 6 = 60 years.

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.

Chapter 2 Linear Equations In One Variable Exercise 2.5

Solve the following linear equations:

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Simplify and solve the following linear equations:

Question 7.
3(t – 3) = 5(2t + 1)
Solution:

Question 8.
15(y – 4) – 2 (y – 9) + 5(y + 6) = 0
Solution:

Question 9.
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Solution:

Question 10.
0.25(4f – 3) = 0.05(10f – 9).
Solution:

Chapter 2 Linear Equations In One Variable Exercise 2.6

Solve the following equations:

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

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.

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 . 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 =
If the numerator is increased by 17 and the denominator is decreased by 1, the number becomes .

Hence, the required rational number = .

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