NCERT Exemplar Solutions Class 8 Maths Chapter 4 Linear Equation In One Variable

# NCERT Exemplar Solutions Class 8 Maths Chapter 4 Linear Equation In One Variable

Subject specialists have designed NCERT Exemplar Solutions for Maths Class 8 Chapter 4 which includes thorough solutions for reference. These solutions are updated according to the latest CBSE syllabus for 2024-25 and are provided in easy language for understanding. Tips and tricks are also provided.

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## NCERT Exemplar Solutions Class 8 Maths Chapter 4 Linear Equation In One Variable

Question. The solution of which of the following equations is neither a fraction nor an integer.

Answer – The given equations are (a) 3x + 2 = 5x + 2, (b) 4x – 18 = 2, (c) 4x + 7 = x + 2, and (d) 5x – 8 = x + 4.

For equation (c), 4x + 7 = x + 2, we need to solve for x.

To solve for x, we need to isolate x on one side of the equation.

First, subtract 7 from both sides of the equation:

4x – 7 = x

Next, subtract x from both sides of the equation:

3x = -5

Finally, divide both sides of the equation by 3:

x = -5/3

The solution of equation (c) is x = -5/3, which is neither a fraction nor an integer.

Question. The solution of the equation ax + b = 0 is

Answer – The given equation is ax + b = 0.

To solve for x, we need to isolate x on one side of the equation.

First, subtract b from both sides of the equation:

ax = -b

Next, divide both sides of the equation by a:

x = -b/a

The solution of the equation ax + b = 0 is x = -b/a.

Question. If 8x – 3 = 25 +17x, then x is

Answer – The given equation is 8x – 3 = 25 + 17x.

To solve for x, we need to isolate x on one side of the equation.

First, subtract 8x from both sides of the equation:

-3 = 25 + 9x

Next, subtract 25 from both sides of the equation:

-28 = 9x

Finally, divide both sides of the equation by 9:

x = -28/9

The solution of the equation 8x – 3 = 25 + 17x is x = -28/9.

Question. The shifting of a number from one side of an equation to the other is called

Answer – The shifting of a number from one side of an equation to the other is called transposition.

Question. If (5x/3) – 4 = (2x/5), then the numerical value of 2x – 7 is

Answer – The given equation is (5x/3) – 4 = (2x/5).

First, we need to solve for x.

To solve for x, we need to isolate x on one side of the equation.

First, multiply both sides of the equation by the least common multiple (LCM) of 3 and 5, which is 15:

15((5x/3) – 4) = 15((2x/5))

Next, distribute the 15 on both sides of the equation:

75x/3 – 60 = 3x

Next, simplify the equation:

25x – 60 = 3x

Next, subtract 3x from both sides of the equation:

22x – 60 = 0

Next, add 60 to both sides of the equation:

22x = 60

Finally, divide both sides of the equation by 22:

x = 60/22

x = 30/11

Next, we need to find the numerical value of 2x – 7.

Substitute x = 30/11 into the expression:

2x – 7 = 2(30/11) – 7

Next, simplify the equation:

2(30/11) – 7 = 60/11 – 7

Next, subtract 7 from both sides of the equation:

60/11 – 7 = 60/11 – 77/11

Next, subtract 60/11 from both sides of the equation:

-13/11

The numerical value of 2x – 7 is -13/11.

Question. The value of x for which the expressions 3x – 4 and 2x + 1 become equal is

Answer – The given expressions are 3x – 4 and 2x + 1.

To find the value of x for which the expressions become equal, we need to solve the equation:

3x – 4 = 2x + 1

To solve for x, we need to isolate x on one side of the equation.

First, subtract 2x from both sides of the equation:

3x – 2x = 1 + 4

Next, simplify the equation:

x = 5

The value of x for which the expressions 3x – 4 and 2x + 1 become equal is 5.

Question. If a and b are positive integers, then the solution of the equation ax = b has to be always

Answer – The given equation is ax = b, where a and b are positive integers.

Since a and b are positive integers, the solution of the equation must be a positive number.

Therefore, the solution of the equation ax = b has to be always a positive number.

Question. Linear equation in one variable has

A linear equation in one variable has only one variable with power 1.

Question. Which of the following is a linear expression:

Answer – The given expressions are (a) x^2 + 1, (b) y + y^2, (c) 4, and (d) 1 + z.

A linear expression is an expression that has the highest power as 1.

Therefore, the linear expression is (d) 1 + z.

Question. A linear equation in one variable has

Answer – A linear equation in one variable has only one solution.

Question. Value of S in (1/3) + S = 2/5

Answer – The given equation is (1/3) + S = 2/5.

To find the value of S, we need to isolate S on one side of the equation.

First, subtract (1/3) from both sides of the equation:

S = 2/5 – 1/3

Next, find the least common multiple (LCM) of 5 and 3, which is 15:

S = (6 – 5)/15

Next, simplify the equation:

S = 1/15

The value of S in (1/3) + S = 2/5 is 1/15.

Question. (-4/3)y = – ¾, then y =

Answer – The given equation is (-4/3)y = -¾.

To find the value of y, we need to isolate y on one side of the equation.

First, divide both sides of the equation by (-4/3):

y = -¾ × -¾

Next, simplify the equation:

y = 9/16

The value of y in (-4/3)y = -¾ is 9/16.

Question. The digit in the tens place of a two-digit number is 3 more than the digit in the units place. Let the digit at the units place be b. Then the number is

Answer – The given information is that the digit in the tens place of a two-digit number is 3 more than the digit in the units place, and the digit at the units place is b.

Therefore, the number is 10(b + 3) + b.

Question. Arpita’s present age is thrice of Shilpa. If Shilpa’s age three years ago was x. Then Arpita’s present age is

Answer – The given information is that Arpita’s present age is thrice of Shilpa, and Shilpa’s age three years ago was x.

Therefore, Shilpa’s present age is x + 3.

Therefore, Arpita’s present age is 3(x + 3).

Question. The sum of three consecutive multiples of 7 is 357. Find the smallest multiple.

Answer – The given information is that the sum of three consecutive multiples of 7 is 357.

Let the three consecutive multiples of 7 be 7x, 7(x + 1), and 7(x + 2).

Therefore, 7x + 7(x + 1) + 7(x + 2) = 357.

Next, simplify the equation:

7x + 7x + 7 + 7x + 14 = 357.

Next, simplify the equation:

21x + 21 = 357.

Next, subtract 21 from both sides of the equation:

21x = 336.

Next, divide both sides of the equation by 21:

x = 16.

Therefore, the smallest multiple of 7 is 7x, which is 112.

Question. In a linear equation, the _________ power of the variable appearing in the equation is one.

Answer – In a linear equation, the highest power of the variable appearing in the equation is one.

Question. The solution of the equation 3x – 4 = 1 – 2 x is .

Answer – The given equation is 3x – 4 = 1 – 2x.

To solve for x, we need to isolate x on one side of the equation.

First, add 2x to both sides of the equation:

5x – 4 = 1.

Next, add 4 to both sides of the equation:

5x = 5.

Next, divide both sides of the equation by 5:

x = 1.

The solution of the equation 3x – 4 = 1 – 2x is 1.

Question. The solution of the equation 2y = 5y – 18/5 is .

Answer – The given equation is 2y = 5y – 18/5.

To solve for y, we need to isolate y on one side of the equation.

First, subtract 2y from both sides of the equation:

-18/5 = 3y.

Next, multiply both sides of the equation by 3:

-18 = 9y.

Next, divide both sides of the equation by 9:

y = -18/9.

y = -2/1.

y = -2.

The solution of the equation 2y = 5y – 18/5 is -2.

Question. Any value of the variable which makes both sides of an equation equal is known as a _________ of the equation.

Answer – Any value of the variable which makes both sides of an equation equal is known as a solution of the equation.

Question. 9x – _________ = –21 has the solution (–2)

Answer – The given equation is 9x – y = -21, and the solution is -2.

To find the value of y, we need to substitute the value of x into the equation.

Substitute x = -2 into the equation:

9(-2) – y = -21.

Next, simplify the equation:

-18 – y = -21.

Next, add 18 to both sides of the equation:

-y = -3.

Next, multiply both sides of the equation by -1:

y = 3.

Therefore, the missing number is 3.

Question. Three consecutive numbers whose sum is 12 are _________, _________ and _________.

Answer – The three consecutive numbers whose sum is 12 are 3, 4, and 5.

Question. The share of A when Rs 25 is divided between A and B so that A gets Rs. 8 more than B is _________.

Answer – The given information is that Rs 25 is divided between A and B so that A gets Rs. 8 more than B.

Let the share of B be x.

Therefore, the share of A is x + 8.

The total share is Rs 25.

Therefore, x + (x + 8) = 25.

Next, simplify the equation:

2x + 8 = 25.

Next, subtract 8 from both sides of the equation:

2x = 17.

Next, divide both sides of the equation by 2:

x = 17/2.

x = 8.5.

The share of B is Rs 8.5.

The share of A is x + 8 = 8.5 + 8 = Rs 16.5.

Therefore, the share of A is Rs 16.5.

Question. The sum of two numbers is 35 and the difference is 9. The larger number is _________.

Answer – The given information is that the sum of two numbers is 35 and the difference is 9.

Let the larger number be x.

Therefore, the smaller number is x – 9.

The sum of the two numbers is 35.

Therefore, x + (x – 9) = 35.

Next, simplify the equation:

2x – 9 = 35.

Next, add 9 to both sides of the equation:

2x = 44.

Next, divide both sides of the equation by 2:

x = 44/2.

x = 22.

Therefore, the larger number is 22.

Question. The sum of two numbers is 25 and the product is 120. The larger number is _________.

Answer – The given information is that the sum of two numbers is 25 and the product is 120.

Let the larger number be x.

Therefore, the smaller number is 25 – x.

The product of the two numbers is 120.

Therefore, x(25 – x) = 120.

Next, simplify the equation:

25x – x^2 = 120.

Next, rewrite the equation in the form of a quadratic equation:

x^2 – 25x + 120 = 0.

(x – 12)(x – 10) = 0.

Next, set each factor equal to zero and solve for x:

x – 12 = 0.

x = 12.

x – 10 = 0.

x = 10.

Therefore, the larger number is 12 or 10.

Question. The sum of two consecutive even numbers is 28. The smaller number is _________.

Answer – The given information is that the sum of two consecutive even numbers is 28.

Let the smaller number be x.

Therefore, the larger number is x + 2.

The sum of the two numbers is 28.

Therefore, x + (x + 2) = 28.

Next, simplify the equation:

2x + 2 = 28.

Next, subtract 2 from both sides of the equation:

2x = 26.

Next, divide both sides of the equation by 2:

x = 26/2.

x = 13.

Therefore, the smaller number is 13.

Question. The sum of two consecutive odd numbers is 44. The smaller number is _________.

Answer – The given information is that the sum of two consecutive odd numbers is 44.

Let the smaller number be x.

Therefore, the larger number is x + 2.

The sum of the two numbers is 44.

Therefore, x + (x + 2) = 44.

Next, simplify the equation:

2x + 2 = 44.

Next, subtract 2 from both sides of the equation:

2x = 42.

Next, divide both sides of the equation by 2:

x = 42/2.

x = 21.

Therefore, the smaller number is 21.

Question. The sum of two consecutive numbers is 30. The smaller number is _________.

Answer – The given information is that the sum of two consecutive numbers is 30.

Let the smaller number be x.

Therefore, the larger number is x + 1.

The sum of the two numbers is 3

Question. A term of an equation can be transposed to the other side by changing its _________.

In an equation, shifting a term to the opposite side involves changing its sign. For instance, in 2x + 3 = 0, moving 3 to the right side transforms it into -3, resulting in 2x = -3 and x = -3/2.

Question. On subtracting 8 from x, the result is 2. The value of x is _________.

Subtracting 8 from x yields 2, making x equal to 10. By moving -8 to the right side, the equation simplifies to x = 2 + 8 = 10.

Question. (x/5) + 30 = 18 has the solution as .

Solving (x/5) + 30 = 18 results in x = -60. By transferring 30 to the right side, the equation becomes (x/5) = -12, leading to x = -12 × 5 = -60.

Question. When a number is divided by 8, the result is –3. The number is _________.

When a number divided by 8 equals -3, the number is -24. This is calculated by solving x/8 = -3, giving x = -3 × 8 = -24.

Question. 9 is subtracted from the product of p and 4, the result is 11. The value of p is _________.

If 9 is subtracted from the product of p and 4, resulting in 11, then p equals 5. By moving -9 to the right side, the equation simplifies to 4p = 20, leading to p = 5.

Question. If (2/5)x – 2 = 5 – (3/5)x, then x = .

If (2/5)x – 2 = 5 – (3/5)x, then x = 7

Given (2/5)x – 2 = 5 – (3/5)x, solving for x gives x = 7. By rearranging terms, the equation simplifies to 5x = 35, resulting in x = 7.

Question. After 18 years, Swarnim will be 4 times as old as he is now. His present age is _________.

Swarnim’s present age is 6 years, as after 18 years, he will be 4 times his current age. By solving x + 18 = 4x, it leads to x = 6, indicating Swarnim’s current age.

Question. Convert the statement adding 15 to 4 times x is 39 into an equation _________.

Converting “Adding 15 to 4 times x is 39” into an equation gives 4x + 15 = 39.

Question. The denominator of a rational number is greater than the numerator by 10. If the numerator is increased by 1 and the denominator is decreased by 1, then the expression for the new denominator is _________.

If the denominator of a rational number exceeds the numerator by 10, then increasing the numerator by 1 and decreasing the denominator by 1 results in the new denominator being x + 9.

Question. The sum of two consecutive multiples of 10 is 210. The smaller multiple is _________.

The sum of two consecutive multiples of 10 totaling 210 reveals the smaller multiple to be 100. By solving 20x + 10 = 210, it leads to x = 10, making the smaller multiple 100.

## Class 8 Maths Chapter 4 Linear Equation In One Variable FAQs

### How to start reading NCERT Exemplar Solutions of this chapter?

At first you can learn and understand Practical Geometry with the help of the Infinity Learn website. Then study the solutions given by Infinity Learn experts at their site. By solving these exercises one can easily go through all the concepts present in these NCERT Exemplar Solutions.

### What are the important topics from an exam perspective present in these NCERT Exemplar Solutions?

The important topics coming under this chapter are 4.1 Introduction 4.2 Constructing a Quadrilateral 4.2.1 When the lengths of four sides and a diagonal are given 4.2.2 When two diagonals and three sides are given 4.2.3 When two adjacent sides and three angles are known 4.2.4 When three sides and two included angles are given 4.3 Some special cases. Students can easily score good marks by practicing these solutions.

### Is it necessary to learn all methods to construct quadrilaterals of NCERT Exemplar Solutions for this chapter?

Yes, it is necessary to know all methods to construct quadrilaterals on different conditions given on these solutions. These topics will continue in higher studies as well as they may come in their finals of Class 8. These concepts are well explained by Infinity Learn. Hence, the main aim of these solutions designed by Infinity Learns experts is to understand maths at its fundamental level, which helps students to understand every concept clearly.

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