Assam SCERT Solutions for Class 7 Maths Chapter 4 Simple Equation Exercise 4.1

Sudev Chandra Das

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Free Solutions for The Assam State School Education Board (ASSEB) Class 7 Maths Chapter 4 Simple Equation Exercise 4.1

Assam SCERT Solutions for Class 7 Maths Chapter 4 Simple Equation – Exercise 4.1 (2025-26) | Download Free PDF

Students can now access SCERT Assam Solutions for Class 7 Maths Chapter 4 – Simple Equation (Exercise 4.1, English Medium), designed as per the latest syllabus for the 2025-26 academic session. These solutions, prepared by subject experts at Digital Pipal Academy, help students develop a strong understanding of algebraic equations and their applications.

Topics Covered in Exercise 4.1:

Understanding simple equations – Formation and solving
Transposing terms – Shifting numbers and variables to simplify equations
Real-life applications of equations – Solving problems using algebra

Key Features of SCERT Assam Class 7 Maths Solutions:

Step-by-step solutions for better clarity
Practice online to strengthen conceptual understanding
Download free PDF for offline study
Exercise-wise solutions to help students score better marks

SCERT Assam Class 7 Maths Chapter 4 Simple Equation – Exercise 4.1

Click on the link below to access detailed solutions for Simple Equation:

SCERT Assam Class 7 Maths Chapter 4 Solutions

SCERT Assam Class 7 Maths Chapter 4 Solutions

Exercise Solution Link
Exercise 4.1 Click Here
Exercise 4.2 Click Here
Exercise 4.3 Click Here

 

 

1. Express the following statement in the form of an equation:

(i) When 5 is added to 6 times of a number the result is 35.

Solution

Let the number be x.

6x + 5 = 35

(ii) One fourth of a number is equal to 9.

Solution

Let the number be x. 

`\frac{1}{4}\times x=9`


(iii) 5 times of a number is equal to 5 more than 20.

Solution

Let the number be x.

5x = 20 + 5

(iv) To get 10 add 3 to 7 times of a number.

Solution

Let the number be x.

7x + 3 = 10

(v) When 4 is subtracted from one fifth of a number the result is 2.

Solution

Let the number be x. 

`\frac{1}{5}\times x-4=2`

(vi) 4 times of p is equal to 20.

Solution

4p = 20

(vii) When 1 is subtracted from 3 times of a number the result is 2.

Solution

Let the number be x.

3x - 1 = 2

(viii) To get 40 divide a number by 10 then subtract 10 from it.

Solution

Let the number be x. 

`\frac{x}{10}-10=40`


 

2. Write the following equation in statement:

(i) 3x - 4 = 5

Solution:   Subtraction of 4 from 3 times of number results 5

(ii) `\frac{m}{3}+6=11`

Solution:  Dividing m by 3, add 6 to get 11

(iii) 7p = 42

Solution:  7 times of a number is equal to 42.

(iv) `\frac{y}{6}=2`

Solution:  Dividing a number 6 will be got 2

(v) 5x + 7 = 2

Solution:  Adding 7 with 5 times of a number will be got 2.

(vi) `\frac{q}{2}-1=4`

Solution:  Substraction of 1 from half of a number results 4.

3. Form equations for each of the following:

(i) Sum of the ages of Anupama, Nirupama and Upoma is 22 years. Anupama is younger to Nirupoma by 1 year, Upoma is older to Nirupoma by 2 years. Write the equation considering the age of Nirupoma.

Solution:   Let the age of Nirupoma be x.

Anupama's age = x - 1

Upoma's age = x + 2

According to question ,

(x - 1) + x + (x + 2) = 22

(ii) Age of Anjan's grandfather is 72 years. Age of grandfather is 2 years more than seven times of the age of Anjan.

Solution:   Let the age of Anjan be x.

According to question,

7x + 2 = 72

(iii) Perimeter of a square is 32 centimeter.

Solution:   Let the side of the square be x.

According to question,

4x = 32

(iv) Romen's father bought potato at the rate of 20 per Kg and onion at the rate of 10 per kg. He paid to the shopkeeper 50 after buying 1kg less onion than potato.

Solution:  Let the weight of potato bought be x.

Weight of onion bought = x - 1

Total cost = 20x + 10(x - 1) = 50

(v) Measure of two angles of a triangle is two times and three times of the smallest angle of the triangle. Sum of the three angles of the triangle is 180°.

Solution:   Let the smallest angle be x. Other angles: 2x, 3x

According to question,

 x + 2x + 3x = 180

 

4. Tell whether the value of variable inserted within the brackets satisfy the given equation or not:

(i) x + 5 = 0, (x = -5)

Solution:  Yes, (-5) + 5 = 0 is satisfy the equation

 

(ii) 2x - 8 = 7, (x = 4)

Solution:  No, 2(4) - 8 ≠ 7 is not satisfy the equation

(iii) `\frac{x}{3}+6=7`, (x = 3)

Solution:  Yes, `\frac{3}{3}+6=7` is satisfy the equation

(iv) `\frac{x}{7}-2=0`, (x = 7)

Solution:  No, `\frac{7}{7}-2=0` is not satisfy the equation

(v) 5x = 35, (x = 7)

Solution:  Yes, 5(7) = 35 is satisfy the equation

(vi) 4x + 8 = 4, (x = -1)

Solution:  Yes, 4(-1) + 8 = 4 is satisfy the equation

(vii) 7x + 2 = 9, (x = 2)

Solution:  No, 7(2) + 2 ≠ 9 is not satisfy the equation

(viii) 2x = 16, (x = 8)

Solution:  Yes, 2(8) = 16 is satisfy the equation

(ix) `\frac{x}{5}=20`, (x = 100)

Solution:  Yes, `\frac{100}{5}=20` is satisfy the equation

(x) `\frac{x}{8}` + 4 = 9, (x = 1)

Solution:  No, `\frac{1}{8}` + 4 ≠ 9 is not satisfy the equation

 

 

5. Examine whether the value of variable inserted within the brackets is the root of the given equation or not:

(i) 4x + 3 = 7, (x = 1)

Solution:  Yes, 4(1) + 3 = 7 is the root of the equation

(ii) `\frac{2x}{3}+5=7`, (x = 3)

Solution:  Yes, `\frac{2(3)}{3}+5=7` is the root of the equation

(iii) x - 4 = 1, (x = 3)

Solution:  No, 3 - 4 ≠ 1 is not the root of the equation

(iv) 6x = 18, (x = 3)

Solution:  Yes, 6(3) = 18 is the root of the equation

(v) 5x - 1 = 7, (x = 2)

Solution:  No, 5(2) - 1 ≠ 7 is not the root of the equation

(vi) x + 9 = 13, (x = 4)

Solution:  Yes, 4 + 9 = 13 is the root of the equation

(vii) 5x - 7 = 8, (x = 3)

Solution:  Yes, 5(3) - 7 = 8 is the root of the equation

(viii) `\frac{y}{3}+5=8`, (y = 9)

Solution:  Yes, `\frac{9}{3}+5=8` is the root of the equation

(ix) `\frac{p}{5}+4=5`, (p = 1)

Solution:  No, `\frac{1}{5}+4\neq5` is not the root of the equation

(x) `\frac{x}{7}=6`, (x = 42)

Solution:  Yes, `\frac{42}{7}=6` is the root of the equation

6. Try equation by changing the value of the variable (Trial and Error method):

  (i) 2x + 5 = 11

 Solution:   Trial 1: Let x = 3

2(3) + 5 = 11

6 + 5 = 11

This is true, so x = 3 is a solution.

Trial 2: Let x = 3

2(3) + 5 = 11

6 + 5 = 11

This is true, so x = 3 is a solution.

(ii) `\frac{x}{5}+5=7`

Solution:   Trial 1: Let x = 10

\frac{10}{5}+5=7

2 + 5 = 7

This is true, so x = 10 is a solution.

(iii) 7x - 4 = 24

Solution:     Trial 1: Let x = 3

7(3) - 4 = 24

21 - 4 = 24

This is not true, so x = 3 is not a solution.

Trial 2: Let x = 4

7(4) - 4 = 24

28 - 4 = 24

This is true, so x = 4 is a solution.

(iv) `\frac{y}{6}=2`

Solution:     Trial 1: Let y = 10

 `\frac{10}{6}=2`

This is not true, so y = 10 is not a solution.

Trial 2: Let y = 12

`\frac{12}{6}=2`

This is true, so y = 12 is a solution.

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Sudev Chandra Das

About the Publisher

Hi! I'm Sudev Chandra Das (B.Sc. Mathematics), the Founder of Digital Pipal Academy. I've dedicated myself to guiding students toward better education. I believe, 'Success comes from preparation, hard work, and learning from failure.' Let’s embark on a journey of growth and digital excellence together!

 


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