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NCERT Solutions

Class 7

Maths

Chapter 4 Simple Equations Exercise 4.1

NCERT Solutions For Class 7 Maths Chapter 4 Simple Equations Exercise 4.1 - 2025-26

Q: 1. What is the correct method to solve a simple equation as per the NCERT Class 7 Maths Chapter 4?

The NCERT textbook for Class 7 Maths introduces the systematic method for solving simple equations. This involves balancing the equation by performing the same mathematical operation (addition, subtraction, multiplication, or division) on both the Left-Hand Side (LHS) and the Right-Hand Side (RHS). The goal is to isolate the variable on one side to find its value. Another key technique covered is transposing, which is a shortcut for moving terms across the equals sign while changing their sign.

Q: 4. Why is it necessary to perform the same operation on both sides of a simple equation?

An equation represents a state of balance, where the value of the Left-Hand Side (LHS) is equal to the value of the Right-Hand Side (RHS). Performing the same operation on both sides ensures that this balance is maintained. Think of it like a weighing scale; if you add 5 kg to one side, you must add 5 kg to the other to keep it level. This principle allows us to manipulate the equation and isolate the variable without changing the fundamental equality, which is the core rule for finding the correct solution.

Q: 7. What is a common mistake students make when solving equations involving negative numbers in Class 7 Maths Chapter 4?

A very common mistake is mishandling signs during transposition. For instance, in an equation like 5x - 3 = 12, students might correctly move '-3' to the other side but forget to change its sign to '+3'. Another frequent error occurs with multiplication or division. In an equation like -2y = 8, students might incorrectly add 2 to both sides instead of dividing by -2. Always remember to apply the correct inverse operation to the entire term, including its sign.

Q: 8. Can a simple equation in one variable have more than one solution? Why or why not, according to the concepts in Chapter 4?

No, a linear or simple equation in one variable, as covered in Class 7 Maths Chapter 4, has only one unique solution. This is because the equation represents a specific point of balance. For example, in the equation x + 5 = 8, only the value x = 3 can satisfy the condition. Any other value for 'x' would make the Left-Hand Side unequal to the Right-Hand Side, thus breaking the equality. More complex equations, which you will study in higher classes, can have multiple solutions.

NCERT Solutions For Class 7 Maths Chapter 4 Simple Equations Exercise 4.1 - 2025-26

Exercise 4.1 Class 7 in Chapter 4 Maths focuses on understanding and solving simple equations. This exercise is essential as it lays the groundwork for solving various types of algebraic equations, a skill crucial for higher-level mathematics. Students will learn how to balance equations and find the value of unknown variables. In NCERT Solutions for Class 7 Maths. It is important to focus on the basic principles of forming and solving equations. Practice is key to mastering this topic, so ensure you work through each problem methodically. This exercise provides a solid foundation for more complex algebraic concepts.

Table of Content

Glance on NCERT Solutions Maths Chapter 4 Exercise 4.1 Class 7 | Vedantu

Class 7th Maths Chapter 4 Exercise 4.1 explains the concept of simple equations, focusing on understanding how to solve them by isolating the variable.
Equations are mathematical statements that show the equality between two expressions.
Variables are symbols that represent unknown values and need to be determined.
Constants are fixed values that do not change within the equation.
Balancing equations involves performing the same operation on both sides to maintain equality.
Solving equations requires manipulating the equation to isolate the variable on one side and the constants on the other.
There are 6 questions in class 7th maths chapter 4 exercise 4.1 which are fully solved by experts at Vedantu.

Access NCERT Solutions for Maths Class 7 Chapter 4 - Simple Equations

Exercise- 4.1

1. Complete the last column of the table:

S.No.	Equation	Value	Say, whether the equation is satisfied. (yes/no)
	$ x+3=0 $	$ x=3 $
	$ x+3=0 $	$ x=0 $
	$ x+3=0 $	$ x=-3 $
	$ x-7=1 $	$ x=7 $
	$ x-7=1 $	$ x=8 $
	$ 5x=25 $	$ x=0 $
	$ 5x=25 $	$ x=5 $
	$ 5x=25 $	$ x=-5 $
	$ \dfrac{m}{3}=2 $	$ m=-6 $
	$ \dfrac{m}{3}=2 $	$ m=0 $
	$ \dfrac{m}{3}=2 $	$ m=6 $

Ans:

S.No.	Equation	Value	Say, whether the equation is satisfied. (yes/no)
	$ x+3=0 $	$ x=3 $	No
	$ x+3=0 $	$ x=0 $	No
	$ x+3=0 $	$ x=-3 $	Yes
	$ x-7=1 $	$ x=7 $	No
	$ x-7=1 $	$ x=8 $	Yes
	$ 5x=25 $	$ x=0 $	No
	$ 5x=25 $	$ x=5 $	Yes
	$ 5x=25 $	$ x=-5 $	No
	$ \dfrac{m}{3}=2 $	$ m=-6 $	No
	$ \dfrac{m}{3}=2 $	$ m=0 $	No
	$ \dfrac{m}{3}=2 $	$ m=6 $	Yes

2. Check whether the value given in the brackets is a solution to the given equation or not:

a. \[n+5=19\left( n=1 \right)\]

Ans: $ n+5=19 $

Putting $ n=1 $ in L.H.S.,

$ 1+5=6 $

$ \because L.H.S.\ne R.H.S.$

$ \therefore n=1 $ is not the solution of given equation.

b. $ 7n+5=19\left( n=-2 \right) $

Ans: $ 7n+5=19 $

Putting $ n=-2 $ in L.H.S.,

$ 7\left( -2 \right)+5=-14+5=-9 $

$ \because L.H.S.\ne R.H.S. $

$ \therefore n=-2 $ is not the solution of given equation.

c. $ 7n+5=19\left( n=2 \right) $

Ans: $ 7n+5=19 $

Putting $ n=2 $ in L.H.S.,

$ 7\left( 2 \right)+5=14+5=19 $

$ \because L.H.S.=R.H.S.$

$ \therefore n=2 $ is the solution of given equation.

d. $ 4p-3=13\left( p=1 \right) $

Ans: $ 4p-3=13 $

Putting $ p=1 $ in L.H.S.,

$ 4\left( 1 \right)-3=4-3=1 $

$ \because L.H.S.\ne R.H.S. $

$ \therefore $ $ p=1 $ is not the solution of given equation.

e. $ 4p-3=13\left( p=-4 \right) $

Ans: $ 4p-3=13 $

Putting $ p=-4 $ in L.H.S.,

$ 4\left( -4 \right)-3=-16-3=-19 $

$ \because L.H.S.\ne R.H.S. $

$ \therefore $ $ p=-4 $ is not the solution of given equation.

f. $ 4p-3=13\left( p=0 \right) $

Ans: $ 4p-3=13 $

Putting $ p=0 $ in L.H.S.,

$ 4\left( 0 \right)-3=0-3=-3 $

$ \because L.H.S.\ne R.H.S. $

$ \therefore $ $ p=0 $ is not the solution of given equation.

3. Solve the following equations by trial and error method:

a. $ 5p+2=17 $

Ans: Putting $ p=-3 $ in L.H.S. $ 5(-3)+2=-15+2=-13 $ $ \because -13\ne 17 $

Therefore, \[p=-3\text{ }\] is not the solution.

Putting $ p=-2 $ in L.H.S. $ 5(-2)+2=-10+2=-8 $ $ \because -8\ne 17\quad $

Therefore, $ p=-2 $ is not the solution.

Putting $ p=-1 $ in L.H.S. $ 5(-1)+2=-5+2=-3 $ $ \because -3\ne 17 $

Therefore, $ p=-1 $ is not the solution.

Putting $ p=0 $ in L.H.S. $ 5(0)+2=0+2=2 $ $ \because 2\ne 17\quad $

Therefore, $ p=0 $ is not the solution.

Putting $ p=1 $ in L.H.S. $ 5(1)+2=5+2=7 $ $ \because 7\ne 17\quad $

Therefore, $ p=1 $ is not the solution.

Putting $ p=2 $ in L.H.S. $ 5(2)+2=10+2=12 $ $ \because 12\ne 17\quad $

Therefore, $ p=2 $ is not the solution.

Putting $ p=3 $ in L.H.S. $ 5(3)+2=15+2=17 $ $ \because 17=17\quad $

Therefore, $ p=3 $ is the solution.

b. $ 3m-14=4 $

Ans: Putting $ m=-2 $ in L.H.S. $ \quad 3(-2)-14=-6-14=-20\,\,\,\because -20\ne 4\quad $

Therefore, $ m=-2 $ is not the solution.

Putting $ m=-1 $ in L.H.S. $ \quad 3(-1)-14=-3-14=-17\,\,\,\,\because -17\ne 4\quad $

Therefore, $ m=-1 $ is not the solution.

Putting $ m=0 $ in L.H.S. $ 3(0)-14-0-14=-14 $ $ \because -14\ne 4\quad $

Therefore, $ m=0 $ is not the solution.

Putting $ m=1 $ in L.H.S. $ 3(1)-14=3-14=-11 $

$ \because -11\ne 4\quad $ Therefore, $ m=1 $ is not the solution.

Putting $ m=2 $ in L.H.S. $ \quad 3(2)-14=6-14=-8 $

$ \because -8\ne 4\quad $ Therefore, $ m=2 $ is not the solution.

Putting $ m=3 $ in L.H.S. $ 3(3)-14=9-14=-5 $

$ \because -5\ne 4\quad $ Therefore, $ m=3 $ is not the solution.

Putting $ m=4 $ in L.H.S. $ \quad 3(4)-14=12-14=-2 $

$ \because -2\ne 4\quad $ Therefore, $ m=4 $ is not the solution.

Putting $ m=5 $ in L.H.S. $ \quad 3(5)-14=15-14=1 $ $ \because 1\ne 4 $

Therefore, $ m=5 $ is not the solution.

4. Write equations for the following statements:

i. The sum of numbers x and 4 is 9.

Ans: $ x+4=9 $

ii. 2 subtracted from y is 8.

Ans: $ y-2=8 $

iii. Ten times a is 70.

Ans: $ 10a=70 $

iv. The number b divided by 5 gives 6.

Ans: $ \dfrac{b}{5}=6 $

v. Three-fourth of t is 15.

Ans: $ \dfrac{3}{4}t=15 $

vi. Seven times m plus 7 gets you 77.

Ans: $ 7m+7=77 $

vii. One-fourth of a number x minus 4 gives 4.

Ans: $ \dfrac{x}{4}-4=4 $

viii. If you take away 6 from 6 times y, you get 60.

Ans: $ 6y-6=60 $

ix. If you add 3 to one-third of z, you get 30.

Ans: $ \dfrac{z}{3}+3=30 $

5. Write the following equations in statement form:

i. $ p+4=15 $

Ans: The sum of numbers p and 4 is 15.

ii. $ m-7=3 $

Ans: 7 subtracted from m is 3

iii. $ 2m=7 $

Ans: Two times m is 7.

iv. $ \dfrac{m}{5}=3 $

Ans: The number m is divided by 5 gives 3.

v. $ \dfrac{3m}{5}=6 $

Ans: Three-fifth of the number m is 6.

vi. $ 3p+4=25 $

Ans: Three times p plus 4 gets 25.

vii. $ 4p-2=18 $

Ans: If you take away 2 from 4 times p, you get 18.

viii. $ \dfrac{p}{2}+2=8 $

Ans: If you added 2 to half is p, you get 8.

6. Set up an equation in the following cases:

i. Irfan says that he has \[\mathbf{7}\] marbles more than five times the marbles Permit has. Irfan has \[\mathbf{37}\] marbles. (Tale m to be the number of Permit’s marbles.)

Ans: Let $ m $ be the number of Permit’s marbles.

$ \therefore \quad 5m+7=37 $

ii. Laxmi’s father is \[\mathbf{49}\] years old. He is \[\mathbf{4}\] years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)

Ans: Let the age of Laxmi be $ y $ years.

$ \therefore \quad 3y+4=49 $

iii. The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus\[\mathbf{7}\]. The highest score is\[\mathbf{87}\]. (Take the lowest score to be l. )

Ans: Let the lowest score be \[l.\]

$ \therefore \quad 2l+7=87 $

iv. In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is \[\mathbf{180}{}^\circ \]

Ans: Let the base angle of the isosceles triangle be \[b,\]so vertex angle \[=2\text{ }b.\] $ \therefore 2b+b+b={{180}^{{}^\circ }}\Rightarrow 4b={{180}^{{}^\circ }}\quad $Angle sum property of a $ \Delta$

Conclusion

Exercise 4.1 of Class 7 Maths Chapter 4 focuses on the foundational concepts of solving simple equations. This exercise is crucial as it introduces students to the process of balancing equations, an essential skill in algebra. By practising these problems, students learn how to isolate variables and perform operations on both sides of an equation to maintain equality. Class 7 Maths Ch 4 Ex 4.1 helps in building a strong understanding of equations, variables, and constants, which are fundamental to more advanced mathematical concepts. Mastery of these basics will greatly benefit students in their future studies.

Class 7 Maths Chapter 4: Exercises Breakdown

Exercises	Number of Questions
Exercise 4.2	4 Questions and Solutions
Exercise 4.3	4 Questions and Solutions

CBSE Class 7 Maths Chapter 4 Other Study Materials

S. No	Important Links for Chapter 4 Simple Equations
1	Class 7 Simple Equations Important Questions
2	Class 7 Simple Equations Revision Notes
3	Class 7 Simple Equations NCERT Exemplar Solution

Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

S.No.	NCERT Solutions Class 7 Chapter-wise Maths PDF
1.	Chapter 1 - Integers Solutions
2.	Chapter 2 - Fractions and Decimals Solutions
3.	Chapter 3 - Data Handling Solutions
4.	Chapter 4 - Simple Equations Solutions
5.	Chapter 5 - Lines and Angles Solutions
6.	Chapter 6 - The Triangle and Its Properties Solutions
7.	Chapter 7 - Comparing Quantities Solutions
8.	Chapter 8 - Rational Numbers Solutions
9.	Chapter 9 - Perimeter and Area Solutions
10.	Chapter 10 - Algebraic Expressions Solutions
11.	Chapter 11 - Exponents and Powers Solutions
12.	Chapter 12 - Symmetry Solutions
13.	Chapter 13 - Visualising Solid Shapes Solutions

Important Related Links for NCERT Class 7 Maths

Access these essential links for NCERT Class 7 Maths, offering comprehensive solutions, study guides, and additional resources to help students master language concepts and excel in their exams.

S.No	Other CBSE Study Materials for Class 7 Maths
1	CBSE Class 7 Maths Revision Notes
2	CBSE Syllabus for Class 7 Maths
3	CBSE Class 7 Maths Important Questions
4	CBSE Class 7 Maths Sample Papers
5	NCERT Books for Class 7 Maths
6.	RS Aggarwal Class 7 Solutions for Math book
7.	RD Sharma Class 7 Maths Solutions
8.	NCERT Class 7 Maths Exemplar Solutions
9.	NCERT Class 7 Maths Formulas

FAQs on NCERT Solutions For Class 7 Maths Chapter 4 Simple Equations Exercise 4.1 - 2025-26

1. What is the correct method to solve a simple equation as per the NCERT Class 7 Maths Chapter 4?

The NCERT textbook for Class 7 Maths introduces the systematic method for solving simple equations. This involves balancing the equation by performing the same mathematical operation (addition, subtraction, multiplication, or division) on both the Left-Hand Side (LHS) and the Right-Hand Side (RHS). The goal is to isolate the variable on one side to find its value. Another key technique covered is transposing, which is a shortcut for moving terms across the equals sign while changing their sign.

2. How do step-by-step NCERT Solutions for Class 7 Maths Chapter 4 help in preparing for exams?

Step-by-step NCERT Solutions are crucial for exam preparation as they:

Demonstrate the correct methodology required by the CBSE pattern, ensuring you don't lose marks for incorrect steps.
Help identify and rectify common errors in solving equations.
Provide a clear path to follow for complex problems, building problem-solving confidence.
Allow for effective self-assessment and revision of all concepts covered in the "Simple Equations" chapter.

3. How do you set up a simple equation from a word problem in Chapter 4?

To set up a simple equation from a statement or word problem, follow these steps:

First, carefully read the problem to understand what is given and what needs to be found.
Identify the unknown quantity and assign a variable to it (like x, y, or t).
Translate the verbal phrases and relationships in the problem into mathematical expressions.
Finally, find the condition that establishes equality and form the equation by setting the Left-Hand Side (LHS) equal to the Right-Hand Side (RHS).

4. Why is it necessary to perform the same operation on both sides of a simple equation?

An equation represents a state of balance, where the value of the Left-Hand Side (LHS) is equal to the value of the Right-Hand Side (RHS). Performing the same operation on both sides ensures that this balance is maintained. Think of it like a weighing scale; if you add 5 kg to one side, you must add 5 kg to the other to keep it level. This principle allows us to manipulate the equation and isolate the variable without changing the fundamental equality, which is the core rule for finding the correct solution.

5. What is the difference between the 'trial and error' method and the 'transposing' method for solving equations in Chapter 4?

The 'trial and error' method involves substituting different values for the variable until you find one that makes the LHS equal to the RHS. It is often inefficient and used for very simple equations. The 'transposing' method, however, is a systematic and efficient technique. It involves moving a term from one side of the equation to the other while changing its sign (e.g., '+' becomes '–', and '×' becomes '÷'). Transposing is a quicker way of applying the balancing method and is the preferred approach for solving equations accurately.

6. How can solving simple equations from the NCERT textbook be applied to real-life problems?

Simple equations are a powerful tool for modeling and solving many real-life problems. For example, you can use them to:

Calculate ages based on given relationships (e.g., "A father's age is three times his son's age...").
Determine quantities or prices (e.g., finding the cost of an item after a discount).
Solve problems related to perimeter, distance, or speed.

Mastering this chapter helps build the foundational skills for applying mathematics to practical situations.

7. What is a common mistake students make when solving equations involving negative numbers in Class 7 Maths Chapter 4?

A very common mistake is mishandling signs during transposition. For instance, in an equation like 5x - 3 = 12, students might correctly move '-3' to the other side but forget to change its sign to '+3'. Another frequent error occurs with multiplication or division. In an equation like -2y = 8, students might incorrectly add 2 to both sides instead of dividing by -2. Always remember to apply the correct inverse operation to the entire term, including its sign.

8. Can a simple equation in one variable have more than one solution? Why or why not, according to the concepts in Chapter 4?

No, a linear or simple equation in one variable, as covered in Class 7 Maths Chapter 4, has only one unique solution. This is because the equation represents a specific point of balance. For example, in the equation x + 5 = 8, only the value x = 3 can satisfy the condition. Any other value for 'x' would make the Left-Hand Side unequal to the Right-Hand Side, thus breaking the equality. More complex equations, which you will study in higher classes, can have multiple solutions.

9. How do Vedantu's NCERT Solutions for Class 7 Maths Chapter 4 ensure correctness for the 2025-26 CBSE session?

Our NCERT Solutions for Class 7 Maths Chapter 4 are curated by subject-matter experts who strictly follow the latest CBSE guidelines and the 2025-26 NCERT textbook. Each solution is meticulously broken down into clear, logical steps that align with the prescribed curriculum, ensuring that you learn the precise methods expected in your examinations. We focus on providing accurate and easy-to-understand solutions for every problem in the chapter.