NCERT Solutions for Class 10 Maths Chapter 1 provide comprehensive, easy-to-understand explanations of every exercise. These solutions cover key concepts, ensuring students are well-prepared for exams. With clear steps and expert-backed guidance, mastering this chapter becomes simpler. Use these solutions for effective learning and better exam results.
NCERT Solutions for Class 10 Maths Chapter 1: Questions and Answers Made Easy
Sample Questions and Answers for NCERT Solutions for Class 10 Maths Chapter 1:
What is the value of 3x + 5 when x = 4?
When x = 4, substitute the value into the equation:
3(4) + 5 = 12 + 5 = 17. So, the value is 17.
Find the area of a triangle with base 6 cm and height 8 cm.
The formula for the area of a triangle is (1/2) × base × height. Substituting the values:
(1/2) × 6 × 8 = 24 cm².
If the perimeter of a rectangle is 24 cm and the length is 7 cm, what is the width?
The formula for the perimeter of a rectangle is 2 × (length + width). So,
2 × (7 + width) = 24.
This simplifies to 7 + width = 12, and the width = 12 - 7 = 5 cm.
Simplify the expression: 4x - 3y + 5x + 2y.
Combining like terms:
(4x + 5x) + (-3y + 2y) = 9x - y.
Find the roots of the equation x² - 5x + 6 = 0.
Factor the quadratic equation:
(x - 2)(x - 3) = 0.
So, x = 2 or x = 3.
Solve the system of equations: 2x + y = 8 and x - y = 1.
From the second equation, x = y + 1. Substitute this into the first equation:
2(y + 1) + y = 8,
2y + 2 + y = 8,
3y = 6,
y = 2.
Now substitute y = 2 into x = y + 1, so x = 3.
So, x = 3 and y = 2.
What is the value of (2x + 3y) when x = 5 and y = 2?
Substitute x = 5 and y = 2 into the expression:
(2(5) + 3(2)) = 10 + 6 = 16.
Find the perimeter of a square with a side length of 4 cm.
The perimeter of a square is 4 × side length. So,
4 × 4 = 16 cm.
What is the area of a circle with radius 3 cm?
The area of a circle is πr². Substituting the radius value:
π × 3² = π × 9 ≈ 28.27 cm².
Simplify the expression: 5x + 2y - 3x + 4y.
Combining like terms:
(5x - 3x) + (2y + 4y) = 2x + 6y.
What is the slope of the line 2x - 3y = 6?
Rewrite the equation in slope-intercept form:
3y = 2x - 6,
y = (2/3)x - 2.
The slope is 2/3.
Find the value of 3x² - 4x + 2 when x = 1.
Substitute x = 1 into the expression:
3(1)² - 4(1) + 2 = 3 - 4 + 2 = 1.
Solve for x in the equation 5x - 7 = 18.
Add 7 to both sides:
5x = 25,
x = 25/5 = 5.
Find the value of x in the equation 3x + 2 = 11.
Subtract 2 from both sides:
3x = 9,
x = 9/3 = 3.
What is the volume of a cube with side length 4 cm?
The volume of a cube is side³:
4³ = 64 cm³.
Solve the equation x² - 4x = 0.
Factor the equation:
x(x - 4) = 0.
So, x = 0 or x = 4.
Find the mean of the numbers 4, 8, 12, and 16.
The mean is the sum of the numbers divided by the total count:
(4 + 8 + 12 + 16)/4 = 40/4 = 10.
What is the difference between the square of 6 and the square of 4?
The square of 6 is 6² = 36, and the square of 4 is 4² = 16.
So, the difference is 36 - 16 = 20.
Find the value of x if 7x = 42.
Divide both sides by 7:
x = 42/7 = 6.
Find the area of a trapezoid with parallel sides of length 6 cm and 8 cm, and height 5 cm.
The area of a trapezoid is (1/2) × (sum of parallel sides) × height.
(1/2) × (6 + 8) × 5 = (1/2) × 14 × 5 = 35 cm².
What is the perimeter of an equilateral triangle with side length 5 cm?
The perimeter of an equilateral triangle is 3 × side length:
3 × 5 = 15 cm.
Solve for x: 4x + 5 = 2x + 9.
Subtract 2x from both sides:
2x + 5 = 9,
Subtract 5 from both sides:
2x = 4,
x = 4/2 = 2.
What is the surface area of a sphere with radius 3 cm?
The surface area of a sphere is 4πr². Substituting the radius:
4π × 3² = 4π × 9 ≈ 113.1 cm².
Simplify the expression: (x² + 2x) + (3x + 4).
Combining like terms:
x² + 5x + 4.
Find the volume of a cylinder with radius 2 cm and height 5 cm.
The volume of a cylinder is πr²h. Substituting the values:
π × 2² × 5 = π × 4 × 5 = 20π ≈ 62.83 cm³.
Find the roots of the equation x² + 3x - 4 = 0.
Factor the quadratic equation:
(x + 4)(x - 1) = 0.
So, x = -4 or x = 1.
What is the length of the diagonal of a square with side length 6 cm?
The diagonal of a square is given by √2 × side length:
√2 × 6 ≈ 8.49 cm.
Find the value of x if x² = 64.
Take the square root of both sides:
x = ±√64 = ±8.
Find the sum of the first 10 even numbers.
The first 10 even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20.
The sum is 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 = 110.
Simplify the expression: 3x² + 5x - x² + 4x.
Combining like terms:
(3x² - x²) + (5x + 4x) = 2x² + 9x.
Solve the equation 2(x - 3) = 4.
Divide both sides by 2:
x - 3 = 2,
Add 3 to both sides:
x = 5.
What is the perimeter of a circle with radius 4 cm?
The perimeter (circumference) of a circle is 2πr. Substituting the radius:
2π × 4 = 8π ≈ 25.13 cm.
What is the sum of the angles in a triangle?
The sum of the angles in any triangle is always 180°.
Find the distance between the points (1, 2) and (4, 6).
Use the distance formula:
√[(x₂ - x₁)² + (y₂ - y₁)²] = √[(4 - 1)² + (6 - 2)²] = √[9 + 16] = √25 = 5.
Solve for x in the equation 5(x + 2) = 30.
Divide both sides by 5:
x + 2 = 6,
Subtract 2 from both sides:
x = 4.
What is the area of a sector with radius 4 cm and angle 60°?
The area of a sector is (θ/360) × πr². Substituting the values:
(60/360) × π × 4² = (1/6) × π × 16 ≈ 8.38 cm².
Find the volume of a cone with radius 3 cm and height 4 cm.
The volume of a cone is (1/3)πr²h. Substituting the values:
(1/3)π × 3² × 4 = (1/3)π × 9 × 4 ≈ 37.7 cm³.
What is the area of a parallelogram with base 6 cm and height 3 cm?
The area of a parallelogram is base × height:
6 × 3 = 18 cm².
Simplify the expression: 2x + 3y - x + y.
Combining like terms:
(2x - x) + (3y + y) = x + 4y.
Find the value of x in the equation 6x - 9 = 27.
Add 9 to both sides:
6x = 36,
x = 36/6 = 6.
What is the surface area of a rectangular prism with length 4 cm, width 5 cm, and height 6 cm?
The surface area of a rectangular prism is 2lw + 2lh + 2wh.
2(4 × 5) + 2(4 × 6) + 2(5 × 6) = 40 + 48 + 60 = 148 cm².
Solve for x: 3x + 4 = 2x + 10.
Subtract 2x from both sides:
x + 4 = 10,
Subtract 4 from both sides:
x = 6.
What is the area of a square with side length 7 cm?
The area of a square is side²:
7² = 49 cm².
Find the roots of the equation x² - 7x + 10 = 0.
Factor the quadratic equation:
(x - 5)(x - 2) = 0.
So, x = 5 or x = 2.
Simplify the expression: 7x + 2y - 3x - 4y.
Combining like terms:
(7x - 3x) + (2y - 4y) = 4x - 2y.
What is the perimeter of a triangle with sides 5 cm, 7 cm, and 10 cm?
The perimeter of a triangle is the sum of its sides:
5 + 7 + 10 = 22 cm.
Find the value of x in the equation 2(x - 5) = 6.
Divide both sides by 2:
x - 5 = 3,
Add 5 to both sides:
x = 8.
What is the surface area of a sphere with radius 2 cm?
The surface area of a sphere is 4πr². Substituting the radius:
4π × 2² = 16π ≈ 50.27 cm².
What is the area of a rhombus with diagonals of length 6 cm and 8 cm?
The area of a rhombus is (1/2) × diagonal₁ × diagonal₂.
(1/2) × 6 × 8 = 24 cm².
Top Indian Books for NCERT Solutions for Class 10 Maths Chapter 1
Mathematics for Class 10 by R.D. Sharma – Published by Dhanpat Rai Publications
This book offers a variety of practice questions to help students understand the basic concepts of algebra. It includes step-by-step solutions and detailed explanations for each topic in Chapter 1, helping students grasp important mathematical principles.
NCERT Solutions for Class 10 Maths by R.S. Aggarwal – Published by S. Chand Publishing
This book provides detailed solutions to all the exercises in the NCERT textbook for Class 10, Chapter 1. It includes simple explanations, important definitions, and multiple variations of questions, including objective, subjective, and application-based problems.
Mathematics Class 10 by M.L. Agarwal – Published by Arya Publishing House
This book covers a wide range of problems related to algebraic expressions, equations, and solutions. It focuses on boosting the problem-solving skills of students with a variety of exercises and multiple sets of problems that align with Chapter 1.
IIT Mathematics for Class 10 by G.S. Rathi – Published by Goyal Brothers Prakashan
This book is ideal for students who aim to excel in mathematics beyond the Class 10 board exams. It features advanced questions that go beyond the NCERT solutions, offering additional challenges and solutions to solidify the knowledge gained in Chapter 1.
Mathematics for Class 10 by S.K. Gupta – Published by Laxmi Publications
A good guide for students aiming to get a deep understanding of algebraic methods, this book breaks down every problem in Chapter 1 into manageable steps. It contains a mixture of solved examples, unsolved questions, and in-depth theory for complete clarity.
Understanding Mathematics for Class 10 by D.K. Singh – Published by New Age International
This book focuses on helping students understand mathematical concepts through examples. Each problem in Chapter 1 is paired with clear explanations and shortcuts to solve problems quickly and effectively, ensuring better preparation for exams.
All in One Mathematics by Rakesh Yadav – Published by R.K. Publishers
Known for its detailed and systematic approach to all topics, this book includes solutions for every problem in Chapter 1, with focus areas on algebraic expressions, equations, and simplifications. It also contains a variety of questions from previous years to help with practice.
Xam Idea Mathematics for Class 10 by Xam Idea – Published by Rachna Sagar
This book includes extensive practice exercises for every chapter of Class 10 Maths. For Chapter 1, the book provides detailed solutions and the application of algebraic expressions to real-life scenarios, helping students with both theoretical knowledge and practical problem-solving.
Target Mathematics for Class 10 by K.K. Agarwal – Published by Target Publications
This book provides a large number of questions for each topic, with clear solutions for better understanding. Chapter 1’s solutions focus on real-world examples, ensuring that students not only understand theory but also develop practical problem-solving skills.
Mathematics for Class 10 by S.C. Sharma – Published by Lakshmi Publications
A well-rounded book offering both solved and unsolved questions. The book provides multiple methods to solve algebraic equations and expressions found in Chapter 1, making it ideal for students who prefer learning through multiple approaches.
Success Mathematics for Class 10 by V.K. Sharma – Published by Kundan Publications
Focused on providing a mix of problem-solving techniques, this book enhances understanding of Chapter 1 concepts with a variety of exercise-based questions. The solutions are designed to encourage conceptual clarity and help students excel in board exams.
Tata McGraw Hill Mathematics for Class 10 – Published by Tata McGraw Hill
This book takes a more application-oriented approach to learning algebra. It provides step-by-step explanations and a variety of questions from Chapter 1 to test knowledge and improve mathematical accuracy.
S.Chand’s Math Complete Class 10 by S. Chand – Published by S. Chand
For students aiming to build strong fundamentals, this book delivers excellent solutions for Chapter 1. It features easy-to-understand explanations and diverse problem sets that ensure comprehensive learning.
Mathematics for Class 10 by M.K. Singhal – Published by Modern Publishers
This book offers clear solutions with an emphasis on problem-solving in algebra. It covers all important aspects of Chapter 1, including linear equations and algebraic simplifications, with plenty of practice questions to help students prepare well.
Creative Mathematics for Class 10 by S. K. Kataria – Published by S.K. Kataria & Sons
This book provides a variety of solved and unsolved problems in Chapter 1, encouraging students to think critically. It also includes tricky problems that help develop the ability to solve complex algebraic equations quickly and efficiently.
Pearson Mathematics for Class 10 – Published by Pearson India
Ideal for students seeking to build strong basics, this book offers detailed explanations and solutions for Chapter 1 problems. It is known for its unique methodology that simplifies complex concepts and prepares students for competitive exams.
Rachna Sagar Class 10 Mathematics by Rachna Sagar – Published by Rachna Sagar
A comprehensive book for Class 10 students, this one presents solutions for every problem in Chapter 1 with adequate practice problems that reinforce learning.
Class 10 Mathematics: A Textbook for Practice by R.K. Gupta – Published by R.K. Gupta
This book provides students with practice problems and solutions designed to focus on core concepts like equations and expressions found in Chapter 1. It also offers tips to solve questions quickly and effectively during exams.
Problems in Algebra for Class 10 by Dinesh K. Gupta – Published by Dinesh Publications
Focused entirely on algebra, this book offers detailed solutions for all algebra-related questions in Chapter 1, helping students enhance their algebra skills through additional problem sets.
Maths: Skill Development for Class 10 by S.K. Soni – Published by Soni Publications
This book focuses on sharpening problem-solving skills, with extra questions designed to improve algebraic thinking. Each problem in Chapter 1 is solved step-by-step to ensure easy comprehension and a deeper understanding.
The first chapter of Class 10 Mathematics covers algebraic expressions and their applications, and it is a crucial foundation for students preparing for their exams. NCERT solutions for this chapter aim to guide students through complex concepts and equations with clarity and precision.
Understanding algebra begins with grasping the basic principles of terms, coefficients, and variables. The NCERT textbook explains how algebraic expressions can be simplified, added, subtracted, multiplied, and divided. The exercises provided help in mastering the fundamental operations with algebraic terms and expressions. The solutions break down each problem to make it easier for students to approach even the most difficult questions step-by-step.
An important aspect of Chapter 1 is the understanding of polynomials. Polynomials are expressions consisting of variables raised to different powers, and learning how to factorize and solve these equations is a key skill for students. The NCERT solutions provide methods to factorize quadratic expressions, as well as techniques for solving them efficiently. These skills are foundational for the problems found later in the curriculum, including those in higher mathematics.
Additionally, the solutions offer a variety of practice problems that help solidify these algebraic skills. From simple linear equations to more complex quadratic equations, the practice problems help students build confidence in their problem-solving abilities. The solutions also provide explanations on how to identify the roots of equations, which is a crucial part of mastering algebra.
Another essential component of this chapter is solving word problems related to algebra. These word problems require students to interpret real-world scenarios and convert them into algebraic expressions. The NCERT solutions guide students on how to break down these problems and solve them by forming and solving equations.
Expert-backed recommendations emphasize consistent practice, which helps in refining mathematical thinking and developing the skills necessary to solve algebraic problems with ease. Whether it’s learning new concepts or revisiting previous ones, practicing regularly with the solutions provided will enhance students' abilities. Additionally, the explanations and step-by-step solutions offered in the NCERT solutions are designed to help students not only learn but also understand the underlying principles of each problem.
Through diligent practice and understanding, students can build a solid foundation in algebra, which will be essential not only for Class 10 but for future mathematical learning.
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