1. What is the perimeter of a rectangle if the length is 8 cm and the breadth is 5 cm?
The perimeter of a rectangle is given by the formula: perimeter = 2 × (length + breadth).
Substitute the values: perimeter = 2 × (8 + 5) = 2 × 13 = 26 cm.
2. How do you find the area of a triangle?
The area of a triangle is given by the formula: area = 1/2 × base × height.
For example, if the base is 6 cm and the height is 4 cm, the area = 1/2 × 6 × 4 = 12 cm².
3. If the sides of a square are 9 cm, what is its area?
The area of a square is given by the formula: area = side².
For a square with side 9 cm, the area = 9² = 81 cm².
4. How do you solve for x in the equation 2x + 5 = 15?
Subtract 5 from both sides: 2x = 15 - 5 = 10.
Now divide by 2: x = 10 / 2 = 5.
5. What is the value of x in the equation 3x - 4 = 11?
Add 4 to both sides: 3x = 11 + 4 = 15.
Now divide by 3: x = 15 / 3 = 5.
6. How do you convert a fraction to a decimal?
Divide the numerator by the denominator.
For example, 3/4 = 3 ÷ 4 = 0.75.
7. What is the volume of a cube with side length 4 cm?
The volume of a cube is given by the formula: volume = side³.
For a cube with side 4 cm, the volume = 4³ = 64 cm³.
8. How do you find the median of the numbers 3, 7, 9, 11, and 15?
The median is the middle number when the data is arranged in order.
The numbers in order: 3, 7, 9, 11, 15. The median is 9.
9. What is the LCM of 4 and 6?
List the multiples of 4 and 6:
Multiples of 4: 4, 8, 12, 16, 20, 24…
Multiples of 6: 6, 12, 18, 24, 30…
The least common multiple is 12.
10. How do you find the HCF of 12 and 18?
List the factors of 12: 1, 2, 3, 4, 6, 12.
List the factors of 18: 1, 2, 3, 6, 9, 18.
The highest common factor is 6.
11. What is the sum of angles in a triangle?
The sum of the interior angles of a triangle is always 180°.
12. What is the formula to calculate the area of a circle?
The area of a circle is given by the formula: area = π × radius².
For a circle with radius 7 cm, the area = π × 7² = 22/7 × 49 ≈ 154 cm².
13. How do you simplify the expression 5x + 3x?
Add the like terms: 5x + 3x = 8x.
14. What is the probability of getting a 5 when rolling a fair die?
There are 6 possible outcomes on a die, and only one of them is 5.
So, the probability is 1/6.
15. What is the difference between a rational and an irrational number?
A rational number can be written as a fraction of two integers, whereas an irrational number cannot.
For example, 1/2 is rational, while √2 is irrational.
16. How do you calculate the area of a parallelogram?
The area of a parallelogram is given by the formula: area = base × height.
For a parallelogram with base 5 cm and height 3 cm, the area = 5 × 3 = 15 cm².
17. What is the sum of 4/5 and 2/5?
Since the denominators are the same, add the numerators:
4/5 + 2/5 = (4 + 2)/5 = 6/5.
18. How do you find the simple interest for a principal amount of 1000, rate 5%, and time 2 years?
Simple Interest = (Principal × Rate × Time) / 100.
SI = (1000 × 5 × 2) / 100 = 100.
19. What is the slope of the line passing through points (1, 2) and (3, 6)?
The slope is given by the formula: slope = (y2 - y1) / (x2 - x1).
Substitute the values: slope = (6 - 2) / (3 - 1) = 4 / 2 = 2.
20. What is the perimeter of a triangle with sides 5 cm, 7 cm, and 9 cm?
The perimeter of a triangle is the sum of the lengths of its sides.
Perimeter = 5 + 7 + 9 = 21 cm.
21. How do you convert a decimal to a fraction?
For example, to convert 0.75 to a fraction, write 75/100 and simplify to 3/4.
22. What is the area of a rectangle with a length of 10 cm and a breadth of 6 cm?
The area of a rectangle is given by the formula: area = length × breadth.
For a rectangle with length 10 cm and breadth 6 cm, the area = 10 × 6 = 60 cm².
23. What is the ratio of the circumference of a circle to its diameter?
The ratio of the circumference to the diameter is always π (approximately 3.14159).
24. How do you find the cube root of 27?
The cube root of 27 is 3, because 3 × 3 × 3 = 27.
25. What is the area of a trapezium with parallel sides 8 cm and 12 cm, and height 5 cm?
The area of a trapezium is given by the formula: area = 1/2 × (sum of parallel sides) × height.
Area = 1/2 × (8 + 12) × 5 = 1/2 × 20 × 5 = 50 cm².
26. How do you find the area of a rhombus?
The area of a rhombus is given by the formula: area = 1/2 × product of diagonals.
For a rhombus with diagonals 6 cm and 8 cm, the area = 1/2 × 6 × 8 = 24 cm².
27. What is the formula for the area of a sector of a circle?
The area of a sector is given by the formula: area = (θ/360) × π × radius², where θ is the angle.
For θ = 90°, the area = (90/360) × π × radius².
28. How do you solve the equation 4x - 7 = 13?
Add 7 to both sides: 4x = 13 + 7 = 20.
Now divide by 4: x = 20 / 4 = 5.
29. What is the volume of a cylinder with radius 3 cm and height 5 cm?
The volume of a cylinder is given by the formula: volume = π × radius² × height.
Volume = π × 3² × 5 = 3.14 × 9 × 5 ≈ 141.3 cm³.
30. How do you calculate the mean of a data set?
The mean is the sum of all numbers divided by the total count of numbers.
For example, for the numbers 2, 4, 6, and 8, the mean = (2 + 4 + 6 + 8) / 4 = 5.
31. What is the perimeter of a regular hexagon with side length 6 cm?
The perimeter of a regular hexagon is 6 times the side length.
Perimeter = 6 × 6 = 36 cm.
32. How do you find the surface area of a sphere?
The surface area of a sphere is given by the formula: surface area = 4π × radius².
For a sphere with radius 3 cm, the surface area = 4π × 3² = 4 × 3.14 × 9 ≈ 113.04 cm².
33. What is the difference between an acute and an obtuse angle?
An acute angle is less than 90°, while an obtuse angle is more than 90° but less than 180°.
34. What is the probability of drawing a red card from a standard deck of cards?
A standard deck has 52 cards, with 26 red cards.
So, the probability of drawing a red card is 26/52 = 1/2.
35. What is the sum of the first 10 natural numbers?
The sum of the first n natural numbers is given by the formula: sum = n(n + 1) / 2.
For n = 10, sum = 10(10 + 1) / 2 = 10 × 11 / 2 = 55.
36. How do you calculate the mean of grouped data?
To calculate the mean of grouped data, multiply the midpoint of each class by the frequency, sum these products, and divide by the total frequency.
37. What is the difference between a prime and a composite number?
A prime number has exactly two distinct positive divisors: 1 and itself.
A composite number has more than two divisors.
38. How do you simplify the expression 2x + 3x - 4?
Combine the like terms: 2x + 3x = 5x.
So the expression simplifies to 5x - 4.
39. What is the area of a square whose diagonal is 10 cm?
The area of a square is given by the formula: area = side².
To find the side, use the relationship between the side and diagonal: side = diagonal/√2.
Side = 10/√2 ≈ 7.07 cm.
Area = (7.07)² ≈ 50 cm².
40. How do you convert a percentage to a fraction?
To convert a percentage to a fraction, write the percentage as a fraction over 100.
For example, 75% = 75/100 = 3/4.
41. What is the area of a rectangle with length 12 cm and width 8 cm?
The area of the rectangle is length × width.
Area = 12 × 8 = 96 cm².
42. How do you calculate the circumference of a circle?
The circumference of a circle is given by the formula: circumference = 2π × radius.
For a circle with radius 7 cm, the circumference = 2π × 7 ≈ 43.96 cm.
43. How do you solve the equation 3x + 4 = 19?
Subtract 4 from both sides: 3x = 19 - 4 = 15.
Now divide by 3: x = 15 / 3 = 5.
44. What is the ratio of the volume of a sphere to its surface area?
The ratio is given by the formula: volume / surface area = radius / 3.
For a sphere with radius 3 cm, the ratio = 3 / 3 = 1.
45. How do you solve the equation 5x - 2 = 13?
Add 2 to both sides: 5x = 13 + 2 = 15.
Now divide by 5: x = 15 / 5 = 3.
46. What is the area of a semicircle with radius 5 cm?
The area of a semicircle is half the area of a full circle.
Area = (1/2) × π × radius².
Area = (1/2) × 3.14 × 5² ≈ 39.25 cm².
47. What is the relationship between the diameter and radius of a circle?
The diameter is twice the length of the radius.
So, diameter = 2 × radius.
48. How do you calculate the surface area of a cylinder?
The surface area of a cylinder is given by the formula: surface area = 2π × radius × height + 2π × radius².
49. What is the total surface area of a cube with side 4 cm?
The total surface area of a cube is given by the formula: surface area = 6 × side².
Surface area = 6 × 4² = 96 cm².
50. What is the sum of the first 20 odd numbers?
The sum of the first n odd numbers is given by the formula: sum = n².
For n = 20, sum = 20² = 400.
Top Indian Books for Class 8th Maths with Author and Publications
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Mathematics for Class 8 by R.S. Aggarwal
Publisher: S. Chand & Co.
This book provides a detailed explanation of each chapter in Class 8 Maths. It includes multiple choice questions, short answer questions, and long answer questions. Each chapter is well-structured with examples that help students grasp the concepts.
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Mathematics for Class 8 by R.D. Sharma
Publisher: Dhanpat Rai & Co.
This book offers a comprehensive explanation of all Class 8 maths topics. It includes a variety of questions, including solved examples, exercises, and worksheets. The book’s simple language and structured approach make it ideal for understanding concepts like algebra and geometry.
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NCERT Mathematics Textbook for Class 8
Publisher: NCERT
The official NCERT textbook for Class 8 Maths offers clear and concise explanations of each topic. It includes practice problems and activities designed to build a strong foundation. The book covers topics such as linear equations, geometry, and mensuration in depth.
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Mathematics for Class 8 by M.L. Agarwal
Publisher: Arya Publications
This book is a great reference for solving problems from the NCERT syllabus. It provides step-by-step solutions to each question, followed by practice exercises to reinforce learning.
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Mathematics for Class 8 by S.K. Gupta
Publisher: Dhanpat Rai & Co.
This book offers a large number of solved examples and exercises. It focuses on building a strong understanding of mathematical concepts like fractions, geometry, and algebra. The easy-to-follow examples help students excel in exams.
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Concepts of Mathematics for Class 8 by M.S. Raghunathan
Publisher: Orient Black Swan
This book provides a more theoretical approach to Class 8 Maths, perfect for students who need to develop a deeper understanding of mathematical concepts. It includes explanation-based questions and challenges students with complex problem-solving exercises.
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Comprehensive Mathematics for Class 8 by S.K. Bansal
Publisher: Laxmi Publications
The book covers all the important topics required in the Class 8 curriculum, including practical questions and examples. Each chapter is followed by a set of exercises to solidify the concepts learned.
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Mathematics Success Class 8 by R.K. Bansal
Publisher: New Saraswati House
This book offers a complete package for Class 8 Maths students, including NCERT solutions, practice sets, and additional exercises. The clear explanations and various question types help improve conceptual clarity.
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Modern’s ABC of Mathematics for Class 8 by A.K. Sharma
Publisher: Modern Publishers
This book is a good resource for revising Class 8 Maths concepts. It includes theory, worked-out examples, and plenty of exercises. It also offers a set of self-assessment questions that help in exam preparation.
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New Learning Mathematics for Class 8 by S. P. Jauhar
Publisher: S. Chand & Co.
The book includes solutions to all questions in the NCERT textbook, along with extra problems for practice. It aims to enhance students' logical reasoning skills and provides a variety of problem-solving techniques.
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Basic Mathematics for Class 8 by A. J. Aggarwal
Publisher: Arihant Publications
This book offers an in-depth study of all the topics included in the NCERT syllabus. It covers fundamental concepts of algebra, geometry, and trigonometry in easy-to-understand language.
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S. Chand’s Mathematics for Class 8 by R. S. Aggarwal
Publisher: S. Chand & Co.
It offers a comprehensive explanation of complex problems, with solutions, practice questions, and examples that help students in mastering the mathematical concepts required in Class 8.
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Mathematics for Class 8 by J.P. Bansal
Publisher: Anand Publications
This book provides solutions to both theoretical and practical problems of Class 8 Maths. It also includes real-life examples and exercises that apply to everyday situations.
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Mathematics for Class 8 by M.K. Agarwal
Publisher: Pustak Mahal
The book is known for its detailed explanations and rich set of exercises. It includes application-based questions and problem-solving strategies.
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Target Maths for Class 8 by Dinesh Khattar
Publisher: Target Publications
This book is designed to provide in-depth knowledge of mathematical concepts. It focuses on question-solving techniques and includes detailed explanations of basic operations, algebra, and geometry.
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Mathematics for Class 8 by J.B. Gupta
Publisher: Goyal Publishers
This book provides a structured approach to solving problems. It includes both subjective and objective type questions, which helps in preparing for exams.
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Advance Learning Mathematics for Class 8 by R.C. Sharma
Publisher: Dhanpat Rai & Co.
This book includes a variety of solved and unsolved problems to practice, along with graphical representations of important concepts. It is suitable for building a strong understanding of both basic and advanced concepts in Class 8 Maths.
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Math Practice Book for Class 8 by S. D. Sharma
Publisher: Vikas Publications
This book provides ample practice questions based on the NCERT syllabus. It includes various types of questions that challenge students and test their knowledge of key topics in algebra, geometry, and measurement.
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Mathematics for Class 8 by P.K. Singhal
Publisher: Oxford Publications
This book offers explanations for difficult topics and breaks them into simpler parts for students to understand easily. The exercises include word problems, reasoning questions, and application-based exercises.
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NCERT Solutions for Class 8 Maths
Publisher: NCERT
This reference guide provides a complete set of solutions for all the exercises in the NCERT textbook. It includes a variety of problems with solutions to strengthen the student's problem-solving ability.
Class 8th Maths is an essential stage in a student’s academic journey. The topics introduced in this level lay the foundation for future mathematical learning. By following the NCERT curriculum, students gain a comprehensive understanding of various mathematical concepts, enhancing their problem-solving abilities.
The NCERT Solutions for Class 8 Maths provide step-by-step solutions to each question in the official textbook. These solutions offer clear explanations and detailed methods for solving problems. As students work through these solutions, they can better grasp concepts like algebra, geometry, mensuration, and statistics. The explanations in these solutions break down the steps, making even the most complex problems easy to understand.
One of the key features of the NCERT solutions is the inclusion of various types of problems. These include direct application questions, word problems, and exercises that encourage logical reasoning. Students are encouraged to approach problems with a structured method that includes identifying known values, applying formulas, and simplifying complex expressions.
Mathematical topics in Class 8, such as linear equations, algebraic expressions, and mensuration are explained in a manner that links them with real-life scenarios. For example, when studying area and perimeter, students can use real-world applications to understand the importance of these concepts. These practical applications help bridge the gap between theoretical knowledge and its practical uses.
The NCERT solutions also emphasize the importance of practice. By regularly solving practice problems, students improve their analytical thinking and calculation skills. This focus on regular practice ensures that students can confidently tackle problems in exams. It also builds their conceptual clarity, which is crucial for mastering complex mathematical topics in higher grades.
Understanding and practicing word problems is another area that the NCERT solutions address in-depth. These types of questions not only assess a student’s understanding of the subject but also their ability to apply their knowledge in real-life situations. Through these problems, students learn how to translate a worded scenario into a mathematical equation and find the solution.
In addition to regular exercises, NCERT solutions include examples that demonstrate how to work through problems systematically. This is particularly beneficial for topics like fractions, ratios, and probability, where a clear understanding of the formula and its application is crucial.
By using NCERT solutions for Class 8 Maths, students are better prepared for their exams, gaining both knowledge and confidence in the subject. The book encourages students to approach problems in an organized manner, promoting critical thinking skills that are vital for mathematical success.