CBSE Class 8 Maths Important Questions: Chapter Wise

CBSE Class 8 Maths Important Questions are essential for every student aiming to build a strong foundation in mathematics and score high in school exams. This set of questions is designed as per the latest CBSE Class 8 syllabus and strictly follows NCERT textbook guidelines. Practicing these chapter-wise important questions helps students master concepts, improve problem-solving skills, and revise effectively before the final exams.

Whether students are looking to strengthen topics like Linear Equations, Algebraic Expressions, Mensuration, or Rational Numbers, this resource covers complete CBSE Class 8 Maths Syllabus comprehensively. The questions are categorized into 1-mark, 2-mark, and 3-mark types, including Higher Order Thinking Skills (HOTS) and application-based problems, which are commonly asked in school-level exams and Olympiads.

Each question comes with a detailed step-by-step solution to enhance conceptual clarity and boost confidence. Infinity Learn has included important formulas, exam tips, and free PDF downloads to support your daily revision and exam preparation. Start solving these CBSE Class 8 Maths chapter-wise important questions to practice smarter, reduce mistakes, and gain the confidence to tackle any question in your exam paper.

CBSE Class 8 Maths Chapter-wise Important Questions

CBSE Class 8 Maths Important Questions are designed to help students master all major topics from the latest syllabus. These chapter-wise important questions and answers cover every concept, formula, and problem type, making revision focused and effective. Practising these questions ensures a thorough understanding and boosts exam confidence.

Benefits of CBSE Class 8 Maths Important Questions

• Focus on key topics and high-weightage chapters for complete exam preparation.
• Strengthen conceptual clarity and problem-solving skills.
• Make revision and practice easier with chapter-wise questions and answers.
• Improve time management and exam performance.

Chapter-wise Important Questions for Class 8 Maths

These CBSE Class 8 Maths Important Questions provide comprehensive practice for each chapter, including Rational Numbers, Linear Equations, Algebraic Expressions, Mensuration, and more. Use these chapter-wise important questions to reinforce your understanding, revise efficiently, and achieve higher marks in your CBSE Class 8 Maths exam.

CBSE Class 8 Maths Imporant Questions

Question 1: Write five rational numbers between 1/2 and 3/4.

Solution:

Step 1: Convert the fractions to have a common denominator.

1/2 = 6/12 and 3/4 = 9/12

Step 2: Find rational numbers between 6/12 and 9/12.

Step 3: To get more numbers, multiply both numerator and denominator by a larger number.

1/2 = 20/40 and 3/4 = 30/40

Step 4: Five rational numbers between 20/40 and 30/40 are:

21/40, 22/40, 23/40, 24/40, 25/40

Answer: 21/40, 22/40, 23/40, 24/40, 25/40

Question 2: State and verify the closure property of rational numbers under addition.

Solution:

Step 1: Statement: The sum of two rational numbers is always a rational number.

Step 2: Verification: Let us take two rational numbers a/b and c/d where b ≠ 0, d ≠ 0

Step 3: Their sum = a/b + c/d = (ad + bc)/(bd)

Step 4: Since a, b, c, d are integers and b ≠ 0, d ≠ 0, we have:

• ad + bc is an integer (sum and product of integers is an integer)

• bd is a non-zero integer (product of non-zero integers is non-zero)

Step 5: Therefore, (ad + bc)/(bd) is a rational number.

Example: 2/3 + 4/5 = (2×5 + 4×3)/(3×5) = (10 + 12)/15 = 22/15 (rational number)

Question 3: Solve: 5t - 3 = 3t - 5

Solution:

Step 1: Given equation: 5t - 3 = 3t - 5

Step 2: Subtract 3t from both sides: 5t - 3t - 3 = 3t - 3t - 5

Step 3: Simplify: 2t - 3 = -5

Step 4: Add 3 to both sides: 2t - 3 + 3 = -5 + 3

Step 5: Simplify: 2t = -2

Step 6: Divide both sides by 2: t = -2/2 = -1

Verification: LHS = 5(-1) - 3 = -5 - 3 = -8; RHS = 3(-1) - 5 = -3 - 5 = -8 ✓

Answer: t = -1

Question 4: If four-fifths of a number is more than three-fourths of the number by 4, find the number.

Solution:

Step 1: Let the number be x.

Step 2: Four-fifths of the number = 4x/5

Step 3: Three-fourths of the number = 3x/4

Step 4: According to the problem: 4x/5 = 3x/4 + 4

Step 5: Subtract 3x/4 from both sides: 4x/5 - 3x/4 = 4

Step 6: Find LCM of 5 and 4 = 20

Step 7: (16x - 15x)/20 = 4

Step 8: x/20 = 4

Step 9: x = 4 × 20 = 80

Verification: 4/5 × 80 = 64; 3/4 × 80 = 60; 64 - 60 = 4 ✓

Answer: The number is 80

Question 5: Define a parallelogram and list its properties.

Solution:

Step 1: Definition: A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

Step 2: Properties of a parallelogram:

1. Opposite sides are parallel and equal in length

2. Opposite angles are equal

3. Adjacent angles are supplementary (sum = 180°)

4. Diagonals bisect each other

5. Each diagonal divides the parallelogram into two congruent triangles

6. The sum of all interior angles is 360°

Question 6: Find the sum of the interior angles of a polygon with 8 sides.

Solution:

Step 1: Formula for sum of interior angles of a polygon = (n - 2) × 180°

Step 2: Here, n = 8 (number of sides)

Step 3: Sum of interior angles = (8 - 2) × 180°

Step 4: = 6 × 180°

Step 5: = 1080°

Answer: 1080°

Question 7: Construct a frequency distribution table for the following data (in kgs): 40, 33, 56, 31, 34, 49, 55, 57, 44, 38, 38, 48, 53, 46, 36, 41, 49, 42, 47, 39 using class intervals 30–35, 35–40, etc.

Solution:

Step 1: Arrange the data in ascending order: 31, 33, 34, 36, 38, 38, 39, 40, 41, 42, 44, 46, 47, 48, 49, 49, 53, 55, 56, 57

Step 2: Create class intervals and count frequencies:

Question 8: The number of students speaking different languages in a class is: Hindi–30, English–12, Marathi–9, Tamil–7, Bengali–4. Display this data in a pie chart.

Solution:

Step 1: Total number of students = 30 + 12 + 9 + 7 + 4 = 62

Step 2: Calculate the central angle for each language:

Central angle = (Frequency/Total) × 360°

Step 3: Calculations:

Step 4: Draw a circle and mark sectors with the calculated angles to create the pie chart.

Question 9: Find the square root of 324 by the prime factorization method.

Solution:

Step 1: Find the prime factorization of 324.

324 = 2 × 162 = 2 × 2 × 81 = 2² × 81

81 = 3 × 27 = 3 × 3 × 9 = 3² × 9

9 = 3 × 3 = 3²

Step 2: So, 324 = 2² × 3² × 3² = 2² × 3⁴

Step 3: √324 = √(2² × 3⁴) = √(2² × (3²)²) = √(2² × 3⁴)

Step 4: = 2 × 3² = 2 × 9 = 18

Verification: 18² = 324 ✓

Answer: √324 = 18

Question 10: Is 105 a perfect square? Give reasons.

Solution:

Step 1: Find the prime factorization of 105.

105 = 3 × 35 = 3 × 5 × 7

Step 2: So, 105 = 3¹ × 5¹ × 7¹

Step 3: For a number to be a perfect square, all prime factors must appear an even number of times.

Step 4: In 105 = 3¹ × 5¹ × 7¹, all prime factors (3, 5, and 7) appear an odd number of times (1 time each).

Step 5: Since the exponents are not even, 105 is not a perfect square.

Answer: No, 105 is not a perfect square because its prime factors 3, 5, and 7 each appear an odd number of times.

Question 11: Find the cube root of 512.

Solution:

Step 1: Find the prime factorization of 512.

512 = 2 × 256 = 2 × 2 × 128 = 2 × 2 × 2 × 64

= 2 × 2 × 2 × 2 × 32 = 2 × 2 × 2 × 2 × 2 × 16

= 2 × 2 × 2 × 2 × 2 × 2 × 8 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 4

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁹

Step 2: ∛512 = ∛(2⁹) = ∛(2³ × 2³ × 2³) = ∛(2³)³ = 2³ = 8

Verification: 8³ = 8 × 8 × 8 = 512 ✓

Answer: ∛512 = 8

Question 12: Write the smallest number by which 72 must be multiplied to make it a perfect cube.

Solution:

Step 1: Find the prime factorization of 72.

72 = 8 × 9 = 2³ × 3²

Step 2: For a number to be a perfect cube, all prime factors must appear in groups of 3.

Step 3: In 72 = 2³ × 3², we have:

• 2 appears 3 times (perfect for cube)

• 3 appears 2 times (needs 1 more to make it 3)

Step 4: To make 72 a perfect cube, we need to multiply by 3¹ = 3.

Step 5: 72 × 3 = 216 = 2³ × 3³ = (2 × 3)³ = 6³

Verification: ∛216 = 6, which is a whole number ✓

Answer: 72 must be multiplied by 3 to make it a perfect cube.

Question 13: Find the compound interest on Rs. 5000 for 2 years at 10% per annum compounded annually.

Solution:

Step 1: Given: Principal (P) = Rs. 5000, Time (T) = 2 years, Rate (R) = 10% per annum

Step 2: Formula: Amount = P(1 + R/100)ᵀ

Step 3: Amount = 5000(1 + 10/100)²

Step 4: = 5000(1 + 0.1)²

Step 5: = 5000(1.1)²

Step 6: = 5000 × 1.21

Step 7: = Rs. 6050

Step 8: Compound Interest = Amount - Principal

Step 9: CI = 6050 - 5000 = Rs. 1050

Answer: Compound Interest = Rs. 1050

Question 14: A shopkeeper purchased 300 bulbs for Rs. 12 each. Five bulbs were fused and thrown away. The remaining bulbs were sold at Rs. 15 each. Find the gain or loss percent.

Solution:

Step 1: Cost price of 300 bulbs = 300 × 12 = Rs. 3600

Step 2: Number of bulbs sold = 300 - 5 = 295 bulbs

Step 3: Selling price of 295 bulbs = 295 × 15 = Rs. 4425

Step 4: Since SP > CP, there is a gain.

Step 5: Gain = SP - CP = 4425 - 3600 = Rs. 825

Step 6: Gain% = (Gain/CP) × 100

Step 7: = (825/3600) × 100

Step 8: = 0.229 × 100 = 22.9%

Answer: Gain percent = 22.9%

Question 15: Classify the following as monomial, binomial, or trinomial: 2x, 5y - 3, x² + 2x + 1.

Solution:

Step 1: Count the number of terms in each expression.

Step 2: Classification rules:

• Monomial: 1 term

• Binomial: 2 terms

• Trinomial: 3 terms

Step 3: Analysis:

• 2x: Has 1 term → Monomial

• 5y - 3: Has 2 terms (5y and -3) → Binomial

• x² + 2x + 1: Has 3 terms (x², 2x, and 1) → Trinomial

Answer: 2x is a monomial, 5y - 3 is a binomial, x² + 2x + 1 is a trinomial.

Question 16: Factorize: p⁴ - 81

Solution:

Step 1: Recognize that 81 = 3⁴

Step 2: So, p⁴ - 81 = p⁴ - 3⁴

Step 3: This is in the form a⁴ - b⁴ = (a² - b²)(a² + b²)

Step 4: p⁴ - 3⁴ = (p² - 3²)(p² + 3²) = (p² - 9)(p² + 9)

Step 5: Further factorize p² - 9 = p² - 3² = (p - 3)(p + 3)

Step 6: p² + 9 cannot be factorized further over real numbers.

Step 7: Therefore, p⁴ - 81 = (p - 3)(p + 3)(p² + 9)

Answer: p⁴ - 81 = (p - 3)(p + 3)(p² + 9)

Question 17: The diagonal of a quadrilateral-shaped field is 24 m, and the perpendiculars dropped from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

Solution:

Step 1: When a diagonal divides a quadrilateral into two triangles, the area of the quadrilateral is the sum of the areas of both triangles.

Step 2: Area of triangle = (1/2) × base × height

Step 3: Here, the diagonal acts as the base for both triangles.

Step 4: Area of first triangle = (1/2) × 24 × 8 = 96 m²

Step 5: Area of second triangle = (1/2) × 24 × 13 = 156 m²

Step 6: Total area of the field = 96 + 156 = 252 m²

Answer: Area of the field = 252 m²

Question 18: Find the area of a rhombus with a side of 6 cm and altitude of 4 cm. If one diagonal is 8 cm, find the length of the other diagonal.

Solution:

Step 1: Finding the area using side and altitude:

Area of rhombus = side × altitude = 6 × 4 = 24 cm²

Step 2: Finding the other diagonal:

Given: One diagonal (d₁) = 8 cm

Let the other diagonal = d₂

Step 3: Area of rhombus = (1/2) × d₁ × d₂

Step 4: 24 = (1/2) × 8 × d₂

Step 5: 24 = 4 × d₂

Step 6: d₂ = 24/4 = 6 cm

Answer: Area = 24 cm², Other diagonal = 6 cm

Question 19: Using the laws of exponents, evaluate: (2)⁻³ × (1/6)⁻³

Solution:

Step 1: Use the law: aᵐ × bᵐ = (ab)ᵐ

Step 2: (2)⁻³ × (1/6)⁻³ = (2 × 1/6)⁻³

Step 3: = (2/6)⁻³ = (1/3)⁻³

Step 4: Use the law: a⁻ᵐ = 1/aᵐ

Step 5: (1/3)⁻³ = 1/(1/3)³ = 1/(1/27) = 27

Alternative method:

Step 1: (2)⁻³ = 1/2³ = 1/8

Step 2: (1/6)⁻³ = (6/1)³ = 6³ = 216

Step 3: (2)⁻³ × (1/6)⁻³ = 1/8 × 216 = 216/8 = 27

Answer: 27

Question 20: Simplify: 3⁴ × 3⁻² ÷ 3³

Solution:

Step 1: Use the law: aᵐ × aⁿ = aᵐ⁺ⁿ

Step 2: 3⁴ × 3⁻² = 3⁴⁺⁽⁻²⁾ = 3⁴⁻² = 3²

Step 3: Now we have: 3² ÷ 3³

Step 4: Use the law: aᵐ ÷ aⁿ = aᵐ⁻ⁿ

Step 5: 3² ÷ 3³ = 3²⁻³ = 3⁻¹

Step 6: 3⁻¹ = 1/3¹ = 1/3

Answer: 1/3