NCERT Solutions for Class 6 Maths: Chapter-Wise Solutions

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Jasmine Grover

Education Journalist | Study Abroad Lead

NCERT Solutions for Class 6 Maths cover detailed explanations and step-by-step solutions to enhance the preparation of a student. Class 6 Maths NCERT Solutions are given in this article along with important concepts covered in this article. The NCERT Solutions for Class 6 Mathematics covers important chapters such as practical geometry, whole numbers, knowing our numbers, etc. 

Class 6 Maths NCERT Solutions

Chapter-wise Class 6 Maths NCERT Solutions are provided as below. Students can also download the PDF of NCERT Solutions for their reference. 

Chapter 1: Knowing Our Numbers


Chapter 2: Whole Numbers


Chapter 3: Playing with Numbers

Additional Resource: Playing Numbers MCQ


Chapter 4: Basic Geometrical Ideas


Chapter 5: Understanding Elementary Shapes


Chapter 6: Integers


Chapter 7: Fractions


Chapter 8: Decimals


Chapter 9: Data Handling


Chapter 10: Mensuration


Chapter 11: Algebra


Chapter 12: Ratio and Proportion


Chapter 13: Symmetry


Chapter 14: Practical Geometry

  • Weightage in Board Exam: Chapter 14 Practical Geometry has a weightage of 10 marks in the Class 6 Maths Examination.
  • Important Topics Covered:​
  • The Class 6 NCERT Solutions chapter exercises for Chapter 14 Practical Geometry are – 

CBSE Study Guides:

CBSE X Related Questions

  • 1.
    In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

      • \(2.5\) cm
      • \(10.5\) cm
      • \(3.5\) cm
      • \(\frac{14}{3}\) cm

    • 2.
      The HCF of 960 and 432 is :

        • 48
        • 54
        • 72
        • 36

      • 3.
        Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)


          • 4.
            If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

              • \(4 \text{ cm}\)
              • \(4\sqrt{2} \text{ cm}\)
              • \(8 \text{ cm}\)
              • \(2\sqrt{2} \text{ cm}\)

            • 5.
              The natural number 1 is :

                • a prime number.
                • a composite number.
                • prime as well as composite.
                • neither prime nor composite.

              • 6.
                In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

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