CBSE Class 8 Mathematics Chapter 1: Rational Numbers NCERT Solutions

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NCERT Solutions PDF Class 8 PDF

This resource provides comprehensive NCERT Solutions for Class 8 Mathematics, Chapter 1 on Rational Numbers. It covers essential concepts like the properties of rational numbers, additive inverses, and verification of identities. The solutions are presented in a clear, step-by-step format, making it easier for students to understand the underlying principles and problem-solving techniques. Each question from the NCERT textbook is addressed with a detailed explanation, ensuring students grasp the methods used to solve problems involving rational numbers. These solutions are ideal for exam preparation, helping students build confidence and accuracy in their answers for the upcoming CBSE examinations.

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Quick info

BoardCBSE
ClassClass 8
SubjectMathematics
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 1: Rational Numbers

Chapter summary

Chapter 1 of the Class 8 NCERT Mathematics textbook focuses on Rational Numbers. This section provides detailed solutions for exercises related to identifying and applying properties of rational numbers, such as the distributive property. It also covers finding the additive inverse of rational numbers and verifying the identity -(-x) = x. The solutions are designed to offer clarity and step-by-step guidance for students to master these fundamental concepts.

Learning outcomes

  • Understand and apply the properties of rational numbers (distributive, associative).
  • Calculate sums and products of rational numbers accurately.
  • Identify and find the additive inverse of given rational numbers.
  • Verify the identity -(-x) = x for rational numbers.
  • Solve problems involving operations on rational numbers.

Topics covered

Paper topics

  • Rational Numbers
  • Properties of Rational Numbers
  • Distributive Property
  • Associative Property
  • Commutative Property
  • Additive Inverse
  • Operations on Rational Numbers
  • Verification of Identities

Important topics

  • Properties of Rational Numbers (Distributive, Associative)
  • Additive Inverse
  • Operations on Rational Numbers
  • Verification of -(-x) = x

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Questions and Solutions

Question 1

Using appropriate properties, find the value of the following expressions:

  1. -\frac{2}{3} \times \frac{3}{5} + \frac{5}{2} - \frac{3}{5} \times \frac{1}{6}
  2. \frac{2}{5} \times \left(\frac{3}{-7}\right) - \frac{1}{6} \times \frac{3}{2} + \frac{1}{14} \times \frac{2}{5}

Solution:

  1. To solve -\frac{2}{3} \times \frac{3}{5} + \frac{5}{2} - \frac{3}{5} \times \frac{1}{6}, we first rearrange the terms using the commutative property of addition to group the terms with the common factor \(\frac{3}{5}\):

    -\frac{2}{3} \times \frac{3}{5} - \frac{3}{5} \times \frac{1}{6} + \frac{5}{2}

    Now, we apply the distributive property, a \times (b + c) = a \times b + a \times c, by factoring out \(\frac{3}{5}\) (or -\frac{3}{5} to be precise, considering the signs):

    =\frac{3}{5}\left(-\frac{2}{3}-\frac{1}{6}\right)+\frac{5}{2}

    Next, we find a common denominator for the fractions inside the parenthesis, which is 6:

    = \frac{3}{5} \left( \frac{-2 \times 2}{3 \times 2} - \frac{1}{6} \right) + \frac{5}{2}

    = \frac{3}{5} \left( \frac{-4}{6} - \frac{1}{6} \right) + \frac{5}{2}

    = \frac{3}{5} \left( \frac{-4-1}{6} \right) + \frac{5}{2}

    = \frac{3}{5} \times \frac{-5}{6} + \frac{5}{2}

    Now, we multiply the fractions:

    = \frac{3 \times -5}{5 \times 6} + \frac{5}{2}

    = \frac{-15}{30} + \frac{5}{2}

    Simplify the fraction \frac{-15}{30} to -\frac{1}{2}:

    = -\frac{1}{2} + \frac{5}{2}

    Finally, add the fractions with the common denominator 2:

    = \frac{-1+5}{2} = \frac{4}{2} = 2

    Thus, the value of the expression is 2.
  2. To solve \frac{2}{5} \times \left(\frac{3}{-7}\right) - \frac{1}{6} \times \frac{3}{2} + \frac{1}{14} \times \frac{2}{5}, we first rearrange the terms using the commutative property to group terms with common factors:

    =\frac{2}{5}\times\left(\frac{-3}{7}\right)+\frac{1}{14}\times\frac{2}{5}-\frac{1}{6}\times\frac{3}{2}

    We can rewrite \frac{1}{14}\times\frac{2}{5} as \frac{2}{5}\times\frac{1}{14} using the commutative property. Then, apply the distributive property by factoring out \frac{2}{5}:

    =\frac{2}{5}\times\left(\frac{-3}{7}+\frac{1}{14}\right)-\frac{1}{6}\times\frac{3}{2}

    Simplify the multiplication \frac{1}{6}\times\frac{3}{2}:

    = \frac{1 \times 3}{6 \times 2} = \frac{3}{12} = \frac{1}{4}

    Now, find a common denominator for the fractions inside the parenthesis, which is 14:

    =\frac{2}{5}\times\left(\frac{-3 \times 2}{7 \times 2}+\frac{1}{14}\right)-\frac{1}{4}

    =\frac{2}{5}\times\left(\frac{-6}{14}+\frac{1}{14}\right)-\frac{1}{4}

    =\frac{2}{5}\times\left(\frac{-6+1}{14}\right)-\frac{1}{4}

    =\frac{2}{5}\times\frac{-5}{14}-\frac{1}{4}

    Multiply the fractions:

    = \frac{2 \times -5}{5 \times 14} - \frac{1}{4}

    = \frac{-10}{70} - \frac{1}{4}

    Simplify \frac{-10}{70} to -\frac{1}{7}:

    = -\frac{1}{7} - \frac{1}{4}

    Find a common denominator, which is 28:

    = \frac{-1 \times 4}{7 \times 4} - \frac{1 \times 7}{4 \times 7}

    = \frac{-4}{28} - \frac{7}{28}

    = \frac{-4-7}{28}

    = \frac{-11}{28}

    Thus, the value of the expression is -\frac{11}{28}.

Question 2

Write the additive inverse of each of the following rational numbers:

  1. \frac{2}{8}
  2. \frac{-5}{9}
  3. \frac{-6}{-5}
  4. \frac{2}{-9}
  5. \frac{19}{-6}

Solution:

The additive inverse of a rational number \frac{a}{b} is \frac{-a}{b}, such that their sum is zero: \frac{a}{b} + \left(\frac{-a}{b}\right) = 0.

  1. The additive inverse of \frac{2}{8} is \frac{-2}{8}.
  2. The additive inverse of \frac{-5}{9} is \frac{-(-5)}{9} = \frac{5}{9}.
  3. First, simplify \frac{-6}{-5} to \frac{6}{5}. The additive inverse of \frac{6}{5} is \frac{-6}{5}.
  4. The additive inverse of \frac{2}{-9} is \frac{-(2)}{-9} = \frac{-2}{-9} = \frac{2}{9}. Alternatively, we can write \frac{2}{-9} as -\frac{2}{9}, so its additive inverse is -\left(-\frac{2}{9}\right) = \frac{2}{9}.
  5. The additive inverse of \frac{19}{-6} is \frac{-(19)}{-6} = \frac{-19}{-6} = \frac{19}{6}. Alternatively, we can write \frac{19}{-6} as -\frac{19}{6}, so its additive inverse is -\left(-\frac{19}{6}\right) = \frac{19}{6}.

Question 3

Verify that -(-x) = x for the following values of x:

  1. x = \frac{11}{15}
  2. x = -\frac{13}{17}

Solution:

  1. Given x = \frac{11}{15}. We need to verify if -(-x) = x. First, find -x:

    -x = -\left(\frac{11}{15}\right) = -\frac{11}{15}

    Now, find -(-x), which is the additive inverse of -x:

    -(-x) = -\left(-\frac{11}{15}\right)

    The additive inverse of -\frac{11}{15} is \frac{11}{15}.

    -(-x) = \frac{11}{15}

    Since x = \frac{11}{15}, we have verified that -(-x) = x.
  2. Given x = -\frac{13}{17}. We need to verify if -(-x) = x. First, find -x:

    -x = -\left(-\frac{13}{17}\right)

    The additive inverse of -\frac{13}{17} is \frac{13}{17}.

    -x = \frac{13}{17}

    Now, find -(-x), which is the additive inverse of -x:

    -(-x) = -\left(\frac{13}{17}\right)

    The additive inverse of \frac{13}{17} is -\frac{13}{17}.

    -(-x) = -\frac{13}{17}

    Since x = -\frac{13}{17}, we have verified that -(-x) = x.

Common mistakes

  • Errors in applying the order of operations (BODMAS/PEMDAS) with fractions.
  • Incorrectly simplifying fractions after multiplication or addition.
  • Confusing additive inverse with multiplicative inverse.
  • Sign errors when dealing with negative rational numbers.

Revision tips

  • Review the properties of rational numbers (commutative, associative, distributive) before attempting problems.
  • Practice identifying common factors for simplification to avoid calculation errors.
  • Ensure you understand the definition of additive inverse and how to find it for any rational number.
  • Work through each step of the verification problems carefully to avoid sign mistakes.

Practice MCQs

Q1. What is the additive inverse of \(\frac{2}{8}\)?

Q2. Which property is used to rearrange \(-\frac{2}{3} \times \frac{3}{5} + \frac{5}{2} - \frac{3}{5} \times \frac{1}{6}\) as \(-\frac{2}{3} \times \frac{3}{5} - \frac{3}{5} \times \frac{1}{6} + \frac{5}{2}\)?

Q3. What is the result of \(\frac{3}{5} \times \frac{-5}{6}\)?

Q4. The additive inverse of \(\frac{-6}{-5}\) is:

Q5. If \(x = \frac{11}{15}\), what is \(-(-x)\)?

Frequently asked questions

What are rational numbers?

Rational numbers are numbers that can be expressed as a fraction \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\). Examples include \(\frac{2}{3}\), \(-5\), and \(0.75\).

What is the additive inverse of a rational number?

The additive inverse of a rational number \(\frac{a}{b}\) is \(\frac{-a}{b}\). When added to the original number, it results in zero. For example, the additive inverse of \(\frac{2}{8}\) is \(\frac{-2}{8}\).

How do the properties of rational numbers help in solving problems?

Properties like the commutative, associative, and distributive laws simplify calculations. For instance, the distributive property allows us to factor out common terms, making complex expressions easier to manage.

How can I verify that -(-x) = x?

To verify \(-(-x) = x\), substitute the given value of \(x\). First, find \(-x\) (the additive inverse of \(x\)), and then find the additive inverse of that result. The final answer should be equal to the original \(x\).

Are these solutions suitable for exam revision?

Yes, these NCERT solutions provide clear, step-by-step explanations for each problem, covering the core concepts of rational numbers. They are excellent for reinforcing understanding and practicing problem-solving techniques for exams.

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