NCERT Solutions for Class 7 Maths Chapter 8: Rational Numbers — Free PDF

Rational Number: Any number that can be expressed in the form $\dfrac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Examples: $\dfrac{3}{5}$, $\dfrac{-7}{2}$, $0 = \dfrac{0}{1}$, $5 = \dfrac{5}{1}$, $-3 = \dfrac{-3}{1}$.

Positive rational numbers: Both numerator and denominator have the same sign. Example: $\dfrac{3}{5}$, $\dfrac{-4}{-7} = \dfrac{4}{7}$.

Negative rational numbers: Numerator and denominator have opposite signs. Example: $\dfrac{-3}{5}$, $\dfrac{4}{-7}$.

Zero is neither positive nor negative. $\dfrac{0}{q} = 0$ for any $q \neq 0$.

Equivalent rational numbers: $\dfrac{p}{q} = \dfrac{p \times k}{q \times k}$ for any non-zero integer $k$. Multiplying (or dividing) both numerator and denominator by the same non-zero number gives an equivalent rational number.

Standard form: A rational number $\dfrac{p}{q}$ is in standard form if:
1. The denominator $q$ is positive.
2. The HCF of $|p|$ and $q$ is $1$ (numerator and denominator are coprime).

Example: $\dfrac{-2}{3}$ is in standard form, but $\dfrac{4}{-6}$ is not (should be $\dfrac{-2}{3}$).

Comparison: To compare two rational numbers, convert them to equivalent fractions with the same positive denominator (use LCM), then compare numerators. The one with the larger numerator is greater.

Number line: Every rational number has a unique position on the number line. Positive rational numbers lie to the right of $0$, negative ones to the left. To plot $\dfrac{-3}{4}$, divide the segment from $-1$ to $0$ into $4$ equal parts and mark the third division from $0$.

Between any two rational numbers, there are infinitely many rational numbers. One method to find a rational number between two given ones is to take their average: the number between $\dfrac{a}{b}$ and $\dfrac{c}{d}$ is $\dfrac{1}{2}\left(\dfrac{a}{b} + \dfrac{c}{d}\right)$.

Addition/Subtraction: Find LCM of denominators, convert to common denominator, then add/subtract numerators.

Multiplication: Multiply numerators and multiply denominators.

Division: Multiply by the reciprocal of the divisor.

Reciprocal: The reciprocal of $\dfrac{p}{q}$ is $\dfrac{q}{p}$ (provided $p \neq 0$). Zero has no reciprocal.

Problem: List five rational numbers between $-1$ and $0$.

Solution:

These all satisfy $-1 < \dfrac{p}{q} < 0$. There are infinitely many such numbers — you can always find more by choosing different denominators or by taking averages.

Problem: Write $\dfrac{-4}{6}$ in standard form.

Solution:
Divide numerator and denominator by their HCF ($2$):

A rational number is in standard form when the denominator is positive and the HCF of $|\text{numerator}|$ and denominator is $1$.

Problem: Which is greater: $\dfrac{-3}{5}$ or $\dfrac{-2}{3}$?

Solution:
Find common denominator (LCM of $5$ and $3$ is $15$):

Since $-9 > -10$, we have $\dfrac{-3}{5} > \dfrac{-2}{3}$.

Remember: on the number line, $\dfrac{-3}{5}$ is to the right of $\dfrac{-2}{3}$, so it is greater.

Problem: Represent $\dfrac{-3}{4}$ on the number line.

Solution:
Divide the segment from $-1$ to $0$ into $4$ equal parts. The third mark from $0$ (towards $-1$) represents $\dfrac{-3}{4}$.

Alternatively, the first mark from $-1$ (towards $0$) represents $\dfrac{-3}{4}$ since $-1 + \dfrac{1}{4} = \dfrac{-3}{4}$.

Problem: Find two equivalent rational numbers for $\dfrac{3}{7}$.

Solution:

Multiply both numerator and denominator by the same non-zero integer ($2$ and $3$ respectively).

Problem: Arrange in ascending order: $\dfrac{-3}{7}, \dfrac{-3}{2}, \dfrac{-3}{4}$.

Solution:
LCM of $7, 2, 4$ is $28$.

Since $-42 < -21 < -12$:

Problem: Add $\dfrac{-3}{5}$ and $\dfrac{2}{3}$.

Solution:
LCM of $5$ and $3$ is $15$.

Problem: Subtract $\dfrac{5}{6}$ from $\dfrac{1}{3}$.

Solution:

Always reduce the answer to standard form.

Problem: Multiply $\dfrac{-3}{7}$ and $\dfrac{5}{4}$.

Solution:

For multiplication: multiply numerators together and denominators together.

Problem: Divide $\dfrac{-2}{3}$ by $\dfrac{4}{5}$.

Solution:

For division: multiply by the reciprocal of the divisor.

Problem: Simplify: $\dfrac{-3}{4} + \dfrac{5}{6} - \dfrac{1}{2}$.

Solution:
LCM of $4, 6, 2$ is $12$.

Problem: Find the product: $\dfrac{-4}{9} \times \dfrac{3}{-16}$.

Solution:

Note: negative divided by negative gives positive.

Problem: Divide $\dfrac{-7}{8}$ by $\dfrac{-3}{4}$.

Solution:

Problem: Find five rational numbers between $\dfrac{1}{4}$ and $\dfrac{1}{2}$.

Solution:
Convert to equivalent fractions with a larger denominator.
$\dfrac{1}{4} = \dfrac{6}{24}$ and $\dfrac{1}{2} = \dfrac{12}{24}$.

Five rational numbers between them: $\dfrac{7}{24}, \dfrac{8}{24}, \dfrac{9}{24}, \dfrac{10}{24}, \dfrac{11}{24}$

Simplified: $\dfrac{7}{24}, \dfrac{1}{3}, \dfrac{3}{8}, \dfrac{5}{12}, \dfrac{11}{24}$.

Problem: The sum of two rational numbers is $\dfrac{-1}{3}$. If one of them is $\dfrac{5}{6}$, find the other.

Solution:
Let the other number be $x$.

Verification: $\dfrac{-7}{6} + \dfrac{5}{6} = \dfrac{-2}{6} = \dfrac{-1}{3}$ ✓

Problem: Verify that $\dfrac{-3}{5} \times \left(\dfrac{2}{7} + \dfrac{4}{9}\right) = \dfrac{-3}{5} \times \dfrac{2}{7} + \dfrac{-3}{5} \times \dfrac{4}{9}$.

Solution:
LHS: $\dfrac{2}{7} + \dfrac{4}{9} = \dfrac{18}{63} + \dfrac{28}{63} = \dfrac{46}{63}$
$\dfrac{-3}{5} \times \dfrac{46}{63} = \dfrac{-138}{315} = \dfrac{-46}{105}$

RHS: $\dfrac{-3}{5} \times \dfrac{2}{7} = \dfrac{-6}{35}$ and $\dfrac{-3}{5} \times \dfrac{4}{9} = \dfrac{-12}{45} = \dfrac{-4}{15}$
$\dfrac{-6}{35} + \dfrac{-4}{15} = \dfrac{-18}{105} + \dfrac{-28}{105} = \dfrac{-46}{105}$

LHS $=$ RHS ✓. The distributive property holds.

Problem: Simplify $\dfrac{-5}{8} + \dfrac{3}{4} \times \dfrac{2}{3} - \dfrac{1}{2}$.

Solution:
Follow BODMAS — multiplication first:
$\dfrac{3}{4} \times \dfrac{2}{3} = \dfrac{6}{12} = \dfrac{1}{2}$

Now: $\dfrac{-5}{8} + \dfrac{1}{2} - \dfrac{1}{2} = \dfrac{-5}{8} + 0 = \dfrac{-5}{8}$

Question: Write $\dfrac{18}{-42}$ in standard form.

Answer: HCF of $18$ and $42$ is $6$. $\dfrac{18}{-42} = \dfrac{-18}{42} = \dfrac{-3}{7}$.

Question: Find $\dfrac{-2}{3} + \dfrac{5}{4} - \dfrac{1}{6}$.

Answer: LCM of $3, 4, 6$ is $12$.
$\dfrac{-8}{12} + \dfrac{15}{12} - \dfrac{2}{12} = \dfrac{-8 + 15 - 2}{12} = \dfrac{5}{12}$.

Question: Find $\dfrac{-5}{9} \times \dfrac{-3}{10}$.

Answer: $\dfrac{(-5)(-3)}{(9)(10)} = \dfrac{15}{90} = \dfrac{1}{6}$.

Question: A rope of length $\dfrac{7}{2}$ m is cut into pieces of length $\dfrac{1}{4}$ m each. How many pieces are there?

Answer: Number of pieces $= \dfrac{7}{2} \div \dfrac{1}{4} = \dfrac{7}{2} \times \dfrac{4}{1} = \dfrac{28}{2} = 14$ pieces.