NCERT Solutions for Class 7 Maths Chapter 8 - Rational Numbers Exercise 8.1

Exercise 8.1

1. List five rational numbers between:

i. $ - 1$ and $0$

Ans: Let us write $ - 1$ and $0$ as rational numbers with denominator $6$.

$ - 1 = \dfrac{{ - 6}}{6}$and$0 = \dfrac{0}{6}$

$\therefore \dfrac{{ - 6}}{6}\langle \dfrac{{ - 5}}{6}\langle \dfrac{{ - 4}}{6}\langle \dfrac{{ - 3}}{6}\langle \dfrac{{ - 2}}{6}\langle \dfrac{{ - 1}}{6}\langle 0$

Thus, $ - 1\langle \dfrac{{ - 5}}{6}\langle \dfrac{{ - 2}}{3}\langle \dfrac{{ - 1}}{2}\langle \dfrac{{ - 1}}{3}\langle \dfrac{{ - 1}}{6}\langle 0$

Therefore, five rational numbers between $ - 1$ and $0$ would be

$ - 1,\dfrac{{ - 5}}{6},\dfrac{{ - 2}}{3},\dfrac{{ - 1}}{2},\dfrac{{ - 1}}{3},\dfrac{{ - 1}}{6}$

ii. $ - 2$ and $ - 1$

Ans: Let us write $ - 2$ and $ - 1$ as rational numbers with denominator $6$

$ - 2 = \dfrac{{ - 12}}{6}$and $ - 1 = \dfrac{{ - 6}}{6}$

$\therefore \dfrac{{ - 12}}{6}\langle \dfrac{{ - 11}}{6}\langle \dfrac{{ - 10}}{6}\langle \dfrac{{ - 9}}{6}\langle \dfrac{{ - 8}}{6}\langle \dfrac{{ - 7}}{6}\langle \dfrac{{ - 6}}{6}$

Thus, \[ - 2\langle \dfrac{{ - 11}}{6}\langle \dfrac{{ - 5}}{3}\langle \dfrac{{ - 3}}{2}\langle \dfrac{{ - 4}}{3}\langle \dfrac{{ - 7}}{6}\langle  - 1\]

Therefore, five rational numbers between $ - 2$ and $ - 1$ would be

\[\dfrac{{ - 11}}{6},\dfrac{{ - 5}}{3},\dfrac{{ - 3}}{2},\dfrac{{ - 4}}{3},\dfrac{{ - 7}}{6}\]

iii. $\dfrac{{ - 4}}{5}$ and $\dfrac{{ - 2}}{3}$

Ans:   Let us write $\dfrac{{ - 4}}{5}$ and$\dfrac{{ - 2}}{3}$ as rational numbers with denominators.

$\dfrac{{ - 4}}{5} = \dfrac{{ - 2}}{3}$and $\dfrac{{ - 2}}{3} = \dfrac{{ - 30}}{{45}}$

$\therefore \dfrac{{ - 36}}{{45}}\langle \dfrac{{ - 35}}{{45}}\langle \dfrac{{ - 34}}{{45}}\langle \dfrac{{ - 33}}{{45}}\langle \dfrac{{ - 32}}{{45}}\langle \dfrac{{ - 31}}{{45}}\langle \dfrac{{ - 30}}{{45}}$

Thus, $\dfrac{{ - 4}}{5}\langle \dfrac{{ - 7}}{9}\langle \dfrac{{ - 34}}{{45}}\langle \dfrac{{ - 11}}{{15}}\langle \dfrac{{ - 32}}{{45}}\langle \dfrac{{ - 31}}{{45}}\langle \dfrac{{ - 2}}{3}$

Therefore, five rational numbers between $\dfrac{{ - 4}}{5}$ and $\dfrac{{ - 2}}{3}$ would be

$\dfrac{{ - 7}}{9},\dfrac{{ - 34}}{{45}},\dfrac{{ - 11}}{{15}},\dfrac{{ - 32}}{{45}},\dfrac{{ - 31}}{{45}}$

iv) $\dfrac{{ - 1}}{2}$ and $\dfrac{2}{3}$

Ans: Let us write $\dfrac{{ - 1}}{2}$ and $\dfrac{2}{3}$ as rational numbers with the same denominator.

$\dfrac{{ - 1}}{2} = \dfrac{{ - 3}}{6}$and $\dfrac{2}{3} = \dfrac{4}{6}$

$\therefore \dfrac{{ - 3}}{6}\langle \dfrac{{ - 2}}{6}\langle \dfrac{{ - 1}}{6}\langle 0\langle \dfrac{1}{6}\langle \dfrac{{ - 2}}{6}\langle \dfrac{{ - 3}}{6}\langle \dfrac{{ - 4}}{6}$

Thus, $\dfrac{{ - 1}}{2}\langle \dfrac{{ - 1}}{3}\langle \dfrac{{ - 1}}{6}\langle 0\langle \dfrac{1}{6}\langle \dfrac{1}{3}\langle \dfrac{1}{2}\langle \dfrac{2}{3}$

Therefore, five rational numbers between $\dfrac{{ - 1}}{2}$ and $\dfrac{2}{3}$ would be

$\dfrac{{ - 1}}{2},\dfrac{{ - 1}}{3},\dfrac{{ - 1}}{6},0,\dfrac{1}{6},\dfrac{1}{3},\dfrac{1}{2},\dfrac{2}{3}$

2. Write four more rational numbers in each of the following patterns: 

i. $\dfrac{{ - 3}}{5},\dfrac{{ - 6}}{{10}},\dfrac{{ - 9}}{{15}},\dfrac{{ - 12}}{{10}},...........$

Ans:  $\dfrac{{ - 3 \times 1}}{{5 \times 1}},\dfrac{{ - 3 \times 2}}{{5 \times 2}},\dfrac{{ - 3 \times 3}}{{5 \times 3}},\dfrac{{ - 3 \times 4}}{{5 \times 4}},...........$

Therefore, the next four rational numbers of this pattern would be,

\[\dfrac{{ - 3 \times 5}}{{5 \times 5}},\dfrac{{ - 3 \times 6}}{{5 \times 6}},\dfrac{{ - 3 \times 7}}{{5 \times 7}},\dfrac{{ - 3 \times 8}}{{5 \times 8}} = \dfrac{{ - 15}}{{25}},\dfrac{{ - 18}}{{30}},\dfrac{{ - 21}}{{35}},\dfrac{{ - 24}}{{40}}\]

ii. $\dfrac{{ - 1}}{4},\dfrac{{ - 2}}{8},\dfrac{{ - 3}}{{12}},...........$

Ans: $\dfrac{{ - 1 \times 1}}{{4 \times 1}},\dfrac{{ - 1 \times 2}}{{4 \times 2}},\dfrac{{ - 1 \times 3}}{{4 \times 3}},...........$

Therefore, the next four rational numbers of this pattern would be,

\[\dfrac{{ - 1 \times 4}}{{4 \times 4}},\dfrac{{ - 1 \times 5}}{{4 \times 5}},\dfrac{{ - 1 \times 6}}{{4 \times 6}},\dfrac{{ - 1 \times 7}}{{4 \times 7}} = \dfrac{{ - 4}}{{16}},\dfrac{{ - 5}}{{20}},\dfrac{{ - 6}}{{24}},\dfrac{{ - 7}}{{28}}\]

iii. $\dfrac{{ - 1}}{6},\dfrac{2}{{ - 12}},\dfrac{3}{{ - 18}},\dfrac{4}{{ - 24}},...........$

Ans: $\dfrac{{ - 1 \times 1}}{{4 \times 1}},\dfrac{{ - 1 \times 2}}{{4 \times 2}},\dfrac{{ - 1 \times 3}}{{4 \times 3}},...........$

Therefore, the next four rational numbers of this pattern would be,

\[\dfrac{{ - 1 \times 4}}{{4 \times 4}},\dfrac{{ - 1 \times 5}}{{4 \times 5}},\dfrac{{ - 1 \times 6}}{{4 \times 6}},\dfrac{{ - 1 \times 7}}{{4 \times 7}} = \dfrac{{ - 4}}{{16}},\dfrac{{ - 5}}{{20}},\dfrac{{ - 6}}{{24}},\dfrac{{ - 7}}{{28}}\]

iv. $\dfrac{{ - 2}}{3},\dfrac{2}{{ - 3}},\dfrac{4}{{ - 6}},\dfrac{6}{{ - 9}},...........$

Ans: $\dfrac{{ - 2 \times 1}}{{3 \times 1}},\dfrac{{2 \times 1}}{{ - 3 \times 1}},\dfrac{{2 \times 2}}{{ - 3 \times 2}},\dfrac{{2 \times 3}}{{ - 3 \times 3}},...........$

Therefore, the next four rational numbers of this pattern would be,

\[\dfrac{{2 \times 4}}{{ - 3 \times 4}},\dfrac{{2 \times 5}}{{ - 3 \times 5}},\dfrac{{2 \times 6}}{{ - 3 \times 6}},\dfrac{{2 \times 7}}{{ - 3 \times 7}} = \dfrac{8}{{ - 12}},\dfrac{{10}}{{ - 15}},\dfrac{{12}}{{ - 18}},\dfrac{{14}}{{ - 21}}\]

3. Give four rational numbers equivalent to:

i. $-\dfrac{2}{7}$

Ans: \[\dfrac{-2\times 2}{7\times 2}=\dfrac{-4}{14},\dfrac{-2\times 3}{7\times 3}=\dfrac{-6}{21},\dfrac{-2\times 4}{7\times 4}=\dfrac{-8}{28},\dfrac{-2\times 5}{7\times 5}=\dfrac{-10}{35}\]

Therefore, four equivalent rational numbers are

\[\dfrac{-4}{14},\dfrac{-6}{21},\dfrac{-8}{28},\dfrac{-10}{35}\]

ii. $\dfrac{5}{-3}$

Ans: \[\dfrac{5\times 2}{-3\times 2}=\dfrac{10}{-6},\dfrac{5\times 3}{-3\times 3}=\dfrac{15}{-9},\dfrac{5\times 4}{-3\times 4}=\dfrac{20}{-12},\dfrac{5\times 5}{-3\times 5}=\dfrac{25}{-15}\]

Therefore, four equivalent rational numbers are

\[\dfrac{10}{-6},\dfrac{15}{-9},\dfrac{20}{-12},\dfrac{25}{-15}\]

iii. $\dfrac{4}{9}$

Ans: \[\dfrac{4\times 2}{9\times 2}=\dfrac{8}{18},\dfrac{4\times 3}{9\times 3}=\dfrac{12}{27},\dfrac{4\times 4}{9\times 4}=\dfrac{16}{36},\dfrac{4\times 5}{9\times 5}=\dfrac{20}{25}\]

Therefore, four equivalent rational numbers are

\[\dfrac{8}{18},\dfrac{12}{27},\dfrac{16}{36},\dfrac{20}{25}\]

4. Draw the number line and represent the following rational numbers on it:

a) $\dfrac{3}{4}$

Ans: Let us take a graphical line,

Consider 1 unit = 4 sub-unit.

Point out the $\dfrac{3}{4}$ point.

b) $\dfrac{{ - 5}}{8}$

Ans: Let us take a graphical line,

Consider 1 unit = 5 sub-units.

Point out the $\dfrac{{ - 5}}{8}$ point.

c) $\dfrac{{ - 7}}{4}$

Ans: Let us take a graphical line,

Consider 1 unit = 5 sub-units.

Point out the $\dfrac{{ - 7}}{4}$ point.

d) $\dfrac{7}{8}$

Ans: Let us take a graphical line,

Consider 1 unit = 8 sub-unit.

Point out the $\dfrac{7}{8}$ point.

The point A is given point.

5. The points P, Q, R, S, T, U, A and B on the number line are such that, TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R and S.

Ans: Each part which is between the two numbers is divided into 3 parts. 

Therefore, \[A = \dfrac{6}{3},P = \dfrac{7}{3},Q = \dfrac{8}{3}\]and $B = \dfrac{9}{3}$

Similarly,\[T = \dfrac{{ - 3}}{3},R = \dfrac{{ - 4}}{3},S = \dfrac{{ - 5}}{3}\]− and $U = \dfrac{{ - 6}}{3}$

Thus, the rational numbers represented \[P,Q,R\]and $S$ are \[\dfrac{7}{3},\dfrac{8}{3},\dfrac{{ - 4}}{3}\] and $\dfrac{{ - 5}}{3}$ respectively.

6. Which of the following pairs represent the same rational numbers?

I. $\dfrac{{ - 7}}{{21}}$ and $\dfrac{3}{9}$

Ans: Converting into the lowest term. We get,

$\dfrac{{ - 7}}{{21}} = \dfrac{{ - 1}}{3}$ and $\dfrac{3}{9} = \dfrac{1}{3}$

$\therefore \dfrac{{ - 1}}{3} \ne \dfrac{1}{3}$

$\because \dfrac{{ - 7}}{{21}} \ne \dfrac{3}{9}$

II. $\dfrac{{ - 16}}{{20}}$ and $\dfrac{{20}}{{ - 25}}$

Ans: Converting into the lowest term. We get,

$\dfrac{{ - 16}}{{20}} = \dfrac{{ - 4}}{5}$ and \[\dfrac{{20}}{{ - 25}} = \dfrac{4}{{ - 5}} = \dfrac{{ - 4}}{5}\]

$\therefore \dfrac{{ - 4}}{5} = \dfrac{{ - 4}}{5}$

$\because \dfrac{{ - 16}}{{20}} = \dfrac{{20}}{{ - 25}}$

III. $\dfrac{{ - 2}}{{ - 3}}$ and $\dfrac{2}{3}$

Ans: Converting into the lowest term. We get,

$\dfrac{{ - 2}}{{ - 3}} = \dfrac{2}{3}$ and \[\dfrac{2}{3} = \dfrac{2}{3}\]

$\therefore \dfrac{2}{3} = \dfrac{2}{3}$

$\because \dfrac{{ - 2}}{{ - 3}} = \dfrac{2}{3}$

IV. $\dfrac{{ - 3}}{5}$ and $\dfrac{{ - 12}}{{20}}$

Ans: Converting into the lowest term. We get,

$\dfrac{{ - 3}}{5} = \dfrac{{ - 3}}{5}$ and \[\dfrac{{ - 12}}{{20}} = \dfrac{{ - 3}}{5}\]

$\therefore \dfrac{{ - 3}}{5} = \dfrac{{ - 3}}{5}$

$\because \dfrac{{ - 3}}{5} = \dfrac{{ - 12}}{{20}}$

V. $\dfrac{8}{{ - 5}}$ and $\dfrac{{ - 24}}{{15}}$

Ans: Converting into the lowest term. We get,

$\dfrac{8}{{ - 5}} = \dfrac{{ - 8}}{5}$ and \[\dfrac{{ - 24}}{{15}} = \dfrac{{ - 8}}{5}\]

$\therefore \dfrac{{ - 8}}{5} = \dfrac{{ - 8}}{5}$

$\because \dfrac{8}{{ - 5}} = \dfrac{{ - 24}}{{15}}$

VI. $\dfrac{1}{3}$ and $\dfrac{{ - 1}}{9}$

Ans: Converting into the lowest term. We get,

$\dfrac{1}{3} = \dfrac{1}{3}$ and \[\dfrac{{ - 1}}{9} = \dfrac{{ - 1}}{9}\]

$\therefore \dfrac{1}{3} \ne \dfrac{{ - 1}}{9}$

$\because \dfrac{1}{3} \ne \dfrac{{ - 1}}{9}$

VII. $\dfrac{{ - 5}}{{ - 9}}$ and $\dfrac{5}{{ - 9}}$

Ans: Converting into the lowest term. We get,

$\dfrac{{ - 5}}{{ - 9}} = \dfrac{5}{9}$ and \[\dfrac{5}{{ - 9}} = \dfrac{5}{{ - 9}}\]

$\therefore \dfrac{5}{9} \ne \dfrac{5}{{ - 9}}$

$\because \dfrac{{ - 5}}{{ - 9}} \ne \dfrac{5}{{ - 9}}$

7. Rewrite the following rational numbers in the simplest form:

a. $\dfrac{{ - 8}}{6}$

Ans: $\dfrac{{ - 8}}{6}$

Dividing numerator and denominator by $2$ we get,

$\dfrac{{\dfrac{{ - 8}}{2}}}{{\dfrac{6}{2}}} = \dfrac{{ - 4}}{3}$

b. $\dfrac{{25}}{{45}}$

Ans: $\dfrac{{25}}{{45}}$

Dividing numerator and denominator by $5$ we get,

$\dfrac{{\dfrac{{25}}{5}}}{{\dfrac{{45}}{5}}} = \dfrac{5}{9}$

c. $\dfrac{{ - 44}}{{72}}$

Ans: $\dfrac{{ - 44}}{{72}}$

Dividing numerator and denominator by $4$ we get,

$\dfrac{{\dfrac{{ - 44}}{4}}}{{\dfrac{{72}}{4}}} = \dfrac{{ - 11}}{{18}}$

d. $\dfrac{{ - 8}}{{10}}$

Ans: $\dfrac{{ - 8}}{{10}}$

Dividing numerator and denominator by $2$ we get,

$\dfrac{{\dfrac{{ - 8}}{2}}}{{\dfrac{{10}}{2}}} = \dfrac{{ - 4}}{5}$

8. Fill in the boxes with the correct symbol out of $\rangle ,\langle $ and =:

i. $\dfrac{{ - 5}}{7}\square \dfrac{2}{3}$

Ans: Since, the positive number is greater than the negative number. 

$\dfrac{{ - 5}}{7}\boxed\langle \dfrac{2}{3}$

ii. $\dfrac{{ - 4}}{5}\square \dfrac{{ - 5}}{7}$

Ans: Since, Both the numbers are negative numbers. Thus, multiple numerator and denominator of the number $\dfrac{{ - 4}}{5}$ with $7$ and of the number $\dfrac{{ - 5}}{7}$with $5$ we get,

$\dfrac{{ - 4 \times 7}}{{5 \times 7}}\square \dfrac{{ - 5 \times 5}}{{7 \times 5}}$

$ = \dfrac{{ - 28}}{{35}}\boxed\langle \dfrac{{ - 25}}{{35}}$

Thus, we can say $\dfrac{{ - 4}}{5}\boxed\langle \dfrac{{ - 5}}{7}$

iii. $\dfrac{{ - 7}}{8}\square \dfrac{{14}}{{ - 16}}$

Ans: Since, Both the numbers are negative numbers. Thus, multiple numerator and denominator of the number $\dfrac{{ - 7}}{8}$with $2$ and of the number $\dfrac{{14}}{{ - 16}}$with $ - 1$ we get,

\[\dfrac{{ - 7 \times 2}}{{8 \times 2}}\square \dfrac{{14 \times ( - 1)}}{{ - 16 \times ( - 1)}}\]

$ = \dfrac{{ - 14}}{{16}}\boxed = \dfrac{{ - 14}}{{16}}$

Thus, we can say $\dfrac{{ - 7}}{8}\boxed = \dfrac{{14}}{{ - 16}}$.

iv. $\dfrac{{ - 8}}{5}\square \dfrac{{ - 7}}{4}$

Ans: Since, Both the numbers are negative numbers. Thus, multiple numerator and denominator of the number $\dfrac{{ - 8}}{5}$with $4$ and of the number $\dfrac{{ - 7}}{4}$with $5$ we get,

$\dfrac{{ - 8 \times 4}}{{5 \times 4}}\square \dfrac{{ - 7 \times 5}}{{4 \times 5}}$

$ = \dfrac{{ - 32}}{{20}}\boxed\rangle \dfrac{{ - 35}}{{20}}$

Thus, we can say $\dfrac{{ - 8}}{5}\boxed\rangle \dfrac{{ - 7}}{4}$

v. $\dfrac{1}{{ - 3}}\square \dfrac{{ - 1}}{4}$

Ans: Since the denominator is greater in the second number. Thus,

$\dfrac{1}{{ - 3}}\boxed\langle \dfrac{{ - 1}}{4}$

vi. $\dfrac{5}{{ - 11}}\square \dfrac{{ - 5}}{{11}}$

Ans:  Since both the numbers have the same denominator and numerator and both of them are negative numbers. Thus, the number is the same.

$\dfrac{5}{{ - 11}}\boxed = \dfrac{{ - 5}}{{11}}$

vii. $0\square \dfrac{{ - 7}}{6}$

Ans: Between these two numbers one is zero and another one is a negative number. And we know that zero is greater than any negative number. Thus, 

$0\boxed\rangle \dfrac{{ - 7}}{6}$

9. Which is greater in each of the following:

i. $\dfrac{2}{3},\dfrac{5}{2}$

Ans: Since, Both the numbers are positive numbers. Thus, multiple numerator and denominator of the number $\dfrac{2}{3}$with $2$ and of the number $\dfrac{5}{2}$with $3$ we get,

$\dfrac{{2 \times 2}}{{3 \times 2}} = \dfrac{4}{6}$and$\dfrac{{5 \times 3}}{{2 \times 3}} = \dfrac{{15}}{6}$

$ = \dfrac{4}{6}\boxed\langle \dfrac{{15}}{6}$

Thus, we can say $\dfrac{2}{3}\boxed\langle \dfrac{5}{2}$

ii. $\dfrac{{ - 5}}{6},\dfrac{{ - 4}}{3}$

Ans: Since, Both the numbers are negative numbers. Thus, multiple numerator and denominator of the number $\dfrac{{ - 5}}{6}$with $1$ and of the number $\dfrac{{ - 8}}{6}$with $2$ we get,

$\dfrac{{ - 5 \times 1}}{{6 \times 1}} = \dfrac{{ - 5}}{6}$and$\dfrac{{ - 4 \times 2}}{{3 \times 2}} = \dfrac{{ - 8}}{6}$

$ = \dfrac{{ - 5}}{6}\boxed\rangle \dfrac{{ - 8}}{6}$

Thus, we can say $\dfrac{{ - 5}}{6}\boxed\rangle \dfrac{{ - 4}}{3}$

ii. $\dfrac{{ - 3}}{4},\dfrac{2}{{ - 3}}$

Ans: Since, Both the numbers are negative numbers. Thus, multiple numerator and denominator of the number $\dfrac{{ - 3}}{4}$with $3$ and of the number $\dfrac{2}{{ - 3}}$with $ - 4$ we get,

$\dfrac{{ - 3 \times 3}}{{4 \times 3}} = \dfrac{{ - 9}}{{12}}$and$\dfrac{{2 \times ( - 4)}}{{ - 3 \times ( - 4)}} = \dfrac{{ - 8}}{{12}}$

$ = \dfrac{{ - 9}}{{12}}\boxed\langle \dfrac{{ - 8}}{{12}}$

Thus, we can say $\dfrac{{ - 3}}{4}\boxed\langle \dfrac{2}{{ - 3}}$

iii. $\dfrac{{ - 1}}{4},\dfrac{1}{4}$

Ans: Since positive numbers are always greater than negative number thus, we can write $\dfrac{{ - 1}}{4}\boxed\langle \dfrac{1}{4}$

iv) $ - 3\dfrac{2}{7}, - 3\dfrac{4}{5}$

Ans: Since, Both the numbers are negative numbers. 

Thus, simplify the number,

$ - 3\dfrac{2}{7} = \dfrac{{ - 23}}{7}$

$ - 3\dfrac{4}{5} = \dfrac{{ - 19}}{5}$

multiple numerator and denominator of the number $\dfrac{{ - 23}}{7}$with $5$ and of the number $\dfrac{{ - 19}}{5}$with $7$ we get,

$\dfrac{{ - 23 \times 5}}{{7 \times 5}} = \dfrac{{ - 115}}{{35}}$and$\dfrac{{ - 19 \times 7}}{{5 \times 7}} = \dfrac{{ - 133}}{{35}}$

$ = \dfrac{{ - 115}}{{35}}\boxed\rangle \dfrac{{ - 133}}{{35}}$

Thus, we can say $ - 3\dfrac{2}{7}\boxed\rangle  - 3\dfrac{4}{5}$

10. Write the following rational numbers in ascending order:

i. $\dfrac{{ - 3}}{5},\dfrac{{ - 2}}{5},\dfrac{{ - 1}}{5}$

Ans: All the numbers have similar type of denominator,

Thus, the ascending order is, $\dfrac{{ - 3}}{5}\langle \dfrac{{ - 2}}{5}\langle \dfrac{{ - 1}}{5}$

ii. $\dfrac{-1}{3},\dfrac{{ - 2}}{9},\dfrac{{ - 4}}{3}$

Ans: To convert all the denominator same, Let us multiply both of numerator and denominator with $3$in the number $\dfrac{1}{3},$and $\dfrac{{ - 4}}{3}$.

Thus, the numbers will be 

$\dfrac{{1 \times 3}}{{3 \times 3}} = \dfrac{3}{9}$ and $\dfrac{{ - 4 \times 3}}{{3 \times 3}} = \dfrac{{ - 12}}{9}$

So, the ascending order will be, $\dfrac{{ - 12}}{9}\langle \dfrac{{ - 2}}{9}\langle \dfrac{3}{9}$ that is $\dfrac{{ - 4}}{3}\langle \dfrac{{ - 2}}{9}\langle \dfrac{1}{3}$.

iii. $\dfrac{{ - 3}}{7},\dfrac{{ - 3}}{2},\dfrac{{ - 3}}{4}$

Ans: As the numerator is the same and all are negative numbers thus, the number having a greater numerator is greater. 

\[\dfrac{{ - 3}}{2}\langle \dfrac{{ - 3}}{4}\langle \dfrac{{ - 3}}{7}\]

Conclusion

NCERT Exercise 8.1 in Chapter 8 of Class 7 Maths is essential for understanding rational numbers. It covers identifying, comparing, and representing rational numbers on a number line. These fundamental skills help students build a solid foundation in mathematics. By practicing these problems, students enhance their computational abilities and logical thinking. Class 7 math exercise 8.1 is crucial for mastering more advanced topics in future classes. Consistent practice with these solutions will ensure a strong grasp of rational numbers and their properties, aiding in overall academic success.

Class 7 Maths Chapter 8: Exercises Breakdown

CBSE Class 7 Maths Chapter 8 Other Study Materials

Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Important Related Links for NCERT Class 7 Maths

Access these essential links for NCERT Class 7 Maths, offering comprehensive solutions, study guides, and additional resources to help students master language concepts and excel in their exams.