NCERT Solutions for Class 7 Maths Chapter 9: Rational Numbers - Exercise 9.1

Exercise 9.1

Refer the page 3 to 11 for the exercise 9.1 in the PDF

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.

(Image Will Be Updated Soon)

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

Ans: Let us take a graphical line,

Consider 1 unit = 5 sub-units.

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

(Image Will Be Updated Soon)

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

Ans: Let us take a graphical line,

Consider 1 unit = 5 sub-units.

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

(Image Will Be Updated Soon)

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

Ans: Let us take a graphical line,

Consider 1 unit = 8 sub-unit.

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

(Image Will Be Updated Soon)

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. 

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

Topics Covered in NCERT Class 7 Maths Exercise 9.1 Rational Numbers

The topics covered in exercise 9.1 are given below. The questions asked in exercise 9.1 of Class 7 NCERT Maths are from these topics. 

Important Points to Remember to Solve Questions of Exercise 9.1 of NCERT Maths Class 7

• Any number that can be expressed in the form of pq is called a rational number. 

• In the number pq, p is termed as the numerator whereas q is called the denominator. p and q both are integers and q ≠ 0. 

• Equivalent rational numbers: They are similar to the equivalent fractions. When after multiplying or dividing any rational number, if we obtain a rational number that is equivalent to another rational number then both the numbers are said to be equivalent rational numbers.

• When either the denominator or numerator is negative, such a rational number is called a negative rational number. Whereas if both numerator and denominator are either positive or negative, then such a rational number is a positive rational number.

• Positive rational numbers are represented on the right of 0 whereas the negative rational numbers are represented on the left of 0.

• To write any rational number in standard form, divide its numerator and denominator by their HCF. Also, any rational number is said to be in standard form when its denominator is a positive integer. 

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Exercise 9.1

Opting for the NCERT solutions for Ex 9.1 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 9.1 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 7 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 7 Maths Chapter 9 Exercise 9.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 7 Maths Chapter 9 Exercise 9.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

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NCERT Solution Class 7 Maths Chapter 9 Other Exercise

NCERT Solutions for Class 7 Maths

• Chapter 1 - Integers

• Chapter 2 - Fractions and Decimals

• Chapter 3 - Data Handling

• Chapter 4 - Simple Equations

• Chapter 5 - Lines and Angles

• Chapter 6 - The Triangle and Its Properties

• Chapter 7 - Congruence of Triangles

• Chapter 8 - Comparing Quantities

• Chapter 9 - Rational Numbers

• Chapter 10 - Practical Geometry

• Chapter 11 - Perimeter and Area

• Chapter 12 - Algebraic Expressions

• Chapter 13 - Exponents and Powers

• Chapter 14 - Symmetry

• Chapter 15 - Visualising Solid Shapes