# NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

Written by Team Trustudies
Updated at 2021-05-07

## NCERT solutions for class 7 Maths Chapter 9 Rational Numbers Exercise 9.1

Q.1 List five rational numbers between:
(i) -1 and 0
(ii) -2 and -1
(iii)$$\frac{-4}{5}$$ and $$\frac{-2}{3}$$
(iv) $$\frac{1}{2}$$ and $$\frac{2}{3}$$

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

(i) The five rational numbers between -1 and 0 are,
-1< $$\frac{-1}{2}$$ < $$\frac{-3}{4}$$ < $$\frac{-4}{5}$$ < $$\frac{-1}{3}$$ < $$\frac{-1}{5}$$ < 0< br> (ii) The five rational numbers between -2 and -1 are,
-2 < $$\frac{-9}{8}$$ , $$\frac{-9}{7}$$ , $$\frac{-9}{6}$$ , $$\frac{-8}{5}$$ , $$\frac{-4}{3}$$ < -1
(iii) The five rational numbers between -4/5 and -2/3 are,
-4/5= $$\frac{-4×6}{5×6}$$ = $$\frac{-24}{30}$$
-2/3 = $$\frac{-2×10}{3×10}$$ = $$\frac{-20}{30}$$
-4/5 < (-21/30) , (-22/30) , (-23/30) , (-16/15) , (-17/16) <-2/3
(iv) The five rational numbers between -1/2 and 2/3 are,
-1/2 < (-1/6) < (0) < (1/3) < (1/2) < (20/36) < 2/3

Q.2 . Write four more rational numbers in each of the following patterns:
(i) -3/5, -6/10, -9/15, -12/20, …..
(ii) -1/4, -2/8, -3/12, …..
(iii) -1/6, 2/-12, 3/-18, 4/-24 …..
(iv) -2/3, 2/-3, 4/-6, 6/-9 …..

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

(i) In the above question, we can observe that the numerator and denominator are the multiples of 3 and 5.
= $$\frac{(-3 × 1)}{(5 × 1)}$$, $$\frac{(-3 × 2)}{(5 × 2)}$$, $$\frac{(-3 × 3)}{(5 × 3)}$$, $$\frac{(-3 × 4)}{ (5 × 4)}$$
Then, next four rational numbers in this pattern are,
= $$\frac{(-3 × 5)}{(5 × 5)}$$, $$\frac{(-3 × 6)}{ (5 × 6)}$$ , $$\frac{(-3 × 7)}{(5 × 7)}, \frac{(-3 × 8)}{(5 × 8)}$$
= -15/25, -18/30, -21/35, -24/40 ….
(ii) In the above question, we can observe that the numerator and denominator are the multiples of 1 and 4.
= (-1 × 1)/ (4 × 1), (-1 × 2)/ (4 × 2), (-1 × 3)/ (1 × 3)
Then, next four rational numbers in this pattern are,
= (-1 × 4)/ (4 × 4), (-1 × 5)/ (4 × 5), (-1 × 6)/ (4 × 6), (-1 × 7)/ (4 × 7)
= -4/16, -5/20, -6/24, -7/28 ….
(iii) In the above question, we can observe that the numerator and denominator are the multiples of 1 and 6.
= (-1 × 1)/ (6 × 1), (1 × 2)/ (-6 × 2), (1 × 3)/ (-6 × 3), (1 × 4)/ (-6 × 4)
Then, next four rational numbers in this pattern are,
= (1 × 5)/ (-6 × 5), (1 × 6)/ (-6 × 6), (1 × 7)/ (-6 × 7), (1 × 8)/ (-6 × 8)
= 5/-30, 6/-36, 7/-42, 8/-48 ….
(iv) In the above question, we can observe that the numerator and denominator are the multiples of 2 and 3.
= (-2 × 1)/ (3 × 1), (2 × 1)/ (-3 × 1), (2 × 2)/ (-3 × 2), (2 × 3)/ (-3 × 3)
Then, next four rational numbers in this pattern are,
= (2 × 4)/ (-3 × 4), (2 × 5)/ (-3 × 5), (2 × 6)/ (-3 × 6), (2 × 7)/ (-3 × 7)
= 8/-12, 10/-15, 12/-18, 14/-21 ….

Q.3 Give four rational numbers equivalent to:
(i) -2/7
(ii) 5/-3
(iii) 4/9

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

(i) $$\frac{-2}{7}$$ = $$\frac{-2×2}{7×2}$$ =$$\frac{-4}{14}$$
$$\frac{-2}{7}$$ =$$\frac{-2×3}{7×3}$$ = $$\frac{-6}{21}$$
$$\frac{-2}{7}$$ =$$\frac{-2×4}{7×4}$$ = $$\frac{-8}{28}$$
$$\frac{-2}{7}$$ =$$\frac{-2×5}{7×5}$$ = $$\frac{-10}{35}$$
(ii) $$\frac{5}{-3}$$ = $$\frac{5×2}{-3×2}$$ =$$\frac{10}{-6}$$
$$\frac{5}{-3}$$ =$$\frac{5×3}{-3×3}$$ = $$\frac{15}{-9}$$
$$\frac{5}{-3}$$ =$$\frac{5×4}{-3×4}$$ = $$\frac{20}{-12}$$
$$\frac{5}{-3}$$ =$$\frac{5×5}{-3×5}$$ = $$\frac{25}{-15}$$
(iii) $$\frac{4}{9}$$ = $$\frac{4×2}{9×2}$$ =$$\frac{8}{18}$$
$$\frac{4}{9}$$ =$$\frac{4×3}{9×3}$$ = $$\frac{12}{27}$$
$$\frac{4}{9}$$ =$$\frac{4×4}{9×4}$$ = $$\frac{16}{36}$$
$$\frac{4}{9}$$ =$$\frac{4×5}{9×5}$$ = $$\frac{20}{45}$$

Q.4 Draw a number line and represent the following rational numbers on it:
(i) 3/4
(ii) -5/8
(iii) -7/4
(iv) 7/8

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

(i)
(ii)
(iii)
(iv)

Q.6 Which of the following pairs represent the same rational number?
(i) $$\frac{-7}{21}$$ and $$\frac{3}{9}$$
(ii) $$\frac{-16}{20}$$ and $$\frac{20}{-25}$$
(iii)$$\frac{-2}{-3}$$ and $$\frac{2}{3}$$
(iv) $$\frac{-3}{5}$$ and $$\frac{-12}{20}$$
(v) $$\frac{8}{-5}$$ and $$\frac{-24}{15}$$
(vi) $$\frac{1}{3}$$ and $$\frac{-1}{9}$$
(vii) $$\frac{-5}{-9}$$ and $$\frac{5}{-9}$$

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

(i) $$\frac{-7}{21}$$ and $$\frac{3}{9}$$
Or, $$\frac{-7×3}{21×3}$$ and $$\frac{3×7}{9×7}$$
Or, $$\frac{-21}{63}$$ and $$\frac{21}{63}$$
Since -21$$\neq$$ 21 so $$\frac{-7}{21}$$ and $$\frac{3}{9}$$do not represent the same rational number
(ii) $$\frac{-16}{20}$$ and $$\frac{20}{-25}$$
Or, $$\frac{-16×5}{20×5}$$ and $$\frac{20×4}{25×4}$$
Or, $$\frac{-80}{100}$$ and $$\frac{80}{100}$$
Since -80$$\neq$$ 80 so $$\frac{-16}{20}$$ and $$\frac{20}{-25}$$do not represent the same rational number
(iii) $$\frac{-2}{-3}$$ and $$\frac{2}{3}$$
Or, $$\frac{2}{3}$$ and $$\frac{2}{3}$$
so$$\frac{2}{3}$$ and $$\frac{2}{3}$$ represent the same rational number
(iv) $$\frac{-3}{5}$$ and $$\frac{-12}{20}$$
Or, $$\frac{-3×4}{5×4}$$ and $$\frac{-12×1}{20×1}$$
Or, $$\frac{-12}{20}$$ and $$\frac{-12}{20}$$
Since $$\frac{-12}{20}$$ = $$\frac{-12}{20}$$
so $$\frac{-3}{5}$$ and $$\frac{-12}{20}$$ represent the same rational number

Q.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.

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

Rational numbers represented by P, Q, R and S. $$\frac{7}{3}$$,$$\frac{8}{3}$$,$$\frac{-4}{3}$$and $$\frac{?-5}{3}$$ respectlvely.

Q.7 Rewrite the following rational numbers in the simplest form:
(i) $$\frac{-8}{6}$$ (ii) $$\frac{-25}{45}$$ (iii) $$\frac{-44}{72}$$

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

(i) $$\frac{-8}{6}$$= $$\frac{-8÷2}{6÷2}$$= $$\frac{-4}{3}$$
(ii) $$\frac{25}{45}$$= $$\frac{25÷5}{45÷5}$$= $$\frac{5}{9}$$
(iii) $$\frac{-44}{72}$$= $$\frac{-44÷4}{72÷4}$$= $$\frac{-11}{18}$$

Q.8 Fill in the boxes with the correct symbol out of >, <
(i) $$\frac{-5}{7}$$ $$\square$$ $$\frac{2}{3}$$
(ii) $$\frac{-4}{5}$$ $$\square$$ $$\frac{-5}{7}$$
(iii) $$\frac{-7}{8}$$ $$\square$$ $$\frac{-14}{16}$$
(iv) $$\frac{-8}{5}$$ $$\square$$ $$\frac{-7}{4}$$
(v) $$\frac{1}{-3}$$ $$\square$$ $$\frac{-1}{4}$$
(vi) $$\frac{5}{-11}$$ $$\square$$ $$\frac{-5}{11}$$
(vii)0 $$\square$$ $$\frac{-7}{6}$$

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

(i) $$\frac{-5}{7}$$ < $$\frac{2}{3}$$
(ii) $$\frac{-4}{5}$$ > $$\frac{-5}{7}$$
(iii) $$\frac{-7}{8}$$ =$$\frac{-14}{16}$$
(iv) $$\frac{-8}{5}$$ > $$\frac{-7}{4}$$
(v) $$\frac{1}{-3}$$ < $$\frac{-1}{4}$$
(vi) $$\frac{5}{-11}$$ = $$\frac{-5}{11}$$
(vii)0 > $$\frac{-7}{6}$$

Q.9 Which is greater in each of the following:
(i) $$\frac{2}{3}$$ , $$\frac{5}{2}$$
(ii) $$\frac{-5}{6}$$ , $$\frac{-4}{3}$$
(iii) $$\frac{-3}{4}$$ , $$\frac{2}{-3}$$
(iv) $$\frac{-1}{4}$$ , $$\frac{1}{4}$$
(v) $$-3\frac{2}{7}$$ , $$-3\frac{4}{5}$$

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

(i) $$\frac{2}{3}$$ = $$\frac{2×2}{3×2}$$=$$\frac{4}{6}$$
$$\frac{5}{2}$$ = $$\frac{5×3}{2×3}$$ = $$\frac{15}{6}$$
Since, $$\frac{2}{3}$$ < $$\frac{5}{2}$$
(ii) $$\frac{-5}{6}$$ = $$\frac{-5×1}{6×1}$$=$$\frac{-5}{6}$$
$$\frac{-4}{3}$$ = $$\frac{-4×2}{3×2}$$ = $$\frac{-8}{6}$$
Since, $$\frac{-5}{6}$$ > $$\frac{-4}{3}$$
(iii) $$\frac{-3}{4}$$ = $$\frac{-3×3}{4×3}$$=$$\frac{-9}{12}$$
$$\frac{2}{-3}$$ = $$\frac{2×4}{-3×4}$$ = $$\frac{8}{-12}$$
Since, $$\frac{-3}{4}$$ < $$\frac{2}{-3}$$
(iv) $$\frac{-1}{4}$$ < $$\frac{1}{4}$$
(v) $$-3\frac{2}{7}$$ = $$\frac{-23}{7}$$ = $$\frac{-23×5}{7×5}$$ = $$\frac{-115}{35}$$
$$-3\frac{4}{5}$$ = $$\frac{-19}{5}$$ =$$\frac{-19×7}{5×7}$$= $$\frac{-133}{35}$$
Since, $$-3\frac{2}{7}$$ > $$-3\frac{4}{5}$$

Q.10 Write the following rational numbers in ascending order:
(i) -3/5, -2/5, -1/5
(ii) -1/3, -2/9, -4/3
(iii) -3/7, -3/2, -3/4

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

(i) -3/5, -2/5, -1/5
Here denominator are same then,
Hence, -3/5 < -2/5 < -1/5
(ii) -1/3, -2/9, -4/3
LCM of denominator = 9
(-1×3)/(3×3), (-2×1)/9×1), (-4×3)/(3×3)
=-2/9 > -3/9 > -12/9
= -2/9>-3/9>-4/3
(iii) -3/7, -3/2, -3/4
LCM of denominator = 28
(-3×4)/(7×4), (-3×14)/(2×14), (-3×7)/(4×7)
= -12/28, -42/28, -21/28
= -12/28 < -21/28 < -42/28
= -3/7< -3/4 < -3/2

## NCERT solutions for class 7 Maths Chapter 9 Rational Numbers Exercise 9.2

Q.1 Find the sum:
(i) $$\frac{5}{4}$$ +( $$\frac{-11}{4}$$)
(ii) $$\frac{5}{3}$$ + $$\frac{3}{5}$$
(iii) $$\frac{-9}{10}$$ + $$\frac{22}{15}$$
(iv) $$\frac{-3}{-11}$$ + $$\frac{5}{9}$$
(v) $$\frac{-8}{-19}$$ + $$\frac{(-2)}{57}$$
(vi) $$\frac{-2}{3}$$ + 0
(vii) $$-2\frac{1}{3}$$ + $$4\frac{3}{5}$$

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

(i) $$\frac{5}{4}$$ +( $$\frac{-11}{4}$$) =$$\frac{5}{4}$$ -$$\frac{11}{4}$$ = $$\frac{5-11}{4}$$= $$\frac{-6}{4}$$
(ii) $$\frac{5}{3}$$ + $$\frac{3}{5}$$ = $$\frac{5×5}{3×5}$$ + $$\frac{3×3}{5×3}$$
= $$\frac{25}{15}$$ + $$\frac{9}{15}$$ = $$\frac{25+9}{15}$$ =$$\frac{25+9}{15}$$ =$$\frac{34 }{15}$$ = $$2\frac{4}{15}$$
(iii) $$\frac{-9}{10}$$ + $$\frac{22}{15}$$ = $$\frac{-9×3}{10×3}$$ + $$\frac{22×2}{15×2}$$=$$\frac{-27}{30}$$ + $$\frac{44}{30}$$ = $$\frac{-27+44}{30}$$ =$$\frac{17}{30}$$
(iv) $$\frac{-3}{-11}$$ + $$\frac{5}{9}$$ = $$\frac{3×9}{11×9}$$ + $$\frac{5×11}{9×11}$$ =$$\frac{27}{99}$$ + $$\frac{55}{99}$$ = $$\frac{27+55}{99}$$ =$$\frac{82}{99}$$
(v) $$\frac{-8}{-19}$$ + $$\frac{(-2)}{57}$$ =$$\frac{8}{19}$$ - $$\frac{2}{57}$$ = $$\frac{8×3}{19×3}$$ - $$\frac{2}{57}$$=$$\frac{24-2}{57}$$ =$$\frac{22}{57}$$
(vi) $$\frac{-2}{3}$$ + 0 = $$\frac{-2}{3}$$
(vii) $$-2\frac{1}{3}$$ + $$4\frac{3}{5}$$= $$\frac{-7}{3}$$ + $$\frac{23}{5}$$
= $$\frac{-7×5}{3×5}$$ + $$\frac{23×3}{5×3}$$ = $$\frac{-35}{15}$$ + $$\frac{69}{15}$$
= $$\frac{-35+69}{15}$$=$$\frac{34}{15}$$=$$2\frac{4}{15}$$

Q.2 Find :
(i) $$\frac{7}{24}$$ - $$\frac{17}{36}$$
(ii) $$\frac{5}{63}$$ - $$(-\frac{6}{21})$$
(iii) $$\frac{-6}{13}$$ - $$(-\frac{7}{15})$$
(iv) $$\frac{-3}{8}$$ - $$\frac{7}{11}$$
(v) $$-2\frac{1}{9}$$ - 6

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

(i) $$\frac{7}{24}$$ - $$\frac{17}{36}$$ = $$\frac{7×3}{24×3}$$ - $$\frac{17×2}{36×2}$$ =$$\frac{21}{72}$$ - $$\frac{34}{72}$$ = $$\frac{21-34}{72}$$ =$$\frac{-13}{72}$$
(ii) $$\frac{5}{63}$$ + $$\frac{6}{21}$$ = $$\frac{5}{63}$$ + $$\frac{6×3}{21×3}$$= $$\frac{5+18}{63}$$= $$\frac{23}{63}$$
(iii) $$\frac{-6}{13}$$ - $$(-\frac{7}{15})$$= $$\frac{-6}{13}$$ + $$\frac{7}{15}$$
= $$\frac{-6×15}{13×15}$$ + $$\frac{7×13}{15×13}$$ = $$\frac{-90}{195}$$ + $$\frac{91}{195}$$ = $$\frac{-90+ 91}{195}$$ = $$\frac{1}{195}$$
(iv) $$\frac{-3}{8}$$ - $$\frac{7}{11}$$ = $$\frac{-3×11}{8×11}$$ - $$\frac{7×8}{11×8}$$
=$$\frac{-33}{88}$$ - $$\frac{56}{88}$$= $$\frac{-33-56}{88}$$ =$$\frac{-89}{88}$$= $$-1\frac{1}{88}$$
(v) $$\frac{-19}{9}$$ - 6 = $$\frac{-19-54}{9}$$ = $$\frac{-73}{9}$$ = $$-8\frac{1}{9}$$

Q.3 Find the product:
(i) $$\frac{9}{2}$$ × $$(-\frac{7}{4} )$$
(ii) $$\frac{9}{2}$$ × (-9 )
(iii) $$\frac{-6}{5}$$× $$\frac{9}{11}$$
(iv) $$\frac{3}{7}$$ × $$(-\frac{2}{5} )$$
(v) $$\frac{3}{11}$$× $$\frac{2}{5}$$
(vi) $$\frac{3}{-5}$$× $$\frac{-5}{3}$$

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

(i) $$\frac{9}{2}$$× $$\frac{-7}{4}$$ = $$\frac{9×-7}{2×4}$$= $$\frac{-63}{8}$$ = $$-7\frac{7}{8}$$
(ii) $$\frac{9}{2}$$ × (-9 )= $$\frac{-81}{2}$$ =$$-40\frac{1}{2}$$
(iii) $$\frac{-6}{5}$$× $$\frac{9}{11}$$ = $$\frac{-6×9}{5×11}$$= $$\frac{-54}{55}$$
(iv) $$\frac{3}{7}$$ × $$(-\frac{2}{5} )$$= $$\frac{-3×2}{7×5}$$ = $$\frac{-6}{35}$$
(v) $$\frac{3}{11}$$× $$\frac{2}{5}$$ = $$\frac{3×2}{11×5}$$=$$\frac{6}{55}$$
(vi) $$\frac{3}{-5}$$× $$\frac{-5}{3}$$ = $$\frac{3×(-5)}{-5×3}$$ =$$\frac{-15}{-15}$$= 1

Q.4 Find the value of:
(i) -4 ÷ $$\frac{2}{3}$$
(ii) $$\frac{-3}{5}$$ ÷ 2
(iii) $$\frac{-4}{5}$$ ÷ (-3)
(iv) $$\frac{-1}{8}$$ ÷ $$\frac{3}{4}$$
(v) $$\frac{-2}{13}$$ ÷ $$\frac{1}{7}$$
(vi) $$\frac{-7}{12}$$ ÷ $$(\frac{-2}{13})$$
(vii) $$\frac{3}{13}$$ ÷ $$(\frac{-4}{65})$$

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

(i) -4 ÷ $$\frac{2}{3}$$= -4 × $$\frac{3}{2}$$ = $$\frac{-12}{2}$$= -6
(ii) $$\frac{-3}{5}$$ ÷ 2 = $$\frac{-3}{5}$$ ×$$\frac{1}{2}$$= $$\frac{-3}{10}$$
(iii) $$\frac{-4}{5}$$ ÷ (-3) = $$\frac{-4}{5}$$× $$\frac{-1}{3}$$ =$$\frac{4}{15}$$
(iv) $$\frac{-1}{8}$$ ÷ $$\frac{3}{4}$$ = $$\frac{-1}{8}$$ × $$\frac{4}{3}$$ = $$\frac{-4}{24}$$ = $$\frac{-1}{6}$$
(v) $$\frac{-2}{13}$$ ÷ $$\frac{1}{7}$$ = $$\frac{-2}{13}$$×7 = $$\frac{-14}{13}$$=$$-1\frac{1}{13}$$
(vi) $$\frac{-7}{12}$$ ÷ $$(\frac{-2}{13})$$= $$\frac{-7}{12}$$ × $$(\frac{13}{-2})$$=$$\frac{-91}{-24}$$ = $$3\frac{19}{24}$$
(vii) $$\frac{3}{13}$$ ÷ $$(\frac{-4}{65})$$ = $$\frac{3}{13}$$ × $$(\frac{65}{-4})$$ = $$\frac{195}{-52}$$
= $$\frac{195÷13}{-52÷13}$$=$$\frac{15}{-4}$$=$$-3\frac{3}{4}$$

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