NCERT solutions for class 7 Maths Chapter 9 Rational Numbers

Solution for 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 \(\frac{2}{3} \)


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Answer :

(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 …..


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Answer :

(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


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Answer :

(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


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Answer :

(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 \(\frac{5}{-9} \)


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Answer :

(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.


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Answer :

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) \(\frac{-44}{72} \)


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Answer :

(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 \(\square \) \(\frac{-7}{6} \)


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Answer :

(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) \(-3\frac{2}{7} \) , \(-3\frac{4}{5} \)


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Answer :

(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


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Answer :

(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

Solution for 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) \(-2\frac{1}{3} \) + \(4\frac{3}{5} \)


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Answer :

(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


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Answer :

(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) \(\frac{3}{-5} \)× \(\frac{-5}{3} \)


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Answer :

(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) \(\frac{3}{13} \) ÷ \((\frac{-4}{65}) \)


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Answer :

(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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