# 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.
= $\frac{\left(?3×1\right)}{\left(5×1\right)}$, $\frac{\left(?3×2\right)}{\left(5×2\right)}$, $\frac{\left(?3×3\right)}{\left(5×3\right)}$, $\frac{\left(?3×4\right)}{\left(5×4\right)}$
Then, next four rational numbers in this pattern are,
= $\frac{\left(?3×5\right)}{\left(5×5\right)}$, $\frac{\left(?3×6\right)}{\left(5×6\right)}$ , $\frac{\left(?3×7\right)}{\left(5×7\right)},\frac{\left(?3×8\right)}{\left(5×8\right)}$
= -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$?$ 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$?$ 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) $\frac{?5}{7}$ $?$ $\frac{2}{3}$
(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 $?$ $\frac{?7}{6}$

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) $\frac{?8}{?19}$ + $\frac{\left(?2\right)}{57}$
(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) $\frac{?8}{?19}$ + $\frac{\left(?2\right)}{57}$ =$\frac{8}{19}$ - $\frac{2}{57}$ = $\frac{8×3}{19×3}$ - $\frac{2}{57}$=$\frac{24?2}{57}$ =$\frac{22}{57}$
(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) $\frac{5}{63}$ - $\left(?\frac{6}{21}\right)$
(iii) $\frac{?6}{13}$ - $\left(?\frac{7}{15}\right)$
(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) $\frac{?6}{13}$ - $\left(?\frac{7}{15}\right)$= $\frac{?6}{13}$ + $\frac{7}{15}$
= $\frac{?6×15}{13×15}$ + $\frac{7×13}{15×13}$ = $\frac{?90}{195}$ + $\frac{91}{195}$ = $\frac{?90+91}{195}$ = $\frac{1}{195}$
(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) $\frac{9}{2}$ × $\left(?\frac{7}{4}\right)$
(ii) $\frac{9}{2}$ × (-9 )
(iii) $\frac{?6}{5}$× $\frac{9}{11}$
(iv) $\frac{3}{7}$ × $\left(?\frac{2}{5}\right)$
(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) $\frac{3}{7}$ × $\left(?\frac{2}{5}\right)$= $\frac{?3×2}{7×5}$ = $\frac{?6}{35}$
(v) $\frac{3}{11}$× $\frac{2}{5}$ = $\frac{3×2}{11×5}$=$\frac{6}{55}$
(vi) $\frac{3}{?5}$× $\frac{?5}{3}$ = $\frac{3×\left(?5\right)}{?5×3}$ =$\frac{?15}{?15}$= 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) $\frac{?7}{12}$ ÷ $\left(\frac{?2}{13}\right)$
(vii) $\frac{3}{13}$ ÷ $\left(\frac{?4}{65}\right)$

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) $\frac{?1}{8}$ ÷