Class 7 Rational Numbers Exercise 4.6

\begin{array}{l} \text {Q1. Draw the number line and represent the following rational numbers on it: } \\ \text {(i) } \quad \frac{2}{3} \quad \text {(ii) } \quad \frac{3}{4} \quad \text {(iii) } \quad \frac{3}{8} \quad \text {(iv) } \quad \frac{-5}{8} \\ \text {(v) } \quad \frac{-3}{16} \quad \text {(vi) } \quad \frac{-7}{3} \quad \text {(vii) } \quad \frac{22}{-7} \quad \text {(viii) } \quad \frac{-31}{3} \end{array}\begin{array}{l} \text {Sol . } \\ \text {(i) We know that } \frac {2}{3} \text { is greater than 2 and less than 3. } \\ \therefore \quad \frac {2}{3} \text { lies between 2 and 3. It can be represented on number line as} \\ \end{array}\begin{array}{cccccccc} \hline -2 & -1 & & 0 & \frac {2}{3} & 1 & & 2 \end{array}\begin{array}{l} \text {(ii) We know that } \frac{3}{4} \text { is greater than 0 and less than 1. } \\ \therefore \quad \frac{3}{4} \text { lies between 0 and 1. It can be represented on number line as } \\ \end{array}\begin{array}{ccccc} \hline -1 & & 0 & \frac{3}{4} & 1 \end{array}\begin{array}{l} \text {(iii) We know that } \frac{3}{8} \text { is greater than 0 and less than 1. } \\ \therefore \quad \frac{3}{8} \text { lies between 0 and 1. It can be represented on number line as} \\ \end{array}\begin{array}{llll} \hline -1 & 0 & \frac{3}{8} & 1 \end{array}\begin{array}{l} \text {(iv) We know that } \frac{-5}{8} \text { is greater than -1 and less than 0. } \\ \therefore \quad \frac{-5}{8} \text { lies between -1 and 0. It can be represented on number line as} \\ \end{array}\begin{array}{llll} \hline -1 & \frac{-5}{8} & 0 & 1 \end{array}\begin{array}{l} \text {(v) We know that } \frac{-3}{16} \text { is greater than -1 and less than 0. } \\ \therefore \quad \frac{-3}{16} \text { lies between -1 and 0. It can be represented on number line as } \\ \end{array}\begin{array}{llll} \hline -1 & \frac{-3}{16} & 0 & 1 \end{array}\begin{array}{l} \text {(vi) We know that } \frac{-7}{3} \text { is greater than -3 and less than -2. } \\ \therefore \quad \frac{-7}{3} \text { lies between -3 and -2. It can be represented on number line as} \\ \end{array}\begin{array}{llll} \hline -3 & \frac{-7}{3} & -2 & -1 \end{array}\begin{array}{l} \text {(vii) We know that } \frac{22}{-7} \text { is greater than -4 and less than -3. } \\ \therefore \quad \frac{22}{-7} \text { lies between -4 and -3. It can be represented on number line as} \\ \end{array}\begin{array}{llll} \hline -4 & \frac{22}{-7} & -3 & -2 \end{array}\begin{array}{l} \text {(viii) We know that } \frac{-31}{3} \text { is greater than -11 and less than -10. } \\ \therefore \quad \frac{-31}{3} \text { lies between -11 and -10. It can be represented on number line as} \\ \end{array}\begin{array}{llll} \hline -11 & \frac{-31}{3} & -10 & -9 \end{array}\begin{array}{l} \text {Q2. Which of the two rational numbers in each of the following pairs of rational numbers is greater? } \\ \text {(i) } \quad \frac{-3}{8}, 0 \quad \text {(ii) } \quad \frac{5}{2}, 0 \quad \text {(iii) } \quad \frac{-4}{11}, \frac{3}{11} \\ \text {(iv) } \quad \frac{-7}{12}, \frac{5}{-8} \quad \text {(v) } \quad \frac{4}{-9}, \frac{-3}{-7} \quad \text {(vi) } \quad \frac{-5}{8}, \frac{3}{-4} \\ \text {(vii) } \quad \frac{5}{9}, \frac{-3}{-8} \quad \text {(viii) } \quad \frac{5}{-8}, \frac{-7}{12} \end{array}\begin{array}{l} \text {Sol. } \\ \text {(i) We know that every positive rational number is greater than zero and } \\ \text {every negative rational number is smaller than zero. } \therefore \quad \frac{-3}{8} \lt 0 \\ \\\text {(ii) Every positive rational number is greater than zero and every negative rational number is smaller than zero. } \\ \therefore \quad \frac{5}{2} \gt 0 \\ \\\text {(iii) As we know that that every positive rational number is greater than zero and every negative rational number} \\ \therefore \quad \frac{-4}{8} \lt \frac{3}{11} \\ \\\text {(iv) LCM of denominators 12 and 8 is 24 } \\ \text {Lets write the rational numbers so that they have common denominators as 24 so we have } \\ \frac{-7}{12}=\frac{-7 \times 2}{12 \times 2}=\frac{-14}{24} \\\text {And } \quad \frac{5}{-8}=\frac{-5 \times 3}{8 \times 3}=\frac{-15}{24} \\ \therefore \quad \frac{-7}{12} \gt \frac{5}{-8} \\ \\\text {(v) } \quad \frac{-3}{-7} = \frac{3}{7} \\ \text {We know that a positive number is greater than a negative number. } \\\therefore \quad \frac{4}{-9} \lt \frac{-3}{-7} \\ \\\text {(vi) LCM of denominators 4 and 8 is 8 } \\ \text {Lets write the rational numbers so that they have common denominators as 8 so we have } \\ \frac{3}{-4}=\frac{-3 \times 2}{4 \times 2}=\frac{-6}{8} \\ \therefore \quad \frac{-5}{8} \gt \frac{3}{-4} \\ \\\text {(vii) LCM of denominators 9 and 8 is 72 } \\ \text {Lets write the rational numbers so that they have common denominators as 72 so we have } \\ \frac{5}{9}=\frac{5 \times 8}{9 \times 8}=\frac{40}{72} \\ \text {And } \quad \frac{-3}{-8}=\frac{3 \times 9}{8 \times 9}=\frac{27}{72} \\ \therefore \quad \frac{5}{9} \gt \frac{-3}{-8} \\ \\\text {(viii) LCM of denominators 12 and 8 is 24 } \\ \text {Lets write the rational numbers so that they have common denominators as 24 so we have } \\ \frac{-7}{12}=\frac{-7 \times 2}{12 \times 2}=\frac{-14}{24} \\\text {And } \quad \frac{5}{-8}=\frac{-5 \times 3}{8 \times 3}=\frac{-15}{24} \\ \therefore \quad \frac{-7}{12} \gt \frac{5}{-8} \\ \end{array}\begin{array}{l} \text {Q3. Which of the two rational numbers in each of the following pairs of rational numbers is smaller? } \text {(i) } \quad \frac{-6}{-13}, \frac{7}{13} \quad \text {(ii) } \quad \frac{16}{-5}, 3 \\ \text {(iii) } \quad \frac{-4}{3}, \frac{8}{-7} \quad \text {(iv) } \quad \frac{-12}{5},-3\end{array}\begin{array}{l} \text {Sol. } \\ \text {(i) Here denomintors of both the rational numbers are same.} \\ \therefore \quad \frac{6}{13} \lt \frac{7}{13} \\ \Rightarrow \quad \frac{-6}{-13} \lt \frac{7}{13} \\ \\\text {(i) Every positive mumber is greater than a negative mumber.} \\ \therefore \quad \frac{16}{-5} \lt 3 \\ \\\text {(iii) LCM of denominators 3 and 7 is 21 } \\ \text {Lets write the rational numbers so that they have common denominators as 21 so we have } \\\frac{-4}{3}=\frac{-4 \times 7}{3 \times 7}=\frac{-28}{21} \\ \text {And } \quad \frac{8}{-7}=\frac{-8 \times 3}{7 \times 3}=\frac{-24}{21} \\ \therefore \quad \frac{-28}{21} \lt \frac{-24}{21} \\ \Rightarrow \quad \frac{-4}{3} \lt \frac{8}{-7} \\ \\\text {(iv) LCM of denominators 5 and 1 is 5 } \\ \text {Lets write the rational numbers so that they have common denominators as 5 so we have } \\-3=\frac{-3 \times 5}{1 \times 5}=\frac{-15}{5} \\\therefore \quad \frac{-12}{5} \gt \frac{-15}{5} \\ \Rightarrow \quad \frac{-12}{5} \gt -3 \\ \end{array}\begin{array}{l} \text {Q4. Fill in the blanks by the correct symbol out of } \gt, = , \lt \\ \text {(i) } \quad \frac{-6}{7} \ldots \ldots \frac{7}{13} \quad \text {(ii) } \quad \frac{-3}{5} \ldots \ldots \frac{-5}{6} \\ \text {(iii) } \quad \frac{-2}{3} \ldots . \frac{5}{-8} \quad \text {(iv) } \quad 0 . . . . \frac{-2}{5} \end{array}\begin{array}{l} \text {Sol. } \\ \text {(i) Every positive mumber is greater than a negative mumber} \\ \therefore \quad \frac{-6}{7} \lt \frac{7}{13} \\ \\\text {(ii) LCM of denominators 5 and 6 is 30 } \\ \text {Lets write the rational numbers so that they have common denominators as 30 so we have } \\\frac{-3}{5}= \frac{-3 \times 6}{5 \times 6} = \frac{-18}{30} \\ \frac{-5}{6}= \frac{-5 \times 5}{6 \times 5} = \frac{-25}{30} \\ \therefore \quad \frac{-18}{30} \gt \frac{-25}{30} \\ \Rightarrow \quad \frac{-3}{5} \gt \frac{-5}{6} \\ \\\text {(iii) LCM of denominators 3 and 8 is 24 } \\ \text {Lets write the rational numbers so that they have common denominators as 24 so we have } \\\frac{-2}{3} = \frac{-2 \times 8}{3 \times 8} = \frac{-16}{24} \\\text {And } \quad \frac{5}{-8} = \frac{5 \times 3}{-8 \times 3} = \frac{15}{-24} =\frac{-15}{24} \\\therefore \quad \frac{-16}{24} \lt \frac{-15}{24} \\ \Rightarrow \quad \frac{-2}{3} \lt \frac{5}{-8} \\ \\\text {(iv) Zero is greater than a negative mumber} \\ \therefore \quad 0 \gt \frac{-2}{5} \end{array}\begin{array}{l} \text {Q5. Arrange the following rational numbers in ascending order: } \\ \text {(i) } \quad \frac{3}{5}, \frac{-17}{-30}, \frac{8}{-15}, \frac{-7}{10} \\ \text {(ii) } \quad \frac{-4}{9}, \frac{5}{-12}, \frac{7}{-18}, \frac{2}{-3} \end{array}\begin{array}{l} \text {Sol. } \\ \text {(i) LCM of denominators 5,30, 15 and 10 is 30 } \\ \text {Lets write the rational numbers so that they have common denominators as 30 so we have } \\\frac{3}{5}=\frac{3 \times 6}{5 \times 6}=\frac{18}{30} \\\frac{-17}{-30}= \frac{17}{30}=\frac{17 \times 1}{30 \times 1}=\frac{17}{30} \\\frac{8}{-15}=\frac{-8 \times 2}{15 \times 2}=\frac{-16}{30} \\\frac{-7}{10}=\frac{-7 \times 3}{10 \times 3}=\frac{-21}{30} \\\text {Lets arrange the numerator of each rational number in ascending order, we have } \\ -21 \lt -16 \lt 17 \lt 18 \\ \therefore \quad \frac{-21}{30} \lt \frac{-16}{30} \lt \frac{17}{30} \lt \frac{18}{30} \\\Rightarrow \quad \frac{-7}{10} \lt \frac{8}{-15} \lt \frac{17}{30} \lt \frac{3}{5} \\ \\\text {(ii) LCM of denominators 9,12,18 and 3 is 36 } \\ \text {Lets write the rational numbers so that they have common denominators as 36 so we have } \\\frac{-4}{9}=\frac{-4 \times 4}{9 \times 4}=\frac{-16}{36} \\\frac{5}{-12}=\frac{-5}{12}=\frac{-5 \times 3}{12 \times 3}=\frac{-15}{36} \\\frac{7}{-18}=\frac{-7}{18}=\frac{-7 \times 2}{18 \times 2}=\frac{-14}{36} \\\frac{2}{-3}=\frac{-2}{3}=\frac{-2 \times 12}{3 \times 12}=\frac{-24}{36} \\\text {Lets arrange the numerator of each rational number in ascending order, we have } \\-24 \lt -16 \lt -15 \lt -14 \\ \therefore \quad \frac{-24}{36} \lt \frac{-16}{36} \lt \frac{-15}{36} \lt \frac{-14}{36} \\ \Rightarrow \quad \frac{2}{-3} \lt \frac{-4}{9} \lt \frac{5}{-12} \lt \frac{7}{-18} \\ \end{array}\begin{array}{l} \text {Q6. Arrange the following rational numbers in descending order: } \\ \text {(i) } \quad \frac{7}{8}, \frac{64}{16}, \frac{36}{-12}, \frac{5}{-4}, \frac{140}{28} \\ \text {(ii) } \quad \frac{-3}{10}, \frac{17}{-30}, \frac{7}{-15}, \frac{-11}{20} \end{array}\begin{array}{l} \text {Sol. } \\ \text {(i) LCM of denominators 8,16,12,4 and 28 is 336 } \\ \text {Lets write the rational numbers so that they have common denominators as 336 so we have } \\\frac{7}{8}=\frac{7 \times 42}{8 \times 42}=\frac{294}{336} \\\frac{64}{16}=\frac{64 \times 21}{16 \times 21}=\frac{1344}{336} \\\frac{36}{-12}=\frac{-36 \times 28}{12 \times 28}=\frac{-1008}{336}\\\frac{5}{-4}=\frac{-5 \times 84}{4 \times 84}=\frac{-420}{336}\\\frac{140}{28}=\frac{140 \times 12}{28 \times 12}=\frac{1680}{336}\\\text {Lets arrange the numerator of each rational number in descending order, we have } \\1680 \gt 1344 \gt 294 \gt -420 \gt -1008 \\ \therefore \quad \frac{1680}{336} \gt \frac{1344}{336} \gt \frac{294}{336} \gt \frac{-420}{336} \gt \frac{-1008}{336} \\\Rightarrow \quad \frac{140}{28} \gt \frac{64}{16} \gt \frac{7}{8} \gt \frac{5}{-4} \gt \frac{36}{-12} \\ \\\text {(ii) LCM of denominators 10,30,15 and 20 is 60. } \\ \text {Lets write the rational numbers so that they have common denominators as 60 so we have } \\\frac{-3}{10}=\frac{-3 \times 6}{10 \times 6}=\frac{-18}{60} \\\frac{17}{-30}=\frac{-17}{30}=\frac{-17 \times 2}{30 \times 2}=\frac{-34}{60}\\\frac{7}{-15}=\frac{-7}{15}=\frac{-7 \times 4}{15 \times 4}=\frac{-28}{60}\\\frac{-11}{20}=\frac{-11 \times 3}{20 \times 3}=\frac{-33}{60} \\\text {Lets arrange the numerator of each rational number in descending order, we have } \\ -18 \gt -28 \gt -33 \gt -34 \\\therefore \quad \frac{-18}{60} \gt \frac{-28}{60} \gt \frac{-33}{60} \gt \frac{-34}{60} \\\Rightarrow \quad \frac{-3}{10} \gt \frac{7}{-15} \gt \frac{-11}{20} \gt \frac{17}{-30} \\ \\\end{array}\begin{array}{l} \text {Q7. Which of the following statements are true: } \\ \text {(i) The rational number } \frac{29}{23} \text { lies to the left of zero on the number line. } \text {(ii) The rational number } \frac{-12}{-17} \text { lies to the left of zero on the number line. } \\ \text {(iii) The rational number } \frac{3}{4} \text { lies to the right of zero on the number line } \\ \text {(iv) The rational numbers } \frac{-12}{-5} \text { and } \frac{-7}{17} \\ \text { are on the opposite side of zero on the number line. } \text {(v) The rational numbers } \frac{-21}{5} \text { and } \frac{7}{-31} \text { are on the opposite side } \\ \text { of zero on the number line. } \\ \text {(vi) The rational number } \frac{-3}{-5} \text { is on the right of } \frac{-4}{7} \text { on the number line.} \end{array}\begin{array}{l} \text {Sol. } \\ \text {(i) False. } \frac{29}{23} \text { lies to the right of zero because it is a positive number.} \\ \\\text {(ii) False. } \frac{-12}{-17} = \frac{12}{17} \text { lies to the right of zero because it is a positive number. } \\ \\\text {(iii) True } \\ \\\text {(iv) True. Because negative and positive numbers are on the opposite side of zero.} \\ \\\text {(v) False. Becuase same sign number lies on same side of zero and both are negative numbers.} \\ \\\text {(vi) True. Positive number is greater than the negative number. } \\ \text {Thus, it is on the right of the negative number. } \\ \end{array}