NCERT Solutions Class 7 Maths Chapter 8 Rational Numbers

5.0NCERT Questions with Solutions for Class 7 Maths Chapter 8 - Detailed Solutions

Exercise: 1.1

• Using appropriate properties find (i) $-\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}$ (ii) $\frac{2}{5}\times \left(-\frac{3}{7}\right)-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}$ Sol. (i) $-\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}=-\frac{2}{3}\times \frac{3}{5}-\frac{3}{5}\times \frac{1}{6}+\frac{5}{2}$ $$\begin{gathered} =\left(-\frac{3}{5}\right)\times \frac{2}{3}+\left(-\frac{3}{5}\right)\times \frac{1}{6}+\frac{5}{2} \\ =\left(-\frac{3}{5}\right)\times \left(\frac{2}{3}+\frac{1}{6}\right)+\frac{5}{2} \\ =\left(-\frac{3}{5}\right)\times \frac{4+1}{6}+\frac{5}{2}=-\frac{3}{5}\times \frac{5}{6}+\frac{5}{2} \\ =-\frac{3\times 5}{5\times 6}+\frac{5}{2}=\frac{-1}{2}+\frac{5}{2}=\frac{-1+5}{2}=\frac{4}{2}=2. \end{gathered}$$ (ii) $\frac{2}{5}\times \left(-\frac{3}{7}\right)-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}$ $=\frac{2}{5}\times \left(-\frac{3}{7}\right)+\frac{1}{14}\times \frac{2}{5}-\frac{1}{6}\times \frac{3}{2}$ $=\frac{2}{5}\times \left(-\frac{3}{7}\right)+\frac{2}{5}\times \frac{1}{14}-\frac{1}{6}\times \frac{3}{2}$ $=\frac{2}{5}\times \left[\left(-\frac{3}{7}\right)+\frac{1}{14}\right]-\frac{1}{6}\times \frac{3}{2}$ $=\frac{2}{5}\times \left(\frac{-6+1}{14}\right)-\frac{3}{12}$ $=\frac{2}{5}\times \frac{-5}{14}-\frac{1}{4}=\frac{-1}{7}-\frac{1}{4}=\frac{-4-7}{28}=\frac{-11}{28}$.
• Write the additive inverse of each of the following (i) $\frac{2}{8}$ (ii) $\frac{-5}{9}$ (iii) $\frac{-6}{-5}$ (iv) $\frac{2}{-9}$ (v) $\frac{19}{-6}$ Sol. (i) The additive inverse of $\frac{2}{8}$ is $\left(\frac{-2}{8}\right)=\frac{-2}{8}$. (ii) The additive inverse of $\frac{-5}{9}$ is $-\left(\frac{-5}{9}\right)=\frac{5}{9}$. (iii) The additive inverse of $\frac{-6}{-5}$ is $\frac{-6}{5}$. (iv) The additive inverse of $\frac{2}{-9}$ is $\frac{2}{9}$. (v) The additive inverse of $\frac{19}{-6}$ is $\frac{19}{6}$.
• Verify that $-(-\mathrm{x})=\mathrm{x}$ for (i) $x=\frac{11}{15}$ (ii) $\mathrm{x}=-\frac{13}{17}$ Sol. (i) $\mathrm{x}=\frac{11}{15}$ $\therefore -(-\mathrm{x})=-\left(-\frac{11}{15}\right)=\frac{11}{15}=\mathrm{x}$ (ii) $\mathrm{x}=\frac{-13}{17}$ $\therefore -(-\mathrm{x})=-\left[-\left(\frac{-13}{17}\right)\right]=-\left[\frac{13}{17}\right]=-\frac{13}{17}$
• Find the multiplicative inverse of the following: (i) -13 (ii) $\frac{-13}{19}$ (iii) $\frac{1}{5}$ (iv) $\frac{-5}{8}\times \frac{-3}{7}$ (v) $-1\times \frac{-2}{7}$ (vi) -1 Sol. (i) The multiplicative inverse of -13 is $(-13{)}^{-1}=\frac{1}{-13}$ (ii) The multiplicative inverse of $\frac{-13}{19}\text{ is }{\left(\frac{-13}{19}\right)}^{-1}=\frac{19}{-13}=\frac{-19}{13}.$ (iii) The multiplicative inverse of $\frac{1}{5}$ is 5 . (iv) We have, $\frac{-5}{8}\times \frac{-3}{7}=\frac{-5\times -3}{8\times 7}=\frac{15}{56}$ The multiplicative inverse of $\frac{15}{56}$ is ${\left(\frac{15}{56}\right)}^{-1}=\frac{56}{15}$ (v) $-1\times \frac{-2}{7}=\frac{2}{7}$ The multiplicative inverse of $\frac{2}{7}=\frac{7}{2}$ (vi) The multiplicative inverse of -1 is -1 .
• Name the property under multiplication used in each of the following (i) $\frac{-4}{5}\times 1=1\times \frac{-4}{5}=-\frac{4}{5}$ (ii) $-\frac{13}{17}\times \frac{-2}{7}=\frac{-2}{7}\times \frac{-13}{17}$ (iii) $\frac{-19}{29}\times \frac{29}{-19}=1$ Sol. (i) Existence of multiplicative identity. (ii) Commutative property of multiplication. (iii) Existence of multiplicative inverse.
• Multiply $\frac{6}{13}$ by the reciprocal of $\frac{-7}{16}$. Sol. $\frac{6}{13}\times ($ the reciprocal of $\frac{-7}{16})$ $=\frac{6}{13}\times {\left(\frac{-7}{16}\right)}^{-1}$ $=\frac{6}{13}\times \frac{16}{-7}=-\frac{96}{91}$.
• Tell what property allows you to compute $\frac{1}{3}\times \left(6\times \frac{4}{3}\right)$ as $\left(\frac{1}{3}\times 6\right)\times \frac{4}{3}$. Sol. Associative property of multiplication over rational numbers allows us to compute : $\frac{1}{3}\times \left(6\times \frac{4}{3}\right)$ as $\left(\frac{1}{3}\times 6\right)\times \frac{4}{3}$.
• Is $\frac{8}{9}$ the multiplicative inverse of $-1\frac{1}{8}$ ? Why or why not? Sol. No, $\frac{8}{9}$ is not the multiplicative inverse of $-1\frac{1}{8}$. Because $\frac{8}{9}\times \left(-1\frac{1}{8}\right)=\frac{8}{9}\times \frac{-9}{8}=-1\ne 1$.
• Is 0.3 the multiplicative inverse of $3\frac{1}{3}$ ? Why or why not? Sol. Yes, 0.3 is multiplicative inverse of $3\frac{1}{3}$. Because $0.3\times 3\frac{1}{3}=\frac{3}{10}\times \frac{10}{3}=1$.
• Write (i) The rational number that does not have a reciprocal. (ii) The rational numbers that are equal to their reciprocals. (iii) The rational number that is equal to its negative. Sol. (i) We know that there is no rational number which when multiplied with 0 , gives 1. Therefore, the rational number 0 has no reciprocal. (ii) We know that the reciprocal of 1 is 1 and the reciprocal of -1 is -1 . Therefore 1 and -1 are the only rational numbers which are equal to their reciprocals. (iii) The rational number 0 is equal to its negative.
• Fill in the blanks (i) Zero has _____ reciprocal. (ii) The numbers ______ and ______ are their own reciprocals. (iii) The reciprocal of - 5 is ______. (iv) Reciprocal of $\frac{1}{x}$, where $x\ne 0$ is ______ (v) The product of two rational numbers is always a _____. (vi) The reciprocal of a positive rational number is _______. Sol. (i) No (ii) $1,-1$ (iii) $\frac{-1}{5}$ (iv) x (v) Rational number (vi) Positive

Exercise: 1.2

• Represent these numbers on the number line. (i) $\frac{7}{4}$ (ii) $\frac{-5}{6}$ Sol. (i) For $7/4$, we make 7 markings of distance 1 / 4 each on the right of zero and starting from 0 . The seventh marking is 7/4.

• The point $P$ represents the rational number $\frac{7}{4}$. (ii) For $\frac{-5}{6}$, we make 5 markings of distance $\frac{1}{6}$ each on the left of zero and starting from 0 . The fifth marking is $\frac{-5}{6}$. The point $P$ represents the rational number $\frac{-5}{6}$.

• Represent $\frac{-2}{11},\frac{-5}{11},\frac{-9}{11}$ on the number line. Sol. For, $\frac{-2}{11},\frac{-5}{11},\frac{-9}{11}$ we make 11 markings of distance $\frac{1}{11}$ each on the left of zero and starting from 0 . The second marking is $\frac{-2}{11}$. The point B represents the rational number $\frac{-2}{11}$.

• The fifth marking is $\frac{-5}{11}$. The point E represents the rational number $\frac{-5}{11}$. The ninth marking is $\frac{-9}{11}$. The point I represents the rational number $\frac{-9}{11}$.
• Write five rational numbers, which are smaller than 2. Sol. Five rational numbers less than 2 may be taken $1,\frac{1}{2},0,-1,-\frac{1}{2}$ There can be many more such rational numbers.
• Find ten rational numbers between $\frac{-2}{5}$ and $\frac{1}{2}$. Sol. Converting the given rational numbers with the same denominators. $\frac{-2}{5}=\frac{-2\times 4}{5\times 4}=\frac{-8}{20}$ and, $\frac{1}{2}=\frac{1\times 10}{2\times 10}=\frac{10}{20}$ We know that $-8<-7<-6\dots <10$ $\Rightarrow \frac{-8}{20}<\frac{-7}{20}<\frac{-6}{20}<\dots <\frac{10}{20}$ Thus, we have the following ten rational number between $\frac{-2}{5}$ and $\frac{1}{2}$ : $\frac{-7}{20},\frac{-6}{20},\frac{-5}{20},\frac{-4}{20},\frac{-3}{20},\frac{-2}{20},\frac{-1}{20},0,\frac{1}{20}\text{ and }\frac{2}{20}$
• Find five rational numbers between (i) $\frac{2}{3}$ and $\frac{4}{5}$ (ii) $\frac{-3}{2}$ and $\frac{5}{3}$ (iii) $\frac{1}{4}$ and $\frac{1}{2}$ Sol. (i) Converting the given rational numbers with the same denominators $\frac{2}{3}=\frac{2\times 5}{3\times 5}=\frac{10}{15}$ and, $\frac{4}{5}=\frac{4\times 3}{5\times 3}=\frac{12}{15}$ also, $\frac{2}{3}=\frac{10}{15}=\frac{10\times 4}{15\times 4}=\frac{40}{60}$ and, $\frac{4}{5}=\frac{12}{15}=\frac{12\times 4}{15\times 4}=\frac{48}{60}$ We know that $40<41<42<43<44<45<46<47<48$ $\Rightarrow \frac{40}{60}<\frac{41}{60}<\frac{42}{60}<\dots <\frac{47}{60}<\frac{48}{60}$ Thus, we have the following five rational numbers between $\frac{2}{3}$ and $\frac{4}{5}$ $\frac{41}{60},\frac{42}{60},\frac{43}{60},\frac{44}{60}$ and $\frac{45}{60}$. Note: We may take any five numbers given above from $\frac{41}{60}$ to $\frac{47}{60}$. (ii) Converting the given rational numbers with the same denominators $\frac{-3}{2}=\frac{-3\times 3}{2\times 3}=\frac{-9}{6}$ and, $\frac{5}{3}=\frac{5\times 2}{3\times 2}=\frac{10}{6}$ We know that $-9<-8<-7<-6<...<0<1<2<$.... $<8<9<10$ $\Rightarrow \frac{-9}{6}<\frac{-8}{6}<\frac{-7}{6}<\frac{-6}{6}<\dots <\frac{0}{6}<\frac{1}{6}<\frac{2}{6}<\dots$ $<\frac{8}{6}<\frac{9}{6}<\frac{10}{6}$. Thus, we have the following five rational numbers between $\frac{-3}{2}$ and $\frac{5}{3}$ : $\frac{-8}{6},\frac{-7}{6},\frac{0}{6},\frac{1}{6}$ and $\frac{2}{6}$ (iii) Converting the given rational numbers with the same denominators $\frac{1}{4}=\frac{1}{4}\times \frac{6}{6}=\frac{6}{24}$ and $\frac{1}{2}=\frac{1}{2}\times \frac{12}{12}=\frac{12}{24}$ We know that $6<7<8<9<10<11<12$ Thus, we have the following five rational numbers between $\frac{6}{24}$ and $\frac{12}{24}$. $\frac{7}{24},\frac{8}{24},\frac{9}{24},\frac{10}{24},\frac{11}{24}$.
• Write five rational numbers greater than $-2$. Sol. Five rational numbers greater than - 2 may be taken as $-\frac{3}{2},-1,\frac{-1}{2},0,\frac{1}{2}$. There can be many more such rational numbers.
• Find ten rational numbers between $\frac{3}{5}$ and $\frac{3}{4}$. Sol. Converting the given rational numbers with the same denominators $\frac{3}{5}=\frac{3\times 20}{5\times 20}=\frac{60}{100}\text{ and }\frac{3}{4}=\frac{3\times 25}{4\times 25}=\frac{75}{100}$ We know that $60<61<62<63<$... $<72<73<74<75$ $\Rightarrow \frac{60}{100}<\frac{61}{100}<\frac{62}{100}<\frac{63}{100}<\dots$ $<\frac{72}{100}<\frac{73}{100}<\frac{74}{100}<\frac{75}{100}$. Thus, we have the following ten rational numbers between $\frac{3}{5}$ and $\frac{3}{4}$; $\frac{61}{100},\frac{62}{100},\frac{63}{100},\frac{64}{100},\frac{65}{100},\frac{66}{100},\frac{67}{100},\frac{68}{100}$, $\frac{69}{100}$ and $\frac{70}{100}$.