Exercise 8.1 (Revised) - Chapter 9 - Rational Numbers - Ncert Solutions class 7 - Maths

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Chapter 8 - Rational Numbers | NCERT Solutions for Class 7 Maths

Ex 8.1 Question 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 $\frac{2}{3}$

Answer:

(i) -1 and 0
Let us write -1 and 0 as rational numbers with denominator 6 .
$
\begin{aligned}
& \Rightarrow-1=\frac{-6}{6} \text { and } 0=\frac{0}{6} \\
& \therefore \frac{-6}{6}<\frac{-5}{6}<\frac{-4}{6}<\frac{-3}{6}<\frac{-2}{6}<\frac{-1}{6}<0 \\
& \Rightarrow-1<\frac{-5}{6}<\frac{-2}{3}<\frac{-1}{2}<\frac{-1}{3}<\frac{-1}{6}<0
\end{aligned}
$

Therefore, five rational numbers between -1 and 0 would be
$
\frac{-5}{6}, \frac{-2}{3}, \frac{-1}{2}, \frac{-1}{3}, \frac{-1}{6}
$
(ii) -2 and -1

Let us write -2 and -1 as rational numbers with denominator 6 .
$
\Rightarrow-2=\frac{-12}{6} \text { and }-1=\frac{-6}{6}
$

$
\begin{aligned}
& \therefore \frac{-12}{6}<\frac{-11}{6}<\frac{-10}{6}<\frac{-9}{6}<\frac{-8}{6}<\frac{-7}{6}<\frac{-6}{6} \\
& \Rightarrow-2<\frac{-11}{6}<\frac{-5}{3}<\frac{-3}{2}<\frac{-4}{3}<\frac{-7}{6}<-1
\end{aligned}
$

Therefore, five rational numbers between -2 and -1 would be

$
\frac{-11}{6}, \frac{-5}{3}, \frac{-3}{2}, \frac{-4}{3}, \frac{-7}{6}
$
(iii) $\frac{-4}{5}$ and $\frac{-2}{3}$

Let us write $\frac{-4}{5}$ and $\frac{-2}{3}$ as rational numbers with the same denominators.
$
\begin{aligned}
& \Rightarrow \frac{-4}{5}=\frac{-36}{45} \text { and } \frac{-2}{3}=\frac{-30}{45} \\
& \therefore \frac{-36}{45}<\frac{-35}{45}<\frac{-34}{45}<\frac{-33}{45}<\frac{-32}{45}<\frac{-31}{45}<\frac{-30}{45} \\
& \Rightarrow \frac{-4}{5}<\frac{-7}{9}<\frac{-34}{45}<\frac{-11}{15}<\frac{-32}{45}<\frac{-31}{45}<\frac{-2}{3}
\end{aligned}
$

Therefore, five rational numbers between $\frac{-4}{5}$ and $\frac{-2}{3}$ would be
$
\frac{-7}{9}, \frac{-34}{45}, \frac{-11}{15}, \frac{-32}{45}, \frac{-31}{45}, \frac{-2}{3}
$
(iv) $\frac{-1}{2}$ and $\frac{2}{3}$

Let us write $\frac{-1}{2}$ and $\frac{2}{3}$ as rational numbers with the same denominators.
$
\begin{aligned}
& \Rightarrow \frac{-1}{2}=\frac{-3}{6} \text { and } \frac{2}{3}=\frac{4}{6} \\
& \therefore \frac{-3}{6}<\frac{-2}{6}<\frac{-1}{6}<0<\frac{1}{6}<\frac{2}{6}<\frac{3}{6}<\frac{4}{6} \\
& \Rightarrow \frac{-1}{2}<\frac{-1}{3}<\frac{-1}{6}<0<\frac{1}{6}<\frac{1}{3}<\frac{1}{2}<\frac{2}{3}
\end{aligned}
$

Therefore, five rational numbers between $\frac{-1}{2}$ and $\frac{2}{3}$ would be $\frac{-1}{3}, \frac{-1}{6}, 0, \frac{1}{6}, \frac{1}{3}$.
Ex 8.1 Question 2.

Write four more rational numbers in each of the following patterns:

(i) $\frac{-3}{5}, \frac{-6}{10}, \frac{-9}{15}, \frac{-12}{20}, \ldots \ldots \ldots$
(ii) $\frac{-1}{4}, \frac{-2}{8}, \frac{-3}{12}, \ldots \ldots \ldots \ldots$
(iii) $\frac{-1}{6}, \frac{2}{-12}, \frac{3}{-18}, \frac{4}{-24}, \ldots \ldots \ldots$.
(iv) $\frac{-2}{3}, \frac{2}{-3}, \frac{4}{-6}, \frac{6}{-9}, \ldots \ldots \ldots$.

Answer:

(i) $\frac{-3}{5}, \frac{-6}{10}, \frac{-9}{15}, \frac{-12}{20}, \ldots \ldots \ldots$.
$
\Rightarrow \frac{-3 \times 1}{5 \times 1}, \frac{-3 \times 2}{5 \times 2}, \frac{-3 \times 3}{5 \times 3}, \frac{-3 \times 4}{5 \times 4}, \ldots \ldots \ldots .
$

Therefore, the next four rational numbers of this pattern would be
$
\frac{-3 \times 5}{5 \times 5}, \frac{-3 \times 6}{5 \times 6}, \frac{-3 \times 7}{5 \times 7}, \frac{-3 \times 8}{5 \times 8}=\frac{-15}{25}, \frac{-18}{30}, \frac{-21}{35}, \frac{-24}{40}
$
(ii) $\frac{-1}{4}, \frac{-2}{8}, \frac{-3}{12}, \ldots \ldots \ldots \ldots$
$
\Rightarrow \frac{-1 \times 1}{4 \times 1}, \frac{-1 \times 2}{4 \times 2}, \frac{-1 \times 3}{4 \times 3}, \ldots \ldots \ldots \ldots
$

Therefore, the next four rational numbers of this pattern would be
$
\frac{-1 \times 4}{4 \times 4}, \frac{-1 \times 5}{4 \times 5}, \frac{-1 \times 6}{4 \times 6}, \frac{-1 \times 7}{4 \times 7}=\frac{-4}{16}, \frac{-5}{20}, \frac{-6}{24}, \frac{-7}{28}
$
(iii) $\frac{-1}{6}, \frac{2}{-12}, \frac{3}{-18}, \frac{4}{-24}, \ldots \ldots \ldots$.
$
\Rightarrow \frac{-1 \times 1}{6 \times 1}, \frac{1 \times 2}{-6 \times 2}, \frac{1 \times 3}{-6 \times 3}, \frac{1 \times 4}{-6 \times 4}, \ldots \ldots . .
$

Therefore, the next four rational numbers of this pattern would be
$
\frac{1 \times 5}{-6 \times 5}, \frac{1 \times 6}{-6 \times 6}, \frac{1 \times 7}{-6 \times 7}, \frac{1 \times 8}{-6 \times 8}=\frac{5}{-30}, \frac{6}{-36}, \frac{7}{-42}, \frac{8}{-48}
$
(iv) $\frac{-2}{3}, \frac{2}{-3}, \frac{4}{-6}, \frac{6}{-9}, \ldots \ldots \ldots$.
$
\Rightarrow \frac{-2 \times 1}{3 \times 1}, \frac{2 \times 1}{-3 \times 1}, \frac{2 \times 2}{-3 \times 2}, \frac{2 \times 3}{-3 \times 3}, \ldots \ldots \ldots .
$

Therefore, the next four rational numbers of this pattern would be
$
\frac{2 \times 4}{-3 \times 4}, \frac{2 \times 5}{-3 \times 5}, \frac{2 \times 6}{-3 \times 6}, \frac{2 \times 7}{-3 \times 7}=\frac{8}{-12}, \frac{10}{-15}, \frac{12}{-18}, \frac{14}{-21}
$

Ex 8.1 Question 3.

Give four rational numbers equivalent to:

(i) $\frac{-2}{7}$
(ii) $\frac{5}{-3}$
(iii) $\frac{4}{9}$

Answer:

(i) $\frac{-2}{7}$
$
\frac{-2 \times 2}{7 \times 2}=\frac{-4}{14}, \frac{-2 \times 3}{7 \times 3}=\frac{-6}{21}, \frac{-2 \times 4}{7 \times 4}=\frac{-8}{28}, \frac{-2 \times 5}{7 \times 5}=\frac{-10}{35}
$

Therefore, four equivalent rational numbers are $\frac{-4}{14}, \frac{-6}{21}, \frac{-8}{28}, \frac{-10}{35}$.

(ii) $\frac{5}{-3}$
$
\frac{5 \times 2}{-3 \times 2}=\frac{10}{-6}, \frac{5 \times 3}{-3 \times 3}=\frac{15}{-9}, \frac{5 \times 4}{-3 \times 4}=\frac{20}{-12}, \frac{5 \times 5}{-3 \times 5}=\frac{25}{-15}
$

Therefore, four equivalent rational numbers are $\frac{10}{-6}, \frac{15}{-9}, \frac{20}{-12}, \frac{25}{-15}$.
(iii) $\frac{4}{9}$
$
\frac{4 \times 2}{9 \times 2}=\frac{8}{18}, \frac{4 \times 3}{9 \times 3}=\frac{12}{27}, \frac{4 \times 4}{9 \times 4}=\frac{16}{36}, \frac{4 \times 5}{9 \times 5}=\frac{20}{45}
$

Therefore, four equivalent rational numbers are $\frac{8}{18}, \frac{12}{27}, \frac{16}{36}, \frac{20}{45}$.
Ex 8.1 Question 4.

Draw the number line and represent the following rational numbers on it:
(i) $\frac{3}{4}$
(ii) $\frac{-5}{8}$
(iii) $\frac{-7}{4}$
(iv) $\frac{7}{8}$

Answer:

(i) $\frac{3}{4}$

$\text { (ii) } \frac{-5}{8}$

$\text { (iii) } \frac{-7}{4}$

$\text { (iv) } \frac{7}{8}$

Ex 8.1 Question 5.

The points P, Q, R, S, T, U, A and B on the number line are such that, $\mathrm{TR}=\mathrm{RS}=\mathrm{SU}$ and $\mathrm{AP}=\mathrm{PQ}=\mathrm{QB}$. Name the rational numbers represented by $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ and $\mathrm{S}$.

Answer:

Each part which is between the two numbers is divided into 3 parts.
Therefore, $\mathrm{A}=\frac{6}{3}, \mathrm{P}=\frac{7}{3}, \mathrm{Q}=\frac{8}{3}$ and $\mathrm{B}=\frac{9}{3}$
Similarly $T=\frac{-3}{3}, R=\frac{-4}{3}, S=\frac{-5}{3}$ and $U=\frac{-6}{3}$
Thus, the rational numbers represented $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ and $\mathrm{S}$ are $\frac{7}{3}, \frac{8}{3}, \frac{-4}{3}$ and $\frac{-5}{3}$ respectively.
Ex 8.1 Question 6.

Which of the following pairs represent the same rational numbers:
(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 $\frac{5}{-9}$

Answer:

(i) $\frac{-7}{21}$ and $\frac{3}{9}$
$\Rightarrow \frac{-7}{21}=\frac{-1}{3}$ and $\frac{3}{9}=\frac{1}{3}$ [Converting into lowest term]
$\because \frac{-1}{3} \neq \frac{1}{3}$

$
\therefore \frac{-7}{21} \neq \frac{3}{9}
$
(ii) $\frac{-16}{20}$ and $\frac{20}{-25}$
$
\Rightarrow \frac{-16}{20}=\frac{-4}{5} \text { and } \frac{20}{-25}=\frac{4}{-5}=\frac{-4}{5}
$
[Converting into lowest term]
$
\begin{aligned}
& \because \frac{-4}{5}=\frac{-4}{5} \\
& \therefore \frac{-16}{20}=\frac{20}{-25}
\end{aligned}
$
(iii) $\frac{-2}{-3}$ and $\frac{2}{3}$
$\Rightarrow \frac{-2}{-3}=\frac{2}{3}$ and $\frac{2}{3}=\frac{2}{3}$ [Converting into lowest term]
$
\begin{aligned}
& \because \frac{2}{3}=\frac{2}{3} \\
& \therefore \frac{-2}{-3}=\frac{2}{3}
\end{aligned}
$
(iv) $\frac{-3}{5}$ and $\frac{-12}{20}$
$\Rightarrow \frac{-3}{5}=\frac{-3}{5}$ and $\frac{-12}{20}=\frac{-3}{5}$ [Converting into lowest term]
$
\begin{aligned}
& \because \frac{-3}{5}=\frac{-3}{5} \\
& \therefore \frac{-3}{5}=\frac{-12}{20}
\end{aligned}
$

$
\begin{aligned}
& \text { (v) } \frac{8}{-5} \text { and } \frac{-24}{15} \\
& \Rightarrow \frac{8}{-5}=\frac{-8}{5} \text { and } \frac{-24}{15}=\frac{-8}{5} \text { [Converting into lowest term] } \\
& \because \frac{-8}{5}=\frac{-8}{5} \\
& \therefore \frac{8}{-5}=\frac{-24}{15}
\end{aligned}
$
(vi) $\frac{1}{3}$ and $\frac{-1}{9}$
$\Rightarrow \frac{1}{3}=\frac{1}{3}$ and $\frac{-1}{9}=\frac{-1}{9}$ [Converting into lowest term]
$
\because \frac{1}{3} \neq \frac{-1}{9}
$

$
\therefore \frac{1}{3} \neq \frac{-1}{9}
$
(vii) $\frac{-5}{-9}$ and $\frac{5}{-9}$
$\Rightarrow \frac{-5}{-9}=\frac{5}{9}$ and $\frac{5}{-9}=\frac{5}{9}$ [Converting into lowest term]
$
\begin{aligned}
& \because \frac{5}{9} \neq \frac{5}{-9} \\
& \therefore \frac{-5}{-9} \neq \frac{5}{-9}
\end{aligned}
$

Ex 8.1 Question 7.

Rewrite the following rational numbers in the simplest form:
(i) $\frac{-8}{6}$
(ii) $\frac{25}{45}$
(iii) $\frac{-44}{72}$
(iv) $\frac{-8}{10}$

Answer:

(i) $\frac{-8}{6}=\frac{-8 \div 2}{6 \div 2}=\frac{-4}{3}$ [H.C.F. of 8 and 6 is 2]
(ii) $\frac{25}{45}=\frac{25 \div 5}{45 \div 5}=\frac{5}{9}$
[H.C.F. of 25 and 45 is 5]
(iii) $\frac{-44}{72}=\frac{-44 \div 4}{72 \div 4}=\frac{-11}{18}$ [H.C.F. of 44 and 72 is 4$]$
(iv) $\frac{-8}{10}=\frac{-8 \div 2}{10 \div 2}=\frac{-4}{5}$ [H.C.F. of 8 and 10 is 2]

Ex 8.1 Question 8.

Fill in the boxes with the correct symbol out of $<,>$ and $=$ :
(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) $0_{\square} \frac{-7}{6}$

Answer:

(i) $\frac{-5}{7} \square<\frac{2}{3}$ Since, the positive number if greater than negative number.
(ii) $\frac{-4 \times 7}{5 \times 7} \square \frac{-5 \times 5}{7 \times 5} \Rightarrow \frac{-28}{35} \square<\frac{-25}{35} \Rightarrow \frac{-4}{5} \square<\frac{-5}{7}$
(iii) $\frac{-7 \times 2}{8 \times 2} \square \frac{14 \times(-1)}{-16 \times(-1)} \Rightarrow \frac{-14}{16} \square \frac{-14}{16} \Rightarrow \frac{-7}{8} \square=\frac{14}{-16}$
(iv) $\frac{-8 \times 4}{5 \times 4} \square \frac{-7 \times 5}{4 \times 5} \Rightarrow \frac{-32}{20} \square>\frac{-35}{20} \Rightarrow \frac{-8}{5} \square>\frac{-7}{4}$
(v) $\frac{1}{-3} \square \frac{-1}{4} \Rightarrow \frac{1}{-3} \square<\frac{-1}{4}$
(vi) $\frac{5}{-11} \square \frac{-5}{11} \Rightarrow \frac{5}{-11} \square=\frac{-5}{11}$
(vii) $0 \square>\frac{-7}{6}$ Since, 0 is greater than every negative number.

Ex 8.1 Question 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) $-3 \frac{2}{7},-3 \frac{4}{5}$

Answer:

(i) $\frac{2 \times 2}{3 \times 2}=\frac{4}{6}$ and $\frac{5 \times 3}{2 \times 3}=\frac{15}{6}$
Since $\frac{4}{6} \square \frac{15}{6}$
Therefore $\frac{2}{3} \square<\frac{5}{2}$
(ii) $\frac{-5 \times 1}{6 \times 1}=\frac{-5}{6}$ and $\frac{-4 \times 2}{3 \times 2}=\frac{-8}{6}$

Since $\frac{-5}{6} \square>\frac{-8}{6}$ Therefore $\frac{-5}{6} \square \frac{-4}{3}$
(iii) $\frac{-3 \times 3}{4 \times 3}=\frac{-9}{12}$ and $\frac{2 \times(-4)}{-3 \times(-4)}=\frac{-8}{12}$

Since $\frac{-9}{12} \square<\frac{-8}{12}$
Therefore $\frac{-3}{4} \square<\frac{2}{-3}$
(iv) $\frac{-1}{4} \square \frac{1}{4}$ Since positive number is always greater than negative number.
(v) $-3 \frac{2}{7}=\frac{-23}{7}=\frac{-23 \times 5}{7 \times 5}=\frac{-115}{35}$ and $-3 \frac{4}{5}=\frac{-19}{5}=\frac{-19 \times 7}{5 \times 7}=\frac{-133}{35}$

Since $\frac{-115}{35} \square>\frac{-133}{35}$
Therefore $-3 \frac{2}{7}>-3 \frac{4}{5}$
Ex 8.1 Question 10.

Write the following rational numbers in ascending order:
(i) $\frac{-3}{5}, \frac{-2}{5}, \frac{-1}{5}$
(ii) $\frac{1}{3}, \frac{-2}{9}, \frac{-4}{3}$
(iii) $\frac{-3}{7}, \frac{-3}{2}, \frac{-3}{4}$

Answer:

(i) $\frac{-3}{5}, \frac{-2}{5}, \frac{-1}{5} \Rightarrow \frac{-3}{5}<\frac{-2}{5}<\frac{-1}{5}$
(ii) $\frac{1}{3}, \frac{-2}{9}, \frac{-4}{3} \Rightarrow \frac{3}{9}, \frac{-2}{9}, \frac{-12}{9}$ [Converting into same denominator]

Now $\frac{-12}{9}<\frac{-2}{9}<\frac{3}{9} \Rightarrow \frac{-4}{3}<\frac{-2}{9}<\frac{1}{3}$
(iii) $\frac{-3}{7}, \frac{-3}{2}, \frac{-3}{4}$

$\Rightarrow \frac{-3}{2}<\frac{-3}{4}<\frac{-3}{7}$