RS Aggarwal solution class 8 chapter 1 Rational Numbers Exercise 1D

Exercise 1D

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Q1 | Ex-1D | Rational Numbers | Class 8 | RS AGGARWAL | Chapter 1 | myhelper

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Question 1:

Find each of the following products:
(i) $\frac{3}{5}\times \frac{-7}{8}$
(ii) $\frac{-9}{2}\times \frac{5}{4}$
(iii) $\frac{-6}{11}\times \frac{-5}{3}$
(iv) $\frac{-2}{3}\times \frac{6}{7}$
(v) $\frac{-12}{5}\times \frac{10}{-3}$
(vi) $\frac{25}{-9}\times \frac{3}{-10}$
(vii) $\frac{5}{-18}\times \frac{-9}{20}$
(viii) $\frac{-13}{15}\times \frac{-25}{26}$
(ix) $\frac{16}{-21}\times \frac{14}{5}$
(x) $\frac{-7}{6}\times 24$
(xi) $\frac{7}{24}\times \left(-48\right)$
(xii) $\frac{-13}{5}\times \left(-10\right)$

Answer 1:

(i)

$$\begin{gathered} \frac{3}{5}\times \frac{-7}{8} \\ =\frac{3\times (-7)}{5\times 8} \\ =-\frac{21}{40} \end{gathered}$$

(ii)

$$\begin{gathered} \frac{-9}{2}\times \frac{5}{4} \\ =\frac{(-9)\times 5}{2\times 4} \\ =\frac{-45}{8} \end{gathered}$$

(iii)

$$\begin{gathered} \frac{-6}{11}\times \frac{-5}{3} \\ =\frac{(-6)\times (-5)}{11\times 3} \\ =\frac{30}{33} \end{gathered}$$

Simplifying the above rational number, we get:

$\frac{30}{33}=\frac{30÷3}{33÷3}=\frac{10}{11}$

(iv)

$$\begin{gathered} \frac{-2}{3}\times \frac{6}{7} \\ =\frac{(-2)\times 6}{3\times 7} \\ =\frac{-12}{21} \end{gathered}$$

Simplifying the above rational number, we get:

$\frac{-12}{21}=\frac{-12÷3}{21÷3}=\frac{-4}{7}$

(v)

$$\begin{gathered} \frac{-12}{5}\times \frac{10}{-3} \\ =\frac{(-12)\times 10}{5\times (-3)} \\ =\frac{-120}{-15} \\ =\frac{120}{15} \end{gathered}$$

Simplifying the above rational number, we get:

$\frac{120}{15}=\frac{120÷3}{15÷3}=\frac{40}{5}=8$

(vi)

$$\begin{gathered} \frac{25}{-9}\times \frac{3}{-10} \\ =\frac{25\times 3}{(-9)\times (-10)} \\ =\frac{75}{90} \end{gathered}$$

Simplifying the above rational number, we get:

$\frac{75}{90}=\frac{75÷15}{90÷15}=\frac{5}{6}$

(vii)

$$\begin{gathered} \frac{5}{-18}\times \frac{-9}{20} \\ =\frac{5\times (-9)}{-18\times 20} \\ =\frac{-45}{-360} \\ =\frac{45}{360} \end{gathered}$$

Simplifying the above rational number, we get:

$\frac{45}{360}=\frac{45÷45}{360÷45}=\frac{1}{8}$

(viii)

$$\begin{gathered} \frac{-13}{15}\times \frac{-25}{26} \\ =\frac{(-13)\times (-25)}{15\times 26} \\ =\frac{325}{390} \end{gathered}$$

Simplifying the above rational number, we get:

$\frac{325}{390}=\frac{325÷5}{390÷5}=\frac{65}{78}=\frac{65÷13}{78÷13}=\frac{5}{6}$

(ix)

$$\begin{gathered} \frac{16}{-21}\times \frac{14}{5} \\ =\frac{16\times 14}{(-21)\times 5} \\ =\frac{224}{-105} \end{gathered}$$

Simplifying the above rational number, we get:

$\frac{224}{-105}=\frac{224÷7}{(-105)÷7}=\frac{32}{-15}=\frac{32\times -1}{-15\times -1}=\frac{-32}{15}$

(x)

$$\begin{gathered} \frac{-7}{6}\times 24 \\ =\frac{(-7)\times 24}{6} \\ =\frac{-168}{6} \end{gathered}$$

Simplifying the above rational number, we get:

$\frac{-168}{6}=\frac{(-168)÷2}{6÷2}=\frac{84}{3}=\frac{-84÷3}{3÷3}=-28$

(xi)

$$\begin{gathered} \frac{7}{24}\times (-48) \\ =\frac{7\times (-48)}{24}=-\frac{336}{24} \end{gathered}$$

Simplifying the above rational number, we get:

$\frac{-336}{24}=\frac{-336÷24}{24÷24}=-14$

(xii)

$$\begin{gathered} \frac{-13}{5}\times (-10) \\ =\frac{(-13)\times (-10)}{5} \\ =\frac{130}{5} \end{gathered}$$

Simplifying the above rational number, we get:

$\frac{130}{5}=\frac{130÷5}{5÷5}=26$

Q2 | Ex-1D | Rational Numbers | Class 8 | RS AGGARWAL | Chapter 1 | myhelper

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Question 2:

Verify each of the following:
(i) $\frac{3}{7}\times \frac{-5}{9}=\frac{-5}{9}\times \frac{3}{7}$
(ii) $\frac{-8}{7}\times \frac{13}{9}=\frac{13}{9}\times \frac{-8}{7}$
(iii) $\frac{-12}{5}\times \frac{7}{-36}=\frac{7}{-36}\times \frac{-12}{5}$
(iv) $-8\times \frac{-13}{12}=\frac{-13}{12}\times \left(-8\right)$

Answer 2:

(i)

$\frac{3}{7}\times \frac{-5}{9}=\frac{-5}{9}\times \frac{3}{7}$

 $$\begin{gathered} \mathrm{LHS}=\frac{3\times (-5)}{7\times 9} \\ =-\frac{15}{63} \\ \text{Simplifying, we get:} \\ -\frac{15}{63}=-\frac{15÷3}{63÷3} \\ =-\frac{5}{21} \end{gathered}$$

  $$\begin{gathered} \mathrm{RHS}=\frac{-5}{9}\times \frac{3}{7} \\ =\frac{(-5)\times 3}{9\times 7} \\ =\frac{-15}{63} \\ \text{Simplifying, we get:} \\ =\frac{-15÷3}{63÷3} \\ =-\frac{5}{21} \end{gathered}$$

LHS = RHS


(ii)

$\frac{-8}{7} \times \frac{13}{9}=\frac{13}{9} \times \frac{-8}{7}$

LHS$=\frac{-8}{7} \times \frac{13}{9}=\frac{(-8) \times 13}{7 \times 9}=-\frac{104}{63}$

RHS$=\frac{13}{9} \times \frac{-8}{7}=\frac{13 \times(-8)}{9 \times 7}=-\frac{104}{63}$

LHS=RHS

(iii)

$\frac{-12}{5}\times \frac{7}{-36}=\frac{7}{-36}\times \frac{-12}{5}$

 $$\begin{gathered} \mathrm{LHS\; =}\frac{-12}{5}\times \frac{7}{-36} \\ =\frac{(-12)\times 7}{5\times (-36)} \\ =\frac{84}{180} \\ \text{Simplifying, we get: } \\ =\frac{84÷12}{180÷12} \\ =\frac{7}{15} \end{gathered}$$

 $$\begin{gathered} \mathrm{RHS}=\frac{7}{-36}\times \frac{-12}{5} \\ =\frac{7\times (-12)}{(-36)\times 5} \\ =\frac{84}{180} \\ \text{Simplifying, we get:} \\ =\frac{84÷12}{180÷12} \\ =\frac{7}{15} \end{gathered}$$

LHS = RHS


(iv)
$-8 \times \frac{-13}{12}=\frac{-13}{12}\times (-8)$

 $$\begin{gathered} \mathrm{LHS\; =}-8 \times \frac{-13}{12} \\ =\frac{(-8)\times (-13)}{12} \\ =\frac{104}{12} \\ \text{Simplifying, we get: } \\ =\frac{104÷4}{12÷4} \\ =\frac{26}{3} \end{gathered}$$

$$\begin{gathered} \mathrm{RHS}=\frac{-13}{12}\times (-8) \\ =\frac{(-13)\times (-8)}{12} \\ =\frac{104}{12} \\ \text{Simplifying, we get: } \\ =\frac{104÷4}{12÷4} \\ =\frac{26}{3} \end{gathered}$$

LHS = RHS

Q3 | Ex-1D | Rational Numbers | Class 8 | RS AGGARWAL | Chapter 1 | myhelper

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Question 3:

Verify each of the following:
(i) $\left(\frac{5}{7}\times \frac{12}{13}\right)\times \frac{7}{18}=\frac{5}{7}\times \left(\frac{12}{13}\times \frac{7}{18}\right)$
(ii) $\frac{-13}{24}\times \left(\frac{-12}{5}\times \frac{35}{36}\right)=\left(\frac{-13}{24}\times \frac{-12}{5}\right)\times \frac{35}{36}$
(iii) $\left(\frac{-9}{5}\times \frac{-10}{3}\right)\times \frac{21}{-4}=\frac{-9}{5}\times \left(\frac{-10}{3}\times \frac{21}{-4}\right)$

Answer 3:

(i)

$\left(\frac{5}{7}\times \frac{12}{13}\right)\times \frac{7}{18}=\frac{5}{7}\times \left(\frac{12}{13}\times \frac{7}{18}\right)$

$$\begin{gathered} \mathrm{LHS}=\left(\frac{5}{7}\times \frac{12}{13}\right)\times \frac{7}{18} \\ =\frac{5\times 12}{7\times 13}\times \frac{7}{18} \\ =\frac{60}{91}\times \frac{7}{18} \\ =\frac{420}{1638} \\ =\frac{10}{39} \end{gathered}$$


$$\begin{gathered} \mathrm{RHS}=\frac{5}{7}\times \left(\frac{12}{13}\times \frac{7}{18}\right) \\ =\frac{5}{7}\times \frac{12\times 7}{13\times 18} \\ =\frac{5}{7}\times \frac{84}{234} \\ =\frac{420}{1638} \\ =\frac{10}{39} \end{gathered}$$

∴ ​$\left(\frac{5}{7}\times \frac{12}{13}\right)\times \frac{7}{18}=\frac{5}{7}\times \left(\frac{12}{13}\times \frac{7}{18}\right)$


(ii)

$\frac{-13}{24}\times \left(\frac{-12}{5}\times \frac{35}{36}\right)=\left(\frac{-13}{24}\times \frac{-12}{5}\right)\times \frac{35}{36}$

 $$\begin{gathered} \mathrm{LHS}=\frac{-13}{24}\times \left(\frac{-12}{5}\times \frac{35}{36}\right) \\ =\frac{-13}{24}\times \frac{(-12)\times 35}{5\times 36} \\ =\frac{-13}{24}\times \frac{-420}{180} \\ =\frac{5460}{4320} \\ =\frac{91}{72} \end{gathered}$$


 $$\begin{gathered} \mathrm{RHS}=\left(\frac{-13}{24}\times \frac{-12}{5}\right)\times \frac{35}{36} \\ =\frac{(-13)\times (-12)}{24\times 5}\times \frac{35}{36} \\ =\frac{156}{120}\times \frac{35}{36} \\ =\frac{156\times 35}{120\times 36} \\ =\frac{5460}{4320} \\ =\frac{91}{72} \end{gathered}$$

 ∴ ​$\frac{-13}{24}\times \left(\frac{-12}{5}\times \frac{35}{36}\right)=\left(\frac{-13}{24}\times \frac{-12}{5}\right)\times \frac{35}{36}$


(iii)

$\left(\frac{-9}{5}\times \frac{-10}{3}\right)\times \frac{21}{-4}=\frac{-9}{5}\times \left(\frac{-10}{3}\times \frac{21}{-4}\right)$

  $$\begin{gathered} \mathrm{LHS}=\left(\frac{-9}{5}\times \frac{-10}{3}\right)\times \frac{21}{-4} \\ =\frac{(-9)\times (-10)}{5\times 3}\times \frac{21}{-4} \\ =\frac{90}{15}\times \frac{21}{-4} \\ =\frac{90\times 21}{15\times (-4)} \\ =-\frac{1890}{60} \\ =-\frac{63}{2} \end{gathered}$$


  $$\begin{gathered} \mathrm{RHS}=\frac{-9}{5}\times \left(\frac{-10}{3}\times \frac{21}{-4}\right) \\ =\frac{-9}{5}\times \frac{(-10)\times 21}{3\times (-4)} \\ =\frac{-9}{5}\times \frac{210}{12} \\ =\frac{(-9)\times 210}{5\times 12} \\ =-\frac{1890}{60} \\ =\frac{-63}{2} \end{gathered}$$

∴ $(\frac{-9}{5}\times \frac{-10}{3})\times \frac{21}{-4}=\frac{-9}{5}\times (\frac{-10}{3}\times \frac{21}{-4})$

Q4 | Ex-1D | Rational Numbers | Class 8 | RS AGGARWAL | Chapter 1 | myhelper

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Question 4:

Fill in the blanks:

(i) $\frac{-23}{17} \times \frac{18}{35}=\frac{18}{35} \times(\ldots \ldots)$

(ii) $-38 \times \frac{-7}{19}=\frac{-7}{19} \times(\ldots \ldots)$

(iii) $\left(\frac{15}{7} \times \frac{-21}{10}\right) \times \frac{-5}{6}=(\ldots \ldots) \times\left(\frac{-21}{10} \times \frac{-5}{6}\right)$

(iv) $\frac{-12}{5} \times\left(\frac{4}{15} \times \frac{25}{-16}\right)=\left(\frac{-12}{5} \times \frac{4}{15}\right) \times(\ldots \ldots)$

Answer 4:

(i) $\frac{-23}{17} \times \frac{18}{35}=\frac{18}{35} \times \boxed{\frac{-23}{17}}$ 

(∵a×b=b×a)

(ii) $-38 \times \frac{-7}{9}=\frac{-7}{9} \times \fbox{-38}$ 

(∵a×b=b×a)

(iii) $\left(\frac{15}{7} \times \frac{-21}{10}\right) \times \frac{-5}{6}=\boxed{\frac{15}{7}} \times\left(\frac{-21}{10} \times \frac{-5}{6}\right)$ 

[∵a×(b×c)=(a×b)×c)]

(iv) $\frac{-12}{5} \times\left(\frac{4}{15} \times \frac{25}{-16}\right)=\left(\frac{-12}{5} \times \frac{4}{15}\right) \times \boxed{\frac{25}{-16}}$ 

[∵a×(b×c)=(a×b)×c]

Q5 | Ex-1D | Rational Numbers | Class 8 | RS AGGARWAL | Chapter 1 | myhelper

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Question 5:

Find the multiplicative inverse (i.e., reciprocal) of:
(i) $\frac{13}{25}$
(ii) $\frac{-17}{12}$
(iii) $\frac{-7}{24}$
(iv) 18
(v) −16
(vi) $\frac{-3}{-5}$
(vii) −1
(viii) $\frac{0}{2}$
(ix) $\frac{2}{-5}$
(x) $\frac{-1}{8}$

Answer 5:

(i) Reciprocal of $\frac{13}{25}$ is $\frac{25}{13}$

(ii) Reciprocal of $\frac{-17}{12}$ is $\frac{12}{-17}$, that is, $\frac{-12}{17}$.

(iii) Reciprocal of $\frac{-7}{24}$ is $\frac{24}{-7}$, that is, $\frac{-24}{7}$.

(iv) Reciprocal of 18 is $\frac{1}{18}$

(v) Reciprocal of -16 is $\frac{1}{-16}$, that is, $\frac{-1}{16}$.

(vi) Reciprocal of $\frac{-3}{-5}$ is $\frac{-5}{-3}$, that is, $\frac{5}{3}$

(vii) Reciprocal of-1 is -1.

(viii) Reciprocal of $\frac{0}{2}$ does not exist as $\frac{2}{0}=\infty$

(ix) Reciprocal of $\frac{2}{-5}$ is $\frac{-5}{2}$

(x) Reciprocal of $\frac{-1}{8}$ is -8.

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Q6 | Ex-1D | Rational Numbers | Class 8 | RS AGGARWAL | Chapter 1 | myhelper

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Question 6:

Find the value of:
(i) ${\left(\frac{5}{8}\right)}^{-1}$
(ii) ${\left(\frac{-4}{9}\right)}^{-1}$
(iii) ${\left(-7\right)}^{-1}$
(iv) ${\left(\frac{1}{-3}\right)}^{-1}$

Answer 6:

We know that  $a^{-1}=\frac{1}{a}$ or $a^{-1}\times a=1$

$$\begin{gathered} \text{(i)} \\ {\left(\frac{5}{8}\right)}^{-1}=\frac{8}{5} \\ \because \frac{5}{8}\times {\left(\frac{5}{8}\right)}^{-1}=1 \\ \text{(ii)} \\ {\left(\frac{-4}{9}\right)}^{-1}=\frac{9}{-4}=\frac{-9}{4} \\ \because \frac{-4}{9}\times {\left(\frac{-4}{9}\right)}^{-1}=1 \\ \text{(iii)} \\ (-7{)}^{-1}=\frac{1}{-7}=\frac{-1}{7} \\ \because -7\times (-7{)}^{-1}=1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right) (-3{)}^{-1} \\ (-3{)}^{-1}=\frac{1}{-3}=\frac{-1}{3} \\ \because (-3{)}^{-1}\times \left(-3\right) = 1 \end{gathered}$$

Q7 | Ex-1D | Rational Numbers | Class 8 | RS AGGARWAL | Chapter 1 | myhelper

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Question 7:

Verify the following:
(i) $\frac{3}{7}\times \left(\frac{5}{6}+\frac{12}{13}\right)=\left(\frac{3}{7}\times \frac{5}{6}\right)+\left(\frac{3}{7}\times \frac{12}{13}\right)$
(ii) $\frac{-15}{4}\times \left(\frac{3}{7}+\frac{-12}{5}\right)=\left(\frac{-15}{4}\times \frac{3}{7}\right)+\left(\frac{-15}{4}\times \frac{-12}{5}\right)$
(iii) $\left(\frac{-8}{3}+\frac{-13}{12}\right)\times \frac{5}{6}=\left(\frac{-8}{3}\times \frac{5}{6}\right)+\left(\frac{-13}{12}\times \frac{5}{6}\right)$
(iv) $\frac{-16}{7}\times \left(\frac{-8}{9}+\frac{-7}{6}\right)=\left(\frac{-16}{7}\times \frac{-8}{9}\right)+\left(\frac{-16}{7}\times \frac{-7}{6}\right)$

Answer 7:

 $$\begin{gathered} \text{(i)} \\ \text{LHS=}\frac{3}{7}\times \left(\frac{5}{6}+\frac{12}{13}\right) \\ =\frac{3}{7}\times \left(\frac{65 +72}{78}\right) \\ =\frac{3}{7}\times \frac{137}{78} \\ =\frac{137}{182} \\ \text{RHS}=\left(\frac{3}{7}\times \frac{5}{6}\right)+\left(\frac{12}{13}\times \frac{3}{7}\right) \\ =\frac{3\times 5}{7\times 6}+\frac{12\times 3}{13\times 7} \\ =\frac{15}{42}+\frac{36}{91} \\ =\frac{195+216}{546} \\ =\frac{411}{546} \\ =\frac{137}{182} \end{gathered}$$

∴ ​$\frac{3}{7}\times (\frac{5}{6}+\frac{12}{13})=(\frac{3}{7}\times \frac{5}{6})+(\frac{3}{7}\times \frac{12}{13})$

$$\begin{gathered} \text{(ii)} \\ \mathrm{LHS}=\frac{-15}{4}\times (\frac{3}{7}+\frac{-12}{5}) \\ =\frac{-15}{4}\times \left(\frac{15-84}{35}\right) \\ =\frac{-15}{4}\times \frac{-69}{35} \\ =\frac{(-15)\times (-69)}{140} \\ =\frac{1035}{140} \\ =\frac{207}{28} \\ \text{RHS}=(\frac{-15}{4}\times \frac{3}{7})+(\frac{-15}{4}\times \frac{-12}{5}) \\ =\frac{(-15)\times 3}{4\times 7}+\frac{(-15)\times (-12)}{4\times 5} \\ =\frac{-45}{28}+\frac{180}{20} \\ =\frac{-225+1260}{140} \\ =\frac{1035}{140} \\ =\frac{207}{28} \\ \therefore \frac{-15}{4}\times (\frac{3}{7}+\frac{-12}{5})=(\frac{-15}{4}\times \frac{3}{7})+(\frac{-15}{4}\times \frac{-12}{5}) \end{gathered}$$

(iii)

$$\begin{gathered} (\frac{-8}{3}+\frac{-13}{12})\times \frac{5}{6}=(\frac{-8}{3}\times \frac{5}{6})+(\frac{-13}{12}\times \frac{5}{6}) \\ \text{LHS}=(\frac{-8}{3}+\frac{-13}{12})\times \frac{5}{6} \\ =\frac{-32-13}{12}\times \frac{5}{6} \\ =\frac{-45}{12}\times \frac{5}{6} \\ =\frac{-45\times 5}{12\times 6} \\ =\frac{-225}{72} \\ =\frac{-225÷9}{72÷9} \\ =-\frac{25}{8} \\ \text{RHS}=(\frac{-8}{3}\times \frac{5}{6})+(\frac{-13}{12}\times \frac{5}{6}) \\ =\frac{-8\times 5}{3\times 6}+\frac{(-13)\times 5}{12\times 6} \\ =\frac{-40}{18}+\frac{-65}{72} \\ =\frac{-160-65}{72} \\ =\frac{-225}{72} \\ =\frac{-225÷9}{72÷9} \\ =\frac{-25}{8} \\ \therefore (\frac{-8}{3}+\frac{-13}{12})\times \frac{5}{6}=(\frac{-8}{3}\times \frac{5}{6})+(\frac{-13}{12}\times \frac{5}{6}) \end{gathered}$$

(iv)

$$\begin{gathered} \frac{-16}{7}\times (\frac{-8}{9}+\frac{-7}{6})=(\frac{-16}{7}\times \frac{-8}{9})+(\frac{-16}{7}\times \frac{-7}{6}) \\ \text{LHS}=\frac{-16}{7}\times (\frac{-8}{9}+\frac{-7}{6}) \\ =\frac{-16}{7}\times \left(\frac{-16-21}{18}\right) \\ =\frac{-16}{7}\times \frac{-37}{18} \\ =\frac{592}{126} \\ =\frac{296}{63} \\ \text{RHS}=(\frac{-16}{7}\times \frac{-8}{9})+(\frac{-16}{7}\times \frac{-7}{6}) \\ =\frac{128}{63}+\frac{112}{42} \\ =\frac{256+336}{126} \\ =\frac{592}{126} \\ =\frac{296}{63} \\ \therefore \frac{-16}{7}\times (\frac{-8}{9}+\frac{-7}{6})=(\frac{-16}{7}\times \frac{-8}{9})+(\frac{-16}{7}\times \frac{-7}{6}) \end{gathered}$$

Q8 | Ex-1D | Rational Numbers | Class 8 | RS AGGARWAL | Chapter 1 | myhelper

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Question 8:

Name the property of multiplication illustrated by each of the following statements:
(i) $\frac{-15}{8}\times \frac{-12}{7}=\frac{-12}{7}\times \frac{-15}{8}$
(ii) $\left(\frac{-2}{3}\times \frac{7}{9}\right)\times \frac{-9}{5}=\frac{-2}{3}\times \left(\frac{7}{9}\times \frac{-9}{5}\right)$
(iii) $\frac{-3}{4}\times \left(\frac{-5}{6}+\frac{7}{8}\right)=\left(\frac{-3}{4}\times \frac{-5}{6}\right)+\left(\frac{-3}{4}\times \frac{7}{8}\right)$
(iv) $\frac{-16}{9}\times 1=1\times \frac{-16}{9}=\frac{-16}{9}$
(v) $\frac{-11}{15}\times \frac{15}{-11}=\frac{15}{-11}\times \frac{-11}{15}=1$
(vi) $\frac{-7}{5}\times 0=0$

Answer 8:

1. Commutative property
2. Associative property
3. Distributive property
4. Property of multiplicative identity
5. Property of multiplicative inverse
6. Multiplicative property of 0

Q9 | Ex-1D | Rational Numbers | Class 8 | RS AGGARWAL | Chapter 1 | myhelper

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Question 9:

Fill in the blanks:
(i) The product of a rational number and its reciprocal is .......
(ii) Zero has ....... reciprocal.
(iii) The numbers ....... and ....... are their own reciprocals.
(iv) zero is ....... the reciprocal of any number.
(v) The reciprocal of a, where a ≠ 0, is .......
(vi) The reciprocal of $\frac{1}{a}$, where a ≠ 0, is .......
(vii) The reciprocal of a positive rational rational number is .......
(viii) The reciprocal of a negative rational number is .......

Answer 9:

(i) 1
(ii) no
(iii) 1; -1
(iv) not
(v) $\frac{1}{a}$
(vi) a
(vii) positive
(viii) negative