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

Exercise 1H

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

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Question 1:

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$\left(\frac{-5}{16}+\frac{7}{12}\right)=?$
(a) $-\frac{7}{48}$
(b) $\frac{1}{24}$
(c) $\frac{13}{48}$
(d) $\frac{1}{3}$

Answer 1:

(c) $\frac{13}{48}$
The denominators of the given rational numbers are 16 and 12, respectively.
LCM of 16 and 12 is $\left(4\times 4\times 3\right), \mathrm{that} \mathrm{is},48$
Now, we have:
$\left(\frac{-5}{16}+\frac{7}{12}\right)=\frac{3\times \left(-5\right)+4\times 7}{48}$

                      $$\begin{gathered} =\frac{\left(-15\right)+28}{48} \\ =\frac{13}{48} \end{gathered}$$

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Question 2:

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$\left(\frac{8}{-15}+\frac{4}{-3}\right)=?$
(a) $\frac{28}{15}$
(b) $\frac{-28}{15}$
(c) $\frac{-4}{5}$
(d) $\frac{-4}{15}$

Answer 2:

(b) $\frac{-28}{15}$
$\frac{8}{-15}=\frac{-8}{15}$ and$\frac{4}{-3}=\frac{-4}{3}$

Now, we have:

$\left(\frac{8}{-15}+\frac{4}{-3}\right)=\left(\frac{-8}{15}+\frac{-4}{3}\right)$

LCM of 15 and 3 is $\left(3\times 5\times 1\right), \mathrm{that} \mathrm{is},15$

$\frac{-8}{15}+\frac{-4}{3}=\frac{1\times \left(-8\right)+5\times \left(-4\right)}{15}$
                   $$\begin{gathered} =\frac{\left(-8\right)+\left(-20\right)}{15} \\ =\frac{-28}{15} \end{gathered}$$

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Question 3:

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$\left(\frac{7}{-26}+\frac{16}{39}\right)=?$
(a) $\frac{11}{78}$
(b) $\frac{-11}{78}$
(c) $\frac{11}{39}$
(d) $\frac{-11}{39}$

Answer 3:

$\frac{7}{-26}=\frac{-7}{26}$

Now, we have:

$\left(\frac{7}{-26}+\frac{16}{39}\right)=\left(\frac{-7}{26}+\frac{16}{39}\right)$

LCM of 26 and 39 is 1014, that is, $\left(29\times 1\times 36\right).$

(a) $\frac{11}{78}$
$\left(\frac{-7}{26}+\frac{16}{39}\right)=\frac{39\times \left(-7\right)+26\times 16}{1014}$
                      $$\begin{gathered} =\frac{\left(-273\right)+416}{1014} \\ =\frac{143}{1014} \\ =\frac{11}{78} \end{gathered}$$

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Question 4:

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$\left(3+\frac{5}{-7}\right)=?$
(a) $\frac{-16}{7}$
(b) $\frac{16}{7}$
(c) $\frac{-26}{7}$
(d) $\frac{-8}{7}$

Answer 4:

(b) $\frac{16}{7}$

$3=\frac{3}{1}$ and $\frac{5}{-7}=\frac{-5}{7}$

Now, we have:
         
$\left(3+\frac{5}{-7}\right)=\left(\frac{3}{1}+\frac{-5}{7}\right)$

LCM of 1 and 7 is 7

$\left(\frac{3}{1}+\frac{-5}{7}\right)=\frac{7\times 3+1\times \left(-5\right)}{7}$
                    $$\begin{gathered} =\frac{21+\left(-5\right)}{7} \\ =\frac{16}{7} \end{gathered}$$

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Question 5:

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$\left(\frac{31}{-4}+\frac{-5}{8}\right)=?$
(a) $\frac{67}{8}$
(b) $\frac{57}{8}$
(c) $\frac{-57}{8}$
(d) $\frac{-67}{8}$

Answer 5:

(d) $\frac{-67}{8}$
 $\frac{31}{-4}=\frac{-31}{4}$

We have:
       
$\left(\frac{31}{-4}+\frac{-5}{8}\right)=\left(\frac{-31}{4}+\frac{-5}{8}\right)$

LCM of 4 and 8 is 8, that is, $\left(4\times 1\times 2\right).$

$\left(\frac{-31}{4}+\frac{-5}{8}\right)=\frac{2\times \left(-31\right)+1\times \left(-5\right)}{8}$
                         $$\begin{gathered} =\frac{\left(-62\right)+\left(-5\right)}{8} \\ =\frac{-67}{8} \end{gathered}$$

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Question 6:

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What should be added to $\frac{7}{12} \mathrm{to} \mathrm{get} \frac{-4}{15}?$
(a) $\frac{17}{20}$
(b) $\frac{-17}{20}$
(c) $\frac{7}{20}$
(d) $\frac{-7}{20}$

Answer 6:

(b) $\frac{-17}{20}$
Let the required number be x.

Now,

$\frac{7}{12}+x=\frac{-4}{15}$

$\Rightarrow x=\left(\frac{-4}{15}+\frac{-7}{12}\right)$

$$\begin{gathered} =\frac{4\times \left(-4\right)+5\times \left(-7\right)}{60} \\ =\frac{\left(-16\right)+\left(-35\right)}{60} \\ =\frac{-51}{60} \\ =\frac{-17}{20} \end{gathered}$$

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Question 7:

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$\left(\frac{2}{3}+\frac{-4}{5}+\frac{7}{15}+\frac{-11}{20}\right)=?$
(a) $\frac{-1}{5}$
(b) $\frac{-4}{15}$
(c) $\frac{-13}{60}$
(d) $\frac{-7}{30}$

Answer 7:

(c) $\frac{-13}{60}$
Using the commutative and associative laws, we can arrange the terms in any suitable manner. Using this rearrangement property, we have:

$\frac{2}{3}+\frac{-4}{5}+\frac{7}{15}+\frac{-11}{20}=\left(\frac{2}{3}+\frac{7}{15}\right)+\left(\frac{-4}{5}+\frac{-11}{20}\right)$
 

                                         $$\begin{gathered} =\frac{(10+7)}{15}+\frac{[\left(-16\right)+\left(-11\right)]}{20} \\ =\left(\frac{17}{15}+\frac{-27}{20}\right) \\ =\frac{[68+\left(-81\right)]}{60} \\ =\frac{-13}{60} \end{gathered}$$

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Question 8:

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The sum of two numbers is $\frac{-4}{3}.$ If one of the numbers is −5, what is the other?
(a) $\frac{-11}{3}$
(b) $\frac{11}{3}$
(c) $\frac{-19}{3}$
(d) $\frac{19}{3}$

Answer 8:

(b) $\frac{11}{3}$
Let the other number be x.

Now,

$x+\left(-5\right)=\frac{-4}{3}$
$$\begin{gathered} \Rightarrow x=\frac{-4}{3}+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of} -5\right) \\ \Rightarrow x=\frac{-4}{3}+5 \end{gathered}$$

       $$\begin{gathered} =\frac{-4}{3}+\frac{5}{1} \\ =\frac{\left(-4\right)+15}{3} \\ =\frac{11}{3} \end{gathered}$$

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Question 9:

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What should be added to $\frac{-5}{7} \mathrm{to} \mathrm{get} \frac{-2}{3}?$
(a) $\frac{-29}{21}$
(b) $\frac{29}{21}$
(c) $\frac{1}{21}$
(d) $\frac{-1}{21}$

Answer 9:

(c) $\frac{1}{21}$
Let the required number be x.

Now,

$\frac{-5}{7}+x=\frac{-2}{3}$
$$\begin{gathered} \Rightarrow x=\frac{-2}{3}+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of} \frac{-5}{7}\right) \\ \Rightarrow x=\left(\frac{-2}{3}+\frac{5}{7}\right) \end{gathered}$$
       $$\begin{gathered} =\frac{\left(-14\right)+15}{21} \\ =\frac{1}{21} \end{gathered}$$

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Question 10:

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What should be subtracted from $\frac{-5}{3} \mathrm{to} \mathrm{get} \frac{5}{6}?$
(a) $\frac{5}{2}$
(b) $\frac{3}{2}$
(c) $\frac{5}{4}$
(d) $\frac{-5}{2}$

Answer 10:

(d) $\frac{-5}{2}$
Let the required number be x.

Now,

$\frac{-5}{3}-x=\frac{5}{6}$
$\Rightarrow x=\left(\frac{-5}{3}-\frac{5}{6}\right)$

       $$\begin{gathered} =\frac{-10-5}{6} \\ =\frac{-15}{6} \\ =\frac{-5}{2} \end{gathered}$$
Thus, the required number is $\frac{-5}{2}$

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Question 11:

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${\left(\frac{-3}{7}\right)}^{-1}=?$
(a) $\frac{7}{3}$
(b) $\frac{-7}{3}$
(c) $\frac{3}{7}$
(d) none of these

Answer 11:

(b) $\frac{-7}{3}$

 ${\left(-\frac{3}{7}\right)}^{-1}\Rightarrow \mathrm{Reciprocal} \mathrm{of}\frac{-3}{7}$

The reciprocal of $\frac{-3}{7}$ is $\frac{7}{-3}, \mathrm{i}.\mathrm{e}., \frac{-7}{3}$

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Question 12:

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The product of two rational numbers is $\frac{-28}{81}$. If one of the numbers is $\frac{14}{27}$ then the other one is
(a) $\frac{-2}{3}$
(b) $\frac{2}{3}$
(c) $\frac{3}{2}$
(d) $\frac{-3}{2}$

Answer 12:

(a) $\frac{-2}{3}$
Let the other number be x.

Now,

$x\times \frac{14}{27}=\frac{-28}{81}$

$\Rightarrow x=\frac{-28}{81}÷\frac{14}{27}$

       $$\begin{gathered} =\frac{-28}{81}\times \frac{27}{14} \\ =\frac{\left(-28\right)\times 27}{81\times 14} \\ =\frac{-\left(28\times 27\right)}{81\times 14} \\ =\frac{-\left(2\times 3\right)}{9\times 1} \\ =\frac{-6}{9} \\ =\frac{-2}{3} \end{gathered}$$
Thus, the other number is $\frac{-2}{3}$

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Question 13:

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The product of two numbers is $\frac{-16}{35}.$ If one of the numbers is $\frac{-15}{14}$, the other is
(a) $\frac{-2}{5}$
(b) $\frac{8}{15}$
(c) $\frac{32}{75}$
(d) $\frac{-8}{3}$

Answer 13:

(c) $\frac{32}{75}$
Let the other number be x.

Now,

$x\times \frac{-15}{4}=\frac{-16}{35}$
$\Rightarrow x=\frac{-16}{35}÷\frac{-15}{14}$

       $$\begin{gathered} =\frac{-16}{35}\times \frac{14}{-15} \\ =\frac{-\left(16\times 14\right)}{-\left(35\times 15\right)} \\ =\frac{16\times 14}{35\times 15}=\frac{224}{525}=\frac{32}{75} \end{gathered}$$

Thus, the other number is $\frac{32}{75}$

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Question 14:

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What should be subtracted from $\frac{-3}{5}$ to get −2?
(a) $\frac{-7}{5}$
(b) $\frac{-13}{5}$
(c) $\frac{13}{5}$
(d) $\frac{7}{5}$

Answer 14:

(d) $\frac{7}{5}$
Let the required number be x. 

Now,

$$\begin{gathered} -\frac{3}{5}-x=-2 \\ \Rightarrow -\frac{3}{5}=-2+x \\ \Rightarrow x=\left(-\frac{3}{5}+2\right) \\ \Rightarrow x =\frac{\left(-3+10\right)}{5} \\ \Rightarrow x=\frac{7}{5} \end{gathered}$$
Thus, the required number is $\frac{7}{5}$
 

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Question 15:

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The sum of two rational numbers is −3. If one of them is $\frac{-10}{3}$ then the other one is
(a) $\frac{-13}{3}$
(b) $\frac{-19}{3}$
(c) $\frac{1}{3}$
(d) $\frac{13}{3}$

Answer 15:

(c) $\frac{1}{3}$
Let the other number be x. 

Now,

$$\begin{gathered} x+\left(-\frac{10}{3}\right)=-3 \\ \Rightarrow x=-3+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of} -\frac{10}{3}\right) \\ \Rightarrow x=\left(-3+\frac{10}{3}\right) \end{gathered}$$
     $$\begin{gathered} =\frac{-3}{1}+\frac{10}{3} \\ =\frac{\left(-9+10\right)}{3} \\ =\frac{1}{3} \end{gathered}$$
Thus, the other number is $\frac{1}{3}$

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Question 16:

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Which of the following numbers is in standard form?
(a) $\frac{-12}{26}$
(b) $\frac{-49}{71}$
(c) $\frac{-9}{16}$
(d) $\frac{28}{-105}$

Answer 16:

(b) $\frac{-49}{71}$ and (c) $\frac{-9}{16}$

The numbers $\frac{-49}{71}$ and $\frac{-9}{16}$ are in the standard form because they have no common divisor other than 1 and their denominators are positive. 

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Question 17:

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$\left(\frac{-9}{16}\times \frac{8}{15}\right)=?$
(a) $\frac{-3}{10}$
(b) $\frac{-4}{15}$
(c) $\frac{-9}{25}$
(d) $\frac{-2}{5}$

Answer 17:

(a) $\frac{-3}{10}$

$\left(\frac{-9}{16}\times \frac{8}{15}\right)=\frac{-9\times 8}{16\times 15}$

                      $$\begin{gathered} =\frac{-72}{240} \\ =\frac{-3}{10} \end{gathered}$$

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Question 18:

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$\left(\frac{-5}{9}÷\frac{2}{3}\right)=?$
(a) $\frac{-5}{2}$
(b) $\frac{-5}{6}$
(c) $\frac{-10}{27}$
(d) $\frac{-6}{5}$

Answer 18:

(b) $\frac{-5}{6}$

$\frac{-5}{9}÷\frac{2}{3}=\frac{-5}{9}\times \frac{3}{2}$

                $$\begin{gathered} =\frac{-5\times 3}{9\times 2} \\ =\frac{-15}{18} \\ =\frac{-5}{6} \end{gathered}$$

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Question 19:

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$\frac{4}{9}÷?=\frac{-8}{15}$
(a) $\frac{-32}{45}$
(b) $\frac{-8}{5}$
(c) $\frac{-9}{10}$
(d) $\frac{-5}{6}$

Answer 19:

(d) $\frac{-5}{6}$

Let $\frac{4}{9}÷\frac{a}{b}=\frac{-8}{15}$

Now,

$$\begin{gathered} \frac{4}{9}\times \frac{b}{a}=\frac{-8}{15} \\ \Rightarrow \frac{b}{a}=\frac{-8}{15}\times \frac{9}{4} \end{gathered}$$

$=\frac{-6}{5}$

$\Rightarrow \frac{a}{b}=\frac{5}{-6}$

$=\frac{-5}{6}$
Hence, the missing number is $\frac{-5}{6}$.

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Question 20:

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Additive inverse of $\frac{-5}{9}$ is
(a) $\frac{-9}{5}$
(b) 0
(c) $\frac{5}{9}$
(d) $\frac{9}{5}$

Answer 20:

(c) $\frac{5}{9}$

Additive inverse of $\frac{-5}{9}$ is $\frac{5}{9}$.

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Question 21:

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Reciprocal of $\frac{-3}{4}$ is
(a) $\frac{4}{3}$
(b) $\frac{3}{4}$
(c) $\frac{-4}{3}$
(d) 0

Answer 21:

(c) $\frac{-4}{3}$
 Reciprocal of $\frac{-3}{4}$ is $\frac{4}{-3}, \mathrm{i}.\mathrm{e}., \frac{-4}{3}.$

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Question 22:

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A rational number between $\frac{-2}{3} \mathrm{and} \frac{1}{4} \mathrm{is}$
(a) $\frac{5}{12}$
(b) $\frac{-5}{12}$
(c) $\frac{5}{24}$
(d) $\frac{-5}{24}$

Answer 22:

(d) $\frac{-5}{24}$
Rational number between $\frac{-2}{3}$ and $\frac{1}{4}$ = $\frac{1}{2}\left(\frac{-2}{3}+\frac{1}{4}\right)$
                                                           $$\begin{gathered} =\frac{1}{2}\left(\frac{-8+3}{12}\right) \\ =\frac{1}{2}\times \frac{-5}{12} \\ =\frac{-5}{24} \end{gathered}$$

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Question 23:

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The reciprocal of a negative rational number
(a) is a positive rational number
(b) is a negative rational number
(c) can be either a positive or a negative rational number
(d) does not exist

Answer 23:

(b) is a negative rational number

The reciprocal of a negative rational number is a negative rational number.