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

Exercise 1C

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

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Question 1:

Add the following rational numbers:

(i) $\frac{-2}{5} \mathrm{and} \frac{4}{5}$
(ii) $\frac{-6}{11} \mathrm{and} \frac{-4}{11}$
(iii) $\frac{-11}{8} \mathrm{and} \frac{5}{8}$
(iv) $\frac{-7}{3} \mathrm{and} \frac{1}{3}$
(v) $\frac{5}{6} \mathrm{and} \frac{-1}{6}$
(vi) $\frac{-17}{15} \mathrm{and} \frac{-1}{15}$

Answer 1:

1. $\frac{-2}{5 }+\frac{4}{5}=\frac{-2+4}{5}=\frac{2}{5}$


2. $\frac{-6}{11}+\frac{-4}{11}=\frac{-6+(-4)}{11}=\frac{-6-4}{11}=\frac{-10}{11}$


3. $\frac{-11}{8}+\frac{5}{8}=\frac{-11+5}{8}=\frac{-6}{8}=\frac{-3\times 2}{4\times 2}=\frac{-3}{4}$


4. $\frac{-7}{3}+\frac{1}{3}=\frac{-7+1}{3}=\frac{-6}{3}=\frac{-3\times 2}{3}=-2$


5. $\frac{5}{6}+\frac{-1}{6}=\frac{5+(-1)}{6}=\frac{4}{6}=\frac{2\times 2}{3\times 2}=\frac{2}{3}$

6.$\frac{-17}{15}+\frac{-1}{15}=\frac{-17+(-1)}{15}$ 

$=\frac{-17-1}{15}=\frac{-18}{15}=\frac{-6 \times 3}{5 \times 3}=\frac{-6}{5}$

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Question 2:

Add the following rational numbers:
(i) $\frac{3}{4} \mathrm{and} \frac{-3}{5}$
(ii) $\frac{5}{8} \mathrm{and} \frac{-7}{12}$
(iii) $\frac{-8}{9} \mathrm{and} \frac{11}{6}$
(iv) $\frac{-5}{16} \mathrm{and} \frac{7}{24}$
(v) $\frac{7}{-18} and \frac{8}{27}$
(vi) $\frac{1}{-12} \mathrm{and} \frac{2}{-15}$
(vii) $-1 \mathrm{and} \frac{3}{4}$
(viii) $2 \mathrm{and} \frac{-5}{4}$
(ix) $0 \mathrm{and} \frac{-2}{5}$

Answer 2:

1. The denominators of the given rational numbers are 4 and 5.

LCM of 4 and 5 is 20.

Now, 

$\frac{3}{4}=\frac{3\times 5}{4\times 5}=\frac{15}{20}$ and $\frac{-3}{5}=\frac{-3\times 4}{5\times 4}=\frac{-12}{20}$

∴ $\frac{3}{4}+\frac{-3}{5}=\frac{15}{20}+\frac{-12}{20}=\frac{15+(-12)}{20}=\frac{15-12}{20}=\frac{3}{20}$

2.​ The denominators of the given rational numbers are 8 and 12.

LCM of 8 and 12 is 24.

Now, 

$\frac{5}{8}=\frac{5\times 3}{8\times 3}=\frac{15}{24}$ and $\frac{-7}{12}=\frac{-7\times 2}{12\times 2}=\frac{-14}{24}$

∴​ $\frac{5}{8}+\frac{-7}{12}=\frac{15}{24}+\frac{-14}{24}=\frac{15+(-14)}{24}=\frac{15-14}{24}=\frac{1}{24}$

3. ​The denominators of the given rational numbers are 9 and 6.

LCM of 9 and 6 is 18.

Now, 

$\frac{-8}{9}=\frac{-8\times 2}{9\times 2}=\frac{-16}{18}$ and $\frac{11}{6}=\frac{11\times 3}{6\times 3}=\frac{33}{18}$

∴​ $\frac{-8}{9}+\frac{11}{6}=\frac{-16}{18}+\frac{33}{18}=\frac{-16+33}{18}=\frac{-16+33}{18}=\frac{17}{18}$

4.​ The denominators of the given rational numbers are 16 and 24.

LCM of 16 and 24 is 48.

Now, 

$\frac{-5}{16}=\frac{-5\times 3}{16\times 3}=\frac{-15}{48}$ and $\frac{7}{24}=\frac{7\times 2}{24\times 2}=\frac{14}{48}$

∴​ $\frac{-5}{16}+\frac{7}{24}=\frac{-15}{48}+\frac{14}{48}=\frac{-15+14}{48}=\frac{-1}{48}$

5. We will first write each of the given numbers with positive denominators.

$\frac{7}{-18}=\frac{7\times (-1)}{-18\times (-1)}=\frac{-7}{18}$

​The denominators of the given rational numbers are 18 and 27.

LCM of 18 and 27 is 54.

Now, 

$\frac{-7}{18}=\frac{-7\times 3}{18\times 3}=\frac{-21}{54}$ and $\frac{8}{27}=\frac{8\times 2}{27\times 2}=\frac{16}{54}$

∴ $\frac{7}{-18}+\frac{8}{27}=\frac{-21}{54}+\frac{16}{54}=\frac{-21+16}{54}=\frac{-5}{54}$

6. ​We will first write each of the given numbers with positive denominators.

$\frac{1}{-12}=\frac{1\times (-1)}{-12\times (-1)}=\frac{-1}{12}$ and $\frac{2}{-15}=\frac{2\times (-1)}{-15\times (-1)}=\frac{-2}{15}$

​The denominators of the given rational numbers are 12 and 15.

LCM of 12 and 15 is 60.

Now, 

$\frac{-1}{12}=\frac{-1\times 5}{12\times 5}=\frac{-5}{60}$ and $\frac{-2}{15}=\frac{-2\times 4}{15\times 4}=\frac{-8}{60}$

∴ $\frac{1}{-12}+\frac{2}{-15}=\frac{-5}{60}+\frac{-8}{60}=\frac{-5+(-8)}{60}=\frac{-5-8}{60}=\frac{-13}{60}$

7. We can write -1 as$\frac{-1}{1}$.

The denominators of the given rational numbers are 1 and 4.

LCM of 1 and 4 is 4.

Now, 

$\frac{-1}{1}=\frac{-1\times 4}{1\times 4}=\frac{-4}{4}$ and $\frac{3}{4}=\frac{3\times 1}{4\times 1}=\frac{3}{4}$

∴ $-1+\frac{3}{4}=\frac{-4}{4}+\frac{3}{4}=\frac{-4+3}{4}=\frac{-1}{4}$

8. ​We can write 2 as$\frac{2}{1}$.

The denominators of the given rational numbers are 1 and 4.

LCM of 1 and 4 is 4.

Now, 

$\frac{2}{1}=\frac{2\times 4}{1\times 4}=\frac{8}{4}$ and $\frac{-5}{4}=\frac{-5\times 1}{4\times 1}=\frac{-5}{4}$

∴ $2+\frac{(-5)}{4}=\frac{8}{4}+\frac{(-5)}{4}=\frac{8+(-5)}{4}=\frac{8-5}{4}=\frac{3}{4}$

9. ​We can write 0 as$\frac{0}{1}$.

The denominators of the given rational numbers are 1 and 5.

LCM of 1 and 5 is 5, that is, (1 $\times$ 5).

Now,
 
$\frac{0}{1}=\frac{0\times 5}{1\times 5}=\frac{0}{5}=0$ and $\frac{-2}{5}=\frac{-2\times 1}{5\times 1}=\frac{-2}{5}$

∴ $0+\frac{(-2)}{5}=\frac{0}{5}+\frac{(-2)}{5}=\frac{0+(-2)}{5}=\frac{0-2}{5}=\frac{-2}{5}$

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Question 3:

Verify the following:

(i) $\frac{-12}{5}+\frac{2}{7}=\frac{2}{7}+\frac{-12}{5}$

(ii) $\frac{-5}{8}+\frac{-9}{13}=\frac{-9}{13}+\frac{-5}{8}$

(iii) $3+\frac{-7}{12}=\frac{-7}{12}+3$

(iv) $\frac{2}{-7}+\frac{12}{-35}=\frac{12}{-35}+\frac{2}{-7}$

Answer 3 :

(i) LHS$=\frac{-12}{5}+\frac{2}{7}$

LCM of 5 and 7 is 35

$\frac{-12 \times 7}{5 \times 7}+\frac{2 \times 5}{7 \times 5}$

$=\frac{-84}{35}+\frac{10}{35}$

$=\frac{-84+10}{35}=\frac{-74}{35}$

RHS$=\frac{2}{7}+\frac{-12}{5}$

LCM of 5 and 7 is 35

$\frac{2 \times 5}{7 \times 5}+\frac{-12 \times 7}{5 \times 7}=\frac{10}{35}+\frac{-84}{35}$

$=\frac{10-84}{35}=\frac{-74}{35}$

∴ $\frac{-12}{5}+\frac{2}{7}=\frac{2}{7}+\frac{-12}{5}$

(2) ​LHS $=\frac{-5}{8}+\frac{-9}{13}$

LCM of 8 and 13 is 104

$\frac{-5 \times 13}{8 \times 13}+\frac{-9 \times 8}{13 \times 8}=\frac{-65}{104}+\frac{-72}{104}$

$=\frac{-65+(-72)}{104}=\frac{-65-72}{104}$

$=\frac{-137}{104}$

RHS$=\frac{-9}{13}+\frac{-5}{8}$

LCM of 13 and 8 is 104

$\frac{-9 \times 8}{13 \times 8}+\frac{-5 \times 13}{8 \times 13}$

$=\frac{-72}{104}+\frac{-65}{104}$

$=\frac{-72-65}{104}=\frac{-137}{104}$

∴ $\frac{-5}{8}+\frac{-9}{13}=\frac{-9}{13}+\frac{-5}{8}$

(3) LHS$=\frac{3}{1}+\frac{-7}{12}$

LCM of 1 and 12 is 12

$\frac{3 \times 12}{1 \times 12}+\frac{-7 \times 1}{12 \times 1}$

$=\frac{36}{12}+\frac{-7}{12}=\frac{36+(-7)}{12}$

$=\frac{36-7}{12}=\frac{29}{12}$

RHS $=\frac{-7}{12}+\frac{3}{1}$

LCM of 12 and 1 is 12

$\frac{-7 \times 1}{12 \times 1}+\frac{3 \times 12}{1 \times 12}$

$=\frac{-7}{12}+\frac{36}{12}$

$=\frac{-7+36}{12}=\frac{29}{12}$

∴ $3+\frac{-7}{12}=\frac{-7}{12}+3$

(4) LHS = ​$\frac{2}{-7}+\frac{12}{-35}$

We will write the given numbers with positive denominators.

$\frac{2}{-7}=\frac{2 \times(-1)}{-7 \times(-1)}=\frac{-2}{7}$ and $\frac{12}{-35}=\frac{12 \times(-1)}{-35 \times(-1)}=\frac{-12}{35}$

LCM of 7 and 35 is 35

$\frac{-2 \times 5}{7 \times 5}+\frac{-12 \times 1}{35 \times 1}=\frac{-10}{35}+\frac{-12}{35}=\frac{-10+(-12)}{35}$ $=\frac{-10-12}{35}=\frac{-22}{35}$

RHS$=\frac{12}{-35}+\frac{2}{-7}$

We will write the given numbers with positive denominators.

$\frac{12}{-35}=\frac{12 \times(-1)}{-35 \times(-1)}=\frac{-12}{35}$ and $\frac{2}{-7}=\frac{2 \times(-1)}{-7 \times(-1)}=\frac{-2}{7}$

LCM of 35 and 7 is 35.

$\frac{-2 \times 5}{7 \times 5}+\frac{-12 \times 1}{35 \times 1}=\frac{-10}{35}+\frac{-12}{35}$ $=\frac{-10+(-12)}{35}=\frac{-10-12}{35}=\frac{-22}{35}$
∴​ $\frac{2}{-7}+\frac{12}{-35}=\frac{12}{-35}+\frac{2}{-7}$

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Question 4:

Verify the following:
(i) $\left(\frac{3}{4}+\frac{-2}{5}\right)+\frac{-7}{10}=\frac{3}{4}+\left(\frac{-2}{5}+\frac{-7}{10}\right)$
(ii) $\left(\frac{-7}{11}+\frac{2}{-5}\right)+\frac{-13}{22}=\frac{-7}{11}+\left(\frac{2}{-5}+\frac{-13}{22}\right)$
(iii) $-1+\left(\frac{-2}{3}+\frac{-3}{4}\right)=\left(-1+\frac{-2}{3}\right)+\frac{-3}{4}$

Answer 4:

(1)
LHS =  $\left\{\left(\frac{3}{4}+\frac{-2}{5}\right)+\frac{-7}{10}\right\}$

$\left\{\left(\frac{15-8}{20}\right)+\frac{-7}{10}\right\}=\left(\frac{7}{20}+\frac{-7}{10}\right)$ $=\left(\frac{7}{20}+\frac{-14}{20}\right)=\left(\frac{7+(-14)}{20}\right)=\frac{-7}{20}$

RHS $=\left\{\frac{3}{4}+\left(\frac{-2}{5}+\frac{-7}{10}\right)\right\}$

$\left\{\frac{3}{4}+\left(\frac{-4}{10}+\frac{-7}{10}\right)\right\}=\left\{\frac{3}{4}+\left(\frac{-4-7}{10}\right)\right\}$ $=\left\{\frac{3}{4}+\left(\frac{-11}{10}\right)\right\}=\left(\frac{3}{4}+\frac{-11}{10}\right)=\left(\frac{15}{20}+\frac{-22}{20}\right)$ $=\left(\frac{15-22}{20}\right)=\frac{-7}{20}$

∴​ $\left(\frac{3}{4}+\frac{-2}{5}\right)+\frac{-7}{10}=\frac{3}{4}+\left({\displaystyle \frac{-2}{5}}+{\displaystyle \frac{-7}{10}}\right)$


(2)
LHS =  $\left\{\left(\frac{-7}{11}+\frac{2}{-5}\right)+\frac{-13}{22}\right\}$

We will first make the denominator positive.

$\left\{\left(\frac{-7}{11}+\frac{2\times (-1)}{-5\times (-1)}\right)+\frac{-13}{22}\right\}=\left\{\left(\frac{-7}{11}+\frac{-2}{5}\right)+\frac{-13}{22}\right\}$

$\left\{\left(\frac{-7}{11}+\frac{-2}{5}\right)+\frac{-13}{22}\right\}=\left\{\left(\frac{-35}{55}+\frac{-22}{55}\right)+\frac{-13}{22}\right\}$ $=\left\{\left(\frac{-35-22}{55}\right)+\frac{-13}{22}\right\}=\left(\frac{-57}{55}+\frac{-13}{22}\right)=\frac{-114}{110}+\frac{-65}{10}$ $=\frac{-114-65}{110}=\frac{-179}{110}$

RHS = $\left\{\frac{-7}{11}+\left(\frac{2}{-5}+{\displaystyle \frac{-13}{22}}\right)\right\}$

We will first make the denominator positive.

$\left\{\left(\frac{-7}{11}+\frac{2 \times(-1)}{-5 \times(-1)}\right)+\frac{-13}{22}\right\}=\left\{\left(\frac{-7}{11}+\frac{-2}{5}\right)+\frac{-13}{22}\right\}$

$\left\{\frac{-7}{11}+\left(\frac{-2}{5}+\frac{-13}{22}\right)\right\}=\left\{\frac{-7}{11}+\left(\frac{-44}{110}+\frac{-65}{110}\right)\right\}$

$=\left\{\frac{-7}{11}+\left(\frac{-44+(-65)}{110}\right)\right\}=\frac{-7}{11}+\frac{-109}{110}$

$=\frac{-70}{110}+\frac{-109}{110}=\frac{-70-109}{110}=\frac{-179}{110}$

∴​ $\left(\frac{-7}{11}+\frac{2}{-5}\right)+\frac{-13}{22}=\frac{-7}{11}+\left({\displaystyle \frac{2}{-5}}+{\displaystyle \frac{-13}{22}}\right)$

(3)
LHS $=-1+\left(\frac{-2}{3}+\frac{-3}{4}\right)$

$\left\{\left(\frac{-7}{11}+\frac{-2}{5}\right)+\frac{-13}{22}\right\}=\left\{\left(\frac{-35}{55}+\frac{-22}{55}\right)+\frac{-13}{22}\right\}$ $=\left\{\left(\frac{-35-22}{55}\right)+\frac{-13}{22}\right\}=\left(\frac{-57}{55}+\frac{-13}{22}\right)$ $=\frac{-114}{110}+\frac{-65}{110}=\frac{-114-65}{110}=\frac{-179}{110}$


RHS $=\left\{\left(-1+\frac{-2}{3}\right)+\frac{-3}{4}\right\}$

$\left\{\left(\frac{-1}{1}+\frac{-2}{3}\right)+\frac{-3}{4}\right\}=\left\{\left(\frac{-3}{3}+\frac{-2}{3}\right)+\frac{-3}{4}\right\}$ $=\left\{\left(\frac{-3-2}{3}\right)+\frac{-3}{4}\right\}=\left\{\left(\frac{-5}{3}\right)+\frac{-3}{4}\right\}=\left(\frac{-5}{3}+\frac{-3}{4}\right)$ $=\left(\frac{-20}{12}+\frac{-9}{12}\right)=\left(\frac{-20-9}{12}\right)=\frac{-29}{12}$

∴ $-1+\left(\frac{-2}{3}+\frac{-3}{4}\right)=\left(-1+\frac{-2}{3}\right)+\frac{-3}{4}$

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Question 5:

Fill in the blanks.
(i) $\left(\frac{-3}{17}\right)+\left(\frac{-12}{5}\right)=\left(\frac{-12}{5}\right)+\left(......\right)$
(ii) $-9+\frac{-21}{8}=\left(......\right)+\left(-9\right)$
(iii) $\left(\frac{-8}{13}+\frac{3}{7}\right)+\left(\frac{-13}{4}\right)=\left(......\right)+\left[\frac{3}{7}+\left(\frac{-13}{4}\right)\right]$
(iv) $-12+\left(\frac{7}{12}+\frac{-9}{11}\right)=\left(-12+\frac{7}{12}\right)+\left(......\right)$
(v) $\frac{19}{-5}+\left(\frac{-3}{11}+\frac{-7}{8}\right)=\left\{\frac{19}{-5}+\left(......\right)\right\}+\frac{-7}{8}$
(vi) $\frac{-16}{7}+......=......+\frac{-16}{7}=\frac{-16}{7}$

Answer 5:

(i) Addition is commutative, that is, $a+b=b+a$.

Hence, the required solution is $\left({\displaystyle \frac{-3}{17}}\right)+\left({\displaystyle \frac{-12}{5}}\right)=\left({\displaystyle \frac{-12}{5}}\right)+\boxed{\left({\displaystyle \frac{-3}{7}}\right)}$.
 
(ii) Addition is commutative, that is, $a+b=b+a$.

Hence, the required solution is $-9+\frac{-21}{8}=\frac{-21}{8}+\boxed{-9}$.

(iii) Addition is associative, that is, $\left(a+b\right)+c=a+\left(b+c\right)$.

Hence, the required solution is $\left(\frac{-8}{13}+\frac{3}{7}\right)+\left(\frac{-13}{4}\right)=\boxed{\left({\displaystyle \frac{-8}{13}}\right)}+\left[{\displaystyle \frac{3}{7}}+\left({\displaystyle \frac{-13}{4}}\right)\right]$.

(iv) Addition is associative, that is, $\left(a+b\right)+c=a+\left(b+c\right)$.

Hence, the required solution is $-12+\left(\frac{7}{12}+\frac{-9}{11}\right)=\left(-12+{\displaystyle \frac{7}{12}}\right)+\frac{-9}{11}$.

(v) Addition is associative, that is, $\left(a+b\right)+c=a+\left(b+c\right)$.

Hence, the required solution is$\frac{19}{-5}+\left({\displaystyle \frac{-3}{11}}+{\displaystyle \frac{-7}{8}}\right)=\left\{\frac{19}{-5}+\boxed{\left({\displaystyle \frac{-3}{11}}\right)}\right\}+\frac{-7}{8}$.

(vi) 0 is the additive identity, that is, $0+a=a$.

Hence, the required solution is $\frac{-16}{7}+\boxed{0}=\boxed{0}+\frac{-16}{7}=\frac{-16}{7}$.

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Question 6:

Find the additive inverse of each of the following:
(i) $\frac{1}{3}$
(ii) $\frac{23}{9}$
(iii) −18
(iv) $\frac{-17}{8}$
(v) $\frac{15}{-4}$
(vi) $\frac{-16}{-5}$
(vii) $\frac{-3}{11}$
(viii) 0
(ix) $\frac{19}{-6}$
(x) $\frac{-8}{-7}$

Answer 6:

The additive inverse of $\frac{a}{b}$ is $\frac{-a}{b}$. Therefore, $\frac{a}{b}+\left(\frac{-a}{b}\right)=0$
(i) Additive inverse of $\frac{1}{3}\mathrm{is}\frac{-1}{3}.$

(ii) Additive inverse of  $\frac{23}{9}\mathrm{is}\frac{-23}{9}.$

(iii) Additive inverse of -18 is 18.

(iv) Additive inverse of $\frac{-17}{8}\mathrm{is}\frac{17}{8}.$

(v) In the standard form, we write $\frac{15}{-4}\mathrm{as}\frac{-15}{4}.$

    Hence, its additive inverse is $\frac{15}{4}$.

(vi) We can write:
 
$\frac{-16}{-5}=\frac{-16\times (-1)}{-5\times (-1)}=\frac{16}{5}$

    Hence, its additive inverse is $\frac{-16}{5}$.

(vii) Additive inverse of $\frac{-3}{11}\mathrm{is}\frac{3}{11}.$

(viii) Additive inverse of 0 is 0.

(ix) In the standard form, we write $\frac{19}{-6}\mathrm{as}\frac{-19}{6}.$

     Hence, its additive inverse is $\frac{19}{6}$.

(x) We can write:
 
$\frac{-8}{-7}=\frac{-8\times (-1)}{-7\times (-1)}=\frac{8}{7}$

Hence, its additive inverse is $\frac{-8}{7}$.

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Question 7:

Subtract:
(i) $\frac{3}{4} \mathrm{from} \frac{1}{3}$
(ii) $\frac{-5}{6} \mathrm{from} \frac{1}{3}$
(iii) $\frac{-8}{9} \mathrm{from} \frac{-3}{5}$
(iv) $\frac{-9}{7} \mathrm{from} -1$
(v) $\frac{-18}{11} \mathrm{from} 1$
(vi) $\frac{-13}{9} \mathrm{from} 0$
(vii) $\frac{-32}{13} \mathrm{from} \frac{-6}{5}$
(viii) $-7 \mathrm{from} \frac{-4}{7}$

Answer 7:

(i) $\left(\frac{1}{3}-\frac{3}{4}\right)=\frac{1}{3}+\left(\right.$ Additive inverse of $\left.\frac{3}{4}\right)$

$=\left(\frac{1}{3}+\frac{-3}{4}\right)=\left(\frac{4}{12}+\frac{-9}{12}\right)$

$=\left(\frac{4-9}{12}\right)=\frac{-5}{12}$


(ii) $\left(\frac{1}{3}-\frac{-5}{6}\right)=\frac{1}{3}+\left(\right.$ Additive inverse of $\left.\frac{-5}{6}\right)$

$=\left(\frac{1}{3}+\frac{5}{6}\right)$ (Because the additive inverse of $\frac{-5}{6}$ is $\frac{5}{6}$ )

$=\left(\frac{2}{6}+\frac{5}{6}\right)=\left(\frac{2+5}{6}\right)=\frac{7}{6}$


(iii) $\left(\frac{-3}{5}-\frac{-8}{9}\right)=\frac{-3}{5}+\left(\text { Additive inverse of } \frac{-8}{9}\right)$

$=\left(\frac{-3}{5}+\frac{8}{9}\right)$ (Because the additive inverse of $\frac{-8}{9}$ is $\frac{8}{9}$ )

$=\left(\frac{-27}{45}+\frac{40}{45}\right)=\left(\frac{-27+40}{45}\right)=\frac{13}{45}$


(iv) $\left(-1-\frac{-9}{7}\right)=-1+\left(\right.$ Additive inverse of $\left.\frac{-9}{7}\right)$

$=\left(\frac{-1}{1}+\frac{9}{7}\right)$ (Because the additive inverse of $\frac{-9}{7}$ is $\frac{9}{7}$ )

$=\left(\frac{-7}{7}+\frac{9}{7}\right)=\left(\frac{-7+9}{7}\right)=\frac{2}{7}$


(v) $\left(1-\frac{-18}{11}\right)=1+\left(\right.$ Additive inverse of $\left.\frac{-18}{11}\right)$

$=\left(\frac{1}{1}+\frac{18}{11}\right)$ (Because the additive inverse of $\frac{-18}{11}$ is $\frac{18}{11}$ )

$=\left(\frac{11}{11}+\frac{18}{11}\right)=\left(\frac{11+18}{11}\right)=\frac{29}{11}$


(vi) $\left(0-\frac{-13}{9}\right)=0+\left(\right.$ Additive inverse of $\left.\frac{-13}{9}\right)$

$=\left(0+\frac{13}{9}\right)$ (Because the additive inverse of $\frac{-13}{9}$ is $\frac{13}{9}$ )

$=\frac{13}{9}$


(vii) $\left(\frac{-6}{5}-\frac{-32}{13}\right)=\frac{-6}{5}+\left(\right.$ Additive inverse of $\left.\frac{-32}{13}\right)$

$=\left(\frac{-6}{5}+\frac{32}{13}\right)$ (Because the additive inverse of $\frac{-32}{13}$ is $\frac{32}{13}$ )

$=\left(\frac{-78}{65}+\frac{160}{65}\right)=\left(\frac{-78+160}{65}\right)=\frac{82}{65}$


(viii) $\left(\frac{-4}{7}-\frac{-7}{1}\right)=\frac{-4}{7}+\left(\right.$ Additive inverse of $\left.\frac{-7}{1}\right)$

$=\left(\frac{-4}{7}+\frac{7}{1}\right)$ (Because the additive inverse of $\frac{-7}{1}$ is $\frac{7}{1}$ )

$=\left(\frac{-4}{7}+\frac{49}{7}\right)=\left(\frac{-4+49}{7}\right)=\frac{45}{7}$

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Question 8:

Using the rearrangement property find the sum:
(i) $\frac{4}{3}+\frac{3}{5}+\frac{-2}{3}+\frac{-11}{5}$
(ii) $\frac{-8}{3}+\frac{-1}{4}+\frac{-11}{6}+\frac{3}{8}$
(iii) $\frac{-13}{20}+\frac{11}{14}+\frac{-5}{7}+\frac{7}{10}$
(iv) $\frac{-6}{7}+\frac{-5}{6}+\frac{-4}{9}+\frac{-15}{7}$

Answer 8:

(i)
 $\left(\frac{4}{3}+\frac{-2}{3}\right)+\left({\displaystyle \frac{3}{5}}+{\displaystyle \frac{-11}{5}}\right)$
= $\left(\frac{4-2}{3}\right)+\left(\frac{3-11}{5}\right)$
$$\begin{gathered} =\left({\displaystyle \frac{2}{3}}+{\displaystyle \frac{-8}{5}}\right) \\ =\left({\displaystyle \frac{10}{15}}+{\displaystyle \frac{-24}{15}}\right) \\ =\left({\displaystyle \frac{10-24}{15}}\right) \\ =\frac{-14}{15} \end{gathered}$$.


(ii)
$\left(\frac{-8}{3}+\frac{-11}{6}\right)+\left({\displaystyle \frac{-1}{4}}+{\displaystyle \frac{3}{8}}\right)$

=$\left({\displaystyle \frac{-16}{6}}+{\displaystyle \frac{-11}{6}}\right)+\left({\displaystyle \frac{-2}{8}}+{\displaystyle \frac{3}{8}}\right)$

=$\left(\frac{-16-11}{6}\right)+\left(\frac{-2+3}{8}\right)$

$$\begin{gathered} =\left(\frac{-27}{6}+\frac{1}{8}\right) \\ =\left({\displaystyle \frac{-108}{24}}+{\displaystyle \frac{3}{24}}\right) \\ =\frac{-105}{24} \end{gathered}$$
=$\frac{35}{8}$


(iii)
$\left(\frac{-13}{20}+\frac{7}{10}\right)+\left({\displaystyle \frac{11}{14}}+{\displaystyle \frac{-5}{7}}\right)$
=$\left({\displaystyle \frac{-13}{20}}+{\displaystyle \frac{14}{20}}\right)+\left({\displaystyle \frac{11}{14}}+{\displaystyle \frac{-10}{14}}\right)$
=$\left(\frac{-13+14}{20}\right)+\left(\frac{11-10}{14}\right)$
$$\begin{gathered} =\left(\frac{1}{20}+\frac{1}{14}\right) \\ =\left({\displaystyle \frac{7}{140}}+{\displaystyle \frac{10}{140}}\right) \\ =\left(\frac{7+10}{140}\right) \\ =\left({\displaystyle \frac{17}{140}}\right) \\ =\frac{17}{140} \end{gathered}$$.


(iv)
$\left(\frac{-6}{7}+\frac{-15}{7}\right)+\left({\displaystyle \frac{-5}{6}}+{\displaystyle \frac{-4}{9}}\right)$

=$\left(\frac{-6}{7}+\frac{-15}{7}\right)+\left({\displaystyle \frac{-15}{18}}+{\displaystyle \frac{-8}{18}}\right)$

=$\left(\frac{-6-15}{7}\right)+\left({\displaystyle \frac{-15-8}{18}}\right)$

$$\begin{gathered} =\left({\displaystyle \frac{-21}{7}}+{\displaystyle \frac{-23}{18}}\right) \\ =\left({\displaystyle \frac{-3}{1}}+{\displaystyle \frac{-23}{18}}\right) \\ =\left({\displaystyle \frac{-54}{18}}+{\displaystyle \frac{-23}{18}}\right) \\ =\left(\frac{-54-23}{18}\right) \\ =\frac{-77}{18} \end{gathered}$$

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

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Question 9:

The sum of two rational numbers is −2. If one of the numbers  is $\frac{-14}{5}$, find the other.

Answer 9:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{be} x. \\ \mathrm{Now}, \\ \Rightarrow x+\frac{-14}{5}=-2 \\ \Rightarrow x-\frac{14}{5}=-2 \\ \Rightarrow x=-2+\frac{14}{5} \\ \Rightarrow x=\frac{(-2)\times 5+14}{5} \\ \Rightarrow x=\frac{-10+14}{5} \\ \Rightarrow x =\frac{4}{5} \end{gathered}$$

Q10 | Ex-1C | Class 8 | RS AGGARWAL | Chapter 1 | Rational Numbers| myhelper

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Question 10:

The sum of two rational numbers is $\frac{-1}{2}$. If one of the numbers is $\frac{5}{6}$, find the other.

Answer 10:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{be} x. \\ \mathrm{Now}, \\ x+\frac{5}{6}=\frac{-1}{2} \\ \Rightarrow x=-\frac{1}{2}-\frac{5}{6} \\ \Rightarrow x=\frac{-3-5}{6} \\ \Rightarrow x=\frac{-8}{6} \\ \Rightarrow x=\frac{-4}{3} \end{gathered}$$

Q11 | Ex-1C | Class 8 | RS AGGARWAL | Chapter 1 | Rational Numbers| myhelper

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Question 11:

What number should be added to $\frac{-5}{8}$ so as to get $\frac{-3}{2}$?

Answer 11:

Let the required number be x.

Now,

$\frac{-5}{8}+x=\frac{-3}{2}$

$\Rightarrow \frac{-5}{6}+x+\frac{5}{8}=\frac{-3}{2}+\frac{5}{6}$ (Adding $\frac{5}{8}$ to both the sides)

$\Rightarrow x=\left(\frac{-3}{2}+\frac{5}{8}\right)$

$\Rightarrow x=\left(\frac{-12}{8}+\frac{5}{8}\right)$

$\Rightarrow x=\left(\frac{-12+5}{8}\right)$

$\Rightarrow x=\frac{-7}{8}$

Hence, the required number is $\frac{-7}{8}$

Q12 | Ex-1C | Class 8 | RS AGGARWAL | Chapter 1 | Rational Numbers| myhelper

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Question 12:

What number should be added to −1 so as to get $\frac{5}{7}$?

Answer 12:

Let the required number be x.
 
Now,

$-1+x=\frac{5}{7}$
$\Rightarrow -1+x+1=\frac{5}{7}+1$     (Adding 1 to both the sides)

$$\begin{gathered} \Rightarrow x=\left({\displaystyle \frac{5+7}{7}}\right) \\ \Rightarrow x=\frac{12}{7} \end{gathered}$$
Hence, the required number is $\frac{12}{7}$.

Q13 | Ex-1C | Class 8 | RS AGGARWAL | Chapter 1 | Rational Numbers| myhelper

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Question 13:

What number should be subtracted from $\frac{-2}{3}$ to get $\frac{-1}{6}$?

Answer 13:

Let the required number be x.
 
Now,

$\frac{-2}{3}-x=\frac{-1}{6}$

$\Rightarrow \frac{-2}{3}-x+x=\frac{-1}{6}+x$  
(Adding x to both the sides)

$\Rightarrow \frac{-2}{3}=\frac{-1}{6}+x$

$\Rightarrow \frac{-2}{3}+\frac{1}{6}=\frac{-1}{6}+x+\frac{1}{6}$ (Adding $\frac{1}{6}$ to both the sides)

$\Rightarrow\left(\frac{-4}{6}+\frac{1}{6}\right)=x$

$\Rightarrow\left(\frac{-4+1}{6}\right)=x$

$\Rightarrow \frac{-3}{6}=x$

$\Rightarrow \frac{-1 \times 3}{2 \times 3}=x$

$\Rightarrow \frac{-1}{2}=x$

Hence, the required number is $\frac{-1}{2}$

Q14 | Ex-1C | Class 8 | RS AGGARWAL | Chapter 1 | Rational Numbers| myhelper

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Question 14:

(i) Which rational number is its own additive inverse?
(ii) Is the difference of two rational numbers a rational number?
(iii) Is addition commutative on rational numbers?
(iv) Is addition associative on rational numbers?
(v) Is subtraction commutative on rational numbers?
(vi) Is subtraction associative on rational numbers?
(vii) What is the negative of a negative rational number?

Answer 14:

1. Zero is a rational number that is its own additive inverse.

2. Yes

Consider $\frac{a}{b}-\frac{c}{d}$ (with a, b, c and d as integers), where b and d are not equal to 0.

$\frac{a}{b}-\frac{c}{d}$ implies $\frac{a d}{b d}-\frac{b c}{b d}$ implies $\left(\frac{a d-b c}{b d}\right)$

Since ad, bc and bd are integers since integers are closed under the operation of multiplication and ad-bc is an integer since integers are closed under the operation of subtraction, then ad-bcbd
since it is in the form of one integer divided by another and the denominator is not equal to 0
Since, b and d were not equal to 0

Thus, $\frac{a}{b}-\frac{c}{d}$ is a rational number.

​3. Yes, rational numbers are commutative under addition. If a and b are rational numbers, then the commutative law under addition is $a+b=b+a$.

4. Yes, rational numbers are associative under addition. If a, b and c are rational numbers, then the associative law under addition is $a+(b+c)=(a+b)+c$.

5. No, subtraction is not commutative on rational numbers. In general, for any two rational numbers, $(a-b) \ne (b - a)$.

6. Rational numbers are not associative under subtraction. Therefore, $a-(b-c)\ne (a-b)-c$.

7. Negative of a negative rational number is a positive rational number.