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#### Question 1:

If $\frac{a}{b}$ is a fraction and $m$ is a non-zero integer, then $\frac{a}{b}=\frac{a×m}{b×m}$.

Now,

(i) $\frac{-3}{5}=\frac{-3×4}{5×4}=\frac{-12}{20}$

(ii) $\frac{-3}{5}=\frac{-3×-6}{5×-6}=\frac{18}{-30}$

(iii)$\frac{-3}{5}=\frac{-3×7}{5×7}=\frac{-21}{35}$

(iv)$\frac{-3}{5}=\frac{-3×-8}{5×-8}=\frac{24}{-40}$

#### Question 2:

If $\frac{a}{b}$ is a fraction and $m$ is a non-zero integer, then $\frac{a}{b}=\frac{a×m}{b×m}$.

Now,

(i) $\frac{-3}{5}=\frac{-3×4}{5×4}=\frac{-12}{20}$

(ii) $\frac{-3}{5}=\frac{-3×-6}{5×-6}=\frac{18}{-30}$

(iii)$\frac{-3}{5}=\frac{-3×7}{5×7}=\frac{-21}{35}$

(iv)$\frac{-3}{5}=\frac{-3×-8}{5×-8}=\frac{24}{-40}$

If $\frac{a}{b}$ is a rational number and $m$ is a common divisor of , then $\frac{a}{b}=\frac{a÷m}{b÷m}$.

∴ $\frac{-42}{98}=\frac{-42÷14}{98÷14}=\frac{-3}{7}$

#### Question 3:

If $\frac{a}{b}$ is a rational number and $m$ is a common divisor of , then $\frac{a}{b}=\frac{a÷m}{b÷m}$.

∴ $\frac{-42}{98}=\frac{-42÷14}{98÷14}=\frac{-3}{7}$

If $\frac{a}{b}$ is a rational integer and $m$ is a common divisor of , then $\frac{a}{b}=\frac{a÷m}{b÷m}$.

∴â€‹ $\frac{-48}{60}=\frac{-48÷12}{60÷12}=\frac{-4}{5}$

#### Question 4:

If $\frac{a}{b}$ is a rational integer and $m$ is a common divisor of , then $\frac{a}{b}=\frac{a÷m}{b÷m}$.

∴â€‹ $\frac{-48}{60}=\frac{-48÷12}{60÷12}=\frac{-4}{5}$

A rational number $\frac{a}{b}$ is said to be in the standard form if $a$ and $b$ have no common divisor other than unity and $b>0$.
Thus,

(i) The greatest common divisor of 12 and 30 is 6.

∴ $\frac{-12}{30}=\frac{-12÷6}{30÷6}=\frac{-2}{5}$ (In the standard form)

(ii)The greatest common divisor of 14 and 49 is 7.

∴ $\frac{-14}{49}=\frac{-14÷7}{49÷7}=\frac{-2}{7}$ (In the standard form)

(iii) $\frac{24}{-64}=\frac{24×\left(-1\right)}{-64×-1}=\frac{-24}{64}$

The greatest common divisor of 24 and 64 is 8.

∴ $\frac{-24}{64}=\frac{-24÷8}{64÷8}=\frac{-3}{8}$ (In the standard form)

(iv) $\frac{-36}{-63}=\frac{-36×\left(-1\right)}{-63×-1}=\frac{36}{63}$

The greatest common divisor of 36 and 63 is 9.

∴ $\frac{36}{63}=\frac{36÷9}{63÷9}=\frac{4}{7}$ (In the standard form)

#### Question 5:

A rational number $\frac{a}{b}$ is said to be in the standard form if $a$ and $b$ have no common divisor other than unity and $b>0$.
Thus,

(i) The greatest common divisor of 12 and 30 is 6.

∴ $\frac{-12}{30}=\frac{-12÷6}{30÷6}=\frac{-2}{5}$ (In the standard form)

(ii)The greatest common divisor of 14 and 49 is 7.

∴ $\frac{-14}{49}=\frac{-14÷7}{49÷7}=\frac{-2}{7}$ (In the standard form)

(iii) $\frac{24}{-64}=\frac{24×\left(-1\right)}{-64×-1}=\frac{-24}{64}$

The greatest common divisor of 24 and 64 is 8.

∴ $\frac{-24}{64}=\frac{-24÷8}{64÷8}=\frac{-3}{8}$ (In the standard form)

(iv) $\frac{-36}{-63}=\frac{-36×\left(-1\right)}{-63×-1}=\frac{36}{63}$

The greatest common divisor of 36 and 63 is 9.

∴ $\frac{36}{63}=\frac{36÷9}{63÷9}=\frac{4}{7}$ (In the standard form)

We know:
(i) Every positive rational number is greater than 0.
(ii) Every negative rational number is less than 0.

Thus, we have:

(i)$\frac{3}{8}$ is a positive rational number.
∴ $\frac{3}{8}>0$

(ii)$\frac{-2}{9}$ is a negative rational number.
∴ $\frac{-2}{9}<0$

(iii) $\frac{-3}{4}$ is a negative rational number.
∴ $\frac{-3}{4}<0$
Also,
$\frac{1}{4}$ is a positive rational number.
∴ $\frac{1}{4}>0$
Combining the two inequalities, we get:
$\frac{-3}{4}<\frac{1}{4}$

(iv)Both $\frac{-5}{7}$ and $\frac{-4}{7}$ have the same denominator, that is, 7.
So, we can directly compare the numerators.

∴ $\frac{-5}{7}<\frac{-4}{7}$

(v)The two rational numbers are $\frac{2}{3}$ and $\frac{3}{4}$.
The LCM of the denominators 3 and 4 is 12.
Now,
$\frac{2}{3}=\frac{2×4}{3×4}=\frac{8}{12}$
Also,
$\frac{3}{4}=\frac{3×3}{4×3}=\frac{9}{12}$
Further
$\frac{8}{12}<\frac{9}{12}$

∴$\frac{2}{3}<\frac{3}{4}$

(vi)The two rational numbers are $\frac{-1}{2}$ and $-1$.
We can write $-1=\frac{-1}{1}$.
The LCM of the denominators 2 and 1 is 2.
Now,
$\frac{-1}{2}=\frac{-1×1}{2×1}=\frac{-1}{2}$
Also,
$\frac{-1}{1}=\frac{-1×2}{1×2}=\frac{-2}{2}$
âˆµ $\frac{-2}{1}<\frac{-1}{1}$
∴ $-1<\frac{-1}{2}$

#### Question 6:

We know:
(i) Every positive rational number is greater than 0.
(ii) Every negative rational number is less than 0.

Thus, we have:

(i)$\frac{3}{8}$ is a positive rational number.
∴ $\frac{3}{8}>0$

(ii)$\frac{-2}{9}$ is a negative rational number.
∴ $\frac{-2}{9}<0$

(iii) $\frac{-3}{4}$ is a negative rational number.
∴ $\frac{-3}{4}<0$
Also,
$\frac{1}{4}$ is a positive rational number.
∴ $\frac{1}{4}>0$
Combining the two inequalities, we get:
$\frac{-3}{4}<\frac{1}{4}$

(iv)Both $\frac{-5}{7}$ and $\frac{-4}{7}$ have the same denominator, that is, 7.
So, we can directly compare the numerators.

∴ $\frac{-5}{7}<\frac{-4}{7}$

(v)The two rational numbers are $\frac{2}{3}$ and $\frac{3}{4}$.
The LCM of the denominators 3 and 4 is 12.
Now,
$\frac{2}{3}=\frac{2×4}{3×4}=\frac{8}{12}$
Also,
$\frac{3}{4}=\frac{3×3}{4×3}=\frac{9}{12}$
Further
$\frac{8}{12}<\frac{9}{12}$

∴$\frac{2}{3}<\frac{3}{4}$

(vi)The two rational numbers are $\frac{-1}{2}$ and $-1$.
We can write $-1=\frac{-1}{1}$.
The LCM of the denominators 2 and 1 is 2.
Now,
$\frac{-1}{2}=\frac{-1×1}{2×1}=\frac{-1}{2}$
Also,
$\frac{-1}{1}=\frac{-1×2}{1×2}=\frac{-2}{2}$
âˆµ $\frac{-2}{1}<\frac{-1}{1}$
∴ $-1<\frac{-1}{2}$

1. The two rational numbers are $\frac{-4}{3}\mathrm{and}\frac{-8}{7}$.

The LCM of the denominators 3 and 7 is 21.

Now,

$\frac{-4}{3}=\frac{-4×7}{3×7}=\frac{-28}{21}$

Also,

$\frac{-8}{7}=\frac{-8×3}{7×3}=\frac{-24}{21}$

Further,

$\frac{-28}{21}<\frac{-24}{21}$

∴ $\frac{-4}{3}<\frac{-8}{7}$

2. â€‹The two rational numbers are $\frac{7}{-9}\mathrm{and}\frac{-5}{8}$.

The first fraction can be expressed as $\frac{7}{-9}=\frac{7×-1}{-9×-1}=\frac{-7}{9}$.

The LCM of the denominators 9 and 8 is 72.

Now,

$\frac{-7}{9}=\frac{-7×8}{9×8}=\frac{-56}{72}$

Also,

$\frac{-5}{8}=\frac{-5×9}{8×9}=\frac{-45}{72}$

Further,

$\frac{-56}{72}<\frac{-45}{72}$

∴â€‹ $\frac{7}{-9}<\frac{-5}{8}$

3. â€‹The two rational numbers are .

$\frac{4}{-5}=\frac{4×-1}{-5×-1}=\frac{-4}{5}$

The LCM of the denominators 3 and 5 is 15.

Now,

$\frac{-1}{3}=\frac{-1×5}{3×5}=\frac{-5}{15}$

Also,

$\frac{-4}{5}=\frac{-4×3}{5×3}=\frac{-12}{15}$

Further,

$\frac{-12}{15}<\frac{-5}{15}$

∴ $\frac{4}{-5}<\frac{-1}{3}$

4. The two rational numbers are $\frac{9}{-13}\mathrm{and}\frac{7}{-12}$.

The LCM of the denominators 13 and 12 is 156.

Now,

$\frac{-9}{13}=\frac{-9×12}{13×12}=\frac{-108}{156}$

Also,

$\frac{-7}{12}=\frac{-7×13}{12×13}=\frac{-91}{156}$

Further,

$\frac{-108}{156}<\frac{-91}{156}$

∴ $\frac{9}{-13}<\frac{7}{-12}$

5. The two rational numbers are .

∴â€‹ $\frac{4}{-5}=\frac{4×-1}{-5×-1}=\frac{-4}{5}$

The LCM of the denominators 5 and 10 is 10.

Now,

$\frac{-4}{5}=\frac{-4×2}{5×2}=\frac{-8}{10}$

Also,

$\frac{-7}{10}=\frac{-7×1}{10×1}=\frac{-7}{10}$

Further,

$\frac{-8}{10}<\frac{-7}{10}$

∴

6. The two rational numbers are
.

The LCM of the denominators is 5.

Now,

$\frac{-3}{1}=\frac{-3×5}{1×5}=\frac{-15}{5}$

Because $\frac{-15}{5}<\frac{-12}{5}$, we can conclude that $-3<\frac{-12}{5}$.

#### Question 7:

1. The two rational numbers are $\frac{-4}{3}\mathrm{and}\frac{-8}{7}$.

The LCM of the denominators 3 and 7 is 21.

Now,

$\frac{-4}{3}=\frac{-4×7}{3×7}=\frac{-28}{21}$

Also,

$\frac{-8}{7}=\frac{-8×3}{7×3}=\frac{-24}{21}$

Further,

$\frac{-28}{21}<\frac{-24}{21}$

∴ $\frac{-4}{3}<\frac{-8}{7}$

2. â€‹The two rational numbers are $\frac{7}{-9}\mathrm{and}\frac{-5}{8}$.

The first fraction can be expressed as $\frac{7}{-9}=\frac{7×-1}{-9×-1}=\frac{-7}{9}$.

The LCM of the denominators 9 and 8 is 72.

Now,

$\frac{-7}{9}=\frac{-7×8}{9×8}=\frac{-56}{72}$

Also,

$\frac{-5}{8}=\frac{-5×9}{8×9}=\frac{-45}{72}$

Further,

$\frac{-56}{72}<\frac{-45}{72}$

∴â€‹ $\frac{7}{-9}<\frac{-5}{8}$

3. â€‹The two rational numbers are .

$\frac{4}{-5}=\frac{4×-1}{-5×-1}=\frac{-4}{5}$

The LCM of the denominators 3 and 5 is 15.

Now,

$\frac{-1}{3}=\frac{-1×5}{3×5}=\frac{-5}{15}$

Also,

$\frac{-4}{5}=\frac{-4×3}{5×3}=\frac{-12}{15}$

Further,

$\frac{-12}{15}<\frac{-5}{15}$

∴ $\frac{4}{-5}<\frac{-1}{3}$

4. The two rational numbers are $\frac{9}{-13}\mathrm{and}\frac{7}{-12}$.

The LCM of the denominators 13 and 12 is 156.

Now,

$\frac{-9}{13}=\frac{-9×12}{13×12}=\frac{-108}{156}$

Also,

$\frac{-7}{12}=\frac{-7×13}{12×13}=\frac{-91}{156}$

Further,

$\frac{-108}{156}<\frac{-91}{156}$

∴ $\frac{9}{-13}<\frac{7}{-12}$

5. The two rational numbers are .

∴â€‹ $\frac{4}{-5}=\frac{4×-1}{-5×-1}=\frac{-4}{5}$

The LCM of the denominators 5 and 10 is 10.

Now,

$\frac{-4}{5}=\frac{-4×2}{5×2}=\frac{-8}{10}$

Also,

$\frac{-7}{10}=\frac{-7×1}{10×1}=\frac{-7}{10}$

Further,

$\frac{-8}{10}<\frac{-7}{10}$

∴

6. The two rational numbers are
.

The LCM of the denominators is 5.

Now,

$\frac{-3}{1}=\frac{-3×5}{1×5}=\frac{-15}{5}$

Because $\frac{-15}{5}<\frac{-12}{5}$, we can conclude that $-3<\frac{-12}{5}$.

(i)We will write each of the given numbers with positive denominators.

One number = $\frac{-3}{7}$
Other number =$\frac{6}{-13}=\frac{6×\left(-1\right)}{-13×\left(-1\right)}=\frac{-6}{13}$

LCM of 7 and 13 = 91

∴ $\frac{-3}{7}=\frac{-3×13}{7×13}=\frac{-39}{91}$

And,

$\frac{-6}{13}=\frac{-6×7}{13×7}=\frac{-42}{91}$$\frac{-6}{13}=\frac{-6×7}{13×7}=\frac{-42}{91}$$\frac{-6}{13}=\frac{-6×7}{13×7}=\frac{-42}{91}$

Clearly,

$-39>-41$

∴ â€‹

Thus,

$\frac{-3}{7}>\frac{6}{-13}$

(ii) We will write each of the given numbers with positive denominators.

One number = $\frac{5}{-13}=\frac{5×\left(-1\right)}{-13×\left(-1\right)}=\frac{-5}{13}$

Other number =$\frac{-35}{91}$

LCM of 13 and 91 = 91

∴ $\frac{-5}{13}=\frac{-5×7}{13×7}=\frac{-35}{91}$ and $\frac{-35}{91}$

Clearly,

$-35=-35$

∴

Thus,

$\frac{-5}{13}=\frac{-35}{91}$

(iii) We will write each of the given numbers with positive denominators.

One number = $-2$

We can write -2 as$\frac{-2}{1}$.
Other number =$\frac{-13}{5}$

LCM of 1 and 5 = 5

∴â€‹ $\frac{-2}{1}=\frac{-2×5}{1×5}=\frac{-10}{5}$ and $\frac{-13}{5}=\frac{-13×1}{5×1}=\frac{-13}{5}$

Clearly,

$-10>-13$

∴ $\frac{-10}{5}>\frac{-13}{5}$

Thus,

$\frac{-2}{1}>\frac{-13}{5}$

$-2>\frac{-13}{5}$

(iv) We will write each of the given numbers with positive denominators.

One number = $\frac{-2}{3}$
Other number =$\frac{5}{-8}=\frac{5×\left(-1\right)}{-8×\left(-1\right)}=\frac{-5}{8}$

LCM of 3 and 8 = 24

∴ â€‹$\frac{-2}{3}=\frac{-2×8}{3×8}=\frac{-16}{24}$ and $\frac{-5}{8}=\frac{-5×3}{8×3}=\frac{-15}{24}$

Clearly,

$-16<-15$

∴ $\frac{-16}{24}<\frac{-15}{24}$

Thus,

$\frac{-2}{3}<\frac{-5}{8}$

$\frac{-2}{3}<\frac{5}{-8}$

(v) $\frac{-3}{-5}=\frac{-3×-1}{-5×-1}=\frac{3}{5}$

$\frac{3}{5}$ is a positive number.

Because every positive rational number is greater than 0, $\frac{3}{5}>0⇒0<\frac{3}{5}$.

(vi) We will write each of the given numbers with positive denominators.

One number = $\frac{-8}{9}$

Other number = $\frac{-9}{10}$

LCM of 9 and 10 = 90

∴â€‹$\frac{-8}{9}=\frac{-8×10}{9×10}=\frac{-80}{90}$ and $\frac{-9}{10}=\frac{-9×9}{10×9}=\frac{-81}{90}$

Clearly,

$-81<-80$

∴â€‹$\frac{-81}{90}<\frac{-80}{90}$

Thus,

$\frac{-9}{10}<\frac{-8}{9}$

#### Question 8:

(i)We will write each of the given numbers with positive denominators.

One number = $\frac{-3}{7}$
Other number =$\frac{6}{-13}=\frac{6×\left(-1\right)}{-13×\left(-1\right)}=\frac{-6}{13}$

LCM of 7 and 13 = 91

∴ $\frac{-3}{7}=\frac{-3×13}{7×13}=\frac{-39}{91}$

And,

$\frac{-6}{13}=\frac{-6×7}{13×7}=\frac{-42}{91}$$\frac{-6}{13}=\frac{-6×7}{13×7}=\frac{-42}{91}$$\frac{-6}{13}=\frac{-6×7}{13×7}=\frac{-42}{91}$

Clearly,

$-39>-41$

∴ â€‹

Thus,

$\frac{-3}{7}>\frac{6}{-13}$

(ii) We will write each of the given numbers with positive denominators.

One number = $\frac{5}{-13}=\frac{5×\left(-1\right)}{-13×\left(-1\right)}=\frac{-5}{13}$

Other number =$\frac{-35}{91}$

LCM of 13 and 91 = 91

∴ $\frac{-5}{13}=\frac{-5×7}{13×7}=\frac{-35}{91}$ and $\frac{-35}{91}$

Clearly,

$-35=-35$

∴

Thus,

$\frac{-5}{13}=\frac{-35}{91}$

(iii) We will write each of the given numbers with positive denominators.

One number = $-2$

We can write -2 as$\frac{-2}{1}$.
Other number =$\frac{-13}{5}$

LCM of 1 and 5 = 5

∴â€‹ $\frac{-2}{1}=\frac{-2×5}{1×5}=\frac{-10}{5}$ and $\frac{-13}{5}=\frac{-13×1}{5×1}=\frac{-13}{5}$

Clearly,

$-10>-13$

∴ $\frac{-10}{5}>\frac{-13}{5}$

Thus,

$\frac{-2}{1}>\frac{-13}{5}$

$-2>\frac{-13}{5}$

(iv) We will write each of the given numbers with positive denominators.

One number = $\frac{-2}{3}$
Other number =$\frac{5}{-8}=\frac{5×\left(-1\right)}{-8×\left(-1\right)}=\frac{-5}{8}$

LCM of 3 and 8 = 24

∴ â€‹$\frac{-2}{3}=\frac{-2×8}{3×8}=\frac{-16}{24}$ and $\frac{-5}{8}=\frac{-5×3}{8×3}=\frac{-15}{24}$

Clearly,

$-16<-15$

∴ $\frac{-16}{24}<\frac{-15}{24}$

Thus,

$\frac{-2}{3}<\frac{-5}{8}$

$\frac{-2}{3}<\frac{5}{-8}$

(v) $\frac{-3}{-5}=\frac{-3×-1}{-5×-1}=\frac{3}{5}$

$\frac{3}{5}$ is a positive number.

Because every positive rational number is greater than 0, $\frac{3}{5}>0⇒0<\frac{3}{5}$.

(vi) We will write each of the given numbers with positive denominators.

One number = $\frac{-8}{9}$

Other number = $\frac{-9}{10}$

LCM of 9 and 10 = 90

∴â€‹$\frac{-8}{9}=\frac{-8×10}{9×10}=\frac{-80}{90}$ and $\frac{-9}{10}=\frac{-9×9}{10×9}=\frac{-81}{90}$

Clearly,

$-81<-80$

∴â€‹$\frac{-81}{90}<\frac{-80}{90}$

Thus,

$\frac{-9}{10}<\frac{-8}{9}$

(i) We will write each of the given numbers with positive denominators.

We have:

$\frac{4}{-9}=\frac{4×\left(-1\right)}{-9×\left(-1\right)}=\frac{-4}{9}$ and$\frac{7}{-18}=\frac{7×\left(-1\right)}{-18×\left(-1\right)}=\frac{-7}{18}$

Thus, the given numbers are

LCM of 9, 12, 18 and 3 is 36.

Now,

$\frac{-4}{9}=\frac{-4×4}{9×4}=\frac{-16}{36}$

$\frac{-5}{12}=\frac{-5×3}{12×3}=\frac{-15}{36}$

$\frac{-7}{18}=\frac{-7×2}{18×2}=\frac{-14}{36}$

$\frac{-2}{3}=\frac{-2×12}{3×12}=\frac{-24}{36}$

Clearly,

$\frac{-24}{36}<\frac{-16}{36}<\frac{-15}{36}<\frac{-14}{36}$

∴ â€‹$\frac{-2}{3}<\frac{-4}{9}<\frac{-5}{12}<\frac{-7}{18}$

That is

$\frac{-2}{3}<\frac{4}{-9}<\frac{-5}{12}<\frac{7}{-18}$

(ii) We  will write each of the given numbers with positive denominators.

We have:

$\frac{5}{-12}=\frac{5×\left(-1\right)}{-12×\left(-1\right)}=\frac{-5}{12}$ and$\frac{9}{-24}=\frac{9×\left(-1\right)}{-24×\left(-1\right)}=\frac{-9}{24}$

Thus, the given numbers are

LCM of  4, 12, 16 and 24 is 48.

Now,

$\frac{-3}{4}=\frac{-3×12}{4×12}=\frac{-36}{48}$

$\frac{-5}{12}=\frac{-5×4}{12×4}=\frac{-20}{48}$

$\frac{-7}{16}=\frac{-7×3}{16×3}=\frac{-21}{48}$

$\frac{-9}{24}=\frac{-9×2}{24×2}=\frac{-18}{48}$

Clearly,

$\frac{-36}{48}<\frac{-21}{48}<\frac{-20}{48}<\frac{-18}{48}$

∴â€‹ $\frac{-3}{4}<\frac{-7}{16}<\frac{-5}{12}<\frac{-9}{24}$

That is

$\frac{-3}{4}<\frac{-7}{16}<\frac{5}{-12}<\frac{9}{-24}$

(iii) We will write each of the given numbers with positive denominators.

We have:

$\frac{3}{-5}=\frac{3×\left(-1\right)}{-5×\left(-1\right)}=\frac{-3}{5}$

Thus, the given numbers are

LCM of 5, 10, 15 and 20 is 60.

Now,

$\frac{-3}{5}=\frac{-3×12}{5×12}=\frac{-36}{60}$

$\frac{-7}{10}=\frac{-7×6}{10×6}=\frac{-42}{60}$

$\frac{-11}{15}=\frac{-11×4}{15×4}=\frac{-44}{60}$

$\frac{-13}{20}=\frac{-13×3}{20×3}=\frac{-39}{60}$

Clearly,

$\frac{-44}{60}<\frac{-42}{60}<\frac{-39}{60}<\frac{-36}{60}$

∴â€‹ $\frac{-11}{15}<\frac{-7}{10}<\frac{-13}{20}<\frac{-3}{5}$.

That is

$\frac{-11}{15}<\frac{-7}{10}<\frac{-13}{20}<\frac{3}{-5}$

(iv) We will write each of the given numbers with positive denominators.

We have:

$\frac{13}{-28}=\frac{13×\left(-1\right)}{-28×\left(-1\right)}=\frac{-13}{28}$

Thus, the given numbers are

LCM of 7, 14, 28 and 42 is 84.

Now,

$\frac{-4}{7}=\frac{-4×12}{7×12}=\frac{-48}{84}$

$\frac{-9}{14}=\frac{-9×6}{14×6}=\frac{-54}{84}$

$\frac{-13}{28}=\frac{-13×3}{28×3}=\frac{-39}{84}$

$\frac{-23}{42}=\frac{-23×2}{42×2}=\frac{-46}{84}$

Clearly,

$\frac{-54}{84}<\frac{-48}{84}<\frac{-46}{84}<\frac{-39}{84}$

∴â€‹ $\frac{-9}{14}<\frac{-4}{7}<\frac{-23}{42}<\frac{-13}{28}$.

That is

$\frac{-9}{14}<\frac{-4}{7}<\frac{-23}{42}<\frac{13}{-28}$

#### Question 9:

(i) We will write each of the given numbers with positive denominators.

We have:

$\frac{4}{-9}=\frac{4×\left(-1\right)}{-9×\left(-1\right)}=\frac{-4}{9}$ and$\frac{7}{-18}=\frac{7×\left(-1\right)}{-18×\left(-1\right)}=\frac{-7}{18}$

Thus, the given numbers are

LCM of 9, 12, 18 and 3 is 36.

Now,

$\frac{-4}{9}=\frac{-4×4}{9×4}=\frac{-16}{36}$

$\frac{-5}{12}=\frac{-5×3}{12×3}=\frac{-15}{36}$

$\frac{-7}{18}=\frac{-7×2}{18×2}=\frac{-14}{36}$

$\frac{-2}{3}=\frac{-2×12}{3×12}=\frac{-24}{36}$

Clearly,

$\frac{-24}{36}<\frac{-16}{36}<\frac{-15}{36}<\frac{-14}{36}$

∴ â€‹$\frac{-2}{3}<\frac{-4}{9}<\frac{-5}{12}<\frac{-7}{18}$

That is

$\frac{-2}{3}<\frac{4}{-9}<\frac{-5}{12}<\frac{7}{-18}$

(ii) We  will write each of the given numbers with positive denominators.

We have:

$\frac{5}{-12}=\frac{5×\left(-1\right)}{-12×\left(-1\right)}=\frac{-5}{12}$ and$\frac{9}{-24}=\frac{9×\left(-1\right)}{-24×\left(-1\right)}=\frac{-9}{24}$

Thus, the given numbers are

LCM of  4, 12, 16 and 24 is 48.

Now,

$\frac{-3}{4}=\frac{-3×12}{4×12}=\frac{-36}{48}$

$\frac{-5}{12}=\frac{-5×4}{12×4}=\frac{-20}{48}$

$\frac{-7}{16}=\frac{-7×3}{16×3}=\frac{-21}{48}$

$\frac{-9}{24}=\frac{-9×2}{24×2}=\frac{-18}{48}$

Clearly,

$\frac{-36}{48}<\frac{-21}{48}<\frac{-20}{48}<\frac{-18}{48}$

∴â€‹ $\frac{-3}{4}<\frac{-7}{16}<\frac{-5}{12}<\frac{-9}{24}$

That is

$\frac{-3}{4}<\frac{-7}{16}<\frac{5}{-12}<\frac{9}{-24}$

(iii) We will write each of the given numbers with positive denominators.

We have:

$\frac{3}{-5}=\frac{3×\left(-1\right)}{-5×\left(-1\right)}=\frac{-3}{5}$

Thus, the given numbers are

LCM of 5, 10, 15 and 20 is 60.

Now,

$\frac{-3}{5}=\frac{-3×12}{5×12}=\frac{-36}{60}$

$\frac{-7}{10}=\frac{-7×6}{10×6}=\frac{-42}{60}$

$\frac{-11}{15}=\frac{-11×4}{15×4}=\frac{-44}{60}$

$\frac{-13}{20}=\frac{-13×3}{20×3}=\frac{-39}{60}$

Clearly,

$\frac{-44}{60}<\frac{-42}{60}<\frac{-39}{60}<\frac{-36}{60}$

∴â€‹ $\frac{-11}{15}<\frac{-7}{10}<\frac{-13}{20}<\frac{-3}{5}$.

That is

$\frac{-11}{15}<\frac{-7}{10}<\frac{-13}{20}<\frac{3}{-5}$

(iv) We will write each of the given numbers with positive denominators.

We have:

$\frac{13}{-28}=\frac{13×\left(-1\right)}{-28×\left(-1\right)}=\frac{-13}{28}$

Thus, the given numbers are

LCM of 7, 14, 28 and 42 is 84.

Now,

$\frac{-4}{7}=\frac{-4×12}{7×12}=\frac{-48}{84}$

$\frac{-9}{14}=\frac{-9×6}{14×6}=\frac{-54}{84}$

$\frac{-13}{28}=\frac{-13×3}{28×3}=\frac{-39}{84}$

$\frac{-23}{42}=\frac{-23×2}{42×2}=\frac{-46}{84}$

Clearly,

$\frac{-54}{84}<\frac{-48}{84}<\frac{-46}{84}<\frac{-39}{84}$

∴â€‹ $\frac{-9}{14}<\frac{-4}{7}<\frac{-23}{42}<\frac{-13}{28}$.

That is

$\frac{-9}{14}<\frac{-4}{7}<\frac{-23}{42}<\frac{13}{-28}$

(i) We will first write each of the given numbers with positive denominators. We have:

$\frac{8}{-3}=\frac{8×\left(-1\right)}{-3×\left(-1\right)}=\frac{-8}{3}$

Thus, the given numbers are

LCM of 1, 6, 3 and 3 is 6

Now,

$\frac{-2}{1}=\frac{-2×6}{1×6}=\frac{-12}{6}$

$\frac{-13}{6}=\frac{-13×1}{6×1}=\frac{-13}{6}$

$\frac{-8}{3}=\frac{-8×2}{3×2}=\frac{-16}{6}$

and

$\frac{1}{3}=\frac{1×2}{3×2}=\frac{2}{6}$

Clearly,Thus,

$\frac{2}{6}>\frac{-12}{6}>\frac{-13}{6}>\frac{-16}{6}$

∴â€‹ $\frac{1}{3}>-2>\frac{-13}{6}>\frac{-8}{3}$. i.e $\frac{1}{3}>-2>\frac{-13}{6}>\frac{8}{-3}$

(ii) We will first write each of the given numbers with positive denominators. We have:

$\frac{7}{-15}=\frac{7×\left(-1\right)}{-15×\left(-1\right)}=\frac{-7}{15}$ and $\frac{17}{-30}=\frac{17×\left(-1\right)}{-30×\left(-1\right)}=\frac{-17}{30}$

Thus, the given numbers are

LCM of 10, 15, 20 and 30 is 60

Now,

$\frac{-3}{10}=\frac{-3×6}{10×6}=\frac{-18}{60}$

$\frac{-7}{15}=\frac{-7×4}{15×4}=\frac{-28}{60}$

$\frac{-11}{20}=\frac{-11×3}{20×3}=\frac{-33}{60}$

and

$\frac{-17}{30}=\frac{-17×2}{30×2}=\frac{-34}{60}$

Clearly,

$\frac{-18}{60}>\frac{-28}{60}>\frac{-33}{60}>\frac{-34}{60}$

∴ $\frac{-3}{10}>\frac{-7}{15}>\frac{-11}{20}>\frac{-17}{30}$. i.e $\frac{-3}{10}>\frac{7}{-15}>\frac{-11}{20}>\frac{17}{-30}$

(iii) We will first write each of the given numbers with positive denominators. We have:

$\frac{23}{-24}=\frac{23×\left(-1\right)}{-24×\left(-1\right)}=\frac{-23}{24}$

Thus, the given numbers are

LCM of 6, 12, 18 and 24 is 72

Now,

$\frac{-5}{6}=\frac{-5×12}{6×12}=\frac{-60}{72}$

$\frac{-7}{12}=\frac{-7×6}{12×6}=\frac{-42}{72}$

$\frac{-13}{18}=\frac{-13×4}{18×4}=\frac{-52}{72}$

and

$\frac{-23}{24}=\frac{-23×3}{24×3}=\frac{-69}{72}$

Clearly,

$\frac{-42}{72}>\frac{-52}{72}>\frac{-60}{72}>\frac{-69}{72}$

∴â€‹ $\frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{-23}{24}$. i.e $\frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{23}{-24}$

(iv) The given numbers are

LCM of 11, 22, 33 and 44 is 132

Now,

$\frac{-10}{11}=\frac{-10×12}{11×12}=\frac{-120}{132}$

$\frac{-19}{22}=\frac{-19×6}{22×6}=\frac{-114}{132}$

$\frac{-23}{33}=\frac{-23×4}{33×4}=\frac{-92}{132}$

and

$\frac{-39}{44}=\frac{-39×3}{44×3}=\frac{-117}{132}$

Clearly,

$\frac{-92}{132}>\frac{-114}{132}>\frac{-117}{132}>\frac{-120}{132}$

∴ $\frac{-23}{33}>\frac{-19}{22}>\frac{-39}{44}>\frac{-10}{11}$

#### Question 10:

(i) We will first write each of the given numbers with positive denominators. We have:

$\frac{8}{-3}=\frac{8×\left(-1\right)}{-3×\left(-1\right)}=\frac{-8}{3}$

Thus, the given numbers are

LCM of 1, 6, 3 and 3 is 6

Now,

$\frac{-2}{1}=\frac{-2×6}{1×6}=\frac{-12}{6}$

$\frac{-13}{6}=\frac{-13×1}{6×1}=\frac{-13}{6}$

$\frac{-8}{3}=\frac{-8×2}{3×2}=\frac{-16}{6}$

and

$\frac{1}{3}=\frac{1×2}{3×2}=\frac{2}{6}$

Clearly,Thus,

$\frac{2}{6}>\frac{-12}{6}>\frac{-13}{6}>\frac{-16}{6}$

∴â€‹ $\frac{1}{3}>-2>\frac{-13}{6}>\frac{-8}{3}$. i.e $\frac{1}{3}>-2>\frac{-13}{6}>\frac{8}{-3}$

(ii) We will first write each of the given numbers with positive denominators. We have:

$\frac{7}{-15}=\frac{7×\left(-1\right)}{-15×\left(-1\right)}=\frac{-7}{15}$ and $\frac{17}{-30}=\frac{17×\left(-1\right)}{-30×\left(-1\right)}=\frac{-17}{30}$

Thus, the given numbers are

LCM of 10, 15, 20 and 30 is 60

Now,

$\frac{-3}{10}=\frac{-3×6}{10×6}=\frac{-18}{60}$

$\frac{-7}{15}=\frac{-7×4}{15×4}=\frac{-28}{60}$

$\frac{-11}{20}=\frac{-11×3}{20×3}=\frac{-33}{60}$

and

$\frac{-17}{30}=\frac{-17×2}{30×2}=\frac{-34}{60}$

Clearly,

$\frac{-18}{60}>\frac{-28}{60}>\frac{-33}{60}>\frac{-34}{60}$

∴ $\frac{-3}{10}>\frac{-7}{15}>\frac{-11}{20}>\frac{-17}{30}$. i.e $\frac{-3}{10}>\frac{7}{-15}>\frac{-11}{20}>\frac{17}{-30}$

(iii) We will first write each of the given numbers with positive denominators. We have:

$\frac{23}{-24}=\frac{23×\left(-1\right)}{-24×\left(-1\right)}=\frac{-23}{24}$

Thus, the given numbers are

LCM of 6, 12, 18 and 24 is 72

Now,

$\frac{-5}{6}=\frac{-5×12}{6×12}=\frac{-60}{72}$

$\frac{-7}{12}=\frac{-7×6}{12×6}=\frac{-42}{72}$

$\frac{-13}{18}=\frac{-13×4}{18×4}=\frac{-52}{72}$

and

$\frac{-23}{24}=\frac{-23×3}{24×3}=\frac{-69}{72}$

Clearly,

$\frac{-42}{72}>\frac{-52}{72}>\frac{-60}{72}>\frac{-69}{72}$

∴â€‹ $\frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{-23}{24}$. i.e $\frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{23}{-24}$

(iv) The given numbers are

LCM of 11, 22, 33 and 44 is 132

Now,

$\frac{-10}{11}=\frac{-10×12}{11×12}=\frac{-120}{132}$

$\frac{-19}{22}=\frac{-19×6}{22×6}=\frac{-114}{132}$

$\frac{-23}{33}=\frac{-23×4}{33×4}=\frac{-92}{132}$

and

$\frac{-39}{44}=\frac{-39×3}{44×3}=\frac{-117}{132}$

Clearly,

$\frac{-92}{132}>\frac{-114}{132}>\frac{-117}{132}>\frac{-120}{132}$

∴ $\frac{-23}{33}>\frac{-19}{22}>\frac{-39}{44}>\frac{-10}{11}$

1. True
A whole number can be expressed as . Thus, every whole number is rational.

2. True
Every integer is a rational number because any integer can be expressed as . Thus, every integer is a rational number.

3. False
Thus, 0 is a rational and whole number.

#### Question 1:

1. True
A whole number can be expressed as . Thus, every whole number is rational.

2. True
Every integer is a rational number because any integer can be expressed as . Thus, every integer is a rational number.

3. False
Thus, 0 is a rational and whole number.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

#### Question 3:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(i) True
A negative number always lies to the left of 0 on the number line.

(ii) False
A negative number always lies to the left of 0 on the number line.

(iii) True
Negative and positive numbers always lie on the opposite sides of 0 on the number line.

(iv) False
The negative sign cancels off and the number becomes $\frac{18}{13}$; it lies to the right of 0 on the number line.

#### Question 1:

(i) True
A negative number always lies to the left of 0 on the number line.

(ii) False
A negative number always lies to the left of 0 on the number line.

(iii) True
Negative and positive numbers always lie on the opposite sides of 0 on the number line.

(iv) False
The negative sign cancels off and the number becomes $\frac{18}{13}$; it lies to the right of 0 on the number line.

1.

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

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

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

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

6. $\frac{-17}{15}+\frac{-1}{15}=\frac{-17+\left(-1\right)}{15}=\frac{-17-1}{15}=\frac{-18}{15}=\frac{-6×3}{5×3}=\frac{-6}{5}$

#### Question 2:

1.

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

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

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

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

6. $\frac{-17}{15}+\frac{-1}{15}=\frac{-17+\left(-1\right)}{15}=\frac{-17-1}{15}=\frac{-18}{15}=\frac{-6×3}{5×3}=\frac{-6}{5}$

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

LCM of 4 and 5 is 20.

Now,

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

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

2.â€‹ The denominators of the given rational numbers are 8 and 12.

LCM of 8 and 12 is 24.

Now,

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

∴â€‹ $\frac{5}{8}+\frac{-7}{12}=\frac{15}{24}+\frac{-14}{24}=\frac{15+\left(-14\right)}{24}=\frac{15-14}{24}=\frac{1}{24}$

3. â€‹The denominators of the given rational numbers are 9 and 6.

LCM of 9 and 6 is 18.

Now,

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

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

4.â€‹ The denominators of the given rational numbers are 16 and 24.

LCM of 16 and 24 is 48.

Now,

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

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

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

$\frac{7}{-18}=\frac{7×\left(-1\right)}{-18×\left(-1\right)}=\frac{-7}{18}$

â€‹The denominators of the given rational numbers are 18 and 27.

LCM of 18 and 27 is 54.

Now,

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

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

6. â€‹We will first write each of the given numbers with positive denominators.

$\frac{1}{-12}=\frac{1×\left(-1\right)}{-12×\left(-1\right)}=\frac{-1}{12}$ and $\frac{2}{-15}=\frac{2×\left(-1\right)}{-15×\left(-1\right)}=\frac{-2}{15}$

â€‹The denominators of the given rational numbers are 12 and 15.

LCM of 12 and 15 is 60.

Now,

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

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

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,

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

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

8. â€‹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,

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

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

9. â€‹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 $×$ 5).

Now,

$\frac{0}{1}=\frac{0×5}{1×5}=\frac{0}{5}=0$ and $\frac{-2}{5}=\frac{-2×1}{5×1}=\frac{-2}{5}$

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

#### Question 3:

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

LCM of 4 and 5 is 20.

Now,

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

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

2.â€‹ The denominators of the given rational numbers are 8 and 12.

LCM of 8 and 12 is 24.

Now,

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

∴â€‹ $\frac{5}{8}+\frac{-7}{12}=\frac{15}{24}+\frac{-14}{24}=\frac{15+\left(-14\right)}{24}=\frac{15-14}{24}=\frac{1}{24}$

3. â€‹The denominators of the given rational numbers are 9 and 6.

LCM of 9 and 6 is 18.

Now,

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

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

4.â€‹ The denominators of the given rational numbers are 16 and 24.

LCM of 16 and 24 is 48.

Now,

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

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

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

$\frac{7}{-18}=\frac{7×\left(-1\right)}{-18×\left(-1\right)}=\frac{-7}{18}$

â€‹The denominators of the given rational numbers are 18 and 27.

LCM of 18 and 27 is 54.

Now,

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

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

6. â€‹We will first write each of the given numbers with positive denominators.

$\frac{1}{-12}=\frac{1×\left(-1\right)}{-12×\left(-1\right)}=\frac{-1}{12}$ and $\frac{2}{-15}=\frac{2×\left(-1\right)}{-15×\left(-1\right)}=\frac{-2}{15}$

â€‹The denominators of the given rational numbers are 12 and 15.

LCM of 12 and 15 is 60.

Now,

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

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

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,

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

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

8. â€‹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,

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

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

9. â€‹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 $×$ 5).

Now,

$\frac{0}{1}=\frac{0×5}{1×5}=\frac{0}{5}=0$ and $\frac{-2}{5}=\frac{-2×1}{5×1}=\frac{-2}{5}$

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

1. LHS = $\frac{-12}{5}+\frac{2}{7}$

LCM of 5 and 7 is 35.

$\frac{-12×7}{5×7}+\frac{2×5}{7×5}=\frac{-84}{35}+\frac{10}{35}=\frac{-84+10}{35}=\frac{-74}{35}$

RHS = $\frac{2}{7}+\frac{-12}{5}$

LCM of 5 and 7 is 35.

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

2. â€‹LHS = $\frac{-5}{8}+\frac{-9}{13}$

LCM of 8 and 13 is 104.

$\frac{-5×13}{8×13}+\frac{-9×8}{13×8}=\frac{-65}{104}+\frac{-72}{104}=\frac{-65+\left(-72\right)}{104}=\frac{-65-72}{104}=\frac{-137}{104}$

RHS = $\frac{-9}{13}+\frac{-5}{8}$

LCM of 13 and 8 is 104.

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

3. â€‹LHS = $\frac{3}{1}+\frac{-7}{12}$

LCM of 1 and 12 is 12.

$\frac{3×12}{1×12}+\frac{-7×1}{12×1}=\frac{36}{12}+\frac{-7}{12}=\frac{36+\left(-7\right)}{12}=\frac{36-7}{12}=\frac{29}{12}$

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

LCM of 12 and 1 is 12.

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

4. LHS = â€‹$\frac{2}{-7}+\frac{12}{-35}$

We will write the given numbers with positive denominators.

$\frac{2}{-7}=\frac{2×\left(-1\right)}{-7×\left(-1\right)}=\frac{-2}{7}$ and $\frac{12}{-35}=\frac{12×\left(-1\right)}{-35×\left(-1\right)}=\frac{-12}{35}$

LCM of 7 and 35 is 35.

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

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

We will write the given numbers with positive denominators.

$\frac{12}{-35}=\frac{12×\left(-1\right)}{-35×\left(-1\right)}=\frac{-12}{35}$ and $\frac{2}{-7}=\frac{2×\left(-1\right)}{-7×\left(-1\right)}=\frac{-2}{7}$

LCM of 35 and 7 is 35.

∴â€‹ $\frac{2}{-7}+\frac{12}{-35}=\frac{12}{-35}+\frac{2}{-7}$

#### Question 4:

1. LHS = $\frac{-12}{5}+\frac{2}{7}$

LCM of 5 and 7 is 35.

$\frac{-12×7}{5×7}+\frac{2×5}{7×5}=\frac{-84}{35}+\frac{10}{35}=\frac{-84+10}{35}=\frac{-74}{35}$

RHS = $\frac{2}{7}+\frac{-12}{5}$

LCM of 5 and 7 is 35.

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

2. â€‹LHS = $\frac{-5}{8}+\frac{-9}{13}$

LCM of 8 and 13 is 104.

$\frac{-5×13}{8×13}+\frac{-9×8}{13×8}=\frac{-65}{104}+\frac{-72}{104}=\frac{-65+\left(-72\right)}{104}=\frac{-65-72}{104}=\frac{-137}{104}$

RHS = $\frac{-9}{13}+\frac{-5}{8}$

LCM of 13 and 8 is 104.

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

3. â€‹LHS = $\frac{3}{1}+\frac{-7}{12}$

LCM of 1 and 12 is 12.

$\frac{3×12}{1×12}+\frac{-7×1}{12×1}=\frac{36}{12}+\frac{-7}{12}=\frac{36+\left(-7\right)}{12}=\frac{36-7}{12}=\frac{29}{12}$

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

LCM of 12 and 1 is 12.

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

4. LHS = â€‹$\frac{2}{-7}+\frac{12}{-35}$

We will write the given numbers with positive denominators.

$\frac{2}{-7}=\frac{2×\left(-1\right)}{-7×\left(-1\right)}=\frac{-2}{7}$ and $\frac{12}{-35}=\frac{12×\left(-1\right)}{-35×\left(-1\right)}=\frac{-12}{35}$

LCM of 7 and 35 is 35.

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

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

We will write the given numbers with positive denominators.

$\frac{12}{-35}=\frac{12×\left(-1\right)}{-35×\left(-1\right)}=\frac{-12}{35}$ and $\frac{2}{-7}=\frac{2×\left(-1\right)}{-7×\left(-1\right)}=\frac{-2}{7}$

LCM of 35 and 7 is 35.

∴â€‹ $\frac{2}{-7}+\frac{12}{-35}=\frac{12}{-35}+\frac{2}{-7}$

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

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

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

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

∴â€‹ $\left(\frac{3}{4}+\frac{-2}{5}\right)+\frac{-7}{10}=\frac{3}{4}+\left(\frac{-2}{5}+\frac{-7}{10}\right)$

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

We will first make the denominator positive.

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

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

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

We will first make the denominator positive.

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

$\left\{\frac{-7}{11}+\left(\frac{-2}{5}+\frac{-13}{22}\right)\right\}=\left\{\frac{-7}{11}+\left(\frac{-44}{110}+\frac{-65}{110}\right)\right\}=\left\{\frac{-7}{11}+\left(\frac{-44+\left(-65\right)}{110}\right)\right\}=\frac{-7}{11}+\frac{-109}{110}=\frac{-70}{110}+\frac{-109}{110}=\frac{-70-109}{110}=\frac{-179}{110}$

∴â€‹ $\left(\frac{-7}{11}+\frac{2}{-5}\right)+\frac{-13}{22}=\frac{-7}{11}+\left(\frac{2}{-5}+\frac{-13}{22}\right)$

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

$\left\{\frac{-1}{1}+\left(\frac{-2}{3}+\frac{-3}{4}\right)\right\}=\left\{\frac{-1}{1}+\left(\frac{-8}{12}+\frac{-9}{12}\right)\right\}=\left\{\frac{-1}{1}+\left(\frac{-8-9}{12}\right)\right\}=\left\{\frac{-1}{1}+\left(\frac{-17}{12}\right)\right\}=\left(\frac{-1}{1}+\frac{-17}{12}\right)=\left(\frac{-1×12}{1×12}+\frac{-17×1}{12×1}\right)=\left(\frac{-12+\left(-17\right)}{12}\right)=\left(\frac{-12-17}{12}\right)=\frac{-29}{12}$

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

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

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

#### Question 5:

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

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

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

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

∴â€‹ $\left(\frac{3}{4}+\frac{-2}{5}\right)+\frac{-7}{10}=\frac{3}{4}+\left(\frac{-2}{5}+\frac{-7}{10}\right)$

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

We will first make the denominator positive.

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

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

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

We will first make the denominator positive.

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

$\left\{\frac{-7}{11}+\left(\frac{-2}{5}+\frac{-13}{22}\right)\right\}=\left\{\frac{-7}{11}+\left(\frac{-44}{110}+\frac{-65}{110}\right)\right\}=\left\{\frac{-7}{11}+\left(\frac{-44+\left(-65\right)}{110}\right)\right\}=\frac{-7}{11}+\frac{-109}{110}=\frac{-70}{110}+\frac{-109}{110}=\frac{-70-109}{110}=\frac{-179}{110}$

∴â€‹ $\left(\frac{-7}{11}+\frac{2}{-5}\right)+\frac{-13}{22}=\frac{-7}{11}+\left(\frac{2}{-5}+\frac{-13}{22}\right)$

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

$\left\{\frac{-1}{1}+\left(\frac{-2}{3}+\frac{-3}{4}\right)\right\}=\left\{\frac{-1}{1}+\left(\frac{-8}{12}+\frac{-9}{12}\right)\right\}=\left\{\frac{-1}{1}+\left(\frac{-8-9}{12}\right)\right\}=\left\{\frac{-1}{1}+\left(\frac{-17}{12}\right)\right\}=\left(\frac{-1}{1}+\frac{-17}{12}\right)=\left(\frac{-1×12}{1×12}+\frac{-17×1}{12×1}\right)=\left(\frac{-12+\left(-17\right)}{12}\right)=\left(\frac{-12-17}{12}\right)=\frac{-29}{12}$

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

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

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

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

Hence, the required solution is $\left(\frac{-3}{17}\right)+\left(\frac{-12}{5}\right)=\left(\frac{-12}{5}\right)+\overline{)\left(\frac{-3}{7}\right)}$.

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

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

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

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

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

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

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

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

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

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

#### Question 6:

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

Hence, the required solution is $\left(\frac{-3}{17}\right)+\left(\frac{-12}{5}\right)=\left(\frac{-12}{5}\right)+\overline{)\left(\frac{-3}{7}\right)}$.

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

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

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

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

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

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

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

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

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

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

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:

$\frac{-16}{-5}=\frac{-16×\left(-1\right)}{-5×\left(-1\right)}=\frac{16}{5}$

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:

$\frac{-8}{-7}=\frac{-8×\left(-1\right)}{-7×\left(-1\right)}=\frac{8}{7}$

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

#### Question 7:

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:

$\frac{-16}{-5}=\frac{-16×\left(-1\right)}{-5×\left(-1\right)}=\frac{16}{5}$

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:

$\frac{-8}{-7}=\frac{-8×\left(-1\right)}{-7×\left(-1\right)}=\frac{8}{7}$

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

(i)

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

(ii)

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

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

(iii)

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

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

(iv)

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

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

(v)

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

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

(vi)

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

=$\frac{13}{9}$

(vii)

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

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

(viii)

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

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

#### Question 8:

(i)

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

(ii)

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

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

(iii)

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

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

(iv)

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

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

(v)

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

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

(vi)

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

=$\frac{13}{9}$

(vii)

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

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

(viii)

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

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

(i)
$\left(\frac{4}{3}+\frac{-2}{3}\right)+\left(\frac{3}{5}+\frac{-11}{5}\right)$
$\left(\frac{4-2}{3}\right)+\left(\frac{3-11}{5}\right)$
$=\left(\frac{2}{3}+\frac{-8}{5}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{10}{15}+\frac{-24}{15}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{10-24}{15}\right)\phantom{\rule{0ex}{0ex}}=\frac{-14}{15}$.

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

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

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

$=\left(\frac{-27}{6}+\frac{1}{8}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{-108}{24}+\frac{3}{24}\right)\phantom{\rule{0ex}{0ex}}=\frac{-105}{24}$
=$\frac{35}{8}$

(iii)
$\left(\frac{-13}{20}+\frac{7}{10}\right)+\left(\frac{11}{14}+\frac{-5}{7}\right)$
=$\left(\frac{-13}{20}+\frac{14}{20}\right)+\left(\frac{11}{14}+\frac{-10}{14}\right)$
=$\left(\frac{-13+14}{20}\right)+\left(\frac{11-10}{14}\right)$
$=\left(\frac{1}{20}+\frac{1}{14}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{7}{140}+\frac{10}{140}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{7+10}{140}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{17}{140}\right)\phantom{\rule{0ex}{0ex}}=\frac{17}{140}$.

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

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

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

$=\left(\frac{-21}{7}+\frac{-23}{18}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{-3}{1}+\frac{-23}{18}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{-54}{18}+\frac{-23}{18}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{-54-23}{18}\right)\phantom{\rule{0ex}{0ex}}=\frac{-77}{18}$

#### Question 9:

(i)
$\left(\frac{4}{3}+\frac{-2}{3}\right)+\left(\frac{3}{5}+\frac{-11}{5}\right)$
$\left(\frac{4-2}{3}\right)+\left(\frac{3-11}{5}\right)$
$=\left(\frac{2}{3}+\frac{-8}{5}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{10}{15}+\frac{-24}{15}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{10-24}{15}\right)\phantom{\rule{0ex}{0ex}}=\frac{-14}{15}$.

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

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

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

$=\left(\frac{-27}{6}+\frac{1}{8}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{-108}{24}+\frac{3}{24}\right)\phantom{\rule{0ex}{0ex}}=\frac{-105}{24}$
=$\frac{35}{8}$

(iii)
$\left(\frac{-13}{20}+\frac{7}{10}\right)+\left(\frac{11}{14}+\frac{-5}{7}\right)$
=$\left(\frac{-13}{20}+\frac{14}{20}\right)+\left(\frac{11}{14}+\frac{-10}{14}\right)$
=$\left(\frac{-13+14}{20}\right)+\left(\frac{11-10}{14}\right)$
$=\left(\frac{1}{20}+\frac{1}{14}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{7}{140}+\frac{10}{140}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{7+10}{140}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{17}{140}\right)\phantom{\rule{0ex}{0ex}}=\frac{17}{140}$.

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

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

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

$=\left(\frac{-21}{7}+\frac{-23}{18}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{-3}{1}+\frac{-23}{18}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{-54}{18}+\frac{-23}{18}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{-54-23}{18}\right)\phantom{\rule{0ex}{0ex}}=\frac{-77}{18}$

#### Question 11:

Let the required number be x

Now,

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

$⇒x=\left(\frac{-3}{2}+\frac{5}{8}\right)\phantom{\rule{0ex}{0ex}}⇒x=\left(\frac{-12}{8}+\frac{5}{8}\right)\phantom{\rule{0ex}{0ex}}⇒x=\left(\frac{-12+5}{8}\right)\phantom{\rule{0ex}{0ex}}⇒x=\frac{-7}{8}$

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

#### Question 12:

Let the required number be x

Now,

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

$⇒x=\left(\frac{-3}{2}+\frac{5}{8}\right)\phantom{\rule{0ex}{0ex}}⇒x=\left(\frac{-12}{8}+\frac{5}{8}\right)\phantom{\rule{0ex}{0ex}}⇒x=\left(\frac{-12+5}{8}\right)\phantom{\rule{0ex}{0ex}}⇒x=\frac{-7}{8}$

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

Let the required number be x.

Now,

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

$⇒x=\left(\frac{5+7}{7}\right)\phantom{\rule{0ex}{0ex}}⇒x=\frac{12}{7}$
Hence, the required number is $\frac{12}{7}$.

#### Question 13:

Let the required number be x.

Now,

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

$⇒x=\left(\frac{5+7}{7}\right)\phantom{\rule{0ex}{0ex}}⇒x=\frac{12}{7}$
Hence, the required number is $\frac{12}{7}$.

Let the required number be x.

Now,

$\frac{-2}{3}-x=\frac{-1}{6}$
$⇒\frac{-2}{3}-x+x=\frac{-1}{6}+x$         (Adding $x$ to both the sides)
$⇒\frac{-2}{3}=\frac{-1}{6}+x$
$⇒\frac{-2}{3}+\frac{1}{6}=\frac{-1}{6}+x+\frac{1}{6}$    (Adding $\frac{1}{6}$ to both the sides)
$⇒\left(\frac{-4}{6}+\frac{1}{6}\right)=x$
$⇒\left(\frac{-4+1}{6}\right)=x$
$⇒\frac{-3}{6}=x\phantom{\rule{0ex}{0ex}}⇒\frac{-1×3}{2×3}=x\phantom{\rule{0ex}{0ex}}⇒\frac{-1}{2}=x$

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

#### Question 14:

Let the required number be x.

Now,

$\frac{-2}{3}-x=\frac{-1}{6}$
$⇒\frac{-2}{3}-x+x=\frac{-1}{6}+x$         (Adding $x$ to both the sides)
$⇒\frac{-2}{3}=\frac{-1}{6}+x$
$⇒\frac{-2}{3}+\frac{1}{6}=\frac{-1}{6}+x+\frac{1}{6}$    (Adding $\frac{1}{6}$ to both the sides)
$⇒\left(\frac{-4}{6}+\frac{1}{6}\right)=x$
$⇒\left(\frac{-4+1}{6}\right)=x$
$⇒\frac{-3}{6}=x\phantom{\rule{0ex}{0ex}}⇒\frac{-1×3}{2×3}=x\phantom{\rule{0ex}{0ex}}⇒\frac{-1}{2}=x$

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

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

2. Yes
Consider

Since  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  $\left(\frac{ad-bc}{bd}\right)$
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.

â€‹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 $a+\left(b+c\right)=\left(a+b\right)+c$.

5. No, subtraction is not commutative on rational numbers. In general, for any two rational numbers, .

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

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

#### Question 1:

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

2. Yes
Consider

Since  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  $\left(\frac{ad-bc}{bd}\right)$
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.

â€‹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 $a+\left(b+c\right)=\left(a+b\right)+c$.

5. No, subtraction is not commutative on rational numbers. In general, for any two rational numbers, .

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

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

(i)

$\frac{3}{5}×\frac{-7}{8}\phantom{\rule{0ex}{0ex}}=\frac{3×\left(-7\right)}{5×8}\phantom{\rule{0ex}{0ex}}=-\frac{21}{40}$

(ii)

$\frac{-9}{2}×\frac{5}{4}\phantom{\rule{0ex}{0ex}}=\frac{\left(-9\right)×5}{2×4}\phantom{\rule{0ex}{0ex}}=\frac{-45}{8}$

(iii)

$\frac{-6}{11}×\frac{-5}{3}\phantom{\rule{0ex}{0ex}}=\frac{\left(-6\right)×\left(-5\right)}{11×3}\phantom{\rule{0ex}{0ex}}=\frac{30}{33}\phantom{\rule{0ex}{0ex}}$

Simplifying the above rational number, we get:

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

(iv)

$\frac{-2}{3}×\frac{6}{7}\phantom{\rule{0ex}{0ex}}=\frac{\left(-2\right)×6}{3×7}\phantom{\rule{0ex}{0ex}}=\frac{-12}{21}$

Simplifying the above rational number, we get:

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

(v)

$\frac{-12}{5}×\frac{10}{-3}\phantom{\rule{0ex}{0ex}}=\frac{\left(-12\right)×10}{5×\left(-3\right)}\phantom{\rule{0ex}{0ex}}=\frac{-120}{-15}\phantom{\rule{0ex}{0ex}}=\frac{120}{15}\phantom{\rule{0ex}{0ex}}$

Simplifying the above rational number, we get:

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

(vi)

$\frac{25}{-9}×\frac{3}{-10}\phantom{\rule{0ex}{0ex}}=\frac{25×3}{\left(-9\right)×\left(-10\right)}\phantom{\rule{0ex}{0ex}}=\frac{75}{90}\phantom{\rule{0ex}{0ex}}$

Simplifying the above rational number, we get:

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

(vii)

$\frac{5}{-18}×\frac{-9}{20}\phantom{\rule{0ex}{0ex}}=\frac{5×\left(-9\right)}{-18×20}\phantom{\rule{0ex}{0ex}}=\frac{-45}{-360}\phantom{\rule{0ex}{0ex}}=\frac{45}{360}\phantom{\rule{0ex}{0ex}}$

Simplifying the above rational number, we get:

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

(viii)

$\frac{-13}{15}×\frac{-25}{26}\phantom{\rule{0ex}{0ex}}=\frac{\left(-13\right)×\left(-25\right)}{15×26}\phantom{\rule{0ex}{0ex}}=\frac{325}{390}\phantom{\rule{0ex}{0ex}}$

Simplifying the above rational number, we get:

$\frac{325}{390}=\frac{325÷5}{390÷5}=\frac{65}{78}=\frac{65÷13}{78÷13}=\frac{5}{6}\phantom{\rule{0ex}{0ex}}$

(ix)

$\frac{16}{-21}×\frac{14}{5}\phantom{\rule{0ex}{0ex}}=\frac{16×14}{\left(-21\right)×5}\phantom{\rule{0ex}{0ex}}=\frac{224}{-105}\phantom{\rule{0ex}{0ex}}$

Simplifying the above rational number, we get:

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

(x)

$\frac{-7}{6}×24\phantom{\rule{0ex}{0ex}}=\frac{\left(-7\right)×24}{6}\phantom{\rule{0ex}{0ex}}=\frac{-168}{6}$

Simplifying the above rational number, we get:

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

(xi)

$\frac{7}{24}×\left(-48\right)\phantom{\rule{0ex}{0ex}}=\frac{7×\left(-48\right)}{24}=-\frac{336}{24}$

Simplifying the above rational number, we get:

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

(xii)

$\frac{-13}{5}×\left(-10\right)\phantom{\rule{0ex}{0ex}}=\frac{\left(-13\right)×\left(-10\right)}{5}\phantom{\rule{0ex}{0ex}}=\frac{130}{5}$

Simplifying the above rational number, we get:

$\frac{130}{5}=\frac{130÷5}{5÷5}=26$

#### Question 2:

(i)

$\frac{3}{5}×\frac{-7}{8}\phantom{\rule{0ex}{0ex}}=\frac{3×\left(-7\right)}{5×8}\phantom{\rule{0ex}{0ex}}=-\frac{21}{40}$

(ii)

$\frac{-9}{2}×\frac{5}{4}\phantom{\rule{0ex}{0ex}}=\frac{\left(-9\right)×5}{2×4}\phantom{\rule{0ex}{0ex}}=\frac{-45}{8}$

(iii)

$\frac{-6}{11}×\frac{-5}{3}\phantom{\rule{0ex}{0ex}}=\frac{\left(-6\right)×\left(-5\right)}{11×3}\phantom{\rule{0ex}{0ex}}=\frac{30}{33}\phantom{\rule{0ex}{0ex}}$

Simplifying the above rational number, we get:

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

(iv)

$\frac{-2}{3}×\frac{6}{7}\phantom{\rule{0ex}{0ex}}=\frac{\left(-2\right)×6}{3×7}\phantom{\rule{0ex}{0ex}}=\frac{-12}{21}$

Simplifying the above rational number, we get:

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

(v)

$\frac{-12}{5}×\frac{10}{-3}\phantom{\rule{0ex}{0ex}}=\frac{\left(-12\right)×10}{5×\left(-3\right)}\phantom{\rule{0ex}{0ex}}=\frac{-120}{-15}\phantom{\rule{0ex}{0ex}}=\frac{120}{15}\phantom{\rule{0ex}{0ex}}$

Simplifying the above rational number, we get:

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

(vi)

$\frac{25}{-9}×\frac{3}{-10}\phantom{\rule{0ex}{0ex}}=\frac{25×3}{\left(-9\right)×\left(-10\right)}\phantom{\rule{0ex}{0ex}}=\frac{75}{90}\phantom{\rule{0ex}{0ex}}$

Simplifying the above rational number, we get:

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

(vii)

$\frac{5}{-18}×\frac{-9}{20}\phantom{\rule{0ex}{0ex}}=\frac{5×\left(-9\right)}{-18×20}\phantom{\rule{0ex}{0ex}}=\frac{-45}{-360}\phantom{\rule{0ex}{0ex}}=\frac{45}{360}\phantom{\rule{0ex}{0ex}}$

Simplifying the above rational number, we get:

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

(viii)

$\frac{-13}{15}×\frac{-25}{26}\phantom{\rule{0ex}{0ex}}=\frac{\left(-13\right)×\left(-25\right)}{15×26}\phantom{\rule{0ex}{0ex}}=\frac{325}{390}\phantom{\rule{0ex}{0ex}}$

Simplifying the above rational number, we get:

$\frac{325}{390}=\frac{325÷5}{390÷5}=\frac{65}{78}=\frac{65÷13}{78÷13}=\frac{5}{6}\phantom{\rule{0ex}{0ex}}$

(ix)

$\frac{16}{-21}×\frac{14}{5}\phantom{\rule{0ex}{0ex}}=\frac{16×14}{\left(-21\right)×5}\phantom{\rule{0ex}{0ex}}=\frac{224}{-105}\phantom{\rule{0ex}{0ex}}$

Simplifying the above rational number, we get:

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

(x)

$\frac{-7}{6}×24\phantom{\rule{0ex}{0ex}}=\frac{\left(-7\right)×24}{6}\phantom{\rule{0ex}{0ex}}=\frac{-168}{6}$

Simplifying the above rational number, we get:

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

(xi)

$\frac{7}{24}×\left(-48\right)\phantom{\rule{0ex}{0ex}}=\frac{7×\left(-48\right)}{24}=-\frac{336}{24}$

Simplifying the above rational number, we get:

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

(xii)

$\frac{-13}{5}×\left(-10\right)\phantom{\rule{0ex}{0ex}}=\frac{\left(-13\right)×\left(-10\right)}{5}\phantom{\rule{0ex}{0ex}}=\frac{130}{5}$

Simplifying the above rational number, we get:

$\frac{130}{5}=\frac{130÷5}{5÷5}=26$

(i)

$\frac{3}{7}×\frac{-5}{9}=\frac{-5}{9}×\frac{3}{7}\phantom{\rule{0ex}{0ex}}$

LHS = RHS

(ii)

(iii)

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

LHS = RHS

(iv)

LHS = RHS

#### Question 3:

(i)

$\frac{3}{7}×\frac{-5}{9}=\frac{-5}{9}×\frac{3}{7}\phantom{\rule{0ex}{0ex}}$

LHS = RHS

(ii)

(iii)

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

LHS = RHS

(iv)

LHS = RHS

(i)

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

$\mathrm{LHS}=\left(\frac{5}{7}×\frac{12}{13}\right)×\frac{7}{18}\phantom{\rule{0ex}{0ex}}=\frac{5×12}{7×13}×\frac{7}{18}\phantom{\rule{0ex}{0ex}}=\frac{60}{91}×\frac{7}{18}\phantom{\rule{0ex}{0ex}}=\frac{420}{1638}\phantom{\rule{0ex}{0ex}}=\frac{10}{39}$

$\mathrm{RHS}=\frac{5}{7}×\left(\frac{12}{13}×\frac{7}{18}\right)\phantom{\rule{0ex}{0ex}}=\frac{5}{7}×\frac{12×7}{13×18}\phantom{\rule{0ex}{0ex}}=\frac{5}{7}×\frac{84}{234}\phantom{\rule{0ex}{0ex}}=\frac{420}{1638}\phantom{\rule{0ex}{0ex}}=\frac{10}{39}$

∴ â€‹$\left(\frac{5}{7}×\frac{12}{13}\right)×\frac{7}{18}=\frac{5}{7}×\left(\frac{12}{13}×\frac{7}{18}\right)$

(ii)

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

$\mathrm{LHS}=\frac{-13}{24}×\left(\frac{-12}{5}×\frac{35}{36}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{-13}{24}×\frac{\left(-12\right)×35}{5×36}\phantom{\rule{0ex}{0ex}}=\frac{-13}{24}×\frac{-420}{180}\phantom{\rule{0ex}{0ex}}=\frac{5460}{4320}\phantom{\rule{0ex}{0ex}}=\frac{91}{72}$

$\mathrm{RHS}=\left(\frac{-13}{24}×\frac{-12}{5}\right)×\frac{35}{36}\phantom{\rule{0ex}{0ex}}=\frac{\left(-13\right)×\left(-12\right)}{24×5}×\frac{35}{36}\phantom{\rule{0ex}{0ex}}=\frac{156}{120}×\frac{35}{36}\phantom{\rule{0ex}{0ex}}=\frac{156×35}{120×36}\phantom{\rule{0ex}{0ex}}=\frac{5460}{4320}\phantom{\rule{0ex}{0ex}}=\frac{91}{72}$

∴ â€‹$\frac{-13}{24}×\left(\frac{-12}{5}×\frac{35}{36}\right)=\left(\frac{-13}{24}×\frac{-12}{5}\right)×\frac{35}{36}$

(iii)

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

$\mathrm{LHS}=\left(\frac{-9}{5}×\frac{-10}{3}\right)×\frac{21}{-4}\phantom{\rule{0ex}{0ex}}=\frac{\left(-9\right)×\left(-10\right)}{5×3}×\frac{21}{-4}\phantom{\rule{0ex}{0ex}}=\frac{90}{15}×\frac{21}{-4}\phantom{\rule{0ex}{0ex}}=\frac{90×21}{15×\left(-4\right)}\phantom{\rule{0ex}{0ex}}=-\frac{1890}{60}\phantom{\rule{0ex}{0ex}}=-\frac{63}{2}$

$\mathrm{RHS}=\frac{-9}{5}×\left(\frac{-10}{3}×\frac{21}{-4}\right)\phantom{\rule{0ex}{0ex}}=\frac{-9}{5}×\frac{\left(-10\right)×21}{3×\left(-4\right)}\phantom{\rule{0ex}{0ex}}=\frac{-9}{5}×\frac{210}{12}\phantom{\rule{0ex}{0ex}}=\frac{\left(-9\right)×210}{5×12}\phantom{\rule{0ex}{0ex}}=-\frac{1890}{60}\phantom{\rule{0ex}{0ex}}=\frac{-63}{2}$

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

#### Question 4:

(i)

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

$\mathrm{LHS}=\left(\frac{5}{7}×\frac{12}{13}\right)×\frac{7}{18}\phantom{\rule{0ex}{0ex}}=\frac{5×12}{7×13}×\frac{7}{18}\phantom{\rule{0ex}{0ex}}=\frac{60}{91}×\frac{7}{18}\phantom{\rule{0ex}{0ex}}=\frac{420}{1638}\phantom{\rule{0ex}{0ex}}=\frac{10}{39}$

$\mathrm{RHS}=\frac{5}{7}×\left(\frac{12}{13}×\frac{7}{18}\right)\phantom{\rule{0ex}{0ex}}=\frac{5}{7}×\frac{12×7}{13×18}\phantom{\rule{0ex}{0ex}}=\frac{5}{7}×\frac{84}{234}\phantom{\rule{0ex}{0ex}}=\frac{420}{1638}\phantom{\rule{0ex}{0ex}}=\frac{10}{39}$

∴ â€‹$\left(\frac{5}{7}×\frac{12}{13}\right)×\frac{7}{18}=\frac{5}{7}×\left(\frac{12}{13}×\frac{7}{18}\right)$

(ii)

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

$\mathrm{LHS}=\frac{-13}{24}×\left(\frac{-12}{5}×\frac{35}{36}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{-13}{24}×\frac{\left(-12\right)×35}{5×36}\phantom{\rule{0ex}{0ex}}=\frac{-13}{24}×\frac{-420}{180}\phantom{\rule{0ex}{0ex}}=\frac{5460}{4320}\phantom{\rule{0ex}{0ex}}=\frac{91}{72}$

$\mathrm{RHS}=\left(\frac{-13}{24}×\frac{-12}{5}\right)×\frac{35}{36}\phantom{\rule{0ex}{0ex}}=\frac{\left(-13\right)×\left(-12\right)}{24×5}×\frac{35}{36}\phantom{\rule{0ex}{0ex}}=\frac{156}{120}×\frac{35}{36}\phantom{\rule{0ex}{0ex}}=\frac{156×35}{120×36}\phantom{\rule{0ex}{0ex}}=\frac{5460}{4320}\phantom{\rule{0ex}{0ex}}=\frac{91}{72}$

∴ â€‹$\frac{-13}{24}×\left(\frac{-12}{5}×\frac{35}{36}\right)=\left(\frac{-13}{24}×\frac{-12}{5}\right)×\frac{35}{36}$

(iii)

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

$\mathrm{LHS}=\left(\frac{-9}{5}×\frac{-10}{3}\right)×\frac{21}{-4}\phantom{\rule{0ex}{0ex}}=\frac{\left(-9\right)×\left(-10\right)}{5×3}×\frac{21}{-4}\phantom{\rule{0ex}{0ex}}=\frac{90}{15}×\frac{21}{-4}\phantom{\rule{0ex}{0ex}}=\frac{90×21}{15×\left(-4\right)}\phantom{\rule{0ex}{0ex}}=-\frac{1890}{60}\phantom{\rule{0ex}{0ex}}=-\frac{63}{2}$

$\mathrm{RHS}=\frac{-9}{5}×\left(\frac{-10}{3}×\frac{21}{-4}\right)\phantom{\rule{0ex}{0ex}}=\frac{-9}{5}×\frac{\left(-10\right)×21}{3×\left(-4\right)}\phantom{\rule{0ex}{0ex}}=\frac{-9}{5}×\frac{210}{12}\phantom{\rule{0ex}{0ex}}=\frac{\left(-9\right)×210}{5×12}\phantom{\rule{0ex}{0ex}}=-\frac{1890}{60}\phantom{\rule{0ex}{0ex}}=\frac{-63}{2}$

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

(i)

(ii)

(iii)

(iv)

(i)

(ii)

(iii)

(iv)

#### Question 6:

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

#### Question 7:

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

∴ â€‹$\frac{3}{7}×\left(\frac{5}{6}+\frac{12}{13}\right)=\left(\frac{3}{7}×\frac{5}{6}\right)+\left(\frac{3}{7}×\frac{12}{13}\right)$

(iii)

(iv)

#### Question 8:

∴ â€‹$\frac{3}{7}×\left(\frac{5}{6}+\frac{12}{13}\right)=\left(\frac{3}{7}×\frac{5}{6}\right)+\left(\frac{3}{7}×\frac{12}{13}\right)$

(iii)

(iv)

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

#### Question 9:

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

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

#### Question 1:

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

$\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\frac{4}{9}÷\frac{-5}{12}\phantom{\rule{0ex}{0ex}}=\frac{4}{9}×\frac{12}{-5}\phantom{\rule{0ex}{0ex}}=\frac{4×12}{9×-5}\phantom{\rule{0ex}{0ex}}=\frac{48}{-45}\phantom{\rule{0ex}{0ex}}=\frac{-48}{45}\phantom{\rule{0ex}{0ex}}=\frac{-16}{15}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}-8÷\frac{-7}{16}\phantom{\rule{0ex}{0ex}}=-8×\frac{16}{-7}\phantom{\rule{0ex}{0ex}}=\frac{8×16}{7}\phantom{\rule{0ex}{0ex}}=\frac{128}{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\phantom{\rule{0ex}{0ex}}\frac{-12}{7}÷\left(-18\right)\phantom{\rule{0ex}{0ex}}=\frac{-12}{7}×\frac{1}{-18}\phantom{\rule{0ex}{0ex}}=\frac{12}{126}\phantom{\rule{0ex}{0ex}}=\frac{12÷3}{126÷3}\phantom{\rule{0ex}{0ex}}=\frac{4}{42}\phantom{\rule{0ex}{0ex}}=\frac{4÷2}{42÷2}\phantom{\rule{0ex}{0ex}}=\frac{2}{21}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iv}\right)\phantom{\rule{0ex}{0ex}}\frac{-1}{10}÷\frac{-8}{5}\phantom{\rule{0ex}{0ex}}=\frac{-1}{10}×\frac{5}{-8}\phantom{\rule{0ex}{0ex}}=\frac{5}{80}\phantom{\rule{0ex}{0ex}}=\frac{5÷5}{80÷5}\phantom{\rule{0ex}{0ex}}=\frac{1}{16}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{v}\right)\phantom{\rule{0ex}{0ex}}\frac{-16}{35}÷\frac{-15}{14}\phantom{\rule{0ex}{0ex}}=\frac{-16}{35}×\frac{14}{-15}\phantom{\rule{0ex}{0ex}}=\frac{224}{525}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{vi}\right)\phantom{\rule{0ex}{0ex}}\frac{-65}{14}÷\frac{13}{7}\phantom{\rule{0ex}{0ex}}=\frac{-65}{14}×\frac{7}{13}\phantom{\rule{0ex}{0ex}}=\frac{-5}{2}$

#### Question 2:

$\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\frac{4}{9}÷\frac{-5}{12}\phantom{\rule{0ex}{0ex}}=\frac{4}{9}×\frac{12}{-5}\phantom{\rule{0ex}{0ex}}=\frac{4×12}{9×-5}\phantom{\rule{0ex}{0ex}}=\frac{48}{-45}\phantom{\rule{0ex}{0ex}}=\frac{-48}{45}\phantom{\rule{0ex}{0ex}}=\frac{-16}{15}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}-8÷\frac{-7}{16}\phantom{\rule{0ex}{0ex}}=-8×\frac{16}{-7}\phantom{\rule{0ex}{0ex}}=\frac{8×16}{7}\phantom{\rule{0ex}{0ex}}=\frac{128}{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\phantom{\rule{0ex}{0ex}}\frac{-12}{7}÷\left(-18\right)\phantom{\rule{0ex}{0ex}}=\frac{-12}{7}×\frac{1}{-18}\phantom{\rule{0ex}{0ex}}=\frac{12}{126}\phantom{\rule{0ex}{0ex}}=\frac{12÷3}{126÷3}\phantom{\rule{0ex}{0ex}}=\frac{4}{42}\phantom{\rule{0ex}{0ex}}=\frac{4÷2}{42÷2}\phantom{\rule{0ex}{0ex}}=\frac{2}{21}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iv}\right)\phantom{\rule{0ex}{0ex}}\frac{-1}{10}÷\frac{-8}{5}\phantom{\rule{0ex}{0ex}}=\frac{-1}{10}×\frac{5}{-8}\phantom{\rule{0ex}{0ex}}=\frac{5}{80}\phantom{\rule{0ex}{0ex}}=\frac{5÷5}{80÷5}\phantom{\rule{0ex}{0ex}}=\frac{1}{16}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{v}\right)\phantom{\rule{0ex}{0ex}}\frac{-16}{35}÷\frac{-15}{14}\phantom{\rule{0ex}{0ex}}=\frac{-16}{35}×\frac{14}{-15}\phantom{\rule{0ex}{0ex}}=\frac{224}{525}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{vi}\right)\phantom{\rule{0ex}{0ex}}\frac{-65}{14}÷\frac{13}{7}\phantom{\rule{0ex}{0ex}}=\frac{-65}{14}×\frac{7}{13}\phantom{\rule{0ex}{0ex}}=\frac{-5}{2}$

#### Question 3:

$\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\left(\frac{5}{9}÷\frac{1}{3}\right)÷\frac{5}{2}=\frac{5}{9}÷\left(\frac{1}{3}÷\frac{5}{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{LHS}\phantom{\rule{0ex}{0ex}}\left(\frac{5}{9}÷\frac{1}{3}\right)÷\frac{5}{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{5}{9}×\frac{3}{1}\right)×\frac{2}{5}\phantom{\rule{0ex}{0ex}}=\frac{5×3×2}{9×1×5}\phantom{\rule{0ex}{0ex}}=\frac{30}{45}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\frac{5}{9}÷\left(\frac{1}{3}÷\frac{5}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{5}{9}÷\left(\frac{1}{3}×\frac{2}{5}\right)\phantom{\rule{0ex}{0ex}}=\frac{5}{9}÷\left(\frac{2}{15}\right)\phantom{\rule{0ex}{0ex}}=\frac{5}{9}×\left(\frac{15}{2}\right)=\frac{75}{18}\phantom{\rule{0ex}{0ex}}=\frac{25}{6}\phantom{\rule{0ex}{0ex}}\mathrm{LHS}\ne \mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{FALSE}$

â€‹(ii)

(iii)
$\left(\frac{-3}{5}÷\frac{-12}{35}\right)÷\frac{1}{14}=\frac{-3}{5}÷\left(\frac{-12}{35}÷\frac{1}{4}\right)\phantom{\rule{0ex}{0ex}}\mathrm{LHS}\phantom{\rule{0ex}{0ex}}=\left(\frac{-3}{5}×\frac{35}{-12}\right)×14\phantom{\rule{0ex}{0ex}}=\frac{\left(-3\right)×35×14}{5×\left(-12\right)}\phantom{\rule{0ex}{0ex}}=\frac{1470}{60}\phantom{\rule{0ex}{0ex}}=\frac{49}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{RHS}\phantom{\rule{0ex}{0ex}}=\frac{-3}{5}÷\left(\frac{-12}{35}÷\frac{1}{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{-3}{5}÷\left(\frac{-12}{35}×\frac{4}{1}\right)\phantom{\rule{0ex}{0ex}}=\frac{-3}{5}÷\left(\frac{-12×4}{35}\right)\phantom{\rule{0ex}{0ex}}=\frac{-3}{5}÷\left(\frac{-48}{35}\right)\phantom{\rule{0ex}{0ex}}=\frac{-3}{5}×\frac{35}{-48}\phantom{\rule{0ex}{0ex}}=\frac{3×35}{5×48}\phantom{\rule{0ex}{0ex}}=\frac{105}{240}\phantom{\rule{0ex}{0ex}}=\frac{7}{16}\phantom{\rule{0ex}{0ex}}\mathrm{LHS}\ne \mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{FALSE}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Question 4:

$\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\left(\frac{5}{9}÷\frac{1}{3}\right)÷\frac{5}{2}=\frac{5}{9}÷\left(\frac{1}{3}÷\frac{5}{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{LHS}\phantom{\rule{0ex}{0ex}}\left(\frac{5}{9}÷\frac{1}{3}\right)÷\frac{5}{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{5}{9}×\frac{3}{1}\right)×\frac{2}{5}\phantom{\rule{0ex}{0ex}}=\frac{5×3×2}{9×1×5}\phantom{\rule{0ex}{0ex}}=\frac{30}{45}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\frac{5}{9}÷\left(\frac{1}{3}÷\frac{5}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{5}{9}÷\left(\frac{1}{3}×\frac{2}{5}\right)\phantom{\rule{0ex}{0ex}}=\frac{5}{9}÷\left(\frac{2}{15}\right)\phantom{\rule{0ex}{0ex}}=\frac{5}{9}×\left(\frac{15}{2}\right)=\frac{75}{18}\phantom{\rule{0ex}{0ex}}=\frac{25}{6}\phantom{\rule{0ex}{0ex}}\mathrm{LHS}\ne \mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{FALSE}$

â€‹(ii)

(iii)
$\left(\frac{-3}{5}÷\frac{-12}{35}\right)÷\frac{1}{14}=\frac{-3}{5}÷\left(\frac{-12}{35}÷\frac{1}{4}\right)\phantom{\rule{0ex}{0ex}}\mathrm{LHS}\phantom{\rule{0ex}{0ex}}=\left(\frac{-3}{5}×\frac{35}{-12}\right)×14\phantom{\rule{0ex}{0ex}}=\frac{\left(-3\right)×35×14}{5×\left(-12\right)}\phantom{\rule{0ex}{0ex}}=\frac{1470}{60}\phantom{\rule{0ex}{0ex}}=\frac{49}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{RHS}\phantom{\rule{0ex}{0ex}}=\frac{-3}{5}÷\left(\frac{-12}{35}÷\frac{1}{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{-3}{5}÷\left(\frac{-12}{35}×\frac{4}{1}\right)\phantom{\rule{0ex}{0ex}}=\frac{-3}{5}÷\left(\frac{-12×4}{35}\right)\phantom{\rule{0ex}{0ex}}=\frac{-3}{5}÷\left(\frac{-48}{35}\right)\phantom{\rule{0ex}{0ex}}=\frac{-3}{5}×\frac{35}{-48}\phantom{\rule{0ex}{0ex}}=\frac{3×35}{5×48}\phantom{\rule{0ex}{0ex}}=\frac{105}{240}\phantom{\rule{0ex}{0ex}}=\frac{7}{16}\phantom{\rule{0ex}{0ex}}\mathrm{LHS}\ne \mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{FALSE}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

.

.

#### Question 9:

$\left(\frac{13}{5}+\frac{-12}{7}\right)÷\left(\frac{-31}{7}×\frac{1}{-2}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{91-60}{35}\right)÷\left(\frac{-31}{-14}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{31}{35}\right)÷\left(\frac{31}{14}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{31}{35}\right)×\left(\frac{14}{31}\right)\phantom{\rule{0ex}{0ex}}=\frac{14}{35}\phantom{\rule{0ex}{0ex}}=\frac{14÷7}{35÷7}\phantom{\rule{0ex}{0ex}}=\frac{2}{5}$

#### Question 10:

$\left(\frac{13}{5}+\frac{-12}{7}\right)÷\left(\frac{-31}{7}×\frac{1}{-2}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{91-60}{35}\right)÷\left(\frac{-31}{-14}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{31}{35}\right)÷\left(\frac{31}{14}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{31}{35}\right)×\left(\frac{14}{31}\right)\phantom{\rule{0ex}{0ex}}=\frac{14}{35}\phantom{\rule{0ex}{0ex}}=\frac{14÷7}{35÷7}\phantom{\rule{0ex}{0ex}}=\frac{2}{5}$

$\left(\frac{65}{12}+\frac{8}{3}\right)÷\left(\frac{65}{12}-\frac{8}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{65}{12}+\frac{32}{12}\right)÷\left(\frac{65}{12}-\frac{32}{12}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\left(\frac{97}{12}\right)÷\left(\frac{33}{12}\right)\phantom{\rule{0ex}{0ex}}=\frac{97}{12}×\frac{12}{33}\phantom{\rule{0ex}{0ex}}=\frac{97}{33}$

#### Question 11:

$\left(\frac{65}{12}+\frac{8}{3}\right)÷\left(\frac{65}{12}-\frac{8}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{65}{12}+\frac{32}{12}\right)÷\left(\frac{65}{12}-\frac{32}{12}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\left(\frac{97}{12}\right)÷\left(\frac{33}{12}\right)\phantom{\rule{0ex}{0ex}}=\frac{97}{12}×\frac{12}{33}\phantom{\rule{0ex}{0ex}}=\frac{97}{33}$

#### Question 12:

â€‹(i)  No, rational numbers are not closed under division in general.

$\frac{a}{0}=\infty$; it is not a rational number.

(ii) No

Therefore, division is not commutative.

(iii) No, rational numbers are not associative under division.

$\frac{a}{b}÷\left(\frac{c}{d}÷\frac{e}{f}\right)\ne \left(\frac{a}{b}÷\frac{c}{d}\right)÷\frac{e}{f}$

(iv) No, we cannot divide 1 by 0. The answer will be$\infty$, which is not defined.

#### Question 1:

â€‹(i)  No, rational numbers are not closed under division in general.

$\frac{a}{0}=\infty$; it is not a rational number.

(ii) No

Therefore, division is not commutative.

(iii) No, rational numbers are not associative under division.

$\frac{a}{b}÷\left(\frac{c}{d}÷\frac{e}{f}\right)\ne \left(\frac{a}{b}÷\frac{c}{d}\right)÷\frac{e}{f}$

(iv) No, we cannot divide 1 by 0. The answer will be$\infty$, which is not defined.

#### Question 7:

We can take any 10 out of these.

#### Question 8:

We can take any 10 out of these.

#### Question 1:

Length of the rope when two pieces of lengths  and are cut off = Total length of the rope - Length of the two cut off pieces
$\therefore 11-\left(2\frac{3}{5}+3\frac{3}{10}\right)$
Now,

LCM of 5 and 10 is 10, i.e., $\left(5×1×2\right)$.

∴â€‹ $2\frac{3}{5}+3\frac{3}{10}=\frac{59}{10}$
Length of the remaining rope $=11-\frac{59}{10}$

Therefore, the length of the remaining rope is .

#### Question 2:

Length of the rope when two pieces of lengths  and are cut off = Total length of the rope - Length of the two cut off pieces
$\therefore 11-\left(2\frac{3}{5}+3\frac{3}{10}\right)$
Now,

LCM of 5 and 10 is 10, i.e., $\left(5×1×2\right)$.

∴â€‹ $2\frac{3}{5}+3\frac{3}{10}=\frac{59}{10}$
Length of the remaining rope $=11-\frac{59}{10}$

Therefore, the length of the remaining rope is .

Weight of rice in the drum = Weight of the drum full of rice - Weight of the empty drum

Therefore, the weight of rice in the drum is .

#### Question 3:

Weight of rice in the drum = Weight of the drum full of rice - Weight of the empty drum

Therefore, the weight of rice in the drum is .

Weight of pears in the basket = Weight of the basket containing three types of fruits - (Weight of apples + Weight of oranges)
=$19\frac{1}{3}-\left(8\frac{1}{9}+3\frac{1}{6}\right)$
Now,

LCM of 9 and 6 is 18, that is, $\left(3×3×2\right)$.

∴â€‹ $8\frac{1}{9}+3\frac{1}{6}=\frac{203}{18}$
Now,
Weight of pears in the basket = $19\frac{1}{3}-\frac{203}{18}$

â€‹Therefore, the weight of the pears in the basket is .

#### Question 4:

Weight of pears in the basket = Weight of the basket containing three types of fruits - (Weight of apples + Weight of oranges)
=$19\frac{1}{3}-\left(8\frac{1}{9}+3\frac{1}{6}\right)$
Now,

LCM of 9 and 6 is 18, that is, $\left(3×3×2\right)$.

∴â€‹ $8\frac{1}{9}+3\frac{1}{6}=\frac{203}{18}$
Now,
Weight of pears in the basket = $19\frac{1}{3}-\frac{203}{18}$

â€‹Therefore, the weight of the pears in the basket is .

Total earning = â‚¹160
Money spent on tea and snacks = â‚¹$26\frac{3}{5}$
Money spent on food = â‚¹$50\frac{1}{2}$
Money spent on repairs = â‚¹$16\frac{2}{5}$
Let the savings be â‚¹x.
Money spent on tea and snacks + Money spent on food + Money spent on repairs + Savings = Total earning
So, $26\frac{3}{5}$ + $50\frac{1}{2}$ + $16\frac{2}{5}$ + x = 160
$⇒26\frac{3}{5}+50\frac{1}{2}+16\frac{2}{5}+x=160\phantom{\rule{0ex}{0ex}}⇒\frac{133}{5}+\frac{101}{2}+\frac{82}{5}+x=160\phantom{\rule{0ex}{0ex}}⇒\frac{266+505+164}{10}+x=160\phantom{\rule{0ex}{0ex}}⇒\frac{935}{10}+x=160\phantom{\rule{0ex}{0ex}}⇒x=160-\frac{935}{10}$
$⇒x=\frac{1600-935}{10}\phantom{\rule{0ex}{0ex}}⇒x=\frac{665}{10}=₹66\frac{1}{2}$
So, the savings are â‚¹$66\frac{1}{2}$.

#### Question 5:

Total earning = â‚¹160
Money spent on tea and snacks = â‚¹$26\frac{3}{5}$
Money spent on food = â‚¹$50\frac{1}{2}$
Money spent on repairs = â‚¹$16\frac{2}{5}$
Let the savings be â‚¹x.
Money spent on tea and snacks + Money spent on food + Money spent on repairs + Savings = Total earning
So, $26\frac{3}{5}$ + $50\frac{1}{2}$ + $16\frac{2}{5}$ + x = 160
$⇒26\frac{3}{5}+50\frac{1}{2}+16\frac{2}{5}+x=160\phantom{\rule{0ex}{0ex}}⇒\frac{133}{5}+\frac{101}{2}+\frac{82}{5}+x=160\phantom{\rule{0ex}{0ex}}⇒\frac{266+505+164}{10}+x=160\phantom{\rule{0ex}{0ex}}⇒\frac{935}{10}+x=160\phantom{\rule{0ex}{0ex}}⇒x=160-\frac{935}{10}$
$⇒x=\frac{1600-935}{10}\phantom{\rule{0ex}{0ex}}⇒x=\frac{665}{10}=₹66\frac{1}{2}$
So, the savings are â‚¹$66\frac{1}{2}$.

Cost of 1 m of cloth = â‚¹$63\frac{3}{4}$
So, cost of $3\frac{2}{5}$ m of cloth
= $63\frac{3}{4}$ × $3\frac{2}{5}$
$=\frac{255}{4}×\frac{17}{5}\phantom{\rule{0ex}{0ex}}=₹216\frac{3}{4}$
So, the cost of $3\frac{2}{5}$ m of cloth is â‚¹$216\frac{3}{4}$.

#### Question 6:

Cost of 1 m of cloth = â‚¹$63\frac{3}{4}$
So, cost of $3\frac{2}{5}$ m of cloth
= $63\frac{3}{4}$ × $3\frac{2}{5}$
$=\frac{255}{4}×\frac{17}{5}\phantom{\rule{0ex}{0ex}}=₹216\frac{3}{4}$
So, the cost of $3\frac{2}{5}$ m of cloth is â‚¹$216\frac{3}{4}$.

Speed =
Time = $6\frac{1}{4}$ h
We know that

Hence, the distance covered in $6\frac{1}{4}$ h is $377\frac{1}{2}\mathrm{km}$.

#### Question 7:

Speed =
Time = $6\frac{1}{4}$ h
We know that

Hence, the distance covered in $6\frac{1}{4}$ h is $377\frac{1}{2}\mathrm{km}$.

Area of the rectangular park = Length of the park $×$ Breadth of the park     (âˆµ Area of rectangle = Length $×$ Breadth)

Therefore, the area of the rectangular park is .

#### Question 8:

Area of the rectangular park = Length of the park $×$ Breadth of the park     (âˆµ Area of rectangle = Length $×$ Breadth)

Therefore, the area of the rectangular park is .

Area of the square plot = Side $×$ Side = ${\left(\mathrm{Side}\right)}^{2}$ = a2  (Because the area of the square is ${a}^{2}$, where a is the side of the square)

Therefore, the area of the square plot is .

#### Question 9:

Area of the square plot = Side $×$ Side = ${\left(\mathrm{Side}\right)}^{2}$ = a2  (Because the area of the square is ${a}^{2}$, where a is the side of the square)

Therefore, the area of the square plot is .

Cost of 1 litre of petrol = â‚¹$63\frac{3}{4}$
Cost of 34 litres of petrol = $63\frac{3}{4}$ × 34 = $\frac{255}{4}×34=₹2167\frac{1}{2}$
So, the cost of 34 litres of petrol is â‚¹$2167\frac{1}{2}$.

#### Question 10:

Cost of 1 litre of petrol = â‚¹$63\frac{3}{4}$
Cost of 34 litres of petrol = $63\frac{3}{4}$ × 34 = $\frac{255}{4}×34=₹2167\frac{1}{2}$
So, the cost of 34 litres of petrol is â‚¹$2167\frac{1}{2}$.

Distance covered by the aeroplane in $4\frac{1}{6}$ hours = $4\frac{1}{6}×1020$

Therefore, the distance covered by the aeroplane is .

#### Question 11:

Distance covered by the aeroplane in $4\frac{1}{6}$ hours = $4\frac{1}{6}×1020$

Therefore, the distance covered by the aeroplane is .

Cost of $3\frac{1}{2}$ m of cloth = â‚¹$166\frac{1}{4}$
So, the cost of 1 m of cloth = $\frac{166\frac{1}{4}}{3\frac{1}{2}}=\frac{\frac{665}{4}}{\frac{7}{2}}=\frac{665}{4}×\frac{2}{7}=\frac{95}{2}=₹47\frac{1}{2}$
Hence, the cost of 1 m of cloth is â‚¹ $47\frac{1}{2}$.

#### Question 12:

Cost of $3\frac{1}{2}$ m of cloth = â‚¹$166\frac{1}{4}$
So, the cost of 1 m of cloth = $\frac{166\frac{1}{4}}{3\frac{1}{2}}=\frac{\frac{665}{4}}{\frac{7}{2}}=\frac{665}{4}×\frac{2}{7}=\frac{95}{2}=₹47\frac{1}{2}$
Hence, the cost of 1 m of cloth is â‚¹ $47\frac{1}{2}$.

Length of each piece of the cord = $71\frac{1}{2}÷26$

Hence, the length of each piece of the cord is .

#### Question 13:

Length of each piece of the cord = $71\frac{1}{2}÷26$

Hence, the length of each piece of the cord is .

Area of a room = Length $×$ Breadth
Thus, we have:
$65\frac{1}{4}=\mathrm{L}\text{ength}×5\frac{7}{16}\phantom{\rule{0ex}{0ex}}\mathrm{L}\text{ength}=65\frac{1}{4}÷5\frac{7}{16}\phantom{\rule{0ex}{0ex}}$

Hence, the length of the room is 12 metres.

#### Question 14:

Area of a room = Length $×$ Breadth
Thus, we have:
$65\frac{1}{4}=\mathrm{L}\text{ength}×5\frac{7}{16}\phantom{\rule{0ex}{0ex}}\mathrm{L}\text{ength}=65\frac{1}{4}÷5\frac{7}{16}\phantom{\rule{0ex}{0ex}}$

Hence, the length of the room is 12 metres.

Let the other fraction be x.

Now, we have:

Hence, the other fraction is $1\frac{1}{55}$.

#### Question 15:

Let the other fraction be x.

Now, we have:

Hence, the other fraction is $1\frac{1}{55}$.

If $\frac{5}{8}$of the students are boys, then the ratio of girls is $1-\frac{5}{8}$, that is, $\frac{3}{8}$.

Now, let x be the total number of students.

Thus, we have:

$=240×\frac{8}{3}\phantom{\rule{0ex}{0ex}}=\frac{240}{1}×\frac{8}{3}\phantom{\rule{0ex}{0ex}}=\frac{240×8}{1×3}\phantom{\rule{0ex}{0ex}}=\frac{1920}{3}\phantom{\rule{0ex}{0ex}}=640$

Hence, the total number of students is 640.
Now,
Number of boys = Total number of students - Number of girls
$=640-240\phantom{\rule{0ex}{0ex}}=400$

Hence, the number of boys is 400.

#### Question 16:

If $\frac{5}{8}$of the students are boys, then the ratio of girls is $1-\frac{5}{8}$, that is, $\frac{3}{8}$.

Now, let x be the total number of students.

Thus, we have:

$=240×\frac{8}{3}\phantom{\rule{0ex}{0ex}}=\frac{240}{1}×\frac{8}{3}\phantom{\rule{0ex}{0ex}}=\frac{240×8}{1×3}\phantom{\rule{0ex}{0ex}}=\frac{1920}{3}\phantom{\rule{0ex}{0ex}}=640$

Hence, the total number of students is 640.
Now,
Number of boys = Total number of students - Number of girls
$=640-240\phantom{\rule{0ex}{0ex}}=400$

Hence, the number of boys is 400.

Ratio of the read book = $\frac{7}{9}$
Ratio of the unread book = $1-\frac{7}{9}$

$=\frac{2}{9}$
Let x be the total number of pages in the book.

Thus, we have:

$=40×\frac{9}{2}\phantom{\rule{0ex}{0ex}}=\frac{40}{1}×\frac{9}{2}\phantom{\rule{0ex}{0ex}}=\frac{40×9}{1×2}\phantom{\rule{0ex}{0ex}}=\frac{360}{2}\phantom{\rule{0ex}{0ex}}=180$

Hence, the total number of pages in the book is 180.

#### Question 17:

Ratio of the read book = $\frac{7}{9}$
Ratio of the unread book = $1-\frac{7}{9}$

$=\frac{2}{9}$
Let x be the total number of pages in the book.

Thus, we have:

$=40×\frac{9}{2}\phantom{\rule{0ex}{0ex}}=\frac{40}{1}×\frac{9}{2}\phantom{\rule{0ex}{0ex}}=\frac{40×9}{1×2}\phantom{\rule{0ex}{0ex}}=\frac{360}{2}\phantom{\rule{0ex}{0ex}}=180$

Hence, the total number of pages in the book is 180.

Amount of money spent on notebooks = $300×\frac{1}{3}$

$=\frac{300}{1}×\frac{1}{3}\phantom{\rule{0ex}{0ex}}=\frac{300}{3}\phantom{\rule{0ex}{0ex}}=100$

∴ Money left after spending on notebooks = $300-100$
$=200$
Amount of money spent on stationery items from the remainder = $200×\frac{1}{4}$
$=\frac{200}{1}×\frac{1}{4}\phantom{\rule{0ex}{0ex}}=\frac{200}{4}\phantom{\rule{0ex}{0ex}}=50$

∴ Amount of money left with Rita = $200-50$

#### Question 18:

Amount of money spent on notebooks = $300×\frac{1}{3}$

$=\frac{300}{1}×\frac{1}{3}\phantom{\rule{0ex}{0ex}}=\frac{300}{3}\phantom{\rule{0ex}{0ex}}=100$

∴ Money left after spending on notebooks = $300-100$
$=200$
Amount of money spent on stationery items from the remainder = $200×\frac{1}{4}$
$=\frac{200}{1}×\frac{1}{4}\phantom{\rule{0ex}{0ex}}=\frac{200}{4}\phantom{\rule{0ex}{0ex}}=50$

∴ Amount of money left with Rita = $200-50$

Amit's income per month = â‚¹32,000
Money spent on food =
Remaining amount = â‚¹32,000 − â‚¹8,000 = â‚¹24,000
Money spent on house rent =
Money left = â‚¹24,000 − â‚¹7,200 = â‚¹16,800
Money spent on education of children = $\frac{5}{21}×₹16,800=₹4,000$
Amount of money still left with him = â‚¹16,800 − â‚¹4,000 = â‚¹12,800

#### Question 19:

Amit's income per month = â‚¹32,000
Money spent on food =
Remaining amount = â‚¹32,000 − â‚¹8,000 = â‚¹24,000
Money spent on house rent =
Money left = â‚¹24,000 − â‚¹7,200 = â‚¹16,800
Money spent on education of children = $\frac{5}{21}×₹16,800=₹4,000$
Amount of money still left with him = â‚¹16,800 − â‚¹4,000 = â‚¹12,800

Let x be the required number.
We know that $\frac{3}{5}$ of the number exceeds its $\frac{2}{7}$ by 44.
That is,

$\frac{3}{5}×x=\frac{2}{7}×x+44$

$\frac{11}{35}×x=44\phantom{\rule{0ex}{0ex}}$
$x=44÷\frac{11}{35}\phantom{\rule{0ex}{0ex}}$

$=44×\frac{35}{11}\phantom{\rule{0ex}{0ex}}=\frac{44}{1}×\frac{35}{11}\phantom{\rule{0ex}{0ex}}=\frac{44×35}{1×11}\phantom{\rule{0ex}{0ex}}=\frac{1540}{11}\phantom{\rule{0ex}{0ex}}=140$

Hence, the number is 140.

#### Question 20:

Let x be the required number.
We know that $\frac{3}{5}$ of the number exceeds its $\frac{2}{7}$ by 44.
That is,

$\frac{3}{5}×x=\frac{2}{7}×x+44$

$\frac{11}{35}×x=44\phantom{\rule{0ex}{0ex}}$
$x=44÷\frac{11}{35}\phantom{\rule{0ex}{0ex}}$

$=44×\frac{35}{11}\phantom{\rule{0ex}{0ex}}=\frac{44}{1}×\frac{35}{11}\phantom{\rule{0ex}{0ex}}=\frac{44×35}{1×11}\phantom{\rule{0ex}{0ex}}=\frac{1540}{11}\phantom{\rule{0ex}{0ex}}=140$

Hence, the number is 140.

Ratio of spectators in the open $=1-\frac{2}{7}\phantom{\rule{0ex}{0ex}}$
$=\frac{5}{7}$
Total number of spectators in the open = x
Then,$\frac{5}{7}×x=15000$
$⇒x=15000÷\frac{5}{7}$

$=15000×\frac{7}{5}\phantom{\rule{0ex}{0ex}}=\frac{15000}{1}×\frac{7}{5}\phantom{\rule{0ex}{0ex}}=\frac{15000×7}{1×5}\phantom{\rule{0ex}{0ex}}=\frac{10500}{5}\phantom{\rule{0ex}{0ex}}=21000$

Hence, the total number of spectators is 21,000

#### Question 1:

Ratio of spectators in the open $=1-\frac{2}{7}\phantom{\rule{0ex}{0ex}}$
$=\frac{5}{7}$
Total number of spectators in the open = x
Then,$\frac{5}{7}×x=15000$
$⇒x=15000÷\frac{5}{7}$

$=15000×\frac{7}{5}\phantom{\rule{0ex}{0ex}}=\frac{15000}{1}×\frac{7}{5}\phantom{\rule{0ex}{0ex}}=\frac{15000×7}{1×5}\phantom{\rule{0ex}{0ex}}=\frac{10500}{5}\phantom{\rule{0ex}{0ex}}=21000$

Hence, the total number of spectators is 21,000

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

$=\frac{\left(-15\right)+28}{48}\phantom{\rule{0ex}{0ex}}=\frac{13}{48}$

#### Question 2:

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

$=\frac{\left(-15\right)+28}{48}\phantom{\rule{0ex}{0ex}}=\frac{13}{48}$

(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

$\frac{-8}{15}+\frac{-4}{3}=\frac{1×\left(-8\right)+5×\left(-4\right)}{15}$
$=\frac{\left(-8\right)+\left(-20\right)}{15}\phantom{\rule{0ex}{0ex}}=\frac{-28}{15}$

#### Question 3:

(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

$\frac{-8}{15}+\frac{-4}{3}=\frac{1×\left(-8\right)+5×\left(-4\right)}{15}$
$=\frac{\left(-8\right)+\left(-20\right)}{15}\phantom{\rule{0ex}{0ex}}=\frac{-28}{15}$

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

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, $\left(29×1×36\right).$

(a) $\frac{11}{78}$
$\left(\frac{-7}{26}+\frac{16}{39}\right)=\frac{39×\left(-7\right)+26×16}{1014}$
$=\frac{\left(-273\right)+416}{1014}\phantom{\rule{0ex}{0ex}}=\frac{143}{1014}\phantom{\rule{0ex}{0ex}}=\frac{11}{78}$

#### Question 4:

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

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, $\left(29×1×36\right).$

(a) $\frac{11}{78}$
$\left(\frac{-7}{26}+\frac{16}{39}\right)=\frac{39×\left(-7\right)+26×16}{1014}$
$=\frac{\left(-273\right)+416}{1014}\phantom{\rule{0ex}{0ex}}=\frac{143}{1014}\phantom{\rule{0ex}{0ex}}=\frac{11}{78}$

(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

$\left(\frac{3}{1}+\frac{-5}{7}\right)=\frac{7×3+1×\left(-5\right)}{7}$
$=\frac{21+\left(-5\right)}{7}\phantom{\rule{0ex}{0ex}}=\frac{16}{7}$

#### Question 5:

(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

$\left(\frac{3}{1}+\frac{-5}{7}\right)=\frac{7×3+1×\left(-5\right)}{7}$
$=\frac{21+\left(-5\right)}{7}\phantom{\rule{0ex}{0ex}}=\frac{16}{7}$

(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, $\left(4×1×2\right).$

$\left(\frac{-31}{4}+\frac{-5}{8}\right)=\frac{2×\left(-31\right)+1×\left(-5\right)}{8}$
$=\frac{\left(-62\right)+\left(-5\right)}{8}\phantom{\rule{0ex}{0ex}}=\frac{-67}{8}$

#### Question 6:

(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, $\left(4×1×2\right).$

$\left(\frac{-31}{4}+\frac{-5}{8}\right)=\frac{2×\left(-31\right)+1×\left(-5\right)}{8}$
$=\frac{\left(-62\right)+\left(-5\right)}{8}\phantom{\rule{0ex}{0ex}}=\frac{-67}{8}$

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

Now,

$\frac{7}{12}+x=\frac{-4}{15}\phantom{\rule{0ex}{0ex}}$

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

$=\frac{4×\left(-4\right)+5×\left(-7\right)}{60}\phantom{\rule{0ex}{0ex}}=\frac{\left(-16\right)+\left(-35\right)}{60}\phantom{\rule{0ex}{0ex}}=\frac{-51}{60}\phantom{\rule{0ex}{0ex}}=\frac{-17}{20}$

#### Question 7:

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

Now,

$\frac{7}{12}+x=\frac{-4}{15}\phantom{\rule{0ex}{0ex}}$

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

$=\frac{4×\left(-4\right)+5×\left(-7\right)}{60}\phantom{\rule{0ex}{0ex}}=\frac{\left(-16\right)+\left(-35\right)}{60}\phantom{\rule{0ex}{0ex}}=\frac{-51}{60}\phantom{\rule{0ex}{0ex}}=\frac{-17}{20}$

(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:

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

$=\frac{\left(10+7\right)}{15}+\frac{\left[\left(-16\right)+\left(-11\right)\right]}{20}\phantom{\rule{0ex}{0ex}}=\left(\frac{17}{15}+\frac{-27}{20}\right)\phantom{\rule{0ex}{0ex}}=\frac{\left[68+\left(-81\right)\right]}{60}\phantom{\rule{0ex}{0ex}}=\frac{-13}{60}$

#### Question 8:

(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:

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

$=\frac{\left(10+7\right)}{15}+\frac{\left[\left(-16\right)+\left(-11\right)\right]}{20}\phantom{\rule{0ex}{0ex}}=\left(\frac{17}{15}+\frac{-27}{20}\right)\phantom{\rule{0ex}{0ex}}=\frac{\left[68+\left(-81\right)\right]}{60}\phantom{\rule{0ex}{0ex}}=\frac{-13}{60}$

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

Now,

$x+\left(-5\right)=\frac{-4}{3}$

$=\frac{-4}{3}+\frac{5}{1}\phantom{\rule{0ex}{0ex}}=\frac{\left(-4\right)+15}{3}\phantom{\rule{0ex}{0ex}}=\frac{11}{3}$

#### Question 9:

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

Now,

$x+\left(-5\right)=\frac{-4}{3}$

$=\frac{-4}{3}+\frac{5}{1}\phantom{\rule{0ex}{0ex}}=\frac{\left(-4\right)+15}{3}\phantom{\rule{0ex}{0ex}}=\frac{11}{3}$

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

Now,

$\frac{-5}{7}+x=\frac{-2}{3}$

$=\frac{\left(-14\right)+15}{21}\phantom{\rule{0ex}{0ex}}=\frac{1}{21}$

#### Question 10:

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

Now,

$\frac{-5}{7}+x=\frac{-2}{3}$

$=\frac{\left(-14\right)+15}{21}\phantom{\rule{0ex}{0ex}}=\frac{1}{21}$

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

Now,

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

$=\frac{-10-5}{6}\phantom{\rule{0ex}{0ex}}=\frac{-15}{6}\phantom{\rule{0ex}{0ex}}=\frac{-5}{2}$
Thus, the required number is $\frac{-5}{2}$

#### Question 11:

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

Now,

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

$=\frac{-10-5}{6}\phantom{\rule{0ex}{0ex}}=\frac{-15}{6}\phantom{\rule{0ex}{0ex}}=\frac{-5}{2}$
Thus, the required number is $\frac{-5}{2}$

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

The reciprocal of $\frac{-3}{7}$ is

#### Question 12:

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

The reciprocal of $\frac{-3}{7}$ is

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

Now,

$x×\frac{14}{27}=\frac{-28}{81}$

$⇒x=\frac{-28}{81}÷\frac{14}{27}$

$=\frac{-28}{81}×\frac{27}{14}\phantom{\rule{0ex}{0ex}}=\frac{\left(-28\right)×27}{81×14}\phantom{\rule{0ex}{0ex}}=\frac{-\left(28×27\right)}{81×14}\phantom{\rule{0ex}{0ex}}=\frac{-\left(2×3\right)}{9×1}\phantom{\rule{0ex}{0ex}}=\frac{-6}{9}\phantom{\rule{0ex}{0ex}}=\frac{-2}{3}$
Thus, the other number is $\frac{-2}{3}$

#### Question 13:

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

Now,

$x×\frac{14}{27}=\frac{-28}{81}$

$⇒x=\frac{-28}{81}÷\frac{14}{27}$

$=\frac{-28}{81}×\frac{27}{14}\phantom{\rule{0ex}{0ex}}=\frac{\left(-28\right)×27}{81×14}\phantom{\rule{0ex}{0ex}}=\frac{-\left(28×27\right)}{81×14}\phantom{\rule{0ex}{0ex}}=\frac{-\left(2×3\right)}{9×1}\phantom{\rule{0ex}{0ex}}=\frac{-6}{9}\phantom{\rule{0ex}{0ex}}=\frac{-2}{3}$
Thus, the other number is $\frac{-2}{3}$

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

Now,

$x×\frac{-15}{4}=\frac{-16}{35}$
$⇒x=\frac{-16}{35}÷\frac{-15}{14}$

$=\frac{-16}{35}×\frac{14}{-15}\phantom{\rule{0ex}{0ex}}=\frac{-\left(16×14\right)}{-\left(35×15\right)}\phantom{\rule{0ex}{0ex}}=\frac{16×14}{35×15}=\frac{224}{525}=\frac{32}{75}$

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

#### Question 14:

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

Now,

$x×\frac{-15}{4}=\frac{-16}{35}$
$⇒x=\frac{-16}{35}÷\frac{-15}{14}$

$=\frac{-16}{35}×\frac{14}{-15}\phantom{\rule{0ex}{0ex}}=\frac{-\left(16×14\right)}{-\left(35×15\right)}\phantom{\rule{0ex}{0ex}}=\frac{16×14}{35×15}=\frac{224}{525}=\frac{32}{75}$

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

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

Now,

Thus, the required number is $\frac{7}{5}$

#### Question 15:

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

Now,

Thus, the required number is $\frac{7}{5}$

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

Now,

$=\frac{-3}{1}+\frac{10}{3}\phantom{\rule{0ex}{0ex}}=\frac{\left(-9+10\right)}{3}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}$
Thus, the other number is $\frac{1}{3}$

#### Question 16:

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

Now,

$=\frac{-3}{1}+\frac{10}{3}\phantom{\rule{0ex}{0ex}}=\frac{\left(-9+10\right)}{3}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}$
Thus, the other number is $\frac{1}{3}$

(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.

#### Question 17:

(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.

(a) $\frac{-3}{10}$

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

$=\frac{-72}{240}\phantom{\rule{0ex}{0ex}}=\frac{-3}{10}$

#### Question 18:

(a) $\frac{-3}{10}$

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

$=\frac{-72}{240}\phantom{\rule{0ex}{0ex}}=\frac{-3}{10}$

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

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

$=\frac{-5×3}{9×2}\phantom{\rule{0ex}{0ex}}=\frac{-15}{18}\phantom{\rule{0ex}{0ex}}=\frac{-5}{6}$

#### Question 19:

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

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

$=\frac{-5×3}{9×2}\phantom{\rule{0ex}{0ex}}=\frac{-15}{18}\phantom{\rule{0ex}{0ex}}=\frac{-5}{6}$

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

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

Now,

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

$⇒\frac{a}{b}=\frac{5}{-6}$

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

#### Question 20:

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

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

Now,

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

$⇒\frac{a}{b}=\frac{5}{-6}$

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

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

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

#### Question 21:

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

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

(c) $\frac{-4}{3}$
Reciprocal of $\frac{-3}{4}$ is

#### Question 22:

(c) $\frac{-4}{3}$
Reciprocal of $\frac{-3}{4}$ is

(d) $\frac{-5}{24}$
Rational number between $\frac{-2}{3}$ and $\frac{1}{4}$ = $\frac{1}{2}\left(\frac{-2}{3}+\frac{1}{4}\right)$
$=\frac{1}{2}\left(\frac{-8+3}{12}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×\frac{-5}{12}\phantom{\rule{0ex}{0ex}}=\frac{-5}{24}$

#### Question 23:

(d) $\frac{-5}{24}$
Rational number between $\frac{-2}{3}$ and $\frac{1}{4}$ = $\frac{1}{2}\left(\frac{-2}{3}+\frac{1}{4}\right)$
$=\frac{1}{2}\left(\frac{-8+3}{12}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×\frac{-5}{12}\phantom{\rule{0ex}{0ex}}=\frac{-5}{24}$

(b) is a negative rational number

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

#### Question 1:

(b) is a negative rational number

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

(i) $\phantom{\rule{0ex}{0ex}}\frac{7}{-10}=\frac{7×-1}{-10×-1}=\frac{-7}{10}\phantom{\rule{0ex}{0ex}}$

#### Question 2:

(i) $\phantom{\rule{0ex}{0ex}}\frac{7}{-10}=\frac{7×-1}{-10×-1}=\frac{-7}{10}\phantom{\rule{0ex}{0ex}}$

#### Question 7:

$\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\frac{-3}{5}×\frac{10}{7}\phantom{\rule{0ex}{0ex}}=\frac{-3×10}{5×7}\phantom{\rule{0ex}{0ex}}=\frac{-30}{35}\phantom{\rule{0ex}{0ex}}=\frac{-6×5}{7×5}\phantom{\rule{0ex}{0ex}}=\frac{-6}{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\left(\frac{-5}{8}{\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\frac{-5}{8}\right)}\phantom{\rule{0ex}{0ex}}=1×\frac{8}{-5}\phantom{\rule{0ex}{0ex}}=\frac{8}{-5}\phantom{\rule{0ex}{0ex}}=\frac{8×-1}{-5×-1}\phantom{\rule{0ex}{0ex}}=\frac{-8}{5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\phantom{\rule{0ex}{0ex}}\left(-6{\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{-6}\phantom{\rule{0ex}{0ex}}=\frac{1×-1}{-6×-1}\phantom{\rule{0ex}{0ex}}=\frac{-1}{6}\phantom{\rule{0ex}{0ex}}$

#### Question 8:

$\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\frac{-3}{5}×\frac{10}{7}\phantom{\rule{0ex}{0ex}}=\frac{-3×10}{5×7}\phantom{\rule{0ex}{0ex}}=\frac{-30}{35}\phantom{\rule{0ex}{0ex}}=\frac{-6×5}{7×5}\phantom{\rule{0ex}{0ex}}=\frac{-6}{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\left(\frac{-5}{8}{\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\frac{-5}{8}\right)}\phantom{\rule{0ex}{0ex}}=1×\frac{8}{-5}\phantom{\rule{0ex}{0ex}}=\frac{8}{-5}\phantom{\rule{0ex}{0ex}}=\frac{8×-1}{-5×-1}\phantom{\rule{0ex}{0ex}}=\frac{-8}{5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\phantom{\rule{0ex}{0ex}}\left(-6{\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{-6}\phantom{\rule{0ex}{0ex}}=\frac{1×-1}{-6×-1}\phantom{\rule{0ex}{0ex}}=\frac{-1}{6}\phantom{\rule{0ex}{0ex}}$

(i) Commutative law of multiplication

(ii) Existence of  multiplicative identity

(iii) Associative law of multiplication

(iv) Multiplicative property of 0

(v) Distributive law of multiplication over addition

#### Question 9:

(i) Commutative law of multiplication

(ii) Existence of  multiplicative identity

(iii) Associative law of multiplication

(iv) Multiplicative property of 0

(v) Distributive law of multiplication over addition

#### Question 10:

(c) $\frac{4}{15}$

Let the number be $x$
Now,

#### Question 11:

(c) $\frac{4}{15}$

Let the number be $x$
Now,

(d) $\frac{-17}{12}$

Let the number be $x$.
Now,

#### Question 12:

(d) $\frac{-17}{12}$

Let the number be $x$.
Now,

(b) $\frac{-4}{5}$

We have:

${\left(\frac{-5}{4}\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\frac{-5}{4}\right)}\phantom{\rule{0ex}{0ex}}=1×\frac{4}{-5}\phantom{\rule{0ex}{0ex}}=\frac{4}{-5}\phantom{\rule{0ex}{0ex}}=\frac{4×-1}{-5×-1}\phantom{\rule{0ex}{0ex}}=\frac{-4}{5}$

#### Question 13:

(b) $\frac{-4}{5}$

We have:

${\left(\frac{-5}{4}\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\frac{-5}{4}\right)}\phantom{\rule{0ex}{0ex}}=1×\frac{4}{-5}\phantom{\rule{0ex}{0ex}}=\frac{4}{-5}\phantom{\rule{0ex}{0ex}}=\frac{4×-1}{-5×-1}\phantom{\rule{0ex}{0ex}}=\frac{-4}{5}$

(a) $\frac{5}{6}$

Let the required number be $x$.
Now,

$\frac{-3}{10}×x=\frac{-1}{4}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1}{4}÷\left(\frac{-3}{10}\right)\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1}{4}×\frac{10}{-3}\phantom{\rule{0ex}{0ex}}⇒x=\frac{10}{12}=\frac{5}{6}$

#### Question 14:

(a) $\frac{5}{6}$

Let the required number be $x$.
Now,

$\frac{-3}{10}×x=\frac{-1}{4}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1}{4}÷\left(\frac{-3}{10}\right)\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1}{4}×\frac{10}{-3}\phantom{\rule{0ex}{0ex}}⇒x=\frac{10}{12}=\frac{5}{6}$

(b)â€‹ $\frac{5}{4}$

We have:
$\left(\frac{-5}{6}÷\frac{-2}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{-5}{6}×\frac{3}{-2}\phantom{\rule{0ex}{0ex}}=\frac{15}{12}\phantom{\rule{0ex}{0ex}}=\frac{5}{4}$

#### Question 15:

(b)â€‹ $\frac{5}{4}$

We have:
$\left(\frac{-5}{6}÷\frac{-2}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{-5}{6}×\frac{3}{-2}\phantom{\rule{0ex}{0ex}}=\frac{15}{12}\phantom{\rule{0ex}{0ex}}=\frac{5}{4}$

(c) $\frac{-8}{15}$

#### Question 16:

(c) $\frac{-8}{15}$

(b) $\frac{-9}{7}$

#### Question 17:

(b) $\frac{-9}{7}$

(b) $\frac{-1}{12}$

#### Question 18:

(b) $\frac{-1}{12}$

(ii)

(iii)

(iv)

#### Question 19:

(ii)

(iii)

(iv)

(i) T

â€‹If  are rational numbers, then $\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}$ is also a rational number because  are all rational numbers.

(ii) F

â€‹Rational numbers are not always closed under division. They are closed under division only if the denominator is non-zero.

(iii) F

cannot be defined.
â€‹
(iv) F

â€‹Let  represent rational numbers.

Now, we have:

$\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}$
$\frac{c}{d}-\frac{a}{b}=\frac{bc-ad}{bd}$

∴ $\frac{a}{b}-\frac{c}{d}\ne \frac{c}{d}-\frac{a}{b}$

(v) T
`
$-\left(\frac{-7}{8}\right)=-1×\left(\frac{-7}{8}\right)=\frac{-1×-7}{8}=\frac{7}{8}$

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