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Rs Aggarwal 2018 Solutions for Class 8 Math Chapter 1 Rational Numbers are provided here with simple step-by-step explanations. These solutions for Rational Numbers are extremely popular among Class 8 students for Math Rational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 8 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

#### Question 1:

Express $\frac{-3}{5}$ as a rational number with denominator
(i) 20
(ii) −30
(iii) 35
(iv) −40

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:

Express $\frac{-42}{98}$ as a rational number with denominator 7.

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:

Express $\frac{-48}{60}$ as a rational number with denominator 5.

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

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

#### Question 4:

Express each of the following rational numbers in standard form:
(i) $\frac{-12}{30}$
(ii) $\frac{-14}{49}$
(iii) $\frac{24}{-64}$
(iv) $\frac{-36}{-63}$

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:

Which of the two rational numbers is greater in the given pair?
(i)
(ii)
(iii)
(iv)
(v)
(vi)

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,
$\frac{-1}{1}=\frac{-1×2}{1×2}=\frac{-2}{2}$
∵ $\frac{-2}{1}<\frac{-1}{1}$
∴ $-1<\frac{-1}{2}$

#### Question 6:

Which of the two rational numbers is greater in the given pair?
(i)
(ii)
(iii)
(iv)
(v)
(vi)

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. 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,

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

∴ $\frac{7}{-9}<\frac{-5}{8}$

3. 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 .

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

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:

Fill in the blanks with the correct symbol out of >, = and <:
(i) $\frac{-3}{7}.....\frac{6}{-13}$
(ii) $\frac{5}{-13}.....\frac{-35}{91}$
(iii)
(iv) $\frac{-2}{3}.....\frac{5}{-8}$
(v)
(vi) $\frac{-8}{9}.....\frac{-9}{10}$

(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,

$-39>-41$

∴

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

∴ $\frac{-2}{1}=\frac{-2×5}{1×5}=\frac{-10}{5}$ 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

∴ $\frac{-2}{3}=\frac{-2×8}{3×8}=\frac{-16}{24}$ 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

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

Clearly,

$-81<-80$

∴$\frac{-81}{90}<\frac{-80}{90}$

Thus,

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

#### Question 8:

Arrange the following rational numbers in ascending order:
(i)
(ii)
(iii)
(iv)

(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,

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

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

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,

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

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

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,

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

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

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,

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

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

That is

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

#### Question 9:

Arrange the following rational numbers in descending order:
(i)
(ii)
(iii)
(iv)

(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,

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

∴ $\frac{1}{3}>-2>\frac{-13}{6}>\frac{-8}{3}$. 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,

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

∴ $\frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{-23}{24}$. 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:

Which of the following statements are true and which are false?
(i) Every whole number is a rational number.
(ii) Every integer is a rational number.
(iii) 0 is a whole number but it is not a rational number.

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:

Represent each of the following numbers on the number line:
(i) $\frac{1}{3}$
(ii) $\frac{2}{7}$
(iii) $1\frac{3}{4}$

(iv) $2\frac{2}{5}$

(v) $3\frac{1}{2}$

(vi) $5\frac{5}{7}$

(vii) $4\frac{2}{3}$

(viii) 8

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

#### Question 2:

Represent each of the following numbers on the number line:
(i) $\frac{-1}{3}$
(ii) $\frac{-3}{4}$
(iii) $-1\frac{2}{3}$
(iv) $-3\frac{1}{7}$
(v) $-4\frac{3}{5}$
(vi) $-2\frac{5}{6}$
(vii) −3
(viii) $-2\frac{7}{8}$

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

#### Question 3:

Which of the following statements are true and which are false?
(i) $\frac{-3}{5}$ lies to the left of 0 on the number line.
(ii) $\frac{-12}{7}$ lies to the right of 0 on the number line.
(iii) The rational numbers $\frac{1}{3}$ and $\frac{-5}{2}$ are on opposite sides of 0 on the number line.
(iv) The rational number $\frac{-18}{-13}$ lies to the left of 0 on the number line.

(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:

Add the following rational numbers:
(i)
(ii)
(iii)
(iv)
(v)
(vi)

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:

Add the following rational numbers:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)

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. 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 $\frac{-7}{12}=\frac{-7×2}{12×2}=\frac{-14}{24}$

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

3. 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 $\frac{11}{6}=\frac{11×3}{6×3}=\frac{33}{18}$

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

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

LCM of 16 and 24 is 48.

Now,

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

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

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

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

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

LCM of 18 and 27 is 54.

Now,

$\frac{-7}{18}=\frac{-7×3}{18×3}=\frac{-21}{54}$ 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. 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 $\frac{2}{-15}=\frac{2×\left(-1\right)}{-15×\left(-1\right)}=\frac{-2}{15}$

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. 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. 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:

Verify the following:
(i) $\frac{-12}{5}+\frac{2}{7}=\frac{2}{7}+\frac{-12}{5}$
(ii) $\frac{-5}{8}+\frac{-9}{13}=\frac{-9}{13}+\frac{-5}{8}$
(iii) $3+\frac{-7}{12}=\frac{-7}{12}+3$
(iv) $\frac{2}{-7}+\frac{12}{-35}=\frac{12}{-35}+\frac{2}{-7}$

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. 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. 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 = $\frac{2}{-7}+\frac{12}{-35}$

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.

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

#### Question 4:

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

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

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

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

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.

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

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

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

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:

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

(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:

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

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:

Subtract:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)

(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:

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

(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:

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

#### Question 10:

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

#### Question 11:

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

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:

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

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:

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

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:

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

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.

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:

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

(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:

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

(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:

Verify each of the following:
(i) $\left(\frac{5}{7}×\frac{12}{13}\right)×\frac{7}{18}=\frac{5}{7}×\left(\frac{12}{13}×\frac{7}{8}\right)$
(ii) $\frac{-13}{24}×\left(\frac{-12}{5}×\frac{35}{36}\right)=\left(\frac{-13}{24}×\frac{-12}{5}\right)×\frac{35}{36}$
(iii) $\left(\frac{-9}{5}×\frac{-10}{3}\right)×\frac{21}{-4}=\frac{-9}{5}×\left(\frac{-10}{3}×\frac{21}{-4}\right)$

(i)

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

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

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

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

(ii)

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

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

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

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

(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:

Fill in the blanks:
(i) $\frac{-23}{17}×\frac{18}{35}=\frac{18}{35}×\left(......\right)$
(ii) $-38×\frac{-7}{19}=\frac{-7}{19}×\left(......\right)$
(iii) $\left(\frac{15}{7}×\frac{-21}{10}\right)×\frac{-5}{6}=\left(......\right)×\left(\frac{-21}{10}×\frac{-5}{6}\right)$
(iv) $\frac{-12}{5}×\left(\frac{4}{15}×\frac{25}{-16}\right)=\left(\frac{-12}{5}×\frac{4}{15}\right)×\left(......\right)$

(i)

(ii)

(iii)

(iv)

#### Question 5:

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

#### Question 6:

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

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

#### Question 7:

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

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

(iii)

(iv)

#### Question 8:

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

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:

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

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

#### Question 1:

Simplify:
(i) $\frac{4}{9}÷\frac{-5}{12}$
(ii) $-8÷\frac{-7}{16}$
(iii) $\frac{-12}{7}÷\left(-18\right)$
(iv) $\frac{-1}{10}÷\frac{-8}{5}$
(v) $\frac{-16}{35}÷\frac{-15}{14}$
(vi) $\frac{-65}{14}÷\frac{13}{7}$

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

#### Question 2:

Verify whether the given statement is true or false:
(i) $\frac{13}{5}÷\frac{26}{10}=\frac{26}{10}÷\frac{13}{5}$
(ii) $-9÷\frac{3}{4}=\frac{3}{4}÷\left(-9\right)$
(iii) $\frac{-8}{9}÷\frac{-4}{3}=\frac{-4}{3}÷\frac{-8}{9}$
(iv) $\frac{-7}{24}÷\frac{3}{-16}=\frac{3}{-16}÷\frac{-7}{24}$

#### Question 3:

Verify whether the given statement is true or false:
(i) $\left(\frac{5}{9}÷\frac{1}{3}\right)÷\frac{5}{2}=\frac{5}{9}÷\left(\frac{1}{3}÷\frac{5}{2}\right)$
(ii) $\left\{\left(-16\right)÷\frac{6}{5}\right\}÷\frac{-9}{10}=\left(-16\right)÷\left\{\frac{6}{5}÷\frac{-9}{10}\right\}$
(iii) $\left(\frac{-3}{5}÷\frac{-12}{35}\right)÷\frac{1}{14}=\frac{-3}{5}÷\left(\frac{-12}{35}÷\frac{1}{4}\right)$

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

(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:

The product of two rational numbers is −9. If one of the numbers is −12, find the other.

#### Question 5:

The product of two rational numbers is $\frac{-16}{9}$. If one of the numbers is $\frac{-4}{3}$, find the other.

.

#### Question 6:

By what rational number should we multiply $\frac{-15}{56}$ to get $\frac{-5}{7}$?

#### Question 7:

By what rational number should $\frac{-8}{39}$ be multiplied to obtain $\frac{1}{26}$?

#### Question 8:

By what number should $\frac{-33}{8}$ be divided to get $\frac{-11}{2}$?

#### Question 9:

Divide the sum of $\frac{13}{5}$ and $\frac{-12}{7}$ by the product of $\frac{-31}{7}$ and $\frac{1}{-2}.$

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

#### Question 10:

Divide the sum of $\frac{65}{12}$ and $\frac{8}{3}$ by their difference.

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

#### Question 11:

Fill in the blanks:
(i) $\frac{9}{8}+\left(...\right)=\frac{-3}{2}$
(ii) $\left(...\right)÷\left(\frac{-7}{5}\right)=\frac{10}{19}$
(iii) $\left(...\right)÷\left(-3\right)=\frac{-4}{15}$
(iv) $\left(-12\right)÷\left(...\right)=\frac{-6}{5}$

#### Question 12:

(i) Are rational numbers always closed under division?
(ii) Are rational numbers always commutative under division?
(iii) Are rational numbers always associative under division?
(iv) Can we divide 1 by 0?

(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:

Find a rational number between

#### Question 2:

Find a rational number between 2 and 3.

#### Question 3:

Find a rational number between

#### Question 4:

Find two rational numbers between −3 and −2.

#### Question 5:

Find three rational numbers between 4 and 5.

#### Question 6:

Find three rational numbers between

#### Question 7:

Find 10 rational numbers between

We can take any 10 out of these.

#### Question 8:

Find 12 rational numbers between −1 and 2.

#### Question 1:

From a rope 11 m long, two pieces of lengths are cut off. What is the length of the remaining rope?

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., $\left(5×1×2\right)$.

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

Therefore, the length of the remaining rope is .

#### Question 2:

A drum full of rice weighs $40\frac{1}{6}$kg. If the empty drum weighs $13\frac{3}{4}$kg, find the weight of rice in the drum.

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:

A basket contains three types of fruits weighing $19\frac{1}{3}\mathrm{kg}$ in all. If $8\frac{1}{9}\mathrm{kg}$ of these be apples, $3\frac{1}{6}\mathrm{kg}$ be oranges and the rest pears, what is the weight of the pears in the basket?

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, $\left(3×3×2\right)$.

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

Therefore, the weight of the pears in the basket is .

#### Question 4:

On one day a rickshaw puller earned Rs 160. Out of his earnings he spent Rs $26\frac{3}{5}$ on tea and snacks, Rs $50\frac{1}{2}$ on food and Rs $16\frac{2}{5}$ on repairs of the rickshaw. How much did he save on that day?

Total earning = ₹160
Money spent on tea and snacks = ₹$26\frac{3}{5}$
Money spent on food = ₹$50\frac{1}{2}$
Money spent on repairs = ₹$16\frac{2}{5}$
Let the savings be ₹x.
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 ₹$66\frac{1}{2}$.

#### Question 5:

Find the cost of $3\frac{2}{5}$ metres of cloth at Rs $63\frac{3}{4}$ per metre.

Cost of 1 m of cloth = ₹$63\frac{3}{4}$
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 ₹$216\frac{3}{4}$.

#### Question 6:

A car is moving at an average speed of $60\frac{2}{5}$ km/hr. How much distance will it cover in $6\frac{1}{4}$ hours?

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:

Find the area of a rectangular park which is $36\frac{3}{5}$ m long and $16\frac{2}{3}$ m broad.

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

Therefore, the area of the rectangular park is .

#### Question 8:

Find the area of a square plot of land whose each side measures $8\frac{1}{2}$ metres.

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:

One litre of petrol costs ₹ $63\frac{3}{4}$ . What is the cost of 34 litres of petrol?

Cost of 1 litre of petrol = ₹$63\frac{3}{4}$
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 ₹$2167\frac{1}{2}$.

#### Question 10:

An aeroplane covers 1020 km in an hour. How much distance will it cover in $4\frac{1}{6}$ hours?

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:

The cost of $3\frac{1}{2}$ metres of cloth is ₹ $166\frac{1}{4}$ . what is the cost of one metre of cloth?

Cost of $3\frac{1}{2}$ m of cloth = ₹$166\frac{1}{4}$
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 ₹ $47\frac{1}{2}$.

#### Question 12:

A cord of length $71\frac{1}{2}$ m has been cut into 26 pieces of equal length. What is the length of each piece?

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

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

#### Question 13:

The area of a room is $65\frac{1}{4}{\mathrm{m}}^{2}.$ If its breadth is $5\frac{7}{16}$ metres, what is its length?

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:

The product of two fractions is $9\frac{3}{5}.$ If one of the fractions is $9\frac{3}{7},$ find the other.

Let the other fraction be x.

Now, we have:

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

#### Question 15:

In a school, $\frac{5}{8}$ of the students are boys. If there are 240 girls, find the number of boys in the school.

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:

After reading $\frac{7}{9}$ of a book, 40 pages are left. How many pages are there in the book?

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:

Rita had Rs 300. She spent $\frac{1}{3}$ of her money on notebooks and $\frac{1}{4}$ of the remainder on stationery items. How much money is left with her?

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:

Amit earns ₹ 32000 per month. He spends $\frac{1}{4}$ of his income of food; $\frac{3}{10}$ of the remainder on house rent and $\frac{5}{21}$ of the remainder on the education of children. How much money is still left with him?

Amit's income per month = ₹32,000
Money spent on food =
Remaining amount = ₹32,000 − ₹8,000 = ₹24,000
Money spent on house rent =
Money left = ₹24,000 − ₹7,200 = ₹16,800
Money spent on education of children = $\frac{5}{21}×₹16,800=₹4,000$
Amount of money still left with him = ₹16,800 − ₹4,000 = ₹12,800

#### Question 19:

If $\frac{3}{5}$ of a number exceeds its $\frac{2}{7}$ by 44, find the number.

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:

At a cricket test match $\frac{2}{7}$ of the spectators were in a covered place while 15000 were in open. Find the total number of spectators.

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:

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

(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:

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

(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:

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

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

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:

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

(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:

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

(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:

Tick (✓) the correct answer
What should be added to
(a) $\frac{17}{20}$
(b) $\frac{-17}{20}$
(c) $\frac{7}{20}$
(d) $\frac{-7}{20}$

(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:

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

(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:

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

(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:

Tick (✓) the correct answer
What should be added to
(a) $\frac{-29}{21}$
(b) $\frac{29}{21}$
(c) $\frac{1}{21}$
(d) $\frac{-1}{21}$

(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:

Tick (✓) the correct answer
What should be subtracted from
(a) $\frac{5}{2}$
(b) $\frac{3}{2}$
(c) $\frac{5}{4}$
(d) $\frac{-5}{2}$

(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:

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

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

The reciprocal of $\frac{-3}{7}$ is

#### Question 12:

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

(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:

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

(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:

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

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

Now,

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

#### Question 15:

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

(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:

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

(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:

Tick (✓) the correct answer
$\left(\frac{-9}{16}×\frac{8}{15}\right)=?$
(a) $\frac{-3}{10}$
(b) $\frac{-4}{15}$
(c) $\frac{-9}{25}$
(d) $\frac{-2}{5}$

(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:

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

(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:

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

(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:

Tick (✓) the correct answer
Additive inverse of $\frac{-5}{9}$ is
(a) $\frac{-9}{5}$
(b) 0
(c) $\frac{5}{9}$
(d) $\frac{9}{5}$

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

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

#### Question 21:

Tick (✓) the correct answer
Reciprocal of $\frac{-3}{4}$ is
(a) $\frac{4}{3}$
(b) $\frac{3}{4}$
(c) $\frac{-4}{3}$
(d) 0

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

#### Question 22:

Tick (✓) the correct answer
A rational number between
(a) $\frac{5}{12}$
(b) $\frac{-5}{12}$
(c) $\frac{5}{24}$
(d) $\frac{-5}{24}$

(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:

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

(b) is a negative rational number

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

#### Question 1:

Find the additive inverse of:
(i) $\frac{7}{-10}$
(ii) $\frac{8}{5}.$

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

Additive inverse of .

(ii) Additive inverse of .

#### Question 2:

The sum of two rational numbers is −4. If one of them is $\frac{-11}{5}$, find the other.

#### Question 3:

What number should be added to

#### Question 4:

What number should be subtracted from $\frac{-3}{4}$ to get $\frac{-1}{2}$?

#### Question 5:

Find the multiplicative inverse of:
(i) $\frac{-3}{4}$
(ii) $\frac{11}{4}.$

#### Question 6:

The product of two numbers is −8. If one of them is −12, find the other.

#### Question 7:

Evaluate:
(i) $\frac{-3}{5}×\frac{10}{7}$
(ii) ${\left(\frac{-5}{8}\right)}^{-1}$
(iii) ${\left(-6\right)}^{-1}$

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

#### Question 8:

Name the property of multiplication shown by each of the following statements:
(i) $\frac{-12}{5}×\frac{3}{4}=\frac{3}{4}×\frac{-12}{5}$
(ii) $\frac{-8}{15}×1=\frac{-8}{15}$
(iii) $\left(\frac{-2}{3}×\frac{7}{8}\right)×\frac{-5}{7}=\frac{-2}{3}×\left(\frac{7}{8}×\frac{-5}{7}\right)$
(iv) $\frac{-2}{3}×0=0$
(v) $\frac{2}{5}×\left(\frac{-4}{5}+\frac{-3}{10}\right)=\left(\frac{2}{5}×\frac{-4}{5}\right)+\left(\frac{2}{5}×\frac{-3}{10}\right)$

(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:

Find two rational numbers lying between

#### Question 10:

Mark (✓) against the correct answer
What should be added to
(a) $\frac{4}{5}$
(b) $\frac{8}{15}$
(c) $\frac{4}{15}$
(d) $\frac{2}{5}$

(c) $\frac{4}{15}$

Let the number be $x$
Now,

#### Question 11:

Mark (✓) against the correct answer
What should be subtracted from $\frac{-2}{3}$ to get $\frac{3}{4}?$
(a) $\frac{-11}{12}$
(b) $\frac{-13}{12}$
(c) $\frac{-5}{4}$
(d) $\frac{-17}{12}$

(d) $\frac{-17}{12}$

Let the number be $x$.
Now,

#### Question 12:

Mark (✓) against the correct answer
${\left(\frac{-5}{4}\right)}^{-1}=?$
(a) $\frac{4}{5}$
(b) $\frac{-4}{5}$
(c) $\frac{5}{4}$
(d) $\frac{3}{5}$

(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:

Mark (✓) against the correct answer
The product of two numbers is $\frac{-1}{4}$. If one of them is $\frac{-3}{10},$ then the other is
(a) $\frac{5}{6}$
(b) $\frac{-5}{6}$
(c) $\frac{4}{3}$
(d) $\frac{-8}{5}$

(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:

Mark (✓) against the correct answer
$\left(\frac{-5}{6}÷\frac{-2}{3}\right)=?$
(a) $\frac{-5}{4}$
(b) $\frac{5}{4}$
(c) $\frac{-4}{5}$
(d) $\frac{4}{5}$

(b) $\frac{5}{4}$

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:

Mark (✓) against the correct answer
$\frac{4}{3}÷?=\frac{-5}{2}$
(a) $\frac{-8}{5}$
(b) $\frac{8}{5}$
(c) $\frac{-8}{15}$
(d) $\frac{8}{15}$

(c) $\frac{-8}{15}$

#### Question 16:

Mark (✓) against the correct answer
Reciprocal of $\frac{-7}{9}$ is
(a) $\frac{9}{7}$
(b) $\frac{-9}{7}$
(c) $\frac{7}{9}$
(d) none of these

(b) $\frac{-9}{7}$

#### Question 17:

A rational number between $\frac{-2}{3}$ and $\frac{1}{2}$ is
(a) $\frac{-1}{6}$
(b) $\frac{-1}{12}$
(c) $\frac{-5}{6}$
(d) $\frac{5}{6}$

(b) $\frac{-1}{12}$

#### Question 18:

Fill in the blanks.
(i) $\frac{25}{8}÷\left(.......\right)=-10.$
(ii) $\frac{-8}{9}×\left(.......\right)=\frac{-2}{3}.$
(iii) $\left(-1\right)+\left(.......\right)=\frac{-2}{9}.$
(iv) $\frac{2}{3}-\left(.......\right)=\frac{1}{15}.$

(ii)

(iii)

(iv)

#### Question 19:

Write 'T' for true and 'F' for false for each of the following:
(i) Rational numbers are always closed under subtraction.
(ii) Rational numbers are aways closed under division.
(iii) 1 ÷ 0 = 0.
(iv) Subtraction is commutative on rational numbers.
(v) $-\left(\frac{-7}{8}\right)=\frac{7}{8}.$

(i) T

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

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.

(iv) F

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

View NCERT Solutions for all chapters of Class 8


text:
Journal article Open Access

# Denoising of EEG Signal using Matlab and SIMULINK Techniques and Estimation of Power Spectral Density of EEG Signal using SIMULINK AR Models

B.Krishna Kumar

### Citation Style Language JSON Export

{
"DOI": "10.35940/ijeat.F8956.129219",
"container_title": "International Journal of Engineering and Advanced Technology (IJEAT)",
"language": "eng",
"title": "Denoising of EEG Signal using Matlab and  SIMULINK Techniques and Estimation of  Power Spectral Density of EEG Signal using  SIMULINK AR Models",
"issued": {
"date-parts": [
[
2019,
12,
30
]
]
},
"abstract": "
The Electroencephalogram (EEG) is the standard technique for investigating the brain’s electrical activity in different psychological and pathological states. Analysis of Electroencephalogram (EEG) signal is a challenging task by reason of the presence of different artifacts such as Ocular Artifacts (OA) and Electromyogram. Normally EEG signals falls in the frequency range of DC to 60 Hz and amplitude of 1-5 µv. Ocular artifacts do have the similar statistical properties of EEG signals, often interfere with EEG signal, thereby making the analysis of EEG signals more complex. In this research paper, removal of artifacts was done using wavelets (matlab coding) as well as using SIMULINK DWT and IDWT blocks and estimated the SNR. In the next stage the output of IDWT block was taken as input to Burg model and Yule walker model to estimate the power spectral density of EEG signal by setting the various parameters of the blocks. The implementation of denoising of EEG signal using SIMULINK DWT and IDWT blocks and estimation of power spectral density of denoised EEG signal using Burg model and Yule walker model was explained in detail in the paper under the methodology heading. In this research paper, the collected EEG signal is normalized and later linearly mixed with the normalized EOG signal resulting in a noisy EEG signal. This noisy EEG signal is decomposed to 4 levels by using different wavelets. This decomposition of EEG signals yields approximate and detail coefficients. Later different thresholding techniques were applied to detail coefficients and estimated the Signal to Noise Ratio of it and estimated the power spectral density of denoised EEG signal obtained from dB4 wavelet as it is providing better SNR than other wavelets mentioned in the results.
",
"author": [
{
"family": "B.Krishna Kumar"
}
],
"page": "418-422",
"volume": "9",
"type": "article-journal",
"issue": "2",
"id": "5602238"
}
22
21
views

format: plain


text:
# nLab Laurent phenomenon

While the composition of birational maps given by Laurent polynomials is in general not a Laurent polynomial there is a number of systematic situations when this is true. In particular, in a cluster algebra, any cluster variable is expressed in terms of any given cluster as a Laurent polynomial with coefficients in the integral group ring $\mathbf{Z}\mathbf{P}$ where $\mathbf{P}$ is the coefficient group. An explanation in the quantum case, from a Hodge-theoretic point of view, has recently being offered by A. Efimov.

Created on January 11, 2012 18:36:42 by Zoran Škoda (161.53.130.104)


text:
You are a guest. Restricted access. Read more

Built-in functions

There are more than 250 built-in functions included with jMathLab. They are implemented as a external MatLab functions (with the extension .m) or as a Java classes. You can look at them using the command path. Look at the reference manual Reference manual.

Here is a simple example. Compute the inverse cosine of an angle:

`x=acos(0.4); printf('%f',x)`

which calls “acos” function. One can get help for this function as:

`help("acos")`

The most common implemented functions are the square root (sqrt(x)), the trigonometric functions (sin(x), cos(x), tan(x)) and inverses (atan(x), atan2(y,x)), and the hyperbolic functions (exp(x), log(x)). A large number of additional functions are available, see the list in chapter 4. Some functions are specific to integers, and also work with arbitrary large numbers: primes(Z) expands Z into prime factors, factorial(Z) calculates the factorial function. Modular division is provided by divide and treated later in the context of polynomials.

`a=log(sqrt(854)); printf('%f',a)         % natural logarithm`

Try also these examples:

```a=0.5*log(854); printf('%f\n',a)
a=gammaln(1234);  printf('%f\n',a)           % log( gamma( x ) )
a=primes(1000000000000000001);  printf('%f\n',a)
a=factorial(35); printf('%f\n',a)```

Let us calculate factorial. We will use “rat” and therefore we specify syms to use it here:

`a=factorial(rat(8));   printf('%f',a)    % to make it exact.`

Standard functions

Here is the table with the rebuild functions for scalar values:

Name(Arguments) Function
float(\$var\$) \$var\$ as floating point number
rat(\$var\$) \$var\$ as exact number
realpart(\$var\$) realpart of \$var\$
imagpart(\$var\$) imaginary part of \$var\$
abs(\$var\$) absolute value of \$var\$
sign(\$var\$) sign of \$var\$
conj(\$var\$) \$var\$ conjugate complex
angle(\$var\$) angle of \$var\$
cfs(\$var\$) [\$var_T\$]) continued fraction expansion of \$var\$ with accuracy \$var_T\$
primes(VAR) VAR decomposed into primes

Built-in unctions

The functions can accept either a value (number), a vector of numbers, or symbolic value. In case of vectors, a function returns a vector. See [jmathlab:vectors]].

Name(Arguments) Function
sqrt(\$var\$) squareroot
exp(\$var\$) exponential
log(\$var\$) natural logarithm
sinh(\$var\$) hyperbolic sine
cosh(\$var\$) hyperbolic cosine
asinh(\$var\$) hyperbolic areasine
acosh(\$var\$) hyperbolic areacosine
sech(\$var\$) hyperbolic secans
csch(\$var\$) hyperbolic cosecans
asech(\$var\$) hyperbolic areasecans
acsch(\$var\$) hyperbolic areacosecans
factorial(N) factorial \$N!\$
nchoosek(N,K) binomial coefficient \$N \choose K\$

Special functions

See the reference reference API

Name(Arguments) Function
erf(\$var\$) error function
inverf(\$var\$) inverse error function
gamma(\$var\$) gamma function
gammaln(\$var\$) logarithm of gamma function
beta(\$var\$) beta function
betaln(\$var\$) logarithm of beta function
j0(\$var\$) Bessel function of the first kind of order 0

as before, the functions can accept values and vectors.