Maths M.C.Q. [1 Marks Each]

 Given, $0.3, 0.03, 0.003$ We need to find sum of all these.
$\therefore$ sum of $0.3 + 0.03 + 0.003 = 0.333$

The correct option is $(c).$
$\frac{14}{9}$ should be added to $\frac{-4}{9}$ to get $1.$
$\text{x}+\Big(\frac{-5}{9}\Big)=1\text{x}$
$=1-\frac{(-5)}{9}=\frac{9+5}{9}=\frac{14}{9}$
Let the required number be $x.$

 If $a$ is reciprocal of $b,$ then the reciprocal of $b$ is $a$
If $a$ is reciprocal of $b,$ then
$⇒ a × b = 1 [$Commutative property is true for multiplication$]$
$⇒ b × a = 1$
Thus reciprocal of $b$ is $a$

 Given numbers are $0, \frac56, 1, \frac24$ are integers and $\frac56, \frac24$ are rational numbers. As, rationals and integers are subset of reals. Thus, all the above numbers are real. we can represent all above numbers on a number line.

 If a is a whole number then $a + 0 = a = 0 + a.$
Therefore, $0$ is the additive identity element for addition of whole number because it does not change the identity or value of the whole number during the operation of addition.
Hence, the correct answer is option $(a).$

 The denominator of $\frac{44}{-77}$ is nagative.
Hence, the correct answer is option $(b).$

 $\frac{-3}{8} = \frac{\text{x}}{24 } \text{ X} =\frac {-3\times-24}{8}\text{ X} = {9}$

 In order to express $\frac{27}{-45}$ as a rational number with denominator $5,$ firstly find a number which gives $5$ when $-45$ is divided by it.
This number is $-45\div5=-9$
Dividing the numerator and denominator of $\frac{27}{-45}$ by $-9,$
We have:
$\frac{27}{-45}=\frac{27\div(-9)}{-45\div(-9)}=\frac{-3}{5}$
Thus, the numerator is $-3.$
Hence, the correct answer is option $(b).$

$\frac{9.826}{10} = \frac{9826}{10000} = {0.9826}$

 $\frac{-7}{13}-\Big(\frac{-8}{15}\Big)$
$=​​\frac{-7}{13}+\frac{8}{15}$ $\Big[-\Big(\frac{-8}{15}\Big)=\frac{8}{15}\Big]$
$=\frac{-7\times15+8\times13}{195} (LCM$ of $13$ and $15 = 195)$
$=\frac{-105+104}{195}$
$=\frac{-1}{195}$
Hence, the correct answer is option $(d).$

The correct option is $(b).$
Multiplicative inverse of $\frac{-2}{3}\text{ is }\frac{-3}{2}$

he rational numbers between the $2$ numbers $a, b$ is given by $\frac{\text{a+b}}{2}$ Here $a = -1, b = 1$ So the rational number between them is $\frac{-1+1}{2} = {0}$

It is given that the rational numbers $\frac{-2}{3}\text{ and }\frac{4}{\text{x}}$ represent a pair of equivalent rational numbers.
We know that the values of two equivalent rational numbers is equal.
$\therefore\frac{-2}{3}=\frac{4}{\text{x}}$
$\Rightarrow-2\times\text{x}=4\times3$ $\Big(\frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}}\Rightarrow\text{ad}=\text{bc}\Big)$
$\Rightarrow-2\text{x}=12$
$\Rightarrow\frac{-2\text{x}}{-2}=\frac{12}{-2} ($Dividing both sides by $-2)$
$\Rightarrow\text{x}=-6$
Hence, the correct answer is option $(b).$

$32 - 27.091 = 4.909$

It is given that the rational numbers $-\frac{3}{8}\text{ and }\frac{\text{x}}{-24}$ are equivalent rational numbers.
We know that the values of two equivalent rational numbers is equal.
$\therefore\frac{\text{x}}{-24}=-\frac{3}{8}$
$\Rightarrow\text{x}\times8=-3\times(-24)$ $\Big(\frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}}\Rightarrow\text{ad}=\text{bc}\Big)$
$\Rightarrow8\text{x}=72$
$\Rightarrow\frac{8\text{x}}{8}=\frac{72}{8}$
$($Dividing both sides by $8)$
$\Rightarrow\text{x}=9 $
Hence, the correct answer is option $(c)$.

$(c)$ We know that, when numerator and denominator of a rational number, both are negative,
it is a positive rational number.
Hence, among the given rational numbers $\frac{-3}{-4}$ is positive.

$\because$ Denominator is $0$, it is not a rational number.

The rational number that does not have a reciprocal $0$ because reciprocal of $0$ is undefined.

All integers are rational number but all rational number are not integer because rational number can be integer, fraction, decimals so $p$ is true and $q$ is false.

Firstly, write $\frac{-3}{7}$ as a rational number with denominator $35.$
Multiplying the numerator and denominator of $\frac{-3}{7}$ by $5,$
We have:
$\frac{-3}{7}=\frac{-3\times5}{7\times5}=\frac{-15}{35}$
$\therefore\frac{-3}{7}=\frac{\text{x}}{35}$
$\Rightarrow\frac{-15}{35}=\frac{\text{x}}{35}$
$\Rightarrow\text{x}=-15$
Hence, the correct answer is option $(c).$

A rational number is a number that can be represented $\frac{\text{a}}{\text{b}}$ where $a$ and $b$ are integers and b is not equal to $0.$ A rational number can also be represented in decimal form and the resulting decimal is a repeating decimal. Also any decimal number that is repeating can be written in the form $\frac{\text{a}}{\text{b}}$ with $b$ not equal to zero so it is a rational number. In the given options, option $D$ is irrational number. option $B$ and $C$ are not lying between $-3$ and $3.$ Only option A lies $-3$ and $3$ and is a rational number.

$ (c)$ By definition, a rational number is said to be in the standard form, if its denominator is a positive integer.

The correct option is $(b).$
$\frac{-5}{12}$ is greater than $\frac{-4}{9}$
$LCM$ of $9$ and $12$ is $36$
$\frac{-5\times3}{12\times3}=\frac{-15}{36}$
$\frac{-4\times4}{12\times4}=\frac{-16}{36}$
$(-15)>(-16)$
$\frac{-5}{12}>\frac{-4}{9}$

$1\div\frac{1}{3}$
$=1\times3$ $\Big(\text{x}\div\text{y}=\text{x}\times\frac{1}{\text{y}}\Big)$
$=3$
Hence, the correct answer is option $(b).$

We know that two rational numbers are equal if they have the same standard form.
The rational number $\frac{-2}{3}$ is in its standard form.
Consider the rational number $\frac{10}{-15}$
This rational numbner can be expressed in standerd form as follows:
$\frac{10}{-15}=\frac{10\times(-1)}{-15\times(-1)}=\frac{-10}{15}$ (Multiplying numerator and denominator by $-1$ to make denominator positive)
$HCF$ of $10$ and $15 = 5$
Dividing the numeator and denominator of $\frac{-10}{15}$ by $5,$
We have:
$\frac{-10}{15}=\frac{-10\div5}{15\div5}=\frac{-2}{3}$
Thus, the standard form of $\frac{-10}{15}$ is $\frac{-2}{3},$ which is same as the given rational number.
So, the rational number equal to $\frac{-2}{3}$ is $\frac{-10}{15}$
Let us check why options $(a)$ and $(c)$ are not correct.
The standard form of $\frac{-10}{25}\text{ is }\frac{-2}{5}$
$HCF$ of $10$ and $25 = 5$
Dividing the numerator and denominator of $=\frac{-10}{25}$ by $5,$
We have:
$\frac{-10}{25}=\frac{-10\div5}{25\div5}=\frac{-2}{5}$
The standard form of $\frac{-9}{6}\text{ is }\frac{-3}{2}$
$HCF$ of $6$ and $9 = 3$
Dividing the numerator and denominator of $\frac{-9}{3}$by $3,$
We have:
$\frac{-9}{6}=\frac{-9\div3}{6\div2}=\frac{-3}{2}$
Hence, the correct answer is option $(b)$

The correct option is $(c).$
Reciprocal of $-6\text{ is }\frac{-1}{6}$

if $p$ and $q$ are perfect squares, then we can writep $= x^2$ and $q = y^2$
$\sqrt{\frac{\text{p}}{\text{q}}} = \sqrt{\frac{\text{x}^{2}}{\text{y}^{2}}} $
$ = \frac{\text{x}}{\text{y}} = {\text{a}}$ rational number. So, the given statement is true

Between two rational numbers there are infinitely many rational number for example.
between $4$ and $5$ there are $4.1, 4.2, .4.22, 4.223.$

$\sqrt{\frac{64}{49}} = {\frac{\sqrt{64}}{\sqrt{49}}} = \frac{8}{7}$ Option $D$ is a rational number. Rest all are irrational numbers.

Product of two rational number is always a rational number Let $a$ and $b$ are two rational number then $a \times b$ will be a rational number.

$​​\frac{\text{x}}{2}+\frac{1}{3}=1$
$\Rightarrow\frac{\text{x}}{2}=1-\frac{1}{3}$
$\Rightarrow\frac{\text{x}}{2}=\frac{3\times1-1}{3}$
$\Rightarrow\frac{\text{x}}{2}=\frac{3-1}{3}$
$\Rightarrow\frac{\text{x}}{2}=\frac{2}{3}$
$\Rightarrow\frac{2\text{x}}{2}=\frac{2\times2}{3}( $Multiplying both sides by $2)$
$\Rightarrow\text{x}=\frac{4}{3}$
Hence, the correct answer is option $(b).$

Since $\frac{-2}{3} < {0}$ it lies on left side of $0$ on the number line.

 The rational numbers $\frac{3}{7}\text{ and }\frac{5}{12}$ are positive rational numbers. We know that every positive rational number is greater than $0,$ so both the rational numbers $\frac{3}{7}\text{ and }\frac{5}{12}$ are represented by points on the right of the zero on the number line.
The rational numbers $-\frac{3}{7}\text{ and }\frac{-5}{12}$ are negative rational numbers. We know that every negative rational number is less than $0,$ so both the rational numbers $\frac{3}{7}\text{ and }\frac{5}{12}$ are represented by points on the left of the zero on the number line.
The rational numbers $\frac{3}{7}$ is a positive rational number whereas the rational number $\frac{-5}{12}$ is a negative rational numbers. We know that every negative rational number is less than $0$ and every positive rational number is greater than $0,$ so the rational number $\frac{3}{7}$ is represented by point on the right of the zero and $\frac{-5}{12}$ is represented by point on the left of the zero on the number line.
Thus, the rational numbers $-\frac{3}{7}\text{ and }\frac{-5}{12}$ are on the opposite side of the zero on the number line.
Hence, the correct answer is option $(c).$

 $3\frac{1}{7}\times1\frac{5}{6}\times1\frac{2}{5}\times1\frac{1}{11}$
$=\frac{22}{7}\times\frac{11}{6}\times\frac{7}5{}\times\frac{12}{11}$
$=\frac{22\times11\times7\times2}{7\times6\times5\times11}$ $\Big(\frac{\text{a}}{\text{b}}\times\frac{\text{c}}{\text{d}}=\frac{\text{a}\times\text{c}}{\text{b}\times\text{d}}\Big)$
$=\frac{44}{5}$
$=\frac{8\times5+4}{5}$
$=8\frac{4}{5}$
Hence, the correct answer is option $(c).$

 Given rational number is $\frac{4}{5},$
So, equivalent rational number $=\frac{4\times4}{5\times4}$
$=\frac{16}{20}[ $Multipying numerator and denominator by $4]$
Note: If the numerator and denominator of a rational number is multiplied/divided by a non-zero integer, then the result we get, is equivalent rational number.

 $(3+\sqrt{23})-\sqrt{23}$
$3+\sqrt{23} - \sqrt{23} = {3}$
Here, $3$ is a rational number.

$ P:$ Every fraction is a rational number: True
$Q:$ Every rational number is a fraction: False

 Let the number to be subtracted be $x$
$\Rightarrow\text{x}=\frac{-2}{3}-\frac{3}{4}$
$LCM$ of $3$ and $4$ is $12$
$=\frac{-8-9}{12}$
$\frac{-17}{12}$

 The correct option is $(a).$
$=\frac{-73}{9}=-8\frac{1}{9}$

 $\frac{-2}{3} = -0.667 - 0.667 < 0$ it will lie to the left side of $0$ on the number line.

 $1\div​​\frac{-5}{7}$
$=1\times\frac{7}{-5}$ $\Big(\text{x}\div\text{y}=\text{x}\times\frac{1}{\text{y}}\Big)$
$=\frac{7}{-5}$
$=\frac{7\times(-1)}{-5\times(-1)}$
$=\frac{-7}{5}$
Hence, the correct answer is option $(d).$

 We need to find value of $\frac {5}{\sqrt {0.0025}}​$
$\therefore \displaystyle \frac{5}{\sqrt{0.0025}} = \frac{5}{0.05} = 100$

 For every non-zero rational number $\frac{\text{a}}{\text{b}}$ there exists a rational number $\frac{\text{b}}{\text{a}}$ such that:
$\frac{\text{a}}{\text{b}}\times\frac{\text{b}}{\text{a}}=1$
Here, the rational number $\frac{\text{b}}{\text{a}}$ is called the multiplicative inverse or reciprocal of $\frac{\text{a}}{\text{b}}$
Thus, if the product of two non-zero rational numbers is $1,$ then they are multiplicative inverse or reciprocal of each other.
Hence, the correct answer is option $(d).$

 The value of $ \frac{18}{6}= 18 \div 6$ as $18$ is divisible by $6 = 3$

 We know that two rational numbers are equal if they have the same standard form.
$\frac{2}{-3}=\frac{2\times(-1)}{-3\times(-1)}=\frac{-2}{3}$
The standard form of $\frac{2}{-3}\text{ is }\frac{-2}{3}$
Consider the rational number $\frac{-6}{9}$
$HCF$ of $6$ and $9 = 3$
Dividing the numerator and denominator of $\frac{-6}{9}$ by $3,$
We have:
$\frac{-6}{9}=\frac{-6\div3}{9\div3}=\frac{-2}{3}$
So, the rational number $\frac{-6}{9}$ is equal to $\frac{2}{-3}$
It can be checked that:
Standard form of $\frac{14}{-18}=\frac{-7}{9}$
Standard form of $\frac{3}{-2}=\frac{-3}{2}$
Hence, the correct answer is option $(b).$

 Consider the rational numbers $\frac{5}{9}\text{ and } \frac{-3}{-8}$
We write the rational number $\frac{-3}{-8}$ with positive denominator.
$\frac{-3}{-8}=\frac{-3\times(-1)}{-8\times(-1)}=\frac{3}{8}$
Now, we write the rational numbers so that they have a common denominator.
$LCM$ of $8$ and $9 = 72$
So, $\frac{5}{9}=\frac{5\times8}{9\times8}=\frac{40}{72}$ and $\frac{3}{8}=\frac{3\times9}{8\times9}=\frac{27}{72}$
Now,
$40>27$
$\Rightarrow\frac{40}{72}>\frac{27}{72}$
$\Rightarrow\frac{5}{9}>\frac{3}{8}$
$\Rightarrow\frac{5}{9}>\frac{-3}{-8}$
Hence the correct option is $(a).$

$ (c)$ There are unlimited numbers between two rational numbers.

 
$H.C.F$ of $102$ and $119$ is $17$
$=\frac{-102\div11}{119\div17}=\frac{-6}{7}$
The standard from of $\frac{-102}{119}\text{ is }\frac{-6}{7}$

 Let the other number to be $x$
$\frac{-5}{8}\times\text{x}=\frac{-1}{6}$
$\Rightarrow\text{x}=\frac{-1}{6}\div\Big(\frac{-5}{8}\Big)$
$=\frac{-1}{6}\times\Big(\frac{8}{-5}\Big)$
$=\frac{-4}{-15}$
$=\frac{4}{15}$

 $?=-2​​\frac{3}{7}+4$
$\Rightarrow?=-\frac{17}{7}+\frac{4}{1}$
$\Rightarrow?=\frac{-17\times1+4\times7}{7}$
$\Rightarrow=\frac{-17+28}{7}$
$\Rightarrow?=\frac{11}{7}$
Hence, the correct answer is option $(b).$

 Given rational number is $\frac{-48}{60}.$
For standrad/ simplest form, divide numerator and denomin by their $HCF$
i.e. $\frac{-48+12}{60+12}=\frac{-4}{5}$
Hence, the standard form of $\frac{-48}{60}$ is $\frac{-4}{5}.$

 $\frac{7}{19}$ can be written as $\frac{{7\times}\text{n}}{{19\times}\text{n}}$ where $n$ is integer.The only equation which satisfies this equation is option $D$ as $\frac{21}{57} = \frac{7\times{3}}{19\times{3}}$ where $n = 3$

 Five rational numbers less than $2$ may be taken $1,\frac{1}{2},\,0,\,-1,\,-\frac{1}{2}$(There can be many more such rational numbers).

$ (b)$ Reciprocal of $\frac{1}{2}=\frac{1}{\frac{1}{2}}=2$

 In the given question numerator is $0$ and $0$ is neither positive nor negative.

 The value of $\frac {1}{(-5)^2}=\dfrac {1}{(-5)(-5)} = \frac{1}{25}$

$ 3 \times 5 \times 2 \times 5 = 150$ The given expression has more than $2$ factors. So, it is a composite number.

The numbers $450$ and $460$ are not divisible by $11.$
Now, both the numbers $451$ and $462$ are divisible by $11.$
Distance between $457$ and $451$ on the number line $= 457 - 451 = 6$
Distance between $457$ and $462$ on the number line $= 462 - 457 = 5$
Thus, the whole number nearest to $457$ and divisible by $11$ is $462.$
Hence, the correct answer is option $(d).$

 $-\frac{3}{4}=\frac{6}{\text{x}}$
$\Rightarrow-3\times\text{x}=6\times4$ $\Big(\frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}}\Rightarrow\text{ad}=\text{bc}\Big)$
$\Rightarrow-3\text{x}=24$
$\Rightarrow\frac{-3\text{x}}{-3}=\frac{24}{-3} ($Dividing both sides by $-3)$
$\Rightarrow\text{x}=-8$
Hence, the correct answer is option $(a).$

 Given, $\frac {5}{13}+ \text{x}= \frac {5}{13}$
$\therefore \text{x}= \frac{5}{13}-\frac{5}{13}$
$\therefore \text{x}= 0$

 $78.12\times1.5 = \frac{7812}{100}\times\frac{15}{10}$
$ = \frac{117180}{1000}$
$= 117.180$
$= 117.18$

 Since $LCM$ of $6$ and $12$ is $12$
$\frac{-5\times2}{6\times}=\frac{-10}{12}$
$\frac{-7\times1}{12\times1}=\frac{-7}{12}$
$\frac{-5}{6}<\frac{-7}{12}$

 Given $ \frac{126}{-196} =\frac{ {-9}\times{14}}{14\times14} = \frac{-9}{14}$

 $(-2)\div\Big(​​-\frac{5}{3}\Big)$
$=-2\times\Big(-\frac{3}{5}\Big)$ $\text{x}\div\text{y}=\text{x}\times\frac{1}{\text{y}}$
$=\frac{-2}{1}\times\frac{(-3)}{5}$
$=\frac{-2\times(-3)}{1\times5}$ $\Big(\frac{\text{a}}{\text{b}}\times\frac{\text{c}}{\text{d}}=\frac{\text{a}\times\text{c}}{\text{b}\times\text{d}}\Big)$
$=\frac{6}{5}$
Hence, the correct answer is option $(c).$

 According to the definition of a rational number, it can be expressed in the form of $\frac{\text{p}}{\text{q}}$ where $p$ and $q$ are an integer and ${\text{q}}\neq{0}$

 The correct option is $(a).$
The value of $x$ is $-14$
$\Big[\text{x}=\frac{7\times6}{-3}=\frac{42^{14}}{-3_1}=-14\Big]$

 The correct option is $(c).$
$\frac{-6}{13}-\frac{[-7]}{15}$
$LCM$ of $13$ and $15$ is $195$
$\frac{-6}{13}-\frac{[-7]}{15}$
$=\frac{-90+91}{195}$
$=\frac{1}{195}$

 Real number is a value that represents a quantity along the number line. Real number includes all rational and irrational numbers. Rational numbers are numbers which can be represented in the form $\dfrac {\text{ p} }{\text{ q} }$ where, ${\text{q}} \neq 0\ p, q $ are integers. rational number is a subset of real number.

$x > y$ and $y > z$
$\therefore x > y > z$
$\Rightarrow x > z$

 The remainder is the integer amount left over after a number is divided by another. The number $7$ goes into $30$ four times, with $2$ left over i.e. $7 \times 4 + 2 = 30,$ so the remainder is $2.$

$\frac{-5}{13}+?=-1$
$\Rightarrow?=-1-\Big(\frac{-5}{13}\Big)$
$\Rightarrow?=-1+\frac{5}{13}$ $\Big[-\Big(\frac{-5}{13}=\frac{5}{13}\Big)\Big]$
$\Rightarrow?=\frac{-1\times13+5}{13}$
$\Rightarrow\frac{-13+5}{13}$
$\Rightarrow=\frac{-8}{13}$
Hence, the correct answer is option $(b).$

Difference of the given number and required number $=\frac{4}{5}$
Given number $=\frac{-2}{3}$
$\therefore$ Required number = Given number - Difference of the numbers
$=\frac{-2}{3}-\frac{4}{5}$
$=\frac{-2}{3}+\Big(\frac{-4}{5}\Big)$
$=\frac{-2\times5+(-4)\times3}{15}$
$=\frac{-10+(-12)}{15}$
$=\frac{-22}{15}$
Hence, the correct answer is option $(b).$

 AIf any number is multiplied by $0,$ the product is $0.$
$\therefore0\times0=0$
If $0$ is divided by any number $(\neq0),$ the quotient is always $0.$
$\therefore\frac{0}{3}\text{ and }\frac{7-7}{3}=\frac{0}{3}=0$
Division of any number by $0$ is meaningless and is not defined.
$\therefore9\div0$ is not defined.
Hence, the correct answer is option $(d).$

An integer is a rational number, so the assertion is true. Whereas root of any integer cant be termed as irrational as $4$ is an integer and a perfect square at the same time, so the root will be rational only.

 given that
$\sqrt { \frac { \text{r} }{\text{ s} } } =\text{s}$ where $S > 0$
squaring both sides
$\Rightarrow \frac{\text{r}}{\text{s}} = \text{s}^{2}$
$⇒ r = s^2× s$
$⇒ r =s^3$

 We know that the multiplicative inverse of the rational number $\frac{\text{a}}{\text{b}}\text{ is }\frac{\text{b}}{\text{a}}$
$\therefore$ Multiplicative inverse of $\frac{4}{-5}=\frac{-5}{4}=\frac{5}{-4}$
Hence, the correct answer is option $(c).$

 As both the digits have equal $(+)$ signs and the operation to be performed is addition, so resultant is $(+12) + (+25) = 37$

We know that if a is a whole number, then $a \times 1 = a = 1 \times a.$
Therefore, $1$ is the multiplicative identity element for multiplication of whole numbers because it does not change the identity or value of the whole number during the operation of multiplication.
Hence, the correct answer is option $(b).$

The correct option is $(b).$
$\frac{-9}{14}+?=-1$
$\therefore?=-1-\frac{(-9)}{14}$
$?=\frac{14+9}{14}$
$?=\frac{-5}{14}$

To reduce a rational number to its standard form, we divide its numerator and denominator by their $HCF.$

We know that $0$ divided by any non-zero rational number is always $0.$
$\therefore0\div\frac{3}{5}=0$
$\Big(0\div\frac{\text{a}}{\text{b}}=0\Big)$
Hence, the correct answer is option $(a).$

The reciprocal of a negative rational number is negative. Let no. be $-a$ its reciprocal is $ \frac{-1}{\text{a}}$ which is a negative number.

$\frac{5}{4}-\frac{7}{6}-\frac{(-2)}{3}$
$LCM$ of $4, 6$ and $3$ is $12$
$=\frac{15-14+8}{12}$
$=\frac{9^3}{12_4}$
$=\frac{3}{4}$

The correct option is $(b).$
Let the number that is to be subtracted be $x.$
$\frac{-3}{4}-\text{x}=\frac{5}{6}$
$\Rightarrow-\text{x}=\frac{5}{6}-\Big(\frac{-3}{4}\Big)$
$\Rightarrow-\text{x}=\frac{5}{6}+\frac{-3}{4}$
$\Rightarrow-\text{x}=\frac{(5\times2)+(3\times3)}{12}$
$\Rightarrow\text{x}=-\frac{19}{12}$
Hence, $\frac{-19}{12}$ should be subtracted from $\frac{-3}{4}$ to get $\frac{5}{6}$

If we divide a positive integer by another positive integer, the resulting number is always a rational number. Though it can be a natural number and an integer only if the denominator is $1.$

Two fractions are equivalent if their cross multiplications are equal.
For example,
$\frac{2}{5} = \frac{2}{5}$
If we cross multiply the above fraction the $2 \times 5 = 10$

Since​ $\frac{1}{0}$ is not rational, the quotient of two integers is not rational.

$\frac{-3}{\text{x}} = \frac{\text{x}}{27}$
$x × x = -3 × 27$
$⇒ x^2= -81$
$⇒ x^2 = -81$
$\text{x}=\sqrt{−81​}$ which is not a rational number.

 
$H.C.F$ of $33$ and $55$ is $11$
$=\frac{-33\div11}{55\div11}=\frac{-3}{5}$

$5.63\div{0.01} = \frac{5.63}{0.01}$
$ = \frac{\frac{563}{100}}{\frac{1}{100}}=\frac{563}{100}\times\frac{100}{1}=563$

Statement - 1: Rational number can even be simply integers which can be further represented as $\frac{\text{p}}{\text{q}}$ form. So statement $1$ is false
Statement - 2: Any number divided by $0$ is not defined. So statement $2$ is true.

values of rational numbers $x$ and $y$ is not given For any two rational numbers all three properties are correct as $x < y$ or $x = y$ or $x > y$

There exists infinite number of rational numbers between any two rational numbers. i.e. in this case between $\frac{2}{5}$ and $\frac{4}{5}$.

We know that the reciprocal of the rational number $\frac{\text{a}}{\text{b}}\text{ is }\Big(\frac{\text{a}}{\text{b}}\Big)^{-1}=\frac{\text{b}}{\text{a}}$
$\therefore$ Reciprocal of $\frac{-3}{4}$
$=\Big(\frac{-3}{4}\Big)^{-1}$
$=\frac{4}{-3}$
$=\frac{4\times(-1)}{-3\times(-1)}$
$=\frac{-4}{3}$
Hence, the correct answer is option $(c).$

$\frac{-18}{5} = -3.6 - 4 < -3.6 < -3 - 3.6$ lies between $-3$ and $-4.$

Difference of $99.999$ and $100$ is $100 - 99.999 = 100.000 - 99.999 = 0.001$

$\frac{5}{4}-\frac{7}{6}-\frac{-2}{3}$
$=\frac{5}{4}+\Big(\frac{-7}{6}\Big)+\frac{2}{3}$ $\Big[-\Big(\frac{-2}{3}\Big)=\frac{2}{3}\Big]$
$=\frac{5\times3+(-7)\times2+2\times4}{12} (LCM$ of $3, 4$ and $6 = 12)$
$=\frac{15-14+8}{12}$
$=\frac{9}{12}$
$=\frac{9\div3}{12\div3} ($Dividing numerator and denominator by $3)$
$=\frac{3}{4}$
Hence, the correct answer is option $(a).$

${125.625}\div{0.5} = \frac{125625}{1000}\times\frac{10}{5}$
$ = \frac{25125}{100} = {251.25}$

 The denominator of the rational number $-\frac{102}{119}$ is positivr.
In order to write the rational number in standerd form, divide its numerator and denominator by the $HCF$ of $102$ and $119.$
$HCF$ of $102$ and $119 = 17$
Dividing the numerator and denominator of $-\frac{102}{119}$ by $17,$
We have:
$-\frac{102}{119}=-\frac{102\div17}{119\div17}=-\frac{6}{7}$
Thus the standard form of $-\frac{102}{119}\text{ is }-\frac{6}{7}$
Hence, the correct answer is option $(a).$

$\frac{2}{3} = {0.67}$ It is clear that $0.67$ lies between $0$ and $1$