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
$\Rightarrow a \times b = 1$ [Commutative property is true for multiplication]
$\Rightarrow b \times a = 1$
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.

$\sqrt{9}$ we can simplify the square root to $3$ which is a natural number, an integer and also can be written as $\frac{3}{1}$ so a rational number.

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

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

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 $(c).$

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

According to the definition, the common factor of numerator and denominator is always $1.$

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 $(c).$

$(c)$ 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 $36$
$\frac{-5\times3}{12\times3}=\frac{-15}{36}$
$\frac{-4\times4}{12\times4}=\frac{-16}{36}$
$(-15)>(-16)$
$\frac{-5}{12}>\frac{-4}{9}$

$(b)$ By definition, in the standard form of a rational number, the common factor of numerator and denominator is always$1.$
Note: Common factor means, a number which divides both the given two numbers.

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

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 bb are two rational number then $a \times b$ will be a rational number.

$\frac { \frac { 1 }{ 3 } }{ \frac { 3 }{ 4 } } = \frac{1}{3} \div \frac{3}{4} = \frac{1}{3}\times\frac{4}{3} = \frac{4}{9}$

Consider given the rational numbers $-\frac{2}{5}$ and $-\frac{1}{5}$ Now, given both rational numbers are negative numbers so the number which lies between them will be negative. so $\frac{3}{10}$ will not lie between them,

Among the following negative rational numbers are $ \frac{-15}{25}$ and $\frac{-3}{5}$

The correct option is $(a).$
$\frac{5\times-1}{-6\times-1}=\frac{-5}{6}$
$LCM$ of $6$ and $12$ is $12$
$\therefore\frac{-5\times2}{6\times2}=\frac{-10}{12}$ and $\frac{-7\times1}{12\times1}=\frac{-7}{12}$
Hence, $\frac{5}{-6}$ is smaller than $\frac{-7}{12}$

$​​\frac{\text{x}}{2}+\frac{1}{3}=1$
$\Rightarrow\frac{\text{x}}{2}=1-\frac{1}{3}$
$\Rightarrow\frac{\text{x}}{2}=\frac{3\times1-1}{3}$
$\Rightarrow\frac{\text{x}}{2}=\frac{3-1}{3}$
$\Rightarrow\frac{\text{x}}{2}=\frac{2}{3}$
$\Rightarrow\frac{2\text{x}}{2}=\frac{2\times2}{3}$ (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 $(c).$

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.

The correct option is $(a).$
$\frac{2\times1}{-3\times-1}=\frac{-2}{3}$
$LCM$ of $3$ and $5$ is $15$
$\therefore\frac{-2\times5}{3\times5}=\frac{-10}{15}$ and $\frac{-4\times3}{5\times3}=\frac{-12}{15}$
Thus $\frac{2}{-3}$ is greater than $\frac{-4}{5}$

$(3+\sqrt{23})-\sqrt{23}$
$3+\sqrt{23} - \sqrt{23} = {3}$
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 $12$
$=\frac{-8-9}{12}$
$\frac{-17}{12}$

$\frac{-2}{3} = -0.667 - 0.667 < 0$ 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 $(d).$

$\frac{-3}{0}$ is undefined. Which means that it is neither a negative rational number nor a positive rational number.

When a number is multiplied by zero, it gives always zero. Then $3.92 \times 0.1 \times 0.0 \times 6.3 = 0$

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 $x$
$\frac{-5}{8}\times\text{x}=\frac{-1}{6}$
$\Rightarrow\text{x}=\frac{-1}{6}\div\Big(\frac{-5}{8}\Big)$
$=\frac{-1}{6}\times\Big(\frac{8}{-5}\Big)$
$=\frac{-4}{-15}$
$=\frac{4}{15}$

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).

$\therefore \sqrt{\frac{16}{36}+\frac{1}{4}}$
$\therefore\sqrt{\frac{16}{36}+\frac{1}{4}}$
$=\sqrt{\frac{16+9}{36}} = \sqrt{\frac{25}{36}} = \frac{5}{6}$

$(b)$ 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.

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

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

$\because$ Both numerator and denominator are negative (i.e., same sign)

Missing number $=(-1)+\frac{9}{14}$
$=\frac{-14+9}{14}$
$=\frac{-5}{14}$

$(\sqrt{2})^{2} = \sqrt{2}\times\sqrt{2} = {2}$
So $(\sqrt{2})^{2}$ is a rational number.

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

$x > y$ 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 $(b).$

If 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
$\Rightarrow \frac{\text{r}}{\text{s}} = \text{s}^{2}$
$\Rightarrow r = s^2\times s$
$\Rightarrow r =s^3$

The correct option is $(b).$
$\frac{2}{3}-1\frac{5}{7}$
$\frac{2}{3}\frac{12}{7}$
LCM of 3 and 7 is 21
$=\frac{14-36}{21}$
$=\frac{-22}{21}$
$=-1\frac{1}{21}$

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

$(b)$ 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 $12$
$=\frac{15-14+8}{12}$
$=\frac{9^3}{12_4}$
$=\frac{3}{4}$

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

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 $100 - 99.999 = 100.000 - 99.999 = 0.001$

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 $(a).$