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 $\frac{-3}{5}$