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.