Maths MCQ

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 $(a).$

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 $(b).$

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 $(b).$

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 $(c).$

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 $(c).$

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)
$\text{HCF}$ 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 $\frac{-10}{25}\text{ is }\frac{-2}{5}$
$\text{HCF}$ 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 $\frac{-9}{6}\text{ is }\frac{-3}{2}$
$\text{HCF}$ 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 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 $(c).$

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 $\frac{-6}{9}$
$\text{HCF}$ 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 $(b).$

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

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 $(d).$