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

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