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

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