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

An integer can be both positive and negative as well as zero.
i.e. $-3, -2, -1, 0, 1, 2, 3,….$

Let $X$ should be subtracted to $-\frac{5}{4}$ to get $-1.$
$-\frac{5}{4} –\text{ x }= -1$
thus $\text{x} = -\frac{5}{4} + 1$
$\text{x} = -5 + \frac{4}{4}$
$\text{x }= -\frac{1}{4}$
Hence, $-\frac{1}{4}$ Should be subtracted to $-\frac{5}{4}$ to get $-1.$

Let $x$ be the rational number that needs to be added,
$\therefore-\frac{5}{16}+\text{x}=\frac{5}{8}$
$\Rightarrow\text{x}=\frac{5}{8}+\frac{5}{16}$
$\Rightarrow\text{x}=\frac{15}{16}$

Let $x$ be subtracted
$=\frac{-5}{3}-\text{x}=\frac{5}{6}$
$=\frac{-5}{3}-\frac{5}{6}=\text{x}$
$=\frac{-10-5}{6}$
$=\frac{-15}{6}$
$=\frac{-15\div3}{6\div3}$
$=\frac{-5}{2}$
Number to be subtracted $=\frac{-5}{2}$

 Converting the mixed number into improper fraction,
$2\frac{2}{4}=\frac{10}{4}$
Multiplicative inverse of $\frac{10}{4}$ is $\frac{4}{10}$
The decimal form of $\frac{4}{10}$ is $0.4$

When we divide $1$ by $4,$ we get $0.25.$

$ 1$ and $-1$ are the only rational numbers which are their own reciprocals. No other rational number is its own reciprocal.
We know that there is no rational number which when multiplied with $0,$ gives $1.$
Therefore, rational number $0$ has no reciprocal or multiplicative inverse.

 Let $x$ is subtracted from $-\frac{2}{3}$
$-\frac{2}{3} – \text{x} = -1$
$ – \text{x} = -1+\frac{2}{3}$
$ – \text{x} = -\frac{2}{3}$
$ \text{x} = \frac{1}{3}$

We know that a number is called in standard form if the numerator and denominator has no common divisor except $1.$
$\frac{-9}{16}$ is in standard form.

$LCM$ of $3, 5, 15$ and $20 = 60$
$\therefore\frac{2}{3}+\frac{-4}{5}+\frac{7}{15}+\frac{-11}{20}$
$=\frac{40+(-48)+28+(-33)}{60}$
$=\frac{40-48+28-33}{60}$
$=\frac{68-81}{60}$
$=\frac{-13}{60}$

We know that, if the product of two rational numbers is $1,$
Then they are multiplicative inverse of each other.
Given number is $-1\frac{1}{7},$ i.e. $\frac{-8}{7}.$
Let the multiplicative inverse of $-\frac{8}{7}$ be $x.$
$\Rightarrow\frac{-8}{7}\times\text{x}=1$
$\Rightarrow\text{x}=1\times\Big(-\frac{7}{8}\Big)$
$=\frac{7}{-8}$
Hence, $\frac{7}{-8}$ is the multiplication inverse of $-\frac{8}{7}.$

Let we should multiply $\frac{8}{21}$ by $x.$
Then according to question, $\text{x}\times\frac{8}{21}=1$
Hence, we should multiply $\frac{8}{21}$ by $\frac{21}{8},$ for getting the product $1.$

We know that, the sum of any rational number and zero $(0)$ is the rational number itself.
Now, $x + 0 = 0 + x = x,$ which is a rational number, then $0$ is called identity for addition of rational numbers.