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

Let the number added be $x$
Then,
$\frac{-3}{5}+\text{x}=\frac{-1}{3}$
$\Rightarrow\text{x}=\frac{1}{3}-\frac{-3}{5}$
$\Rightarrow\text{x}=\frac{-1\times5-(-3)\times3}{15}$
$\Rightarrow\text{x}=\frac{-5+9}{15}$
$\Rightarrow\text{x}=\frac{4}{15}$

A number Which can be experssed as $\frac{\text{p}}{\text{q}},$ where $p$ and $q$ are integers $\text{q}\neq0$ is a rational number.

$\frac{3}{4} = 0.75$ and $\frac{7}{4} = 1.75$
Here we see $1.25$ lies between $0.75$ and $1.75.$

The reciprocal of $1$ is the number itself.

One $(1)$ is the identity for multiplication of rational numbers.
That means,
If $a$ is a rational number.
Then, $a - 1 = 1 - a = a$
Note: One $(1)$ is the multiplication identity for integers and whole number also.

We know that, the distributive property of multiplication over addition for rational numbers can be expressed as $a \times (b + c) = ab + ac,$ where $a, b$ and $c$ are rational numbers.
Here, $-\frac{1}{4}\times\bigg\{\frac{2}{3}+\Big(\frac{-4}{7}\Big)\bigg\}=\Big[-\frac{1}{4}\times\frac{2}{3}\Big]+\bigg[-\frac{1}{4}\times\Big(\frac{-4}{7}\Big)\bigg]$
Is the example of distributive property of multiplication over addition for rational numbers.

Let $X$ should be added to $-\frac{5}{4}$ to get $-1.$
$-\frac{5}{4}+\text{x}=-1$
$\text{x}=-1+\frac{5}{4}$
To make the denominator same multiply and divide first term by $4.$
$\text{x}=-\frac{4}{4}+\frac{5}{4}$
$\text{x}=\frac{-4+5}{4}$
$\text{x}=\frac{1}{4}$

We can get reciprocal of a number just be interchanging its numerator and denominator. Thus, the reciprocal of $a$ is $\frac{1}{\text{a}}$​.

 Any number added to zero is equal to the number itself.
$5 + 0 = 5$
Therefore, $0$ is the additive identity of rational numbers.

Let $x$ be the required number
Then,
$\frac{-15}{14}\times\text{x}=\frac{-16}{35}$
$\Rightarrow\text{x}=\frac{-16}{35}\div\frac{-15}{14}$
$\Rightarrow\text{x}=\frac{-16}{35}\times\frac{14}{-15}$
$\Rightarrow\frac{-224}{-525}=\frac{224}{525}$
$=\frac{224\div7}{525\div7}$
$=\frac{32}{75}$

If $x$ is any rational number,
Then $x + 0 = x [0$ is the additive identity$]$

We can write $\frac{3}{4}$ as $\frac{30}{40}$ and $1$ as $\frac{40}{40}$.
Hence the rational numbers between them are:
$\frac{31}{40}, \frac{32}{40}, \frac{33}{40}, \frac{34}{40}, \frac{35}{40}, \frac{36}{40}, \frac{37}{40}, \frac{38}{40}, \frac{39}{40}.$
Note: There are countless rational numbers between any two rational numbers.

Let the number be $x$
Now,
$\frac{-2}{3}-\text{x}=\frac{3}{4}$
$\Rightarrow-1\times\Big(\frac{2}{3}+\text{x}\Big)=\frac{3}{4}$
$\Rightarrow\frac{2}{3}+\text{x}=\frac{-3}{4}$
$\Rightarrow\text{x}=\frac{-3}{4}+\Big(\text{Additive inverse of }\frac{2}{3}\Big)$
$\Rightarrow\text{x}=\frac{-3}{4}-\Big(\frac{-2}{3}\Big)$
$\Rightarrow\text{x}=\frac{-3}{4}+\frac{2}{3}$
$\Rightarrow\text{x}=\frac{-3\times3}{4\times3}+\frac{2\times4}{3\times4}$
$\Rightarrow\text{x}=\frac{-9}{12}+\frac{-8}{12}$
$\Rightarrow\text{x}=\frac{-17}{12}$

 The reciprocal of any rational number $\frac{\text{p}}{\text{q}},$
Where $p$ and $q$ are integers and $\text{q}\neq0$ is $\frac{\text{q}}{\text{p}}.$

The additive identity element is $0.$
We know that the sum of $0$ and any number is the number itself. Let the number be $x.$
$\therefore x + 0 = 0$
$⇒ x = 0$

 A number of the form $\frac{\text{p}}{\text{q}}$ is said to be a rational number, if $p$ and $q$ are integers.

Any number multiplied by $1$ is equal to the number itself.
$5 × 1 = 5$
Therefore, $1$ is the multiplicative identity of rational numbers.

Given,
$\frac{3}{4}+\frac{5}{6}+\frac{2}{7}$
$LCM$ of $4, 6$ and $7 = 84$
Now, making the denominators the same and adding the given rational numbers, we get;
$\Rightarrow\frac{63}{84}+\frac{70}{84}+\frac{24}{84}$
$\Rightarrow\frac{157}{84}$