Free Rational Numbers 01 Practice Test - 8th Grade 

A number which can be represented in the form of $\frac{p}{q},q\ne 0$ where p and q are integers is called a rational number.

For example, $\frac{2}{3},5,$ $-4$ etc are all rational numbers as they can be expressed in $\frac{p}{q}$ form with $p$ and $q$ as integers and $q\ne 0$.

If 'a' and 'b' are any two whole numbers, their product will also be a whole number. We know that the multiplication of any two numbers can be expressed as repeated addition of one number. So, if we are adding any whole number to itself again and again, then we will surely get a whole number.
Let's assume $a=2,b=3$.
$a\times b=2\times 3$ which is same as adding '2' three times to itself, which will result in a whole number.

Rational numbers can be compared by making their denominators equal.

Here, we need to compare all given rational numbers $\frac{25}{4},\frac{22}{3},\frac{13}{2},\frac{15}{2},\frac{17}{2}$ and $\frac{8}{1}$.

We know LCM of 1, 2, 3 and 4, i.e., the denominators of the above fractions, is 12.

Converting all the above fractions into like fractions, we get

$\frac{25}{4}=\frac{25\times 3}{4\times 3}=\frac{75}{12}$

$\frac{22}{3}=\frac{22\times 4}{3\times 4}=\frac{88}{12}$

$\frac{13}{2}=\frac{13\times 6}{2\times 6}=\frac{78}{12}$

$\frac{15}{2}=\frac{15\times 6}{2\times 6}=\frac{90}{12}$

$\frac{17}{2}=\frac{17\times 6}{2\times 6}=\frac{102}{12}$

$8=\frac{8}{1}=\frac{8\times 12}{1\times 12}=\frac{96}{12}$

Arranging these fractions in an ascending order, we have
$\frac{75}{12},\frac{78}{12},\frac{88}{12},\frac{90}{12},\frac{96}{12},\frac{102}{12}$

Thus, $\frac{13}{2}$ lies between $\frac{25}{4}$ and $\frac{22}{3}$.

An operation is commutative if changing the order of operands does not change the result.

Addition and multiplication are commutative for whole numbers.

$\text{For example,}$ $2+3=5,$  $3+2=5$.

$\text{Also,}$ $3\times 4=12,$  $4\times 3=12$

The reciprocal of any non-zero number 'a' is $\frac{1}{a}$.

​​​​​​The reciprocal of '-10' = $\frac{1}{(-10)}=-0.1$

Reciprocal of $\frac{5}{11}$ is $\frac{11}{5}$.

$\frac{11}{5}÷\frac{-55}{18}$

$=\frac{11\times 18}{5\times (-55)}$

$=\frac{-18}{25}$

Let the rational number be $x$.

$\frac{22}{7}÷x$  = $\frac{-11}{14}$ (given)

$\Rightarrow \frac{22}{7}=x\times \frac{-11}{14}$

$\Rightarrow \frac{22}{7}÷\frac{-11}{14}=x$

$\Rightarrow \frac{22}{7}\times \frac{14}{-11}=x$

$\Rightarrow x=-4$

Therefore, $\frac{22}{7}$ should be divided by -4 to get $\frac{-11}{14}$.

$\frac{-12}{x}=3$

$\Rightarrow \frac{-12}{3}=x$

Or
$x=-4$

In the fractions $\frac{1}{3}$ and  $\frac{5}{3}$, the denominators are same.

So, considering numerator values, we can find the required rational number.

4 lies between 1 and 5.

So, keep 4 in the numerator and keep the same denominator as it is.

The required rational number is $\frac{4}{3}$.

Let's first evaluate LHS.

$\left(\frac{59}{13}\right)\times \frac{1}{52}=\frac{59}{676}$

Now, RHS

$59\times \left(\frac{52}{13}\right)$

$=\frac{3068}{13}$

We see that both LHS and RHS are not equal.