Free Rational Numbers 03 Practice Test - 8th Grade 

Natural numbers start from 1 and goes on as 2, 3, 4, ...

$\frac{3}{10}=0.3\text{and}\frac{7}{10}=0.7$

Thus, there are no natural numbers between 0.3 and 0.7.

There are infinite rational numbers between any two rational numbers. Let's consider an example to understand this better.

Rational numbers between $\frac{-1}{10}$ and $\frac{3}{10}$:

$\frac{0}{10}$, $\frac{1}{10}$, $\frac{2}{10}$ are the obvious rational numbers between them.

But $\frac{-1}{10}$ can be written as $\frac{-10000}{100000}$ and $\frac{3}{10}$ = $\frac{30000}{100000}$

So, rational numbers between $\frac{-1}{10}$ and $\frac{3}{10}$ are $\frac{-9999}{100000}$, $\frac{-9998}{100000}$ ,…..$\frac{29998}{100000}$, $\frac{29999}{100000}$
Thus, we can find infinite rational numbers between any two rational numbers.

All real numbers can be represented on a number line and rational numbers are also real numbers. Thus rational numbers can also be represented on the number line.

Zero is the additive identity for rational numbers, integers and whole numbers, as adding zero to these numbers does not affect its value.

$Eg:a+0=0+a=a.$

There are only 4 natural numbers between 15 and 20 which are 16, 17, 18 and 19.

Multiplication of any number with its reciprocal results in 1.
$\text{For example},\frac{2}{5}\times \frac{5}{2}=1$

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

Division of whole numbers is not commutative. $\frac{a}{b}\ne \frac{b}{a}$

For example: a = 2, b = 3; $\frac{2}{3}\ne \frac{3}{2}$

Let the number be $y$.

$(y-\frac{1}{2})\times \left(\frac{1}{2}\right)=\left(\frac{3}{8}\right)$

$(y-\frac{1}{2})=\left(\frac{3}{4}\right)$

$y=\frac{3}{4}+\frac{1}{2}=\frac{3}{4}+\frac{2}{4}=\frac{5}{4}=1.25$

Raju's total monthly income = ₹ 34000

Raju’s expenses on food = $\frac{1}{4}\times 34000$

                                         = ₹ 8500

Money left with Raju = 34000 - 8500
                                   = ₹ 25500

House rent = $\frac{3}{10}\times 25500$ 
                   = ₹ 7650

$\text{Now, money left with Raju =}25500–7650$ 
                                                = ₹ 17850

Children's education = $\frac{5}{21}\times 17850$
                                   = ₹ 4250

Money left with Raju = 17850 – 4250
                                   = ₹13600

Let the  total number of pages in the book be $x$.

 $(\frac{4}{9}\times x)\text{pages are read}$

$\text{Remaining pages}=x–\frac{4x}{9}$
$=\frac{5x}{9}=30$$\Rightarrow 5x=30\times 9$
$x=54\text{pages}$