How many fifths are in $2\dfrac{3}{5}$ ?

Hint: First, we have to convert into rational or mixed-fraction form. That means we have to convert it in the form $\dfrac{p}{q}$ . Then there will be a number in the numerator and $5$ as the denominator. The number in the numerator is our answer.

Formula used: $a + \dfrac{b}{c} = \dfrac{{(a \times c) + b}}{c}$

Complete step-by-step solution:
We have $2\dfrac{3}{5}$
This can be written as $2 + \dfrac{3}{5}$ .
Then we have to convert it into $\dfrac{p}{q}$ form.
To convert it we have to add these two numbers.
$2$ is a natural number and $\dfrac{3}{5}$ is a rational number.
The formula for adding a natural number and a rational number is;
Let, $a$ is the natural number and $\dfrac{b}{c}$ is a rational number.
Then $a + \dfrac{b}{c} = \dfrac{{(a \times c) + b}}{c}$
So, $2 + \dfrac{3}{5} = \dfrac{{(2 \times 5) + 3}}{5}$
After multiplication we get;
$\Rightarrow 2 + \dfrac{3}{5} = \dfrac{{10 + 3}}{5}$
After addition we get;
$\Rightarrow 2 + \dfrac{3}{5} = \dfrac{{13}}{5}$

$\therefore$ There are $\;13$ fifths in $2\dfrac{3}{5}$.

Note: The numbers which are in the form $\dfrac{p}{q}$ , where $p$ and $q$ can be natural number can be integers but $q \ne 0$ are called rational numbers. The numbers like $1$ , $2$ , $3$ etc. are natural numbers. But remember $0$ is not the natural number. Integers can be two types positive integer and negative integer. Like $1$ , $2$ , $3$ etc. are positive integers and $- 1$ , $- 2$ , $- 3$ etc. are negative integers.
Similar example:
How many sevenths are in $4\dfrac{1}{7}$ ?
We will apply the formula $a + \dfrac{b}{c} = \dfrac{{(a \times c) + b}}{c}$ and get;
$4\dfrac{1}{7} = \dfrac{{(4 \times 7) + 1}}{7}$
That will be nothing but equal to $\dfrac{{28 + 1}}{7}$ .
From here we will easily get;
$4\dfrac{1}{7} = \dfrac{{29}}{7}$
So, we can see that there are $\;29$ seventh in this example.