Question

How do you write $\dfrac{15}{6}$ as decimal?

Answer 1

Hint: We first try to explain the improper fraction and the representation in mixed fraction. We use variables to express the condition between those representations. Then we apply long division to express the improper fraction in mixed fraction where we convert the proper fraction part into decimal.

Complete step-by-step answer:
The given fraction $\dfrac{15}{6}$ is an improper fraction. Improper fractions are those fractions who has greater value in numerator than the denominator. We need to convert it into decimal.
We need to convert it into mixed fraction which is representation in the form sum of an integer and a proper fraction. Then we change the proper fraction into decimal to add with the integer.
We express the process in the form of variables.
Let the fraction be $\dfrac{a}{b}$ where $a>b$. Now we express it in the form of mixed fraction. Let’s assume the integer we get is $x$ and the proper fraction is $\dfrac{c}{b}$.
Then the equational condition will be $\dfrac{a}{b}=x+\dfrac{c}{b}$. The representation of the mixed fraction will be $x\dfrac{c}{b}$. We convert the $\dfrac{c}{b}$ part into decimal.
Now we solve our fraction $\dfrac{15}{6}$.
We express it in regular long division processes. The denominator is the divisor. The numerator is the dividend. The quotient will be the integer of the mixed fraction. The remainder will be the numerator of the proper fraction.
$6\overset{2}{\overline{\left){\begin{align}
  & 15 \\
 & \underline{12} \\
 & 3 \\
\end{align}}\right.}}$
Therefore, the proper fraction is $\dfrac{3}{6}=\dfrac{1}{2}$. The integer is 3.
Now we find the decimal form of $\dfrac{1}{2}$. We get
$2\overset{0.5}{\overline{\left){\begin{align}
  & 10 \\
 & \underline{10} \\
 & 0 \\
\end{align}}\right.}}$
The addition will give $3+0.5=3.5$.
The decimal of $\dfrac{15}{6}$ is $3.5$.
So, the correct answer is “$3.5$”.

Note: We need to remember that the denominator in both cases of improper fraction and the mixed fraction will be the same. The only change happens in the numerator. The relation being the equational representation of $a=bx+c$.