Question

Express $1.\overline{62}$ in the $\dfrac{p}{q}$ form, where $q\ne 0$, p, q are integers.

Answer 1

Hint: We start solving this question by realising that the given decimal is a recurring decimal. Then we assume the decimal to be n. Then we multiply it with 100 and we can see that the numbers after the decimal of n and 100n are the same. Then we subtract n from 100n to convert it into an integer. Then we can find the value of the decimal in the $\dfrac{p}{q}$ form.

Complete step-by-step answer:
The decimal we are given is $1.\overline{62}$.
It is recurring decimal. A recurring decimal is a decimal that keeps occurring and that does not end. We represent the recurring part of the decimal by keeping a symbol of the bar on it.
For example,
$4.\overline{17}=4.17171717.......$
$5.37\overline{43}=5.37434343.........$
Now, let us consider the decimal $1.\overline{62}$.
Let us assume that this decimal is equal to n. So, we get
$n=1.\overline{62}$
As it is a recurring decimal, we can write it as,
$n=1.6262626262.........$
Now, let us multiply it with 100. Then we get,
$100n=162.6262626262.........$
Now, let us subtract n from 100n. Then we get,
$\begin{align}
  & 100n=162.6262626262......... \\
 & \text{ }n\text{ }=1.6262626262......... \\
\end{align}$
$\begin{align}
  & 99n=161.0000000 \\
 & n=\dfrac{161}{99} \\
\end{align}$
So, we can write the decimal $1.\overline{62}$ in the $\dfrac{p}{q}$ form as $\dfrac{161}{99}$.
Hence, the answer is $\dfrac{161}{99}$.

Note: The common mistake that one does while solving this type of problem is one might not realize that the given decimal is a recurring decimal and assume that it a normal decimal and write it as,
$\begin{align}
  & n=1.62 \\
 & 100n=162 \\
 & n=\dfrac{162}{100}=\dfrac{81}{50} \\
\end{align}$
So, they might write the answer as $\dfrac{81}{50}$. But one should remember that when a bar is given over the decimal it is a recurring decimal.