Question

Convert the decimal numbers into the form \[\dfrac{p}{q}\].
1) \[18.\overline {48} \]
2) \[32.12\overline {35} \]

Answer 1

Hint:
We are required to convert the given decimal numbers into the \[\dfrac{p}{q}\] form. We know that \[\dfrac{p}{q}\] form is the rational form. We will write our numbers as equations. We will then multiply them with multiples of 10 equal to the number of recurring digits, and then subtract them with the original equation to get the required rational numbers.

Complete step by step solution:
Rational numbers are the numbers which are expressed in the form of \[\dfrac{p}{q}\], where \[p\] and \[q\] are integers and \[q \ne 0\].
1) We will now convert the decimal number \[18.\overline {48} \] into the form \[\dfrac{p}{q}\].
Let us assume \[x = 18.\overline {48} \].
\[x = 18.484848\]……………..\[\left( 1 \right)\]
Since there are two recurring decimal places here, so we will multiply equation (1) with 100. On doing so we get,
\[100x = 1848.484848\]…………………\[\left( 2 \right)\]
Now we will subtract equation (1) from equation (2). On doing so we get,
\[\begin{array}{l}100x - x = 1848.484848..... - 18.484848.....\\ \Rightarrow \left( {100 - 1} \right)x = 1848 - 18\end{array}\]
On subtracting the above terms, we get
\[99x = 1830\]
On dividing both sides of the above equation with 99, we get
\[\begin{array}{l}\dfrac{{99x}}{{99}} = \dfrac{{1830}}{{99}}\\ \Rightarrow x = \dfrac{{1830}}{{99}}\end{array}\]

Thus, the required \[\dfrac{p}{q}\] form of the decimal number \[18.\overline {48} \] is \[\dfrac{{1830}}{{99}}\].

2) We will now convert the decimal number \[32.12\bar 3\bar 5\] into the form \[\dfrac{p}{q}\].
Let us assume \[x = 32.12\overline {35} \].
\[x = 32.12353535\]………………\[\left( 1 \right)\]
Since there are two recurring decimal places here, so we will multiply equation (1) with 100. On doing so we get,
\[100x = 3212.35353535\]…………….\[\left( 2 \right)\]
Now we will subtract equation (1) from equation (2). On doing so we get,
\[\begin{array}{l}100x - x = 3212.35353535..... - 32.12353535.....\\ \Rightarrow \left( {100 - 1} \right)x = 3212.35 - 32.12\end{array}\]
On subtracting the above terms, we get
\[99x = 3180.23\]
On dividing both sides of the above equation with 99, we get
\[\begin{array}{l}\dfrac{{99x}}{{99}} = \dfrac{{3180.23}}{{99}}\\ \Rightarrow x = \dfrac{{3180.23}}{{99}}\end{array}\]
Multiply the numerator and denominator with 100. On doing so we get,
\[\begin{array}{l}x = \dfrac{{3180.23 \times 100}}{{99 \times 100}}\\ \Rightarrow x = \dfrac{{318023}}{{9900}}\end{array}\]

Thus, the required \[\dfrac{p}{q}\] form of the decimal number
\[32.12\overline {35} \] is \[\dfrac{{318023}}{{9900}}\].

Note:
The bar over a number indicates a recurring, non-terminating, infinite decimal. The decimal conversions of the rational numbers are of two types. These are –
Finite, terminating decimals or Rational numbers – These are the rational numbers that upon division terminate after a finite number of digits. This means that they eventually give the remainder 0.
Example - \[5.3456\]
Infinite, Non-terminating decimals or Rational numbers – These are the rational numbers that upon division do not terminate after a finite number of digits. This means that they do not give the remainder 0, and the division keeps on continuing. They are further of three kinds –
Non-recurring – In these decimals, the numbers do not repeat and do not form any pattern.
Example - \[4.678291538502715183940.....\]
Recurring – In these decimals, the numbers of the decimal part either repeat continuously as a single digit or the set. A bar is used to denote the recurring number or the set.
Example - \[4.666\]……..or \[4.\bar 6\], \[8.93939\]……….or \[8.\overline {93} \].
Mixed recurring – In these decimals, some numbers do not repeat and some of them repeat. The bar is used to denote the repeating of the recurring terms.
Example - \[2.35666\]……..or \[7.16767\]……….or \[{\rm{7}}.1\overline {67} \].