Answer

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**Hint:**In this type of question we have to use the concept of conversion of decimal numbers into fraction. Here, we consider the given decimal number equal to some variable and call it as equation 1. Then we multiplied both sides by 100 and called it as equation 2. When we perform multiplication by 100, the number of zeros is equal to the number of digits after the decimal point. Then by performing subtraction of equation 1 from equation 2 we can obtain the result.

**Complete step by step answer:**

Here, we have to express \[1.\overline{81}\] in the form of \[\dfrac{p}{q}\] where p and q are integers and q is not equal to zero.

Let us suppose,

\[\Rightarrow x=1.\overline{81}=1.818181\cdots \cdots \text{ ---}\left( 1 \right)\]

Now, multiplying both sides by 100 we get,

\[\begin{align}

& \Rightarrow 100\times x=100\times 1.\overline{81} \\

& \Rightarrow 100\times x=181.8181\cdots \cdots \text{ ---}\left( 2 \right) \\

\end{align}\]

By subtracting equation (1) from equation (2), we can write,

\[\begin{align}

& \Rightarrow 99x=180 \\

& \Rightarrow x=\dfrac{180}{99} \\

\end{align}\]

Dividing both numerator and denominator by the highest common divisor that is 9, we get

\[\Rightarrow x=\dfrac{20}{11}\]

Hence, \[1.\overline{81}\] can be expressed as \[\dfrac{20}{11}\] in the form of \[\dfrac{p}{q}\] where p and q are integers and q is not equal to zero.

**Note:**In this type of question students have to take care when they perform multiplication by 100. Students have to note that the number of zeros over 1 is equal to the number of digits after the decimal point. Also students have to note that in \[1.\overline{81}\] the number 81 gets repeated infinitely many times hence we can write \[1.\overline{81}\] as 1.818181……

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