Question

The value of \[0.4\dot{2}\dot{3}\] is
A. $\dfrac{419}{990}$
B. $\dfrac{419}{999}$
C. $\dfrac{417}{990}$
D. $\dfrac{417}{999}$

Answer 1

Hint: In this question, we are to find the sum of the repeated terms in the given non-terminating decimal number. By writing the non-terminating decimal number in the form of a series, we are able to find the sum of the series using the appropriate formula, and then adding with the remaining non-repeating term we get the required value.

Formula Used: Here the non-terminating values will form a series by writing them repeatedly. So, the series represents a geometric series since the next term in the series is multiplied by the previous term.
Thus, the sum of the infinite terms is calculated by
 ${{S}_{n}}=\dfrac{a}{1-r}$ where $r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{n}}$ is the sum of the n terms of the series; $n$ is the number of terms; $a$ is the first term in the series, and $r$ is the common ratio of the terms in the series.

Complete step by step solution: The given value is \[0.4\dot{2}\dot{3}\]
Expanding the given non-terminating decimal, we get
\[0.4\dot{2}\dot{3}=0.4232323.....\]
On simplifying the series,
\[\begin{align}
  & 0.4\dot{2}\dot{3}=0.4232323... \\
 & \text{ }=0.4+0.023+0.00023+0.0000023+... \\
 & \text{ }=\dfrac{4}{10}+\dfrac{23}{1000}+\dfrac{23}{100000}+\dfrac{23}{10000000}+... \\
 & \text{ }=\dfrac{4}{10}+23\left[ \dfrac{1}{{{10}^{3}}}+\dfrac{1}{{{10}^{5}}}+\dfrac{1}{{{10}^{7}}}+... \right]\text{ }...\text{(1)} \\
\end{align}\]
Since the series formed in the equation (1) is a geometric series, the common ratio is
$r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
\[\begin{align}
  & \Rightarrow r=\dfrac{{{a}_{2}}}{{{a}_{1}}} \\
 & \text{ =}\dfrac{\dfrac{1}{{{10}^{5}}}}{\dfrac{1}{{{10}^{3}}}} \\
 & \text{ =}\dfrac{1}{{{10}^{5}}}\times \dfrac{{{10}^{3}}}{1} \\
 & \text{ =}\dfrac{1}{{{10}^{2}}} \\
\end{align}\]
Thus, the sum of the infinite terms of the obtained geometric series is calculated by the formula,
${{S}_{\infty }}=\dfrac{a}{1-r}$
On substituting $a=\dfrac{1}{{{10}^{3}}};r=\dfrac{1}{{{10}^{2}}}$, we get
\[\begin{align}
  & {{S}_{\infty }}=\dfrac{\dfrac{1}{1000}}{1-\dfrac{1}{100}} \\
 & \text{ }=\dfrac{\dfrac{1}{1000}}{\dfrac{100-1}{100}} \\
 & \text{ }=\dfrac{1}{1000}\times \dfrac{100}{99} \\
 & \text{ }=\dfrac{1}{990} \\
\end{align}\]
Then, substituting the obtained sum in the equation (1) we get,
\[\begin{align}
  & 0.4\dot{2}\dot{3}=\dfrac{4}{10}+23\left[ \dfrac{1}{{{10}^{3}}}+\dfrac{1}{{{10}^{5}}}+\dfrac{1}{{{10}^{7}}}+... \right] \\
 & \text{ }=\dfrac{4}{10}+23\left( \dfrac{1}{990} \right) \\
 & \text{ }=\dfrac{4}{10}+\dfrac{23}{990} \\
 & \text{ }=\dfrac{419}{990} \\
\end{align}\]

Option ‘A’ is correct

Note: Here the given value is a non-terminating decimal number. So, on expanding the decimal, a geometric series is formed. So, we can find the common ratio from this easily and we can use it for finding the sum of infinite terms. On substituting these values in the expansion of the given decimal we get the required value of the decimal in the rational form.