Answer
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Hint: Problems on addition, subtraction, multiplication and division of repeating decimals can be easily solved by converting the repeating decimals into fractions and then solving them as per the requirements. We assume the decimal of the number to be a variable $x$ and then we multiply it with ${{10}^{n}}$ ($n$ is any positive integer selected as per the requirements). Now, we subtract expressions of $x$ having different numbers of zeros and further simplifying we get the fraction.
Complete step by step answer:
To add, subtract, divide and multiply repeating decimals we first convert the decimal numbers into fractions. After doing so the numbers can be added, subtracted, multiplied with each other or can be divided as needed following the rules of simple fractions.
We solve some examples to have a basic idea about solving this type of problem.
Let’s take the number \[6.\bar{2}\bar{3}\bar{5}\] and convert it to fraction as shown below
Assuming, $x=0.\bar{2}\bar{3}\bar{5}$
As, there are three digits repeating in the number we multiply it with $1000$
$\Rightarrow 1000x=235.\bar{2}\bar{3}\bar{5}$
Now, we subtract $x$ from the above expression as shown below
$\Rightarrow 1000x-x=235.\bar{2}\bar{3}\bar{5}-0.\bar{2}\bar{3}\bar{5}$
$\Rightarrow 999x=235.00$
$\Rightarrow x=\dfrac{235}{999}$
Also, the number \[6.\bar{2}\bar{3}\bar{5}\] can also be written as
\[=6+.\bar{2}\bar{3}\bar{5}\]
\[=6+x\]
\[=6+\dfrac{235}{999}\]
\[=\dfrac{6229}{999}\]
Therefore, the number \[6.\bar{2}\bar{3}\bar{5}\] can be written as \[\dfrac{6229}{999}\] .
Also, we take another number \[8.2\bar{3}\bar{5}\bar{9}\]
Assuming, $x=0.2\bar{3}\bar{5}\bar{9}$
We multiply $x$ with \[10000\] and get
$10000x=2359.\bar{3}\bar{5}\bar{9}.....\left( 1 \right)$
Also, we multiply $x$ with $10$ and get
$10x=2.\bar{3}\bar{5}\bar{9}....\left( 2 \right)$
Now, we subtract $\left( 2 \right)$ from $\left( 1 \right)$ as shown below
$\Rightarrow 10000x-10x=2359.\bar{3}\bar{5}\bar{9}-2.\bar{3}\bar{5}\bar{9}$
$\Rightarrow 9990x=2357.00$
$\Rightarrow x=\dfrac{2357}{9990}$
Also, the number \[8.2\bar{3}\bar{5}\bar{9}\] can also be written as
$=8+0.2\bar{3}\bar{5}\bar{9}$
$=8+x$
$=8+\dfrac{2357}{9990}$
$=\dfrac{82277}{9990}$
Therefore, the number \[8.2\bar{3}\bar{5}\bar{9}\] can be written as $\dfrac{82277}{9990}$ .
In this way any number with repeating decimals can be converted into fractions.
After converting we can do any type of addition, subtraction, multiplication and division following the rules of fractional numbers.
Note: We must carefully pick the value of the power of $10$ that is to be multiplied with $x$ as incorrect power of $10$ will make the problem complicated and mistakes will be unavoidable. Also, we must properly check which of the digits are repeating after the decimal point and which are not so that the rest of the calculation can be done accordingly.
Complete step by step answer:
To add, subtract, divide and multiply repeating decimals we first convert the decimal numbers into fractions. After doing so the numbers can be added, subtracted, multiplied with each other or can be divided as needed following the rules of simple fractions.
We solve some examples to have a basic idea about solving this type of problem.
Let’s take the number \[6.\bar{2}\bar{3}\bar{5}\] and convert it to fraction as shown below
Assuming, $x=0.\bar{2}\bar{3}\bar{5}$
As, there are three digits repeating in the number we multiply it with $1000$
$\Rightarrow 1000x=235.\bar{2}\bar{3}\bar{5}$
Now, we subtract $x$ from the above expression as shown below
$\Rightarrow 1000x-x=235.\bar{2}\bar{3}\bar{5}-0.\bar{2}\bar{3}\bar{5}$
$\Rightarrow 999x=235.00$
$\Rightarrow x=\dfrac{235}{999}$
Also, the number \[6.\bar{2}\bar{3}\bar{5}\] can also be written as
\[=6+.\bar{2}\bar{3}\bar{5}\]
\[=6+x\]
\[=6+\dfrac{235}{999}\]
\[=\dfrac{6229}{999}\]
Therefore, the number \[6.\bar{2}\bar{3}\bar{5}\] can be written as \[\dfrac{6229}{999}\] .
Also, we take another number \[8.2\bar{3}\bar{5}\bar{9}\]
Assuming, $x=0.2\bar{3}\bar{5}\bar{9}$
We multiply $x$ with \[10000\] and get
$10000x=2359.\bar{3}\bar{5}\bar{9}.....\left( 1 \right)$
Also, we multiply $x$ with $10$ and get
$10x=2.\bar{3}\bar{5}\bar{9}....\left( 2 \right)$
Now, we subtract $\left( 2 \right)$ from $\left( 1 \right)$ as shown below
$\Rightarrow 10000x-10x=2359.\bar{3}\bar{5}\bar{9}-2.\bar{3}\bar{5}\bar{9}$
$\Rightarrow 9990x=2357.00$
$\Rightarrow x=\dfrac{2357}{9990}$
Also, the number \[8.2\bar{3}\bar{5}\bar{9}\] can also be written as
$=8+0.2\bar{3}\bar{5}\bar{9}$
$=8+x$
$=8+\dfrac{2357}{9990}$
$=\dfrac{82277}{9990}$
Therefore, the number \[8.2\bar{3}\bar{5}\bar{9}\] can be written as $\dfrac{82277}{9990}$ .
In this way any number with repeating decimals can be converted into fractions.
After converting we can do any type of addition, subtraction, multiplication and division following the rules of fractional numbers.
Note: We must carefully pick the value of the power of $10$ that is to be multiplied with $x$ as incorrect power of $10$ will make the problem complicated and mistakes will be unavoidable. Also, we must properly check which of the digits are repeating after the decimal point and which are not so that the rest of the calculation can be done accordingly.
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