How do you convert $3.123$ ( $123$ being repeated) to a fraction?
Answer
627k+ views
Hint: We will consider the fraction to be a variable $x$ . Then, since there is repetition, we will multiply the fraction to remove the $3$ decimal places and then subtract the initial value and simplify to get the required value of $x$ .
Complete step-by-step solution:
We have the number given to us as:
$3.123$ ( $123$ being repeated)
We will consider the number to be $x$ . Therefore, it can be written as:
$x = 3.123$($123$ being repeated).
Now to remove the recurring decimal place, we will multiply both the sides of the equation by $1000$ .
On multiplying, we get:
$\Rightarrow$$1000x = 3123.123$
Now on subtracting the term $x$from both the sides, we get:
$\Rightarrow$$1000x - x = 3123.123 - x$
Now on substituting the value of $x$in the right-hand side, we get:
$\Rightarrow$$1000x - x = 3123.123 - 3.123$
Now on subtracting the value, we get:
$\Rightarrow$$999x = 3120$
Now on transferring the term $999$ from the left-hand side to the right-hand side, we get:
$\Rightarrow$$x = \dfrac{{3120}}{{999}}$
But, we assumed $x = 3.123$ .
Hence, $3.123 = \dfrac{{3120}}{{999}}$ is the required solution.
Note: It is to be remembered that whenever a value is added, subtracted, multiplied or divided on both the sides of the equation, the value of the equation does not change.
It is to be remembered that when a term which is in multiplication is transferred across the $ = $ sign, it has to be written as division. Same rule applies for addition and subtraction.
In the above question, we have a smaller number in the denominator and a larger number in the numerator. These types of fractions are called improper fractions. There also exists proper fractions which are the opposite of it.
Complete step-by-step solution:
We have the number given to us as:
$3.123$ ( $123$ being repeated)
We will consider the number to be $x$ . Therefore, it can be written as:
$x = 3.123$($123$ being repeated).
Now to remove the recurring decimal place, we will multiply both the sides of the equation by $1000$ .
On multiplying, we get:
$\Rightarrow$$1000x = 3123.123$
Now on subtracting the term $x$from both the sides, we get:
$\Rightarrow$$1000x - x = 3123.123 - x$
Now on substituting the value of $x$in the right-hand side, we get:
$\Rightarrow$$1000x - x = 3123.123 - 3.123$
Now on subtracting the value, we get:
$\Rightarrow$$999x = 3120$
Now on transferring the term $999$ from the left-hand side to the right-hand side, we get:
$\Rightarrow$$x = \dfrac{{3120}}{{999}}$
But, we assumed $x = 3.123$ .
Hence, $3.123 = \dfrac{{3120}}{{999}}$ is the required solution.
Note: It is to be remembered that whenever a value is added, subtracted, multiplied or divided on both the sides of the equation, the value of the equation does not change.
It is to be remembered that when a term which is in multiplication is transferred across the $ = $ sign, it has to be written as division. Same rule applies for addition and subtraction.
In the above question, we have a smaller number in the denominator and a larger number in the numerator. These types of fractions are called improper fractions. There also exists proper fractions which are the opposite of it.
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