How do you convert $0.23\bar 4$ (with $4$ repeating) as a fraction
Answer
620.4k+ views
Hint: In this question, we need to convert $0.23\bar 4$ (with $4$ repeating) into fraction. Here, we will consider $0.23\bar 4$ as x. So, to bring the repeating entity immediately after the decimal point, we multiply and divide the given decimal $0.23\bar 4$ by $100$. Then, as there is only $1$ digit being repeated. So, we multiply and divide the decimal by $10$.
Complete step-by-step solution:
In this question, we need to convert $0.23\bar 4$ to a fraction.
Let x be that fraction.
Here, consider the given value as $x = 0.23\bar 4$.
Now, let us multiply and divide $0.23\bar 4$ by $100$, we have,
$x = 0.23\bar 4 \times \dfrac{{100}}{{100}}$
Then, $100x = 0.23\bar 4 \times 100$
$ \Rightarrow 100x = 23.\bar 4$
Hence, $100x = 23.4444....$
Let us consider this as the equation $\left( 1 \right)$.
Now, let us multiply and divide $23.\bar 4$ by $10$, we have,
$100x = 23.\bar 4 \times \left( {\dfrac{{10}}{{10}}} \right)$
Then, $1000x = 23.\bar 4 \times 10$
$ \Rightarrow 1000x = 234.\bar 4$
Hence, $1000x = 234.\bar 4$
Let us consider this as equation \[\left( 2 \right)\].
Now, we will subtract equation $\left( 1 \right)$ from equation \[\left( 2 \right)\].
Therefore, we have,
$1000x - 100x = \left( {234.\bar 4 - 23.\bar 4} \right)$
Hence, $900x = \left( {234.4444.... - 23.4444.....} \right)$
$ \Rightarrow 900x = 211$
\[ \Rightarrow x = \dfrac{{211}}{{900}}\]
Therefore, \[x = \dfrac{{211}}{{900}}\]
Hence, the converted value of $0.23\bar 4$ to a fraction is \[\left( {\dfrac{{211}}{{900}}} \right)\].
Note: In this question it is important to note that, here we have multiplied and divided $0.23\bar 4$ firstly by $100$ and then by $10$ respectively, then subtracted both the equations to determine the value of x as in this question we have a repetition of a repetition of $4$ in $0.23\bar 4$. The scenario may be different in each question depending on the situation as the decimal may have more number of digits as its repeating entity.
Complete step-by-step solution:
In this question, we need to convert $0.23\bar 4$ to a fraction.
Let x be that fraction.
Here, consider the given value as $x = 0.23\bar 4$.
Now, let us multiply and divide $0.23\bar 4$ by $100$, we have,
$x = 0.23\bar 4 \times \dfrac{{100}}{{100}}$
Then, $100x = 0.23\bar 4 \times 100$
$ \Rightarrow 100x = 23.\bar 4$
Hence, $100x = 23.4444....$
Let us consider this as the equation $\left( 1 \right)$.
Now, let us multiply and divide $23.\bar 4$ by $10$, we have,
$100x = 23.\bar 4 \times \left( {\dfrac{{10}}{{10}}} \right)$
Then, $1000x = 23.\bar 4 \times 10$
$ \Rightarrow 1000x = 234.\bar 4$
Hence, $1000x = 234.\bar 4$
Let us consider this as equation \[\left( 2 \right)\].
Now, we will subtract equation $\left( 1 \right)$ from equation \[\left( 2 \right)\].
Therefore, we have,
$1000x - 100x = \left( {234.\bar 4 - 23.\bar 4} \right)$
Hence, $900x = \left( {234.4444.... - 23.4444.....} \right)$
$ \Rightarrow 900x = 211$
\[ \Rightarrow x = \dfrac{{211}}{{900}}\]
Therefore, \[x = \dfrac{{211}}{{900}}\]
Hence, the converted value of $0.23\bar 4$ to a fraction is \[\left( {\dfrac{{211}}{{900}}} \right)\].
Note: In this question it is important to note that, here we have multiplied and divided $0.23\bar 4$ firstly by $100$ and then by $10$ respectively, then subtracted both the equations to determine the value of x as in this question we have a repetition of a repetition of $4$ in $0.23\bar 4$. The scenario may be different in each question depending on the situation as the decimal may have more number of digits as its repeating entity.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

