Convert the recurring decimal \[0.22222...\] into a vulgar fraction
A.\[\dfrac{2}{9}\]
B.\[\dfrac{2}{{99}}\]
C.\[\dfrac{{22}}{{99}}\]
D.\[\dfrac{{222}}{{999}}\]
Answer
576k+ views
Hint: Here, we will denote the recurring decimal by a variable to form an equation. Then we will multiply by 10 on both sides to form another equation. We will then subtract the first equation from the second and solve it further to get the required vulgar fraction.
Complete step-by-step answer:
The recurring decimal given to us is \[0.22222...\]
Let us denote this by a variable \[x\].
\[x = 0.22222...\]…………………………………….\[\left( 1 \right)\]
In the decimal part of \[0.22222...\], there is only one digit that is recurring i.e., 2.
In this case, we will multiply both sides of equation \[\left( 1 \right)\] by 10 to get a new recurring decimal. Thus, we get,
\[10x = 2.22222...\] …………………………………\[\left( 2 \right)\]
Now, we will subtract equation \[\left( 1 \right)\] from equation \[\left( 2 \right)\] i.e., we will subtract the original recurring decimal from the new recurring decimal. Therefore, we get
\[10x - x = \left( {2.22222...} \right) - \left( {0.22222...} \right)\]
Subtracting the terms, we get
\[ \Rightarrow 9x = 2\]
Taking 9 on the LHS to the denominator in RHS, we get,
\[ \Rightarrow x = \dfrac{2}{9}\] …………………………\[\left( 3 \right)\]
\[ \Rightarrow x = 0.22222... = \dfrac{2}{9}\]
Therefore, the recurring decimal \[0.22222...\] is represented as \[\dfrac{2}{9}\] in the fractional form.
Thus the correct option is A.
Note: Here, we have converted the recurring or non terminating decimal to the vulgar fraction. A decimal number is said to be non terminating if the digits after the decimal number are infinite or repeat themselves. Similarly, a decimal number is said to be terminating if the digits after the decimal point are finite or do not repeat themselves. A fraction is called a vulgar fraction if the numerator and denominator consist of only integers.
Complete step-by-step answer:
The recurring decimal given to us is \[0.22222...\]
Let us denote this by a variable \[x\].
\[x = 0.22222...\]…………………………………….\[\left( 1 \right)\]
In the decimal part of \[0.22222...\], there is only one digit that is recurring i.e., 2.
In this case, we will multiply both sides of equation \[\left( 1 \right)\] by 10 to get a new recurring decimal. Thus, we get,
\[10x = 2.22222...\] …………………………………\[\left( 2 \right)\]
Now, we will subtract equation \[\left( 1 \right)\] from equation \[\left( 2 \right)\] i.e., we will subtract the original recurring decimal from the new recurring decimal. Therefore, we get
\[10x - x = \left( {2.22222...} \right) - \left( {0.22222...} \right)\]
Subtracting the terms, we get
\[ \Rightarrow 9x = 2\]
Taking 9 on the LHS to the denominator in RHS, we get,
\[ \Rightarrow x = \dfrac{2}{9}\] …………………………\[\left( 3 \right)\]
\[ \Rightarrow x = 0.22222... = \dfrac{2}{9}\]
Therefore, the recurring decimal \[0.22222...\] is represented as \[\dfrac{2}{9}\] in the fractional form.
Thus the correct option is A.
Note: Here, we have converted the recurring or non terminating decimal to the vulgar fraction. A decimal number is said to be non terminating if the digits after the decimal number are infinite or repeat themselves. Similarly, a decimal number is said to be terminating if the digits after the decimal point are finite or do not repeat themselves. A fraction is called a vulgar fraction if the numerator and denominator consist of only integers.
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