
What is the value of $2.\overline{6}-1.\overline{9}$ .
A. $0.\overline{6}$
B. $0.\overline{9}$
C. $0.\overline{7}$
D. 0.7
Answer
616.5k+ views
Hint: First, convert $2.\overline{6}\text{ and 1}\text{.}\overline{\text{9}}$ to their respective $\dfrac{p}{q}$ form and then subtract to get the answer. Do not forget to convert the answer back to the decimal form.
Complete step-by-step answer:
Before moving to the solution, let us first discuss the meaning of the notation used. The bar above the decimal part of the two numbers given in the question represents recurring decimals. They can be written as:
$2.\overline{6}=2.666666.............$
$1.\overline{9}=1.9999999..........$
To start with the question, let $2.\overline{6}\text{ and 1}\text{.}\overline{\text{9}}$ be x and y, respectively.
Now let us convert both the numbers to the $\dfrac{p}{q}$ form. First, let us start with x.
$\therefore x=2.66666......$ ……………(i)
So, on multiplying both sides of the equation by 10, our equation becomes:
$10x=10\times 2.66666......$
$\Rightarrow 10x=26.6666......$ ……………(ii)
Now, if we subtract equation (i) from equation (ii) we will get rid of the recurring decimal part of the right-hand side of the equations. So, subtracting equation (i) from equation (ii), we get
$10x-x=26.66666.....-2.66666.....$
$\Rightarrow 9x=24$
$\Rightarrow x=\dfrac{24}{9}$
Similarly, let us convert y to the $\dfrac{p}{q}$ form.
$\therefore y=1.99999......$ ………………(iii)
As we did for x, here also, we will multiply equation (iii) by 10. So, equation (iii) becomes:
$10y=10\times 1.9999........$
$\Rightarrow 10y=19.9999.........$ ……………(iv)
Now subtracting equation (iii) from equation (iv) to get rid of the recurring decimal part, we get
$10y-y=19.9999.....-1.99999.....$
$\Rightarrow 9y=18$
$\Rightarrow y=\dfrac{18}{9}$
Now solving the expression given in the question, we get
$2.\overline{6}-1.\overline{9}$
$=x-y$
Here if we put the values of x and y in $\dfrac{p}{q}$ form, we get
$\dfrac{24}{9}-\dfrac{18}{9}$
$=\dfrac{24-18}{9}$
\[=\dfrac{6}{9}\]
\[=\dfrac{2}{3}=0.66666......=0.\overline{6}\]
So, the answer is option (a) $0.\overline{6}$ .
Note: Whenever you come across a question consisting of a recurring decimal, the first thing you should do is to convert the recurring decimal to its $\dfrac{p}{q}$ form as you cannot deal with a recurring number in its decimal form.
Complete step-by-step answer:
Before moving to the solution, let us first discuss the meaning of the notation used. The bar above the decimal part of the two numbers given in the question represents recurring decimals. They can be written as:
$2.\overline{6}=2.666666.............$
$1.\overline{9}=1.9999999..........$
To start with the question, let $2.\overline{6}\text{ and 1}\text{.}\overline{\text{9}}$ be x and y, respectively.
Now let us convert both the numbers to the $\dfrac{p}{q}$ form. First, let us start with x.
$\therefore x=2.66666......$ ……………(i)
So, on multiplying both sides of the equation by 10, our equation becomes:
$10x=10\times 2.66666......$
$\Rightarrow 10x=26.6666......$ ……………(ii)
Now, if we subtract equation (i) from equation (ii) we will get rid of the recurring decimal part of the right-hand side of the equations. So, subtracting equation (i) from equation (ii), we get
$10x-x=26.66666.....-2.66666.....$
$\Rightarrow 9x=24$
$\Rightarrow x=\dfrac{24}{9}$
Similarly, let us convert y to the $\dfrac{p}{q}$ form.
$\therefore y=1.99999......$ ………………(iii)
As we did for x, here also, we will multiply equation (iii) by 10. So, equation (iii) becomes:
$10y=10\times 1.9999........$
$\Rightarrow 10y=19.9999.........$ ……………(iv)
Now subtracting equation (iii) from equation (iv) to get rid of the recurring decimal part, we get
$10y-y=19.9999.....-1.99999.....$
$\Rightarrow 9y=18$
$\Rightarrow y=\dfrac{18}{9}$
Now solving the expression given in the question, we get
$2.\overline{6}-1.\overline{9}$
$=x-y$
Here if we put the values of x and y in $\dfrac{p}{q}$ form, we get
$\dfrac{24}{9}-\dfrac{18}{9}$
$=\dfrac{24-18}{9}$
\[=\dfrac{6}{9}\]
\[=\dfrac{2}{3}=0.66666......=0.\overline{6}\]
So, the answer is option (a) $0.\overline{6}$ .
Note: Whenever you come across a question consisting of a recurring decimal, the first thing you should do is to convert the recurring decimal to its $\dfrac{p}{q}$ form as you cannot deal with a recurring number in its decimal form.
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