Question

What is the sum of $0.\overline{6}$ and $0.\overline{7}$ ?
$\begin{align}
  & \text{(A) 1}\text{.}\overline{\text{3}} \\
 & \text{(B) 1}\text{.3} \\
 & \text{(C) 1}\text{.}\overline{\text{4}} \\
 & \text{(D) An irrational number} \\
\end{align}$

Answer 1

Hint: We will convert non terminating recurring decimals $0.\overline{6}$ and $0.\overline{7}$ to fraction and then we will add those two fractions and at last we will convert the fraction into decimal to get the desired result.

Complete step-by-step answer:

We will convert non terminating recurring decimals to fractions by assuming x=$0.\overline{6}$ and y=$0.\overline{7}$
x can also be written as 0.66666……. and y can also be written as 0.77777……
First, we will convert $0.\overline{6}$ to fraction.
We assumed x=0.66666….. and let this be equation 1.
Now we multiply x with 10, multiplying x with 10 we will get,
$\Rightarrow 10x=6.6666.....$ and let this be equation 2.
Subtracting equation 1 from equation 2 we will get 9x=6 and x can be written as $x=\dfrac{6}{9}$ .
So, $x=\dfrac{6}{9}$.
Now, we will convert $0.\overline{7}$ to fraction.
We assumed y=0.77777….. and let this be equation 3.
Now we multiply y with 10, multiplying y with 10 we will get,
$\Rightarrow 10y=7.7777.....$ and let this be equation 4.
Subtracting equation 3 from equation 4 we will get 9y=7 and x can be written as $y=\dfrac{7}{9}$.
So, $y=\dfrac{7}{9}$.
Now, according to the question we have to add x and y.
So, we will add x and y, adding x and y we will get $\Rightarrow \dfrac{6}{9}+\dfrac{7}{9}=\dfrac{13}{9}$ .
Now we will convert $\dfrac{13}{9}$ to decimal, converting $\dfrac{13}{9}$ to decimal we will get 1.44444…….
1.4444…… can be written as $1.\overline{4}$ .
So, the sum of $0.\overline{6}$ and $0.\overline{7}$ is $1.\overline{4}$ .
Hence the correct answer is (C).

Note: We cannot directly add the non terminating recurring decimals to get answer for the above question instead we have to change the non terminating recurring decimals to their respective fractions and has to be added.