Question

The value of $ 0.\overline {63} + 0.\overline {37} $ is

A. $ 1 $

B. $ \dfrac{{100}}{{99}} $

C. $ \dfrac{{100}}{{90}} $

D. None of these

Answer 1

Hint: In this problem, we need to find the sum of two rational numbers whose decimal expansion is non-terminating and repeating. First we will assume the given number as $ x $ and then we will multiply it by $ 100 $. After simplification, we will get the required answer.

Complete step-by-step answer:

$0.\overline {63} + 0.\overline {37} $ . 

We know that $ 0.\overline {63} $ can be written as $ 0.636363 \ldots $ and $ 0.\overline {37} $ can be written as $0.373737 \ldots $ .

First we will find the value of $ 0.\overline {63} $ . For this, let us assume that $ x = 0.\overline {63} = 0.636363 \ldots \cdots \cdots \left( 1 \right) $.

 Here we can see that two digits are repeated. So, we will multiply by $ 100 $ on both sides of equation $ \left( 1 \right) $ . Therefore, we get

$  100x = 100 \times \left( {0.636363 \ldots } \right) $

$  \Rightarrow 100x = 63.636363 \ldots \cdots \cdots \left( 2 \right) $

Let us subtract equation $ \left( 1 \right) $ from equation $ \left( 2 \right) $. 

Therefore, we get

$  100x - x = \left( {63.636363 \ldots } \right) - \left( {0.636363 \ldots } \right) $

$   \Rightarrow 99x = 63 $

$   \Rightarrow x = \dfrac{{63}}{{99}} $

$   \Rightarrow 0.\overline {63} = \dfrac{{63}}{{99}} \cdots \cdots \left( 3 \right) $

Now we will find the value of $ 0.\overline {37} $ . 

For this, let us assume that $ y = 0.\overline {37} = 0.373737 \ldots \cdots \cdots \left( 4 \right) $.

Now we will multiply by $ 100 $ on both sides of equation $ \left( 4 \right) $ . Therefore, we get

$  100y = 100 \times \left( {0.373737 \ldots } \right) $

$   \Rightarrow 100y = 37.373737 \ldots \cdots \cdots \left( 5 \right)   $

Let us subtract equation $ \left( 4 \right) $ from equation $ \left( 5 \right) $ . Therefore, we get

 $  100y - y = \left( {37.373737 \ldots } \right) - \left( {0.373737 \ldots } \right) $

 $  \Rightarrow 99y = 37 $

 $  \Rightarrow y = \dfrac{{37}}{{99}} $

 $ \Rightarrow 0.\overline {37} = \dfrac{{37}}{{99}} \cdots \cdots \left( 6 \right) $

Let us add two numbers $ 0.\overline {63} $ and $ 0.\overline {37} $ . So, from equations $ \left( 3 \right) $ and $ \left( 6 \right) $ , we can write

$  0.\overline {63} + 0.\overline {37} = \dfrac{{63}}{{99}} + \dfrac{{37}}{{99}} $

$   \Rightarrow 0.\overline {63} + 0.\overline {37} = \dfrac{{63 + 37}}{{99}} $

$   \Rightarrow 0.\overline {63} + 0.\overline {37} = \dfrac{{100}}{{99}}  $

Therefore, the value of $ 0.\overline {63} + 0.\overline {37} $ is $ \dfrac{{100}}{{99}} $ .

So, the correct answer is “Option B”.

Note: If we want to find the value of $ 5.\overline 2 $ then first we will write $ 5.\overline 2 $ as $ 5.222222 \ldots $ . Here we can see that only one digit is repeated. Therefore, we will multiply by $ 10 $ on both sides of equation $ x = 5.222222 \ldots $ . Then, we will apply the same procedure as we discussed in the given problem. If a digit or a sequence of digits is repeated infinitely in a decimal then that decimal is called non-terminating repeating or recurring decimal. The recurring decimal can be written by putting a bar over the repeated digit.