Question

The value of $0.\overline{1}+0.0\overline{1}+0.00\overline{1}$is equal to:
(a)$\dfrac{407}{3300}$
(b)$\dfrac{37}{3300}$
(c)$\dfrac{111}{900}$
(d)$\dfrac{1343}{3300}$

Answer 1

Hint: Individually find the value of $0.\overline{1}$,$0.0\overline{1}$ and $0.00\overline{1}$. Equate each number with x call this equation as eq. (1) then multiply 10 on both the sides and subtract this new equation with eq. (1). Then solve the value of x. This is how we have found the value of these three numbers and then add them.

Complete step-by-step answer:
In the expression $0.\overline{1}+0.0\overline{1}+0.00\overline{1}$, we can see the bar on a number. Bar on a number means trail of the same number which is not ending.
Let us assume $0.\overline{1}=x$. We can also write:
x = 0.111111111111………..Eq. (1)
Multiplying 10 on both the sides we get,
10x = 1.1111111111……………...Eq. (2)
Subtracting eq. (1) from eq. (2) we get,
9x = 1.000000000000
$\Rightarrow x=\dfrac{1}{9}$
Let us assume $0.0\overline{1}=y$. We can also write:
y = 0.01111111111…………..Eq. (3)
Multiplying 10 on both the sides we get,
10y = 0.111111111111………………Eq. (4)
Subtracting eq. (3) from eq. (4) we get,
9y = 0.10000
$\Rightarrow y=\dfrac{1}{90}$
Let us assume $0.00\overline{1}=z$. We can also write:
z =0.0011111111…………….Eq. (5)
Multiplying 10 on both the sides we get,
10z = 0.011111111……………….Eq. (6)
Subtracting eq. (5) from eq. (6) we get,
9z = 0.010000000000
$z=\dfrac{1}{900}$
Adding x, y and z will give:
$x+y+z=0.\overline{1}+0.0\overline{1}+0.00\overline{1}$
Substituting values of x, y and z in the above equation we get,
$\begin{align}
  & \dfrac{1}{9}+\dfrac{1}{90}+\dfrac{1}{900}=0.\overline{1}+0.0\overline{1}+0.00\overline{1} \\
 & \Rightarrow \dfrac{100+10+1}{900}=0.\overline{1}+0.0\overline{1}+0.00\overline{1} \\
 & \Rightarrow \dfrac{111}{900}=0.\overline{1}+0.0\overline{1}+0.00\overline{1} \\
\end{align}$
Hence, the value of $0.\overline{1}+0.0\overline{1}+0.00\overline{1}$is$\dfrac{111}{900}$.
Hence, the correct option is (c).

Note: We might think of adding the three repeating numbers given in the question as:
$\begin{align}
  & \text{ }0.111111111...... \\
 & +0.011111111...... \\
 & \dfrac{+0.001111111......}{0.1233333333..} \\
\end{align}$
But then again we are getting the repeating terms so the method that we have discussed above is suitable for solving the question.