# Ex.1.3 Q3 Number System Solution - NCERT Maths Class 9

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## Question

Express the following in the form of \begin{align}\frac{\mathrm{p}}{\mathrm{q}},\end{align} where $$p$$ and $$q$$ are integers and $$q≠ 0$$.

(i) \begin{align}0.\overline 6\end{align}

(ii) \begin{align}0.4\overline 7\end{align}

(iii) \begin{align} 0.\overline {001} .\end{align}

Video Solution
Number Systems
Ex 1.3 | Question 3

## Text Solution

Steps:

\begin{aligned} \text { (i) } &0 . \overline{6} \\ & \text { Let } \mathrm{x} =0 . \overline{6} \\&\qquad \mathrm{x} =0.666 \ldots \qquad \qquad \dots(1) \end{aligned}

Since one digit is repeating. We should multiply both sides of ($$1$$) by $$10$$. We get,

\begin{align}10\,{\rm{ x}} &= {\rm{ }}6.666...\\&= {\rm{ }}6{\rm{ }} + {\rm{ }}0.666...\\10\;{\rm{ x}}& = {\rm{ }}6{\rm{ }} + {\rm{ }}x\\10{\rm{\,x }}{\rm{ }}-{\rm{x }}{\rm{ }} &= {\rm{ }}6\\9\,{\rm{x }} &= {\rm{ }}6\\\rm{x} &= \frac{6}{9}\\&= \frac{2}{3}\\\end{align}

Hence\begin{align}0.\overline 6 = \frac{2}{3}\end{align}

\begin{align}{\rm{(ii}})\quad 0.4\overline 7 \end{align}

\begin{align}\text{Let x = 0.4777} \qquad…… (1) \end{align}

Here the repetition starts after the first decimal place and one digit is repeated.

\begin{align}\text{10 x} = 4.777 \qquad…... (2)\end{align}

\begin{align}(2) – (1)\; \text{gives}\end{align}

\begin{align} 10 \mathrm{x}-\mathrm{x} &=4.777 \ldots-0.4777 \ldots \\ 9 \mathrm{x} &=4.3 \\ 9 \mathrm{x} &=\frac{43}{10} \\ ∴ \mathrm{x} &=\frac{43}{90} \end{align}

Hence \begin{align}0.4 \overline{7}=\frac{43}{90}\end{align}

\begin{align}\rm(iii) \quad 0.\overline {001} \end{align}

\begin{align}\text{Let x }= 0.001001…… \qquad(1)\end{align}

Since $$3$$ digits are repeated multiply both the sides of ($$1$$) by $$1000$$

\begin{align}1000\;\mathrm{x}&=1.001001 \\ 1000\;\mathrm{x}&=1+0.001001 \\ 1000 \;\mathrm{x}&=1+\mathrm{x} \\ 1000\;\mathrm{x}-\mathrm{x}&=1 \\ 999 \;\mathrm{x}&=1 \\ ∴ \;\mathrm{x}&=\frac{1}{999}\end{align}

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