Question

express the following in the p/q , where p and q are integers and q is not equal to 0 (1)
1.327 bar is on 2and 7​

Answer 1

$\large\underline{\sf{Solution-}}$

Given rational number is

$\sf \: 1.3 \overline{27}$

Let assume that

$\sf \: x = 1.3 \overline{27}$

Multiply by 10 on both sides, we get

$\sf \: 10x = 1.3 \overline{27} \times 10$

$\sf \: 10x = 13. \overline{27} - - - (1)$

$\sf \: 10x = 13.272727... \: \:$

Multiply by 100 on both sides, we get

$\sf \: 1000x = 13.272727... \times 100 \: \:$

$\sf \: 1000x = 1327.2727... \: \:$

can be rewritten as

$\sf \: 1000x = 1327.\overline{27} - - - - (2) \: \:$

On Subtracting equation (1) from equation (2), we get

$\sf \: 990x = 1314$

$\sf \: x = \dfrac{1314}{990}$

$\sf \: \sf \: \implies \: x = \dfrac{657}{445}$

Hence,

$\sf \: \bf \: \implies \: 1.3\overline{27} = \dfrac{657}{445}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$