Question

Express 2.13(bar on 3) and 0.123( bar on 123) in p/q form...class 9 chapter 1 Number System​

Answer 1

Answer :-

Required to find :-

• p/q form of the given non - terminating recurring decimals

Solution :-

Given :-

2.13 ( bar on 3 )

$\rm{ 2.1 \overline{3}}$

So,

Let x = 2.13333 - - - - - $\rightarrowtail{\text{Equation 1 }}$

$\boxed{\begin{minipage}{4cm} \\ \sf{ period = 3 } \\ \\ \sf{ Periodicity = 1 } \end{minipage}}$

Since,

Periodicity is 1

Multiply equation 1 with 10 on both sides

So,

10 ( x ) = 10 ( 2.13333 - - - - )

10x = 21.3333 - - - - $\rightarrowtail{\text{Equation 2 }}$

Consider this as equation 2

Subtract equation 1 from equation 2

So,

10x = 21.3333 - - - - -

1x = 2.1333 - - - -

$\rule{100}{1}$

9x = 19.2000 - - - -

$\rule{100}{1}$

This implies ,

9x = 19.2

$\tt{ x = \dfrac{19.2}{ 9 } }$

$\tt{ x = \dfrac{6.4}{3} }$

Hence,

$\large{\leadsto{\boxed{\tt{ 2.1 \overline{3} = \dfrac{6.4}{3} }}}}$

Similarly ,

Given :-

0.123 ( bar on 123 )

$\rm{ 0.\overline{123}}$

So,

Let x = 0.123123123 - - - -$\rightarrowtail{\text{Equation 1 }}$

$\boxed{\begin{minipage}{7cm} \\ \sf{ period = 123 } \\ \\ \sf{ Periodicity = 3 } \end{minipage}}$

Since,

periodicity = 3

Multiply equation 1 with 1000 on bith sides

So,

1000 ( x ) = 1000 ( 0.123123123 - - - )

1000x = 123.123123 - - - - $\rightarrowtail{\text{Equation 2 }}$

Consider this as equation 2

So,

Subtract equation 1 from equation 2

1000x = 123.123123123 - - - -

x = 0.123123123 - - - -

$\rule{100}{1}$

999x = 123.000000000 - -

$\rule{100}{1}$

This implies ,

999x = 123

$\tt{ x = \dfrac{123}{999} }$

$\tt{ x = \dfrac{ 41 }{ 333}}$

Hence,

$\large{\leadsto{\boxed{\tt{ 0. \overline{123} = \dfrac{ 41}{333} }}}}$

✅ Hence solved !