Question

can someone pls give me the solution of this possess..... pls????

Answer 1

Let denominator of fraction be X .



Given that : Numerator of fraction is 6 Less than the denominator of fraction .


So,

Numerator of fraction = ( X - 6 ).



Fraction = Numerator / Denominator = ( X - 6 ) / X.






Also given that : I f 3 is added to Numerator then fraction is equal to 2/3.


New Numerator = ( X - 6 + 3 ) = ( X - 3 ).


Denominator = X


New fraction = New Numerator / Denominator = ( X - 3 ) / X.




2/3 = ( X - 3 ) / X



2X = 3 ( X - 3 )


2X = 3X - 9


3X - 2X = 9


X = 9



Denominator of fraction = X = 9


And,

Numerator of fraction = X - 6 = 9 -6 = 3.


Fraction = Numerator/Denominator = 3/9 = 1/3.

Answer 2

Let the denominator of the original fraction be x and the numerator of the original fraction be 6 less than denominator that is ( x - 6 ),




So, original fraction = numerator / denominator


$\bold{Original \: \: fraction }= \frac{x - 6}{x}$






Given in the question that if 3 is added to the numerator of the original fraction, the new fraction becomes ( 2 / 3 ) ,





According to the question :

$\frac{numerator + 3}{denominator \: } = \frac{2}{3} \\ \\ \\ = > \frac{x - 6 + 3}{x} = \frac{2}{3} \\ \\ \\ = > \frac{x - 3}{x} = \frac{2}{3} \\ \\ \\ = > 3(x - 3) = 2x \\ \\ \\ = > 3x - 9= 2x \\ \\ \\ = > 3x - 2x = 9 \\ \\ \\ = > x = 9$






Hence,
Required original fraction = numerator / denominator


Required original fraction = ( x - 6 ) / ( x )

$Required \: original \: fraction = \frac{9 - 6}{9} \\ \\ \\ \boxed{ \bold{ \underline{Required \: original \: fraction }= \frac{3}{9} }}$