Question

the numerator of a fraction is 6 less than the denominator . if 3 is added to the numerator,the fraction becomes 2/3.Find the original fraction​

Answer 1

Answer:

Required fraction is 3 / 9.

Step-by-step explanation:

Let the denominator of the required fraction be b and numerator of that same fraction be a.

So the fraction should be a / b.

Given,

Numerator of a fraction is 6 less than the denominator. It means if the denominator of fraction a / b is d then its numerator should be 6 less than b. { now a = b - 6 }

Thus,

Required fraction should be ( b - 6 ) / b

Also,

If 3 is added to the numerator, the fraction becomes 2 / 3.

It means : ( Original numerator + 3 ) / ( original denominator ) = 2 / 3

= > [ ( b - 6 ) + 3 ] / b = 2 / 3

= > [ b - 6 + 3 ] / b = 2 / 3

= > ( b - 3 ) / b = 2 / 3

= > 3( b - 3 ) = 2b

= > 3b - 9 = 2b

= > 3b - 2b = 9

= > b = 9

Hence :

• Required fraction : ( b - 6 ) / b i.e. ( 9 - 6 ) / 9 i.e. 3 / 9.

Hence the required fraction is 3 / 9.

Answer 2

Answer:

$question.$

The numerator of a fraction is 6 less than the denominator if 3 is added to the numerator the fraction becomes 2/3 find the original fraction .

$to \: find \: .$

The original fraction .

$answer.$

Suppose the denominator of fraction = b.

and the numerator of fraction = a .

The fraction = a/b.

According to question the numerator of fraction is 6 less than the denominator .The denominator be = b .

so , it fraction be = a(b - 6)

= (b - 6)/6

Formula :

( Original numerator / Original denominator )

If 3 added to the numerator , the fraction becomes 2/3.

So it can ( original numerator + 3 / ( original denominator ) = 2/3 .

$solution \: .$

= [( b - 6) + 3]/b = 2/3.

= [b - 6 + 3]/b = 2/3.

= (b - 3))b = 2/3.

= 3(b - 3)= 2b.

= 3b - 9 = 2b .

= 3b - 2b = 9

= b = 9

The Required fraction :

= (b - 6),(9 - 6)9 , (3/9).

Thus the required fraction :

= 3/9.