Answer:

The required fraction is 8 / 20.

Step-by-step-explanation:

Let the numerator of fraction be x.

And denominator of fraction be y.

∴ The fraction = x / y

From the first condition,

The denominator of a fraction is 12 more than the numerator.

∴ y = x + 12 - - - ( 1 )

From the second condition,

If 16 is added to both the mumerator and the denominator, the fraction obtained is 2 / 3.

∴ ( x + 16 ) / ( y + 16 ) = 2 / 3

⇒ 3 ( x + 16 ) = 2 ( y + 16 )

⇒ 3x + 48 = 2y + 32

⇒ 3x - 2y = 32 - 48

⇒ 3x - 2y = - 16

⇒ 3x - 2 ( x + 12 ) = - 16 - - - [ From ( 1 ) ]

⇒ 3x - 2x - 24 = - 16

⇒ x = - 16 + 24

⇒ x = 8

∴ Numerator = 8

By substituting x = 8 in equation ( 1 ),

y = x + 12 - - - ( 1 )

⇒ y = 8 + 12

⇒ y = 20

∴ Denominator = 20

Now,

The fraction = Numerator / Denominator

⇒ The fraction = 8 / 20

∴ The required fraction is 8 / 20.

Answer:

Given :-

• The denominator of a fraction is 12 more than the numerator.
• If 16 is added to both the numerator and the denominator, then the fraction obtained is 2/3.

To Find :-

• What is the original fraction.

Solution :-

Let,

[tex]\mapsto \bf Numerator =\: x\\[/tex]

[tex]\bigstar[/tex] The denominator of a fraction is 12 more than the numerator.

[tex]\mapsto \bf Denominator =\: x + 12\\[/tex]

So, the original fraction will be :

[tex]\leadsto \sf Original\: Fraction =\: \dfrac{Numerator}{Denominator}\\[/tex]

[tex]\leadsto \sf\bold{Original\: Fraction =\: \dfrac{x}{x + 12}}\\[/tex]

According to the question by using the formula we get :

[tex]\bigstar[/tex] If 16 is added to both the numerator and the denominator, then the fraction obtained is 2/3.

So,

[tex]\implies \sf\boxed{\bold{\bigg\{\dfrac{Numerator + 16}{Denominator + 16}\bigg\} =\: \bigg\{\dfrac{2}{3}\bigg\}}}\\[/tex]

[tex]\implies \sf \dfrac{x + 16}{x + 12 + 16} =\: \dfrac{2}{3}[/tex]

[tex]\implies \sf \dfrac{x + 16}{x + 28} =\: \dfrac{2}{3}[/tex]

By doing cross multiplication we get,

[tex]\implies \sf 3(x + 16) =\: 2(x + 28)[/tex]

[tex]\implies \sf 3x + 48 =\: 2x + 56[/tex]

[tex]\implies \sf 3x - 2x =\: 56 - 48[/tex]

[tex]\implies \sf\bold{x =\: 8}[/tex]

Hence, the required original fraction will be :

[tex]\dag[/tex] Original Fraction :

[tex]\dashrightarrow \sf Original\: Fraction =\: \dfrac{x}{x + 12}\\[/tex]

[tex]\dashrightarrow \sf Original\: Fraction =\: \dfrac{8}{8 + 12}\\[/tex]

[tex]\dashrightarrow \sf\bold{\underline{Original\: Fraction =\: \dfrac{8}{20}}}\\[/tex]

[tex]\sf\boxed{\bold{\therefore\: The\: original\: fraction\: is\: \dfrac{8}{20}\: .}}\\[/tex]