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.
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]
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Answers & Comments
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 :-
To Find :-
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]