Question

Denominator of fraction is 3 more than thrice it's numerator. Denominator becomes 12times the numerator and the denominator is reduced by 3 .Find the fraction ​

Answer 1

Appropriate Question :

• Denominator of fraction is 3 more than thrice it's numerator. Denominator becomes 12 times the numerator and the denominator and denominator is reduced by 3 .Find the fraction.

Given :

• Denominator of fraction is 3 more than thrice it's numerator.
• Denominator becomes 12 times the numerator and the denominator and numerator are reduced by 3.

To Find :

• The Fraction.

Solution :

Let the numerator of the fraction be x.

Let the denominator of the fraction be y.

Fraction = $\bold{\dfrac{x}{y}}$

Case 1 :

The denominator is 3 more than thrice the numerator.

Equation :

$\sf{\longrightarrow{y=3x+3}}$

$\sf{\longrightarrow{y-3=3x}}$

$\sf{\longrightarrow{\dfrac{y-3}{3}=x\:\:\:\:\:(1)}}$

Case 2 :

Denominator becomes 12 times of numerator, denominator is reduced by 3.

Numerator = (x - 3)

Denominator = (y - 3)

Equation :

$\sf{\longrightarrow{(y-3)=12(x-3)}}$

$\sf{\longrightarrow{y-3=12x-36}}$

$\sf{\longrightarrow{y-3=12\:\Big(\dfrac{y-3}{3}\Big)-36}}$

$\bold{\big[From\:equation\:(1)\:x\:=\:\dfrac{y-3}{3}\big]}$

$\sf{\longrightarrow{y-3=\dfrac{12y-36}{3}-36}}$

$\sf{\longrightarrow{y-3=\dfrac{12y-36-108}{3}}}$

$\sf{\longrightarrow{3(y-3) =12y-36-108}}$

$\sf{\longrightarrow{3y-9=12y-36-108}}$

$\sf{\longrightarrow{3y-9=12y-144}}$

$\sf{\longrightarrow{3y-12y=-144+9}}$

$\sf{\longrightarrow{-9y=-135}}$

$\sf{\longrightarrow{y=\dfrac{-135}{-9}}}$

$\sf{\longrightarrow{y=15}}$

Substitute, y = 15 in equation (1),

$\sf{\longrightarrow{x=\dfrac{y-3}{3}}}$

$\sf{\longrightarrow{x=\dfrac{15-3}{3}}}$

$\sf{\longrightarrow{x=\dfrac{12}{3}}}$

$\sf{\longrightarrow{x=4}}$

$\large{\boxed{\bold{Numerator\:of\:fraction\:=\:x\:=\:4}}}$

$\large{\boxed{\bold{Denominator\:of\:fraction\:=\:y\:=\:15}}}$

$\large{\boxed{\bold{Fraction\:=\:\dfrac{x}{y}=\dfrac{4}{15}}}}$

Answer 2

$\small \mathsf{let \: the \: x \: be \: numertor \: and \: y \: be \: denominator \: } \\ \large \therefore \mathsf{fraction \: = \frac{x}{y} } \\ \small \mathsf{according \: to \: 1st \: condition} \\ \large \therefore \mathsf \: {y = 3x+ 3} \\ \mathsf{ 3x - y = - 3 - - - - (i)} \\ \small \mathsf{according \: to \: 2nd \: condition} \\ \therefore \mathsf{(y - 3) = 12(x - 3)} \\\therefore \mathsf{y - 3 = 12x - 36} \\ \mathsf{12x - y = 33 - - - - (ii)} \\ \small \mathsf{subtracting \: equation\: (i)\: from \: (ii) \: we \: get \: } \\ 12x - y = 33 \\ 3x - y \: \: \: \: = - 3 \\ - \: \: \: \: \: \: \: \: + \: \: \: \: \: \: + \: \: \: \: \: \: \: \: \: \: \: \: \\ 9x = 36 \\ \large \mathsf \therefore \boxed{ \: x = 4} \\ \mathsf{put \: x = 4 \: in \: equation \: (i)} \\ 3(4) - y = - 3 \\ 12 - y = - 3 \\ - y = - 3 - 12 \\ - y = - 15 \\ \large \mathsf \therefore \boxed{y = 15} \\ \therefore \mathsf {fraction \: = \frac{x}{y} = } \: \boxed {\frac{4}{15} }$

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