Question

the sum of the numerator and the denomenter of a fraction is 4.for more than twice numerator if 4 is add to each of the numerator and denomator ratio become 2:3 find the fraction​

Answer 1

To Find:-

• Find the fraction.

Given:-

• The sum of the numerator and the denomenter of a fraction is 4.for more than twice numerator if 4 is add to each of the numerator and denomator ratio become 2 : 3.

Solution:-

Let the numerator be " x "

Denominator be " y "

According to the Question

x + y = 2x + 4

y = x + 4 . . . . . ( 1 )

Condition 2

$\tt\implies \: \dfrac { x + 3 } { y + 3 } = \dfrac { 2 } { 3 }$

$\tt\implies \: 3 ( x + 3 ) = 2 ( y + 3 )$

$\tt\implies \: 3x + 9 = 2y + 6$

$\tt\implies \: 3x - 2y = - 3$

Substitute " y " in equation ( 1 )

$\tt\implies \: 3x - 2 ( x + 4 ) = - 3$

$\tt\implies \: 3x - 2x - 8 = - 3$

$\tt\implies \: x = 5$

• Numerator is 5

• Denominator is x + 4 = 9

Fraction is $\tt \dfrac { 5 } { 9 }$

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

$\begin{gathered}\begin{gathered}\bf \: Let\: - \begin{cases} &\sf{numerator \: = \: x} \\ &\sf{denominator \: = y} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: So, \: fraction \: is - \begin{cases} &\sf{\dfrac{x}{y} } \\ \end{cases}\end{gathered}\end{gathered}$

$\bigstar \: \: \red{ \rm \: According \: to \: statement }$

The sum of the numerator and the denomenter of a fraction is 4 more than twice numerator.

$\rm : \implies \:x + y = 4 + 2x$

$\rm : \implies \:y = 4 + 2x - x$

$\boxed{ \pink{ \implies \bf \: y \: = 4 + x \: }} - - - (i)$

$\bigstar \: \: \green{ \rm \: According \: to \: statement }$

Now,

Numerator and denominator both are increased by 4.

$\begin{gathered}\begin{gathered}\bf \: So\: - \begin{cases} &\sf{numerator \: = \: x + 4} \\ &\sf{denominator \: = y + 4} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: So, \: fraction \: is - \begin{cases} &\sf{\dfrac{x + 4}{y + 4} } \\ \end{cases}\end{gathered}\end{gathered}$

$\rm : \implies \:\dfrac{x + 4}{y + 4} = \dfrac{2}{3}$

$\rm : \implies \:\dfrac{x + 4}{x + 4 + 4} = \dfrac{2}{3} \: \: ( \because \: of \: using \: (i) \: )$

$\rm : \implies \:\dfrac{x + 4}{x + 8} = \dfrac{2}{3}$

$\rm : \implies \:3x + 12 = 2x + 16$

$\bigstar \: \: \boxed{ \pink{ \rm : \implies \:x \: = \: 4}}$

On substituting the value of x, in equation (i), we get

$\rm : \implies \:y \: = \: 4 + 4$

$\bigstar \: \: \boxed{ \pink{ \rm : \implies \:y \: = \: 8}}$

$\begin{gathered}\begin{gathered}\bf \: So, \: fraction \: is - \begin{cases} &\sf{\dfrac{4}{8} } \\ \end{cases}\end{gathered}\end{gathered}$

Verification

• Numerator = 4

• Denominator = 8

First condition

• Sum of numerator and denominator is 4 more than twice the numerator.

• 4 + 8 = 4 + 2 × 4

• 12 = 12 (verified).

Second condition

• Numerator = 4 + 4 = 8

• Denominator = 8 + 4 = 12

• Ratio of numerator to denominator is 8 : 12 = 2 : 3.

Hence, verified.