Question

The denominator of a fraction exceed the numerator by 4, and fraction formed by Squaring both numerator and
denomination is equal to 4 by 9 find the fraction
​

Answer 1

AnswEr

The required fraction is 8/12

Given

• The denominator of a fraction exceed the numerator by 4
• Fraction formed by squaring both numerator and denominator is equal to 4/9

To Find

• The fraction

Solution

Let us consider the numerator and denominator be x and y respectively

•Therefore, the fraction is x/y

Applying the first condition

$y = x + 4$_______(1)

And from the second condition

$\frac{ ({x})^{2} }{ {(y)}^{2} } = \frac{4}{9} \\ \implies (\frac{x}{y} {)}^{2} = \frac{4}{9} \\ \implies (\frac{x }{x + 4} ) {}^{2} = \frac{4}{9} \\ \implies \frac{ {x}^{2} }{(x + 4) {}^{2} } = \frac{4}{9} \\ \implies 9 {x}^{2} = 4( {x + 4)}^{2} \\ \implies9 {x}^{2} = 4( {x}^{2} + 16 + 8x) \\ \implies9x {}^{2} = 4 {x}^{2} + 64 + 32x \\ \implies5x {}^{2} - 32x - 64 = 0 \\ \implies 5{x}^{2} - 40x + 8x - 64 = 0 \\ \implies5x(x - 8) + 8(x - 8) = 0 \\ \implies(5x + 8)(x - 8) = 0$

Now

x - 8 = 0 and 5x + 8 = 0

⇒ x = 8 and ⇒ x = -8/5

But since x is a numerator so it couldn't be a fraction itself . Thus x ≠ -8/5

Now putting the value of x in (1) we have

$y = 8 + 4 \\ \implies y = 12$

So the fraction is

$\huge{ \bold{\frac{8}{12}}}$

Answer 2

Answer:

Let the Numerator be n and Denominator be (n + 4) of the Fraction respectively.

$\underline{\bigstar\:\:\textsf{According to the Question :}}$

$:\implies\sf\bigg(\dfrac{Numerator}{Denominator}\bigg)^2=\dfrac{4}{9}$

$:\implies\sf\bigg(\dfrac{n}{(n+4)}\bigg)^2=\dfrac{4}{9}$

$:\implies\sf\dfrac{n^2}{(n+4)^2}=\dfrac{4}{9}$

• By Cross Multiplication

$:\implies\sf$ 9n² = 4(n + 4)²

$:\implies\sf$ 9n² = 4(n² + 16 + 8n)

$:\implies\sf$ 9n² = 4n² + 64 + 32n

$:\implies\sf$ 9n² – 4n² – 64 – 32n = 0

$:\implies\sf$ 5n² – 32n – 64 = 0

$:\implies\sf$ 5n² – (40 – 8)n – 64 = 0

$:\implies\sf$ 5n² – 40n + 8n – 64 = 0

$:\implies\sf$ 5n(n – 8) + 8(n – 8) = 0

$:\implies\sf$ (n – 8)(5n + 8) = 0

$:\implies\sf$ n = 8⠀or,⠀n = $\sf{}^{-8}\!/{}_{5}$

• But Acc. to Situation n ≠ $\sf{}^{-8}\!/{}_{5}$

$\rule{150}{1}$

$\bf{\dag}\:\boxed{\sf Original\: Fraction=\dfrac{n}{(n+4)}=\dfrac{\textsf{\textbf{8}}}{\textsf{\textbf{12}}}}$