Question

The numerator of a certain fraction is 8 less than the denominator. If 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes 3/4. Find the original fraction?

Answer 1

$\large \dag$ Question :-

The numerator of a certain fraction is 8 less than the denominator. If 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes 3/4. Find the original fraction.

$\large \dag$ Answer :-

$\red\dashrightarrow\underline{\underline{\sf  \green{The   \: Original \:  Fraction  \: is \:  \frac{3}{11} }} }$

$\large \dag$ Step by step Explanation :-

Let Numerator and Denominator of Original Fraction be :

Numerator = x

As Per the question numerator of the fraction is 8 less than the denominator so,

Denominator should be = x + 8

So ,

$\text{Original Fraction = } \frac{\text x}{\text x + 8}$

❒ When 3 is added to the numerator and 3 is subtracted from the denominator :

$\rm \text{Fraction Becomes } : \frac{x + 3}{x + 5}$

⏩ According To Question :

$\large \blue \bigstar \: \red{ \bf \frac{x + 3}{x + 5} = \frac{3}{4} }$

$:\longmapsto \rm 4(x + 3) = 3(x + 5 )$

$:\longmapsto \rm 4x + 12 = 3x + 15$

$:\longmapsto \rm 4x - 3x = 15 - 13$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf x = 3} }}}$

$\blue\dashrightarrow\underline{\underline{\sf  \orange{Numerator  \: of \:  Original  \: Fraction = 3 }} }$

☆ As numerator of the fraction is 8 less than the denominator

$\rm\therefore \: Denominator = 3 + 8$

$\blue\dashrightarrow\underline{\underline{\sf  \orange{Denominator  \: of \:  Original  \: Fraction = 11 }} }$

Therefore,

$\large\underline{\pink{\underline{\frak{\pmb{ Original  \: Fraction  =  \dfrac{3}{11} }}}}}$

Answer 2

Answer:-

Let the fraction be x/y.

Given:-

The numerator of a fraction is 8 less than the denominator.

That is,

⟹ Numerator = Denominator - 8

⟹ x = y - 8 -- equation (1).

Also given that,

If 3 is added to numerator and 3 is subtracted from the denominator, the fraction becomes 3/4.

$\implies \sf \: \dfrac{x + 3}{y - 3} = \dfrac{3}{4}$

Substitute the value of x from equation (1).

$\begin{gathered} \implies \sf \: \frac{y - 8 + 3}{y - 3} = \frac{3}{4} \\ \\ \\ \implies \sf \: \frac{y - 5}{y - 3} = \frac{3}{4} \\ \\ \\ \implies \sf \:4(y - 5) = 3(y - 3) \\ \\ \\ \implies \sf \:4y - 20 = 3y - 9 \\ \\ \\ \implies \sf \:4y - 3y = - 9 + 20 \\ \\ \\ \implies \boxed{\sf \:y = 11}\end{gathered}$

Substitute y = 11 in equation (1).

⟹ x = y - 8

⟹ x = 11 - 8

⟹ x = 3

∴ The required fraction x/y =$\large \frac{3}{11}$