Question

${\large{\pink{\bf {For \: Best \: Users }}}}$
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

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).

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

Substitute y = 11 in equation (1).

⟹ x = y - 8

⟹ x = 11 - 8

⟹ x = 3

∴ The required fraction x/y = 3/11.

Answer 2

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.​

EXPLANATION-:

Let the denominator be x

Then numerator will be x-8

Fraction so formed-:

$\bf\; \frac{x-8}{x}$

After adding 3 to numerator-:

Numerator=x-8+3

Numerator=x-5

After subtracting 3 from the denominator-:

Denominator=x-3

Fraction so formed-:

$\bf\; \frac{x-5}{x-3}$

A.T.Q,

$\rightarrow \bf\; \frac{x-5}{x-3}=\frac{3}{4}\\\\\rightarrow \bf\; 4x-20=3x-9\\\\\dashrightarrow \underline{\boxed{\bigstar \tt x=11}}$

So -:

Denominator=11

Numberator=11-3

Numerator=8

Original fraction-:

$\underline{\underline{\tt \frac{3}{11}}}$