Question

the numerator of a rational number is less than that in its denominator by 3 if the numerator becomes three times then the denominator is increased by 20 the new number becomes 1 by 8 find the original number​

Answer 1

Answer :-

Let the denominator be x.

Numerator = x - 3

Fraction = $\dfrac{x - 3}{x}$

New numerator = 3 (x - 3)

Denominator = x + 20

New Fraction = $\dfrac{3 (x - 3)}{x + 20}$

According to Question :-

$\dfrac{3 (x - 3)}{x + 20}$ = $\dfrac{1}{8}$

$\implies$ $\dfrac{3x - 9 }{x + 20}$ = $\dfrac{1}{8}$

By cross Multiplication :-

$\implies$ (3x - 9) × 8 = x + 20

$\implies$ 24x - 72 = x + 20

$\implies$ 24x - x = 20 + 72

$\implies$ 23x = 92

$\implies$ x = $\dfrac{92}{23}$

$\implies$ x = 4

So, the Numerator = x - 3

= 4 - 3 = 1

Denominator = 4

So, the fraction is :-

=) $\dfrac{1}{4}$

Answer 2

Answer :-

Let the denominator = x

Numerator = x - 3

Fraction = $\dfrac{x - 3}{x}$

New numerator = 3 (x - 3)

Denominator = x + 20

New Fraction = $\dfrac{3 (x - 3)}{x + 20}$

According to Question :-

=> $\dfrac{3 (x - 3)}{x + 20}$ =$\dfrac{1}{8}$

=> $\dfrac{3x - 9 }{x + 20}$ = $\dfrac{1}{8}$

By cross Multiplication :-

=> (3x - 9) × 8 = x + 20

=> 24x - 72 = x + 20

=> 24x - x = 20 + 72

=> 23x = 92

=> x = $\dfrac{92}{23}$

=> x = 4

So, the Numerator = x - 3

= 4 - 3 = 1

Denominator = 4

So, the fraction is :-

=> $\boxed{\dfrac{1}{4}}$