Question

The numerator of a fraction is 6 less than the

denominator. If 3 is added to thenumerator,the fraction is equal to ⅔. Find the original fraction.​

Answer 1

Answer :

The required original fraction = 3/9

Step-by-step explanation :

Given :

• The numerator of a fraction is 6 less than the  denominator.
• If 3 is added to the numerator, the fraction is equal to ⅔

To find :

the original fraction

Solution :

Let the denominator be x

Since the numerator is 6 less than the denominator,

numerator = x - 6

  $\sf fraction=\dfrac{numerator}{denominator}$

The original fraction is  $\sf \dfrac{x-6}{x}$

Now, 3 is added to the numerator.

new numerator = (x - 6) + 3 = x - 3

 new fraction =  $\sf \dfrac{x-3}{x}$

 According to the question,

   $\sf \dfrac{x-3}{x}=\dfrac{2}{3} \\\\ \sf 3(x-3)=2x \\\\ \sf 3x-9=2x \\\\ \sf 3x-2x=9 \\\\ \sf x=9$

Therefore,

denominator = 9

numerator = 9 - 6 = 3

The original fraction = 3/9

       [ In simplest form it's 1/3 ]

Verification :

• numerator of the fraction = 3
• denominator of the fraction = 9

Condition : If 3 is added to the numerator, the fraction is equal to ⅔

   $\sf \dfrac{3+3}{9}=\dfrac{2}{3} \\\\ \sf \dfrac{6}{9}=\dfrac{2}{3} \\\\ \sf \dfrac{3 \times 2}{3 \times 3}=\dfrac{2}{3} \\\\ \sf \dfrac{2}{3}=\dfrac{2}{3}$

 LHS = RHS

Hence verified!