Question

if 1 is added to the numerator of a certain fraction it's value becomes 1/2 and if 1 is added to its denominator 1/3.find the original fraction by simultaneous equation​

Answer 1

Answer :

The required fraction is 3/8

Given :

• If 1 is added to the numerator of a certain fraction its value becomes 1/2
• If 1 is added to its denominator the fraction becomes 1/3

To Find :

• The fraction

Solution :

Let us consider the numerator be x and denominator be y

Therefore, the fraction is :

$\sf \dfrac{x}{y}$

Now according to question

$\sf \implies \dfrac{x + 1}{y} = \dfrac{1}{2} \\\\ \sf \implies 2 (x + 1) = y \\\\ \sf \implies y = 2x + 2......... (1)$

Again by the question we have :

$\sf \implies \dfrac{x}{y+1}= \dfrac{1}{3} \\\\ \sf \implies 3x = y + 1........ (2)$

Again putting the value of y from (1) in (2)

$\sf \implies 3x = 2x + 2 + 1\\\\ \sf \implies 3x-2x = 3\\\\ \implies \sf x = 3$

Now using the value of x in in (1)

$\sf \implies y = 2 (3) + 2 \\\\ \sf \implies y = 6 + 2 \\\\ \sf \implies y = 8$

Thus the fraction is :

$\huge{\bf \dfrac{3}{8}}$

Answer 2

Answer:

Let the Numerator be N and Denominator be D of the Fraction.

$\underline{\bigstar\:\textsf{According to the first statement :}}$

$:\implies\sf \dfrac{Numerator+1}{Denominator}=\dfrac{1}{2}$

$:\implies\sf \dfrac{N+1}{D}=\dfrac{1}{2}$

• By Cross Multiplication

$:\implies$ (N + 1) × 2 = 1 × D

$:\implies$ 2N + 2 = D⠀⠀⠀⠀—eq. ( I )

$\rule{150}{1}$

$\underline{\bigstar\:\textsf{According to the second statement :}}$

$:\implies\sf \dfrac{Numerator}{Denominator+1}=\dfrac{1}{3}$

$:\implies\sf \dfrac{N}{D+1}=\dfrac{1}{3}$

• By Cross Multiplication

$:\implies$ N × 3 = 1 × (D + 1)

$:\implies$ 3N = D + 1

• Putting the value of D from eq. ( I )

$:\implies$ 3N = 2N + 2 + 1

$:\implies$ 3N – 2N = 3

$:\implies$ N = 3

$\rule{120}{1}$

$\underline{\bigstar\:\textsf{Putting value of N in eq. ( I ) :}}$

$\dashrightarrow\:\:$ D = 2N + 2

$\dashrightarrow\:\:$ D = 2(3) + 2

$\dashrightarrow\:\:$ D = 6 + 2

$\dashrightarrow\:\:$ D = 8

$\rule{170}{2}$

$\bf{\dag}\:\boxed{\sf Fraction=\dfrac{Numerator}{Denominator}=\dfrac{N}{D}=\dfrac{\textsf{\textbf{3}}}{\textsf{\textbf{8}}}}$