Question

the denominator of a fraction is 1 less than twice its numerator. if 3 is added to the numerator and 4 is added to the denominator the fraction becomes 3/5. find the fraction.​

Answer 1

$Let \: \: the \: \: numerator \: \: be \: \: x \\ \\ denominator = 2x-1 \\ \\ Therefore \: \: the \: \: fraction \: \: is \: \: \frac{x}{2 x- 1} \\ \\ = \frac{x + 3}{2x - 1 + 4} = \frac{3}{5}$

$⇒ \frac{x + 3}{2x + 3} = \frac{3}{5} \\ \\ ⇒ 5x + 15 = 6x + 9 \\ \\ ⇒ \: 15 - 9 = 6x - 5x \\ \\ ⇒ \: x = 6$

$Therefore \: \: the \: \: fraction \: \: is \: \: = \frac{x}{2x - 1} = \frac{6}{2 \times 6 - 1} = \frac{6}{11}$

$Hence \: \: the \: \: required \: \: fraction \: \: is \: \: \frac{6}{11}$

Answer 2

❒ Question :-

The denominator of a fraction is 1 less than twice its numerator. If 3 is added to the numerator and 4 is added to the denominator the fraction becomes 3/5. find the fraction.

❒ Answer :-

Read the question carefully , The denominator of a fraction is 1 less than twice its numerator. That means :-

• If the numerator is X . Then,
• Denominator is 2X - 1

Now , as we know when Numerator and Denominator is there that forms a Fraction .

So, fraction :-

$\boxed{ \rm \: Hence , the \: fraction \: is \: = \: \dfrac{ Numerator}{Denominator}}$

$\rm \: Fraction \: \: = \: \dfrac{ x}{2x - 1}$

Now , after that it is given if 3 is added to the Numerator and 4 is added to the Denominator the fraction becomes 3/5 . Therefore , fraction becomes :-

$\rm \: Fraction \: \: = \: \dfrac{ x + 3}{2x - 1 + 4} = \dfrac{3}{5}$

So , solving it :-

$\hookrightarrow\rm \: \dfrac{ x +3 }{2x - 1 + 4} = \dfrac{3}{5}$

$\hookrightarrow\rm \: \dfrac{ x +3 }{2x + 3} = \dfrac{3}{5}$

Now doing cross Multiplication ,

$\hookrightarrow \rm \: 5(x + 3) = 3(2x + 3)$

$\hookrightarrow \rm \: 5 x + 15 \: = 6x + 9$

Sign changes as it sides changes

$\hookrightarrow \rm \: 15 - 9 \: = 6x - 5x$

$\hookrightarrow \rm \: 6 \: = x$

$\boxed{ \rm \: \: Fraction \: \: = \: \dfrac{ Numerator}{Denominator}}$

Hence , the fraction

$\rm \dfrac{6}{2 \times 6 - 1} = \bf \dfrac{6}{11} \big \star$

$\rm \bf \: Hence , the \: required \: fraction \: is \: \dfrac{6}{11}$