Question

The sum of the denominator and numerator of a fraction is 40. If 1 is added to both the numerator and denominator ,the fraction become
$\frac{3}{4}$
.find the fraction
​

Answer 1

AnswEr :

• Given :

⇒ Denominator + Numerator = 40

⇒ D + N = 40

⇒ D = (40 - N)

$\rule{150}{1}$

• According to the Question Now :

$:\implies \sf \dfrac{Numerator+1}{Denominator+1} = \dfrac{3}{4}\\\\\\:\implies \sf \dfrac{N+1}{D+1} = \dfrac{3}{4}\\\\ \qquad \scriptsize{\bf{ \dag}\texttt{ But ; D = (40 - N)}}\\\\:\implies \sf \dfrac{N+1}{(40 - N)+1} = \dfrac{3}{4}\\\\\\:\implies \sf \dfrac{N+1}{41 - N} = \dfrac{3}{4}\\\\\\:\implies \sf 4(N+1) = 3(41 - N)\\\\\\:\implies \sf 4N + 4 = 123 - 3N\\\\\\:\implies \sf 4N + 3N = 123 - 4\\\\\\:\implies \sf 7N = 119\\\\\\:\implies \sf N = \cancel\dfrac{119}{7} \\\\\\:\implies \sf N =17$

$\rule{200}{2}$

$\longrightarrow\tt Original \:Fraction =\dfrac{N}{D} \\\\\longrightarrow\tt Original \:Fraction=\dfrac{N}{(40-N)} \\\\\longrightarrow\tt Original \:Fraction=\dfrac{17}{(40-17)} \\\\\\\therefore\quad\boxed{ \red{\tt Original \:Fraction=\dfrac{17}{23}}}$

Answer 2

Answer:

$\huge{\boxed{\bf\green{Answer=\frac{17}{23}}}}$

_______________________

Let's take numerator be X and the denominator be Y.

Now,

$\bold {X + Y = 40 \: \: \: \: \:(eq.1)} \\ \\ \bold{ \frac{X+1}{Y+1} = \frac{3}{4} \: \: \: \: \: \:(eq.2)}$

$\tt From\:(eq.2)\: we\:get,$

$\bold{4(X+1) = 3(Y+1)}$

$\bold{\rightarrow{4X + 4=3Y + 3}}$

$\bold{\rightarrow{4X = 3Y + 3 - 4}}$

$\bold{\rightarrow{4X = 3Y - 1\: \: \: \: \:(eq.3)}}$

$\tt From\:(eq.1)\:we\:get,$

$\bold{X = 40 - Y}$

$\tt Putting\:the\:value\:of\:X \:into\:(eq.3)\:we\:get,$

$\bold{4(40-Y) = 3Y - 1} \\ \bold{\rightarrow{160 - 4Y = 3Y - 1}} \\ \bold{\rightarrow{-4Y - 3Y = - 1 - 160 }} \\ \bold{\rightarrow{- 7Y = -161}} \\ \bold{\rightarrow{Y =\frac{-161}{-7} }} \\ \bold{\rightarrow{Y = 23}}$

$\therefore\tt So,\:denominator\:is\:23$

Now,Putting the value of Y into (eq.1) we get,

$\bold{X + 23 = 40} \\ \bold{\rightarrow{X = 40 - 23}} \\ \bold{\rightarrow{X = 17}}$

$\therefore\tt So,\:numerator\:is\:17$

${\boxed{\bf\red{Fraction=\frac{17}{23}}}}$