Question

The sum of numerator and denominator of a fraction is 3 less than twice the denominator.if each numerator and denominator is decreased by 1, the fraction becomes 1/2. find the fraction

Answer 1

$\huge {\red {\mathfrak{Question :-}}}$

• The sum of numerator and denominator of a fraction is 3 less than twice the denominator. If each numerator and denominator is decreased by 1, the fraction becomes $\dfrac {1}{2}$. Find the fraction.

$\huge {\red {\mathfrak{Solution:-}}}$

$\underline{\bold{To\:Find:}}$

• The fraction .

Let the numerator be x.

Then, the denominator be y.

Sum of numerator and denominator of the fraction is 3 less than twice the denominator.

$\therefore {\sf{x+y=2y-3}}$

$\hookrightarrow x + y = 2y - 3 \\ \\ \hookrightarrow x = 2y - 3 - y \\ \\ \hookrightarrow x = y - 3\:\:\:\:\:\: ....(i)$

If each numerator and denominator is decreased by 1 , the fraction becomes $\bf{\frac{1}{2}}$

$\therefore{\sf{\dfrac {x-1}{y-1}=\dfrac{1}{2}}}$

$\hookrightarrow \frac{x - 1}{y - 1} = \frac{1}{2} \\ \\ \hookrightarrow 2(x - 1) = 1(y - 1) \\ \\ \hookrightarrow 2x - 2 = y - 1 \\ \\ \sf{By \:substituting \: the \: value \: of \: equation \: (i)} : \\\hookrightarrow 2x - 2 = y - 1 \\ \\ \hookrightarrow 2(y - 3)-2= y - 1 \\ \\ \hookrightarrow \ 2y - 6-2 = y - 1 \\ \\\hookrightarrow 2y - 8= y - 1 \\ \\\hookrightarrow 2y-y=-1+8 \\ \\\hookrightarrow y = 7$

Now we find the numerator (Value of x) :

$\hookrightarrow x = y - 3 \\ \\ \hookrightarrow x =7- 3 \\ \\ \hookrightarrow x = 4$

$\green{\therefore{\tt {The\:fraction\:is\:\dfrac {4}{7}.}}}$

$\rule {197}{2}$

Answer 2

Given :

• The sum of numerator and denominator of a fraction is 3 less than twice the denominator.
• If each numerator and denominator is decreased by 1, the fraction becomes 1/2.

To Find :

• The fraction.

Solution :

Let the numerator of the fraction be x.

Let the denominator of the fraction be y.

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

Case 1 :

The sum of the numerator and denominator is 3 less than twice the denominator.

Equation :

$\longrightarrow$ $\sf{x+y=2y-3}$

$\longrightarrow$ $\sf{x=2y-y-3}$

$\sf{x=y-3\:\:\:(1)}$

Case 2 :

The numerator and denominator both are decreased by 1, then the fraction becomes 1/2.

Numerator = (x-1)

Denominator = (y-1)

Equation :

$\longrightarrow$ $\sf{\dfrac{(x-1)}{(y-1)}\:=\:\dfrac{1}{2}}$

$\longrightarrow$ $\sf{2(x-1) =1(y-1) }$

$\longrightarrow$ $\sf{2x-2=y-1}$

$\longrightarrow$ $\sf{2x-y=-1+2}$

$\longrightarrow$ $\sf{2(y-3)-y=1}$

$\longrightarrow$ $\sf{2y-6-y=1}$

$\longrightarrow$ $\sf{2y-y=1+6}$

$\longrightarrow$ $\sf{y=7}$

Substitute, y = 7 in equation (1),

$\longrightarrow$ $\sf{x=y-3}$

$\longrightarrow$ $\sf{x=7-3}$

$\longrightarrow$ $\sf{x=4}$

$\large{\boxed{\bold{Numerator\:=\:x\:=\:4}}}$

$\large{\boxed{\bold{Denominator\:=\:y\:=\:7}}}$

$\large{\boxed{\bold{Fraction\:=\:\dfrac{x}{y}\:=\:\dfrac{4}{7}}}}$