Question

The sum of numerator and denominator of a fraction is greater by 1 than thrice numerator. If the numerator is decreased by 1 than the fraction reduces to 1/3. Find the fraction.​

Answer 1

$\underline{ \underline{ \Large \pmb{\sf { {Given:}} }} }$

• The sum of numerator and denominator of a fraction is greater by 1 than thrice numerator.

• If the numerator is decreased by 1 than the fraction reduces to $\sf {\dfrac{1}{3} }$.

$\underline{ \underline{ \Large \pmb{\sf { {To \: find:}} }} }$

• The fraction.

$\underline{ \underline{ \Large \pmb{\sf { {Calculation:}} }} }$

✰ Here, we are given that the sum of numerator and denominator of a fraction is greater by 1 than thrice numerator and if the numerator is decreased by 1 than the fraction reduces to $\sf {\dfrac{1}{3} }$. We have to find the original fraction, at first we'll assume the numerator and the denominator as variables and then we'll algebraic equations then by linking one equation with other, we'll find the value of the variables.

⠀⠀⠀⠀⠀_____________

Let the numerator and the denominator be p and q respectively. So, the original fraction,

$\to \sf {\dfrac{p \longrightarrow Numerator}{q\longrightarrow Denominator} }$

As per the given question,

» The sum of numerator and denominator of a fraction is greater by 1 than thrice numerator.

$\sf {\longrightarrow p + q = 3p + 1 }$

Let it be the equation (i).

Also,

» If the numerator is decreased by 1 than the fraction reduces to $\sf {\dfrac{1}{3} }$.

$\longrightarrow \sf {\dfrac{(p +1)}{q} = \dfrac{1}{3} }$

Let it be the equation (ii).

⠀⠀⠀⠀⠀_____________

From the equation (i), we have :

$\dashrightarrow \bf { q = 3p + 1 - p }$

$\dashrightarrow \bf { q = 2p + 1 }$

Henceforth, value of q is 2p + 1.

⠀⠀⠀⠀⠀_____________

Substituting the value of q in the equation (ii).

$\longrightarrow \sf {\dfrac{(p +1)}{(2p + 1)} = \dfrac{1}{3} }$

By cross multiplication,

$\longrightarrow \sf { 3(p-1) = 1(1 + 2p) } \\ \\ \\ \longrightarrow \sf { 3p - 3 =1 + 2p } \\ \\ \\ \longrightarrow \sf { 3p - 2p = 1 + 3 } \\ \\ \\ \dashrightarrow \boxed{ \bf { p = 4 }}$

Also, from the equation (i) :

$\longrightarrow \sf { q = 2p + 1} \\ \\ \\ \longrightarrow \sf { q = 2(4) + 1} \\ \\ \\ \longrightarrow \sf { q = 8 + 1} \\ \\ \\ \dashrightarrow \boxed{ \bf {q= 9 }}$

Therefore,

$\dashrightarrow \underline{\boxed { \bf{ Fraction = \dfrac{4}{9} }}} \: \: \bigstar$

Answer 2

GIVEN:

The sum of the numerator and

denominator of a fraction is greater by 1 than thrice the numerator.

If the numerator is decreased by 1 then the fraction reduces to 1/3

TO FIND:

What is the fraction ?

SOLUTION:

Let the numerator of the fraction be 'x' and the denominator of the fraction be 'y'.

So, the fraction will become

$\bf{\star \: Fraction = \dfrac{Numerator}{Denominator} = \dfrac{x}{y}}$

CASE:- 1

The sum of the numerator and denominator of a fraction is greater by 1 than thrice the numerator.

$\rm{\hookrightarrow x + y = 3x + 1 }$

$\rm{\hookrightarrow -1 = 3x - x -y }$

$\bf{\hookrightarrow -1 = 2x - y...1) }$

$\rm{\hookrightarrow y = 2x + 1}$

CASE:- 2

If the numerator is decreased by 1 then the fraction reduces to 1/3.

$\rm{\hookrightarrow \dfrac{x - 1}{y} = \dfrac{1}{3} }$

$\rm{\hookrightarrow 3(x-1) = y }$

$\rm{\hookrightarrow 3x - 3 = y }$

$\bf{\hookrightarrow 3x - y = 3...2) }$

Put the value of 'y' from equation 1) in equation 2)

$\rm{\hookrightarrow 3x -(2x+1) = 3}$

$\rm{\hookrightarrow 3x - 2x - 1 = 3}$

$\rm{\hookrightarrow 3x - 2x = 3 + 1 }$

$\bf{\hookrightarrow x = 4 }$

Put the value of 'x' in equation 1)

$\rm{\hookrightarrow 2 \times 4 - y = -1 }$

$\rm{\hookrightarrow 8 - y = -1}$

$\rm{\hookrightarrow -y = -1 -8 }$

$\bf{\hookrightarrow y = 9 }$

FRACTION = x/y

FRACTION = 4/9

______________________