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

Questions

June 2021 1 52 Report

The sum of 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 Find the fraction.

Answers & Comments

itzsecretagent

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

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 $\sf {\dfrac{1}{3} }$

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

The fraction.

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

✰ 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 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 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,

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

Also, from the equation (i) :

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