Question

The sum of the numerator and denominator of a fraction is 15 if the denominator is increased by 1 the fraction becomes 1 by 3 find the fraction

Answer 1

$\huge{\underline{\red{\purple{Solution:}}}}$

$\large{\underline{\red{\rm{Let:}}}}$

• $\rm{The\: numerator\:of\: fraction=x}$

• $\rm{The\: Denominator\:of\: fraction=y}$

• $\rm{The\: fraction=\dfrac{x}{y}}$

$\large{\underline{\purple{\rm{To\: Find:}}}}$

• $\rm{The\: fraction=?}$

$\small{\underline{\orange{\rm{As\:per\:the\: Question:}}}}$

$\small{\underline{\red{\rm{Given\:in\:the\:1st\:case:}}}}$

• $\rm{The\:sum\:N\:and\:D=15}$

$\rm{\implies x+y=15}$

$\rm\green{\implies x+y=15-----(i)}$

$\small{\underline{\red{\rm{Given\:in\:the\:2nd\:case:}}}}$

• $\rm{If\:the\:D\:is\: increased\:by\:1\:the\: fraction\: become\:\frac{1}{3}}$

$\rm{\implies \dfrac{x}{y+1}=\dfrac{1}{3}}$

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

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

$\rm\green{\implies 3x-y=1-----(ii)}$

• $\rm{Now,\:in\:eq\:1st\: multiplying\:by\:3\:than\: subtract\:in\:eq\:2nd}$

$\rm{\implies 3x+3y=45}$

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

• $\rm{By\: solving\: equation\:we\:get\:here}$

$\rm{\implies \cancel{4}y=\cancel{44}}$

$\rm\purple{\implies y=11}$

• $\rm{putting\: the\: value\:of\:y=12\:in\:eq\:1st}$

$\rm{\implies x+y=15}$

$\rm{\implies x+11=15}$

$\rm{\implies x=15-11}$

$\rm\purple{\implies x=4}$

Hence,

$\rm\orange\implies{The\: fraction=\dfrac{x}{y}=\dfrac{4}{11}}$

Answer 2

Let the denominator of the fraction be $x.$

Then, since the sum of numerator and denominator is $15,$ the numerator will be $15-x.$

Hence our fraction will be $\dfrac{15-x}{x}.$

If denominator is increased by $1,$ then our fraction will be equivalent to $\dfrac{1}{3}.$

$\longrightarrow \dfrac{15-x}{x+1}=\dfrac{1}{3}$

By cross multiplication.

$\longrightarrow 3(15-x)=x+1$

$\longrightarrow 45-3x=x+1$

$\longrightarrow 4x=44$

$\longrightarrow x=11$

This is the denominator. The numerator will be,

$\longrightarrow 15-x=4$

Hence our fraction is $\bf{\dfrac{4}{11}.}$