Question

A fraction becomes 1/2 when 1 is added to the numerator and it becomes 1/3 when 1 is subtracted from the numerator and 2 is added to the denominator .find the fraction . Also find the numerator obtained when 5 is added to numerator and 4 ia subtracted from the denominator​

Answer 1

let numerator be x and denominatior be y.

Answer 2

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{1. \ Fraction \ is \ \frac{7}{16}}$

$\sf{2. \ Numenator \ obtained \ is \ 1.}$

$\sf\orange{Given:}$

$\sf{\implies{A \ fraction \ becomes \ \frac{1}{2} \ when}}$

$\sf{1 \ is \ added \ to \ the \ numenator.}$

$\sf{\implies{It \ becomes \ \frac{1}{3} \ when \ 1 \ is}}$

$\sf{subracted \ from \ the \ numenator \ and \ 2 \ is}$

$\sf{added \ to \ the \ denominator.}$

$\sf\pink{\underline{\underline{To \ find:}}}$

$\sf{1. \ The \ fraction.}$

$\sf{2. \ The \ numenator \ of \ a \ fraction}$

$\sf{obtained \ when \ 5 \ is \ added \ to \ numerator}$

$\sf{and \ 4 \ is \ subtracted \ from \ the \ denominator.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{1.}$

$\sf{Let \ numenator \ of \ a \ fraction \ be \ x }$

$\sf{and \ it's \ denominator \ be \ y.}$

$\sf{According \ to \ first \ condition.}$

$\sf{\frac{x+1}{y}=\frac{1}{2}}$

$\sf{\therefore{(2(x+1)=y}}$

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

$\sf{According \ to \ second \ condition.}$

$\sf{\frac{x-1}{y+2}=\frac{1}{3}}$

$\sf{\therefore{3(x-1)=y+2}}$

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

$\sf{\therefore{3x-y=5...(2)}}$

$\sf{Subtract \ equation (1) \ from \ equation (2)}$

$\sf{3x-y=5}$

$\sf{-}$

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

__________________

$\boxed{\sf{\therefore{x=7}}}$

$\sf{Substitute \ x=7 \ in \ equation (1)}$

$\sf{2(7)-y=-2}$

$\sf{14-y=-2}$

$\sf{-y=-2-14}$

$\sf{-y=-16}$

$\boxed{\sf{\therefore{y=16}}}$

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

$\sf\purple{\tt{\therefore{Fraction \ is \ \frac{7}{16}}}}$

___________________________________

$\sf{2.}$

$\sf{According \ to \ given \ condition.}$

$\sf{\frac{x+5}{y-4}=\frac{7+5}{16-4}}$

$\sf{=\frac{12}{12}}$

$\sf{=\frac{1}{1}}$

$\sf\purple{\tt{\therefore{Numenator \ obtained \ is \ 1.}}}$