Question

The denominator of fraction is 12 more than the numberator . If 16 8s added to both the numberator and the denominator ,the new fraction so obtained is 10/ 13 .Find the original value .​

Answer 1

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

$\sf{The \ original \ value \ of \ fraction \ is \ \frac{7}{10}.}$

$\sf\orange{Given:}$

$\sf{\implies{The \ denominator \ of \ fraction \ is \ 12}}$

$\sf{more \ than \ numerator.}$

$\sf{\implies{If \ 16 \ is \ added \ to \ both \ the}}$

$\sf{numerator \ and \ denominator, \ the \ new}$

$\sf{fraction \ so \ obtained \ is \ \frac{10}{13}.}$

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

$\sf{The \ original \ value \ of \ fraction.}$

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

$\sf{Let \ the \ numerator \ of \ the \ fraction \ be \ x}$

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

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

$\sf{y-x=12}$

$\sf{\therefore{-x+y=12...(1)}}$

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

$\sf{\frac{x+12}{y+12}=\frac{10}{13}}$

$\sf{13(x+12)=10(y+12)}$

$\sf{13x+156=10y+120}$

$\sf{13x-10y=120-156}$

$\sf{\therefore{13x-10y=-36...(2)}}$

$\sf{Multiply \ equation (1) \ throughout \ by \ 13}$

$\sf{-13x+13y=156...(3)}$

$\sf{Add \ equations \ (2) \ and \ (3)}$

$\sf{13x-10y=-36}$

$\sf{+}$

$\sf{-13x+13y=156}$

_____________________

$\sf{3y=120}$

$\sf{y=\frac{120}{3}}$

$\boxed{\sf{\therefore{y=40}}}$

$\sf{Substitute \ y=40 \ in \ equation (1)}$

$\sf{-x+40=12}$

$\sf{-x=12-40}$

$\sf{-x=-28}$

$\boxed{\sf{\therefore{x=28}}}$

$\sf{Required \ fraction \ is \ \frac{x}{y}}$

$\sf{i.e. \ \frac{28}{40}=\frac{7}{10}}$

$\sf\purple{\tt{\therefore{The \ original \ value \ of \ fraction \ is \ \frac{7}{10}.}}}$

Answer 2

★$\color{darkblue}\underline{\underline{\sf Question:}}$

$\rm{The \ denominator \ of \ fraction \ is \ 12 \ more }$

$\rm{than \ the \ numerator. \ if \ 16 \ is \ added \ to \ both }$

$\rm{the \ numerator \ and \ denominator , \ the \ new}$

$\rm{fraction \ so \ obtained \ is \ \frac{10}{13} .}$

$\rm{Find \ the \ original \ value}$

___________________________________________

★$\color{pink}\underline{\underline{\sf Answer:}}$

$\bf\orange{The \ original \ value \ of \ fraction \ is }$

$\bf\orange{ \frac{7}{10}.}$

___________________________________________

$\sf\purple{Given:}$

$\rm{The \ denominator \ of \ fraction \ is \ 12}$

$\rm{more \ than \ numerator.}$

$\rm{if \ 16 \ is \ added \ to \ both \ the}$

$\rm{numerator \ and \ denominator, \ the \ new}$

$\rm{fraction \ so \ obtained \ is }$

$\rm{\frac{10}{13}}$

___________________________________________

$\sf\purple{To \ find:}$

$\rm{The \ original \ value \ of \ fraction}$

___________________________________________

★$\color{red}\underline{\underline{\sf Solution:}}$

$\rm{Let \ us \ consider \ the \ numerator \ of \ the }$

$\rm{fraction \ is \ m \ it's \ denominator \ is \ n}$

$\rm{By \ the \ first \ condition.}$

$\rm{we \ have,}$

$\rm{m \ - \ n \ = \ 12}$

$\rm{\implies{- \ m \ + \ n \ = \ 12 \ ............(a)}}$

$\rm{By \ the \ second \ condition....}$

$\rm{we \ have,}$

$\rm{\frac{m+12}{n+12} \ = \ \frac{10}{13}}$

$\rm{13(m+12) \ = \ 10(n+12)}$

$\rm{13m \ + \ 156 \ = \ 10n \ + \ 120}$

$\rm{13m \ - \ 10n \ = \ 120 \ - \ 156}$

$\rm{\implies{13m \ - \ 10n=-36..................(b)}}$

$\rm{Multiply \ equation \ (a) \ by \ 13}$

$\rm{we \ get}$

$\rm{- \ 13m \ + \ 13n \ = \ 156 \ .................(c)}$

$\rm{Add \ eq^s \ (b) \ and \ (c)}$

$\rm{we \ get ,}$

$\rm{3n=120}$

$\sf{\implies{\frac{120}{3}}}$

$\boxed{\star{\bf{\orange{n=40}}}}$

$\rm{put \ n=40 \ in \ eq^n (a)}$

$\rm{-m+40=12}$

$\rm{-m=12-40}$

$\rm{-m=-28}$

$\boxed{\star{\bf{\orange{m=28}}}}$

$\rm{Required \ fraction \ is \ \frac{m}{n}}$

$\rm{\ \frac{28}{40}=\frac{7}{10}}$

$\sf\orange{\bf{\boxed{The \ original \ value \ of \ fraction \ is \ \frac{7}{10}.}}}$