Question

The digit in the ten's place of a two digit number is one less than the digit in the one's place. If we add this number to the number obtained by reversing its digits, the result is 55, find the number.

Answer 1

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

$\sf{The \ number \ is \ 23.}$

$\sf\orange{Given:}$

$\sf{\implies{The \ digit \ in \ the \ ten's \ place \ of}}$

$\sf{a \ two \ digit \ number \ is \ one \ less \ than \ the}$

$\sf{digit \ in \ the \ one's \ place }$

$\sf{\implies{The \ sum \ of \ number \ and \ the}}$

$\sf{number \ obtained \ by \ reversing \ the \ digits}$

$\sf{is \ 55.}$

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

$\sf{The \ numbers.}$

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

$\sf{Let \ the \ digit \ in \ the \ ten's \ place \ be \ x}$

$\sf{and \ the \ digit \ in \ the \ unit's \ place \ be \ y.}$

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

$\sf{y-x=1}$

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

$\sf{Original \ number=10x+y}$

$\sf{Number \ with \ reversed \ digits=10y+x}$

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

$\sf{(10x+y)+(10y+x)=55}$

$\sf{11x+11y=55}$

$\sf{11(x+y)=55}$

$\sf{x+y=\frac{55}{11}}$

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

$\sf{Add \ equations \ (1) \ and \ (2)}$

$\sf{-x+y=1}$

$\sf{+}$

$\sf{x+y=5}$

_______________________

$\sf{2y=6}$

$\sf{\therefore{y=\frac{6}{2}}}$

$\boxed{\sf{\therefore{y=3}}}$

$\sf{Substitute \ y=3 \ in \ equation \ 1}$

$\sf{x+3=5}$

$\sf{x=5-3}$

$\boxed{\sf{\therefore{x=2}}}$

$\sf{Original \ number=10(2)+3=23}$

$\sf\purple{\tt{\therefore{The \ number \ is \ 23.}}}$

Answer 2

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

$\rm{The \ digit \ in \ the \ ten's \ place \ of \ a}$

$\rm{two \ digit \ number \ is \ one \ less \ than \ the }$

$\rm{digit \ in \ the \ one's \ place, \ If \ we \ add \ this }$

$\rm{number \ to \ obtained \ by \ reversing \ it's \ digit,}$

$\rm{the \ result \ is \ 55 , \ find \ the \ number.}$

_________________________________________

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

$\rm\orange{The \ number \ is \ 23.}$

_________________________________________

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

$\rm{The \ digit \ in \ the \ ten's \ place \ of}$

$\rm{a \ two \ digit \ number \ is \ one \ less \ than \ the}$

$\rm{digit \ in \ the \ one's \ place }$

$\rm{The \ sum \ of \ number \ and \ the}$

$\rm{number \ obtained \ by \ reversing \ the \ digits}$

$\rm{is \ 55.}$

_________________________________________

$\sf\underline\red{To \ Find:}$

$\rm{The \ numbers.}$

_________________________________________

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

$\rm{Let \ the \ digit \ in \ the \ ten's \ place \ be \ m}$

$\rm{and \ the \ digit \ in \ the \ unit's \ place \ be \ n.}$

$\rm{According \ to \ the \ first \ condition}$

$\bf{n-m=1}$

$\bf{\implies{- \ m \ + \ n \ = \ 1 \ ............[a]}}$

$\rm{Original \ number \ = \ 10 \ m \ + \ n}$

$\rm{Number \ with \ reversed \ digits \ = \ 10 \ n \ + \ m}$

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

$\rm{( \ 10m \ + \ n \ ) \ + \ ( \ 10n \ + \ m \ ) \ = \ 55}$

$\rm{11m \ + \ 11y \ = \ 55}$

$\rm{11(m+n) \ = \ 55}$

$\rm{m \ + \ n \ = \ \frac{55}{11}}$

$\rm{\implies{m \ + \ n \ = \ 5 \ ..............[b]}}$

$\sf{Add \ equations \ [a] \ and \ [b]}$

$\rm{2n=6}$

$\rm{\implies{m=\frac{6}{2}}}$

$\star\pink{\boxed{\bf{\implies{m \ = \ 3}}}}$

$\rm{putting \ this \ value \ in \ equation \ [a]}$

$\rm{m+3=5}$

$\rm{m=5-3}$

$\star\pink{\boxed{\bf{\implies{m \ = \ 2}}}}$

$\rm{Original \ number=10(2)+3=23}$

$\bf\orange{\sf{\implies{The \ number \ is \ 23.}}}$

_________________________________________

$\rm\pink{Hope \ it \ helps \ :))}$