Question

Consider a two digit number such that when its digits are reversed the new number obtained is 6 more than three times the original number also one digit is five times that of the other find the original number

Answer 1

$\sf\small\underline\blue{Let:-}$

$\sf{\implies The\:ones\:digit=R}$

$\sf{\implies The\:tens\: digit=S}$

$\sf{\implies The\: orginal\: number=10S+R}$

$\sf\small\underline\blue{To\: Find:-}$

$\sf{\implies The\: orginal\: number=?}$

$\sf\small\underline\blue{Solution:-}$

Here we will solve this question by setting up equation then solve the the value of R and S. After that by applying value to calculate the original number.

$\sf\small\underline\blue{Given\:in\:case-(i):-}$

when its digits are reversed the new number obtained is 6 more than three times the original number.

$\sf{\implies (Orginal\: number)=3(New\: number)+6}$

$\sf{\implies (10R+S)=3(10S+R)+6}$

$\sf{\implies 10R+S=30S+3R+6}$

$\sf{\implies 10R-3R+S-30S=6}$

$\sf{\implies 7R-29S=6----(i)}$

$\sf\small\underline\blue{Given\:in\:case-(ii):-}$

One digit is five times that of the other digit.

$\sf{\implies Ones\: digit=5(other\: digit)}$

$\sf{\implies R=5S------(ii)}$

• by Solving equation (i) and (ii) we get:-

$\sf{\implies 7R - 29S=6}$

• Putting the value of R=5S:-

$\sf{\implies 7(5S)-29S=6}$

$\sf{\implies 35S-29S=6}$

$\sf{\implies6S=6}$

$\sf\red{\implies S=1}$

• Putting the value of S = in eq (ii):-

$\sf{\implies R=5S}$

$\sf{\implies R=5(1)}$

$\sf\red{\implies R=5}$

$\sf\large{Hence,}$

$\sf{\implies The\: orginal\: number=10S+R}$

$\sf{\implies The\: orginal\: number=10(1)+5}$

$\bf{\implies The\: orginal\: number=15}$

Answer 2

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$\underline{\textsf{\textbf{Question\::}}}$

Consider a two digit number such that when its digits are reversed the new number obtained is 6 more than three times the original number also one digit is five times that of the other find the original number.

$\underline{\textsf{\textbf{To\:find\::}}}$

• Original number ?

$\underline{\textsf{\textbf{Let\::}}}$

• Ones digit = x
• Tens digit = y
• Original number = 10y + x

$\underline{\textsf{\textbf{Solution\::}}}$

¤ First case :

¤ Given :

• When its digits are reversed the new number obtained is 6 more than three times the original number.

$\dashrightarrow\:\tt New\:number = 3(Original\:number) + 6$

$\dashrightarrow\:\tt 10x + y = 3(10y + x) + 6$

$\dashrightarrow\:\tt 10x + y = 30y + 3x + 6$

$\dashrightarrow\:\tt 10x - 3x + y - 30y = 6$

$\dashrightarrow\:\tt 7x - 29y = 6 \qquad\qquad\lgroup 1 \rgroup$

¤ Second case :

¤ Given :

• One digit is five times that of the other digit.

$\longrightarrow\:\tt One\:digit\:=\:5(Other\:digit)$

$\longrightarrow\:\tt x\:=\:5y \qquad\qquad\qquad \lgroup 2 \rgroup$

Now,

¤ From (2) put in (1) :

$\:\:\:\dashrightarrow\qquad\tt 7(5y) - 29y = 6$

$\:\:\:\dashrightarrow\qquad\tt 35y - 29y = 6$

$\:\:\:\dashrightarrow\qquad\tt 6y = 6$

$\:\:\:\dashrightarrow\qquad\tt y = \dfrac{6}{6}$

$\:\:\:\dashrightarrow\qquad\tt y = \dfrac{\cancel{6}}{\cancel{6}}$

$\:\:\:\dashrightarrow\qquad\bf { y = \red{1}}$

¤ Putting value of y in (2) :

$\:\:\:\longrightarrow\qquad\tt x = 5(1)$

$\:\:\:\longrightarrow\qquad\bf {x = \red{5}}$

Therefore,

$\dashrightarrow\:\tt Original\:number = 10y + x$

¤ Putting values of y and x :

$\dashrightarrow\:\tt Original\:number = 10(1) + 5$

$\dashrightarrow\:\tt Original\:number = 10 + 5$

$\dashrightarrow\:\bf {Original\:number = \red{ 15}}$

This is the required answer.

$\large\underline{\boxed{\sf{Original\:number\:\leadsto\:\rm\purple{15}}}}$

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