Question

The digits in the ten's place of a two-digit number is three times that in the units place. If the digits are reversed, the new number will be 36 less then the original number. Find the original number

Answer 1

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Question:-

The digits in the ten's place of a two-digit number is three times that in the units place. If the digits are reversed, the new number will be 36 less then the original number. Find the original number

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Answer:-

Let the one's digit be y and tens digit be x,

$\bold{Number = 10x + y}$

$\bold{Then,x=3y⋯(i)}$

Reversed number = 10y + x

Reversed number = 10y + xA.t.Q :- (10x+y)−(10y+x)=36 Put x = 3y in eq. (i)

$\bold{⇒9x−9y=36}$

$\bold{⇒x−y=4⋯(ii)}$

$\bold{⇒3y−y=4}$

$\bold{∴2y=4 x=3y ∴x=6}$

$\bold{y=2}$

$\large \blue{ \boxed{ \red{ \bf ∴ Number = 62}}}$

$\huge\blue{\ddot\smile}$

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Answer 2

$\huge\boxed{\fcolorbox{black}{pink}{Answer}}</p><p>$

Let the one's digit be y and tens digit be x,

Number = 10x + y Number=10x+y

Then,x=3y⋯(i) Then,x=3y⋯(i)

Reversed number = 10y + x

Reversed number = 10y + xA.t.Q :- (10x+y)−(10y+x)=36 Put x = 3y in eq. (i)

⇒9x−9y=36 ⇒9x−9y=36

⇒x−y=4⋯(ii) ⇒x−y=4⋯(ii)

⇒3y−y=4 ⇒3y−y=4

∴2y=4 x=3y ∴x=6 ∴2y=4x=3y∴x=6

y=2 y=2

Number = 62