the digit in the tens place of a 2- digit number is twice the unit digit if the ...

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the digit in the tens place of a 2- digit number is twice the unit digit if the digits are reversed the new number is 18 less than the original number find the original number

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Given :

$\sf\purple{Digit \:in\: tens\: place\: is\: twice \:the \:unit's \:digit.}$

$\sf\purple{If \:digits \:reversed, new \:number\: is\: 18\: less\: than\: }$ $\sf\purple{the\: original \:number.}$

To find :

$\sf\orange{Original \:number}$

Solution :

Let the digit at unit place be y and digit at ten's place be 2y [∵ Digit at tens place = twice the digit at units place]

$\sf\green{∴ Original \:number = y + 10(2y) = y + 20y}$ $\sf\green{ = 21y}$

Now if digits are reversed, then :

$\sf\red{∴ \:New\: number = 2y + 10y = 12y}$

Now atq,

$\sf\pink{⇒ 12y = 21y - 18}$

$\sf\pink{⇒ 12y - 21y = - 18}$

$\sf\pink{⇒ -9y = -18}$

$\sf\pink{⇒ y ={\frac{ -18}{-9}}}$

$\sf\pink{⇒ y = 2}$

$\sf\blue{∴ Digit\: at \:units \:place \:= y = 2}$

$\sf\blue{∴ Digit \:at\: tens \:place \:= 2y = 2(2) = 4}$

∴ Original number = 21y

= 21(2)

= 42

Therefore,

$\sf\red{Original \:number = 42}$

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