Question

The sum of two digits number 12 if the digits are reserved the new number is less 12 than twice the orginal number find the orginal number

Answer 1

$\Large{\underline{\underline{\tt{\purple{Given}}}}}$

The sum of two digits number 12 if the digits are reserved the new number is less 12 than twice the orginal number.

$\Large{\underline{\underline{\tt{\purple{Find\:out}}}}}$

Find the original number

$\Large{\underline{\underline{\tt{\purple{Solution}}}}}$

★ Let the ten's digit be x and one's

digit be y

• Original number = 10x + y

✞ According to the given condition

✰Sum of two digits number 12

• x + y = 12 ---(i)

✰If the digits are reserved the new number is less 12 than twice the orginal number

• Reversed number = 10y + x

• 2(10x + y) - (10y + x) = 12

➟ 20x + 2y - 10y - x = 12

➟ 19x - 8y = 12 ---(ii)

✞ Multiply (i) by 8 and (ii) by 1

• 8x + 8y = 96
• 19x - 8y = 12

✞ Add both the equations

➟ (8x + 8y) + (19x - 8y) = 96 + 12

➟ 8x + 8y + 19x - 8y = 108

➟ 27x = 108

➟ x = 108/27

➟ x = 4

✞ Put the value of x in eqⁿ (i)

➟ x + y = 12

➟ 4 + y = 12

➟ y = 12 - 4

➟ y = 8

$\large{\underline{\boxed{\tt{\red{Hence,\:x=4\:and\:y=8}}}}}$

$\large{\underline{\boxed{\tt{\red{So,Original\:number= 10x + y = 48}}}}}$

Answer 2

Let the two digits number be " xy "

i.e., 10 x + y

If the digits are reversed new number will be " 10 y + x  "

A/c " The sum of two digits number 12 "

⇒ x + y = 12 ... (1)

A/c " if the digits are reserved the new number is less 12 than twice the original number "

⇒ 2 ( 10 x + y ) - ( 10 y + x ) = 12  

⇒ 19 x - 8 y = 12 ... (2)

Now solve 8(1) + (2) ,

⇒ 8 ( x + y ) + ( 19 x - 8 y ) = 96 + 12

⇒ 8 x + 8 y + 19 x - 8 y = 108

⇒ 27 x = 108

⇒ x = 4

Sub. x value in (1) , we  get ,

⇒ y = 12 - 4

⇒ y = 8

Hence the original number is xy = 48