Question

in the 2 digit number in unit place is twice the digit in ten place If 27 is added to it digits are reversed find the number​

Answer 1

Answer:

Step-by-step explanation:

Given :-

In a two digit number digit in units place is twice the digit in the tens place.

If 27 is added to its digits are reversed.

To Find :-

The Number

Solution :-

Let ten's digit = x  

Unit's digit = 2x

Required Number = 10x + 2x = 12x

On Interchanging the Digit's Number = 10 (2x) + x = 21x

As per the Question,

$\sf\implies 12x + 27 = 21x$

$\sf\implies27 = 21x-12x$

$\sf\implies27=9x$

$\sf\implies\dfrac{27}{9}=x$

$\bf\implies x=3$

Required Number = 12 × 3 = 36

Hence, the number​ required number is 36.

Answer 2

AnswEr :

Let the Unit Digit be y and, Ten's Digit be x.

◗ Original Number : (10x + y)

◗ Reversed Number : (10y + x)

• Given :

↦ Unit Digit = Twice the Ten's Digit

↦ y = 2x ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀—eq.( I )

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• According to the Question Now :

⇒ Original No. + 27 = Reversed No.

⇒ (10x + y) + 27 = (10y + x)

⇒ 10x + y - 10y - x + 27 = 0

⇒ 9x - 9y + 27 = 0

⇒ 27 = 9y - 9x

⇒ 27 = 9(y - x)

• Dividing Both term by 3

⇒ 3 = y - x

• putting the value of y from eq.( I )

⇒ 3 = 2x - x

⇒ x = 3

• Using the Value of x in eq.( I ) :

⇒ y = 2x

⇒ y = 2(3)

⇒ y = 6

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• Original Number formed will be :

↠ Original No. = (10x + y)

↠ Original No. = (10(3) + 6)

↠ Original No. = (30 + 6)

↠ Original No. = 36

∴ Hence, Original Number will be 36.