Question

■In a two digit number, digit at the ten's place is twice the digit at units's place. If the number obtained by interchanging the digit is added to the original number, the sum is 66. Find the number.
PLZ......
■GIVE ME STEP BY STEP ANSWER ​

Answer 1

HeY MatE !!

HerE IS YouR AnsweR ----

Let the digit at ones place = x

So, the digit at tens place = 2x

therefore, original no. =

$= (x\times 1) + (2x \times 10) \\ \\ = x + 20x \\ \\ = 21x$

On interchanging the digits ,

New no.

$= (x \times 10) + (2x \times 1) \\ \\ = 10x + 2x \\ \\ = 12x$

acc. to the given condition----

$= > new \: no. \: + original \: no. = 66 \\ \\ = > 12x + 21x = 66 \\ \\ = > 33x = 66 \\ \\ = > x = \frac{66}{33} \\ \\ = > x = 2$

therefore, original no.= 21x= 21×2 = 42

and , new no. = 24

Answer 2

happy friendship day dear...god bless you.. and have a glorious day.... my lovely friend........