Question

digit of a two digit number is twice the tens digit when the number formed by reversing the digit is added to original number the sum is 99, find the original number?​

Answer 1

Answer

Let the two digit number is xy

y is digit at unit's place

x is digit at ten's place

⇒ Number can be written as = 10x+y

According to the conditions : y=2x ___ (1)

and

(10x+y)+(10y+x)=99

⇒11x+11y=99

⇒x+y=9 (dividing by 11)

Now x+2x=9 (from (1))

⇒3x=9

⇒x=9/3=3

y=2x=6

⇒ Number is 36

$o$

Answer 2

Answer:

The original number is 12 times 3 = 36.

Step-by-step explanation:

Let's assume that, the tens digit is x

According to the problem, the units digit is 2x

Therefore, the original number is 10x + 2x = 12x

When the digits are reversed, the number becomes 10(2x) + x = 21x

According to the problem:

12x + 21x = 99

=> 33x = 99

∴ x = 3

∴ The original number is 12 times 3 = 36.